QUESTION:
People are worried that Earth is slowly
loosing its moon, because Earth will go
crazy. Mercury and Venus don't have moons
and their orbits and rotations are stable.
What gives?
ANSWER:
Funny, I have never met a single person
who is "…worried that Earth is slowly
loosing [sic] its moon…". Do you
know the rate at which this is happening?
1.5 inches per year! So tell all those
worried people to stop worrying. There will
be no noticable change for hundreds of
generations. Why do you think "…Earth will
go crazy"? And what does that mean, anyway?
If you are interested in why the moon is
receding rather than coming closer, see an
earlier answer. There is also a
nice video which explains that the moon
will never actually leave the earth.
QUESTION:
My assumtion: Things that get too hot
eventually glow. Question: Does coolder
matter have some way of binding photons? If
so, What force weakeneds within matter that
it's grip on those escaping photons when the
matter heats up the the point of glowing and
emiting light? I do get that light is always
shinning through different spectra, is it
just the rapidly vibrating hotter matter
that speeds up the wave form of an emited
photon?
ANSWER:
Everything radiates regardless of its
temperature. Most objects are pretty well
described as
black bodies. Also see a
earlier answer of mine. The intensity of
the radiation as a function of the
wavelength of the radiation is shown in the
figure for three temperatures. At 5000 K,
approximately the temperature of the surface
of the sun, the most ratiation is in our
visible range; this is a wonderful example
of evolution, showing that we evolved such
that our eyes are most sensitive to the
wavelengths which illuminate our world. As
the temperature gets cooler, the peak shifts
toward longer (redder) wavelengths; At 3000
K, roughly room temperature, nearly all the
radiation is in the infrared part of the
spectrum where our eyes cannot detect it but
infrared cameras and glasses can. You are
right that as you increase temperature the
radiation in the visible region will become
intense enough to see it. But your idea that
all those emitted photons have been sitting
around waiting to be radiated is totally
wrong. Photons are created when electric
charges are accelerated. The electrons in
the atoms are vibrating about and get
created at the time of emission.
QUESTION:
What does it mean when 'if a charge is
taken through a potential difference'?
ANSWER:
It means that the charge moves in a
volume where there is an electric field. It
means that work is done (positive or
negative) on the charge as it moves from one
point to another; the work done is
independent of the path between the two
points.
QUESTION:
How is it that half the surface of a
sphere [the Moon] can be illuminated with
the same intensity of sunlight rather than a
gradient?
ANSWER:
I don't understand the question.
REPLY:
A computergenerated lambertian model
illuminated with collimated light (as from
the sun) contrasted with the illumination of
the moon:
ANSWER:
It turns out that the reason is quite
simple: The moon is not welldescribed as
Lambertian surface for numerous reasons. A
good explanation can be seen at this
link.
QUESTION:
We now know that protons and neutrons
are comprised of 3 quarks. Is there any
reason to believe that is it? Quarks are the
most basic so called 'building blocks' of
particles or...., is it possible quarks are
comprised of some other as of yet
undiscovered particles and so on and so
on...?
ANSWER:
It is generally believed that, like
electrons, quarks are indivisible particles;
no selfrespecting scientist would insist
that it is not possible that they have some
underlying structure. But, the notion that
protons are comprised simply of 3 quarks is
a big simplification. See an earlier
answer for a discussion of this.
QUESTION:
Hi, as I was surfing the internet I
became interested in the concept of
renormalization and found something related
to it's history quite interesting.
Renormalization is usually credited to three
men, Richard Feymann (the famous one),
Julian Schwinger and Shin'ichirō Tomonaga,
they even received the Nobel, Tomonaga seems
to have published his first paper about
renormalization in 1943, in japanese
(https://academic.oup.com/ptp/article/1/2/27/1877120?login=false).
Is there some reason he is not given a
pioneering role in places like wikipedia and
physics books? I thought the discovery was
made almost simultaneous, but he just was
obstructed by war and language barrier. As
I'm not expert in the subject at hand, does
the first Tomonaga paper lacks something
that makes it not a renormalization paper?
ANSWER:
I do not understand what is bothering
you so much. All received three were awarded
the Nobel Prize in Physics 1965 for their
contributions to quantum electrodynamics;
you can't get any better recognition than
that. The Wikepedia article on
renormalization credits all three along with
Kramer, Bethe, and Dyson for making
important contributions to renormalization
theory (Tomonaga has the most references).
QUESTION:
Hopefull I’m not breaking ground rules.
Related to climat change which I beleive is
real.. but the recent finding of a hiker
lost on a gkazier 4 decades ago seems to say
the glazier is at the same point it was at
40 years ago…. So assuming glazier are not
normall ever growing and they seem to ebb
snd flow(is this right). Then this
particular glazier seems to be at the same
point it was at 40 years ago… how does this
knowledge fit in to the obvious climate
changes we seem to be experiencing?
ANSWER:
I do not see how you can glean any
information from his being found. First,
since he was lost and never found, we have
no idea where he was when he died. Second,
there is a reason why glaciers are called
"rivers of ice"—they are always moving. So
you neither know where this hiker died nor
where that location has moved to over the
intervening time.
QUESTION:
Distance between 2 planets (A & B) is 10
light years. One spaceship (1) leaves planet
A and travels to planet B at an average
velocity of 0.25c. A second spaceship (2)
leaves planet A 20 years later and travels
to planet B at an average velocity of 0.5c.
Do they arrive at planet B at the same time?
ANSWER:
It depends on whose clocks you used.
Let's first use clocks on planet A: It
obviously takes a time T_{1}=40
years for #1 and T_{2}=20
years for #2, so, as observed by A, they
arrive simultaneously. Now, #1 would
measure, because of length contraction, the
distance between A and B to be L'_{1}=L_{1}√(10.25^{2})=9.68
ly; similarly, #2 would measure that
distance to be L'_{2}=L_{2}√(10.5^{2})=8.7
ly. The corresponding times (T=L/v)
would be T'_{1}=38.7 yr and
T'_{2}=17.3 yr. Now we need
to find the speed of #2 relative to #1; [v_{21}=0.5c+0.125c]/[1+(0.5x0.25]=0.286c.
We can now calculate how long it takes
for #2 to catch up with #1, (L_{2}/2)/v_{21}=16.9
years; but it takes #1 T'_{1}/2=19.4
years to arrive at B, so #2 arrives at B 2.5
years earlier than #1. What is simultaneous
in one frame is not necessarily simultaneous
in another.
QUESTION:
Hello, I am very interested in how
standing, walking etc in a pool affects the
location your standing center of gravity 
normally just in front of S1S2. My gut
feeling is that it raises is a tiny bit, but
I'm not at all sure. Do you have the answer?
ANSWER:
There is, of course, no exact answer to
this question—bodies are different,
distribution of mass is highly nonuniform,
the depth of the water matters, etc.
Center of gravity (cog) is different than
center of mass (com) because com depends on
how mass is distributed whereas cog depends
on how apparent weight is
distributed. Because there is a buoyant
force on the body in the water, his apparent
weight of that section is less. The figure
shows a man of height H standing in
water of a depth D. I was a little
surprised to find for the man not in water,
as the questioner implies, that the cog is
near his waist, about H/2; then I
realized that the heaviest bones (legs and
pelvis) are below the waist so this was
reasonable. Being in the middle like this
suggested that the mass is fairly uniformly
distributed so I chose to model the man by a
uniform cylinder of mass M and
volume V=AH; this is a very rough
estimate but is the best I can do and it
will give an orderofmagnitude of how far
the cog will move upward. Before the water
is added the cog of the cylinder is at H/2
and the total weight is Mg where
g=9.8 m/s^{2} is the
acceleration due to gravity. The total
weight is Mg=ρ_{man}Vg,
where ρ_{man}=985 kg/m^{3}
is a typical mass density for a man*.
Now let's add the water. The apparent weight
of the submerged part of the cylinder is
W_{1}=(ρ_{water}ρ_{man})DAg
and the cog is y_{1}=D/2
above the bottom because the mass is assumed
to be uniformly distributed. The weight of
the unsubmerged volume is W_{2}=ρ_{man}(HD)Ag
and its position is at y_{2}=(H+D)/2.
So we can now write the position of the new
cog, y_{cog}=(W_{1}y_{1}+W_{2}y_{2})/(W_{1}+W_{2}).
The density of water is 997 kg/m3, so y_{cog}=[6D^{2}+493(H^{2}D^{2}]/[12D+985(HD)].
As a numerical example I will look at an
H=2 m tall person standing in water of
depth D=1.3 m; this would be
roughly the scale which is in the figure I
have drawn. Doing the arithmetic I find that
y_{cog}=1.63 m. This is
just barely below the cog of the unsubmerged
portion of the cylinder or man which is at
1.65 m. Note that this is by no means a
"tiny bit" as you speculated. The reason for
this is easy to understand—because the
density of water and the density of the man
are so close to each other, the effective
mass of the submerged mass is very nearly
zero.
*As a
touchpoint, a 200 lb man has a mass of about
90 kg, so V=90/985=0.1 m^{3}.
Since V=AH where A is the
area of our model cylinder, A=V/H=0.05
m^{2}.
QUESTION:
When light moves upward against gravity
it redshifts (and blue shifts if it were
relected back down). Let's say that the
shift is causes blue visible light to
actually go red. How does this work in a
photonic sense. Do blue photons turn into
red photons? If so, what has happened to the
energy, did it disappear? Energy can't just
disappear, photons have no mass, so it
hasn't become potential energy, where did
the energy go?
ANSWER:
When you throw a ball straight up in the
air it slows down, eventually stops, and
falls back down. In physics terms, the ball
loses kinetic energy on the way up and gains
it on the way down. Usually we "invent" a
potential energy function and consider it
the result of work done by the gravitational
force, negative work going up, positive work
falling down. In this case the ball's lost
(gained) energy is gained (lost) by the by
the gravitational field. In the case of a
photon, again energy of the photon is lost
on the way up and gained on the way down.
Again, the energy which is lost is gained by
the field and vice versa. A photon
cannot slow down or speed up because that is
a law of nature; its energy is given by
E=hf where h is Planck's
constant and f is the frequency of
the corresponding electromagnetic field of
which the photon is a "member". Thus the
red/blue shifts. To make this a little more
concrete, you probably know that light
cannot escape from a black hole so as it
moves away it eventually disappears
altogether; so the field has gained all the
energy of that photon and this results in
the source of the field, the black hole,
increasing its mass by the amount Δm=hf/c^{2}.
QUESTION:
Hi, my question centres around the
motorcycle community and that “loud pipes
saves lives”. For context if you are not a
biker, the premise is that if your bike is
very loud, you are more likely to be heard
by cars/drivers sooner and therefore
reducing your risk of them not seeing you
and crashing into you. The counter argument
to this is, as the exhausts are at the rear
of the bike the sound is lost out of the
back and therefore they won’t hear you until
you have reached or gone past the car. This
then is further argued that for this to
apply you will need to be travelling faster
than the speed of sound. I believe the
Doppler effect sits in this somewhere too
but I’m not knowledgeable enough to know. So
my question is; Can a motorcycle travelling
down a road at Xmph be heard by a vehicle
ahead of it and at what distance does the
vehicle ahead begin to hear the motorcycle?
I realise there are a number of variables
that would need to be factored in to make
here, (noise inside the car, speed
travelling, etc., but I guess that’s why I’m
asking the question
ANSWER:
I found a
website where this question was
addressed scientifically, not just
somebody's opinion. I could do no better.
The bottom line is that loud pipes do not
save lives. I might add that loud pipes do
annoy people who value a quiet environment!
I might also add to that gasoline leaf
blowers, lawn mowers, loud cars, etc.
QUESTION:
If I were travelling at 100mph in a
sealed unit ie a train with no air movement
and I dropped a cricket ball from directly
above a cross marked on the floor, would the
ball hit the cross? As the drop is approx. 8
feet would the result be the same if the
sealed unit/train was 100 feet tall. I
maintain the once the ball has left my hand
it no longer has direct contact with the
movement force and it would very slowly
loose forward momentum as while the
contained air is still moving at 100mph it
would not have the density/power/whatever to
overcome the weight of the cricket ball and
it's forward momentum would gradually
decrease. I ask this as someone says that no
matter from what height the ball was dropped
the forward momentum world remain exactly
100mph and therefore hit the cross. I am 72
and back when I was at school we did not do
this sort of maths(?).
ANSWER:
This is interesting because it is such a
common gedanken used to demonstrate the
Galilean relativity of twodimensional
kinematics. But never is it discussed what
the conditions are for it to be absolutely
true. Since the speeds involved are all much
less than the speed of light, I answer this
question considering only Newtonian
classical mechanics. The answer, as I will
show for the real world, is no, even from
8'. The conditions are:

The experiment must be performed in an
inertial frame of reference. That is,
the frame of reference must not be
accelerating and we are assuming that
the frame is the train itself.

The experiment must be performed in a
uniform gravitational field. The
direction and magnitude of the
gravitational force must be constant in
magnitude every everywhere.
These
are well approximated in the real world as
long as the distances aren't too large.
First, referring to #1, the train is on the
surface of the earth which is a sphere so
the train is moving with some speed v
in a circle which has the radius of the
earth R so it has a centripetal
acceleration v^{2}/R.
But, it is really more complicated than that
because this circular train track is
attached to the earth and the earth is
rotating on its axis. I won't go any
farther, but we could also say that the
earth is revolving around the sun, the sun
is revolving around the center of the
galaxy, etc. ad nauseum. The point
is that the frame of the train is not an
inertial frame; it is an excellent
approximation for most everyday situations.
Next
I will give examples of how these two
accelerating frames (circular track and
rotating earth) can affect where the dropped
ball lands. I have drawn a figure which
shows the train car and the ball on the
ceiling, the car being h meters above the
floor. But, if the car is traversing a
circle, the ceiling is moving faster than
the floor as shown in the figure by a factor
of (1+(h/R). The ball will
take some time to fall and during that time
the floor will move forward a distance v_{floor}t
and the ball v_{floor}t(1+(h/R))
so the ball lands v_{floor}th/R
farther forward than the X initially under
it. Of course for a normal train car h/R
is an incredibly small number and you would
be hardpressed to observe it. But if you
dropped it from an altitude of R,
the ball would go twice as far as the X on
the floor. A much larger effect, though, is
from the rotation of the earth and is called
the Coriolis force. It is beyond the scope
of this site to get into this force, but the
effect turns out to be that the falling ball
will be deflected in an easterly direction
by the amount d=(ωcosφ/3)√[8h^{3}/g]
where φ is the latitude where the
ball is dropped, ω is the angular
velocity of the earth's rotation (7.27x10^{5}
s^{1}), and g is the
acceleration due to gravity (9.8 m/s^{2}).
If you do the calculation for h=100
m, φ=45° you find that d=1.55x10^{2}
m=1.55 cm. Compare this with the deflection
of the ball dropping from 100 m and the
train going 100 mph=0.028 m/s, d=0.2
mm.
The
central part of your question, "…I
maintain…gradually decrease." is totally
wrong. You, on the train, see the ball with
zero horizontal velocity falling in
perfectly still air. An observer outside
sees the ball with some initial horizontal
speed the same as the horizontal speed of
the air. Air drag will have an effect on the
falling ball, increasing the fall time.
Finally, condition #2 that the experiment
must be conducted in a uniform gravitational
field. This is an exellent approximation
near the surface, but going away from the
earth the field decreases like 1/r^{2}.
I haven't thought about the impact this
would have the falling ball experiment, but
I think I have adequately shown that the
"falling on the X" part is wrong, so I will
save that for another day!
QUESTION:
In photoelectric effect, why do we deal
with "maximum" kinetic energy?
ANSWER:
When a photon strikes the surface of
metal it may interact with an electron in
the metal. Suppose that an energy W
is required to remove an electron from the
surface. Then, if a photon of energy E
interacts with an electron on the surface,
the electron is ejected with a kinetic
energy K and the electron will have
a kinetic energy K=E(E'+W)
where E' is the energy which the
photon has after the interaction. Now,
suppose that the photon gives all
its energy to the electron, E'=0,
and this will be the largest possible
kinetic energy the electron to have, K_{max}=EW.
QUESTION:
Hello!! I have a question that is
related to space but deals mostly with the
physics of orbital mechanics and free fall
objects. If you were to have a space station
that maintains a constant altitude above a
planet or stellar body, would you still have
gravity? Would that gravity be the same as
the gravity on the surface of the body or
would it be a portion of that gravity?
ANSWER:
Gravity is a force field which is caused
by the presence of mass. Because it is a
very weak force, a very large amount of mass
is necessary for us to be able to perceive
its presence. It is a field which, if caused
by a massive spherical object (planet, star, etc.), decreases as you get farther
away. Isaac Newton first discovered the
force F on an object of mass m,
a distance r from the center of a
spherical object of mass M and
radius R is F=GMm/r^{2
}where G is a constant. If
d is the altitude, d=rR, so
r=(R+d) and F=GMm/(R+d)^{2}.
I suppose that answers your questions—yes,
you "would you still have gravity"; no,
gravity would not "…be the same as the
gravity on the surface…" because it
decreases like 1/r^{2}.
There is, however, an important thing to
keep in mind—although there is gravity
there, you would not feel it because when in
orbit you are in free fall as gravity
provides the centripetal force to keep you
in orbit. Discussions of astronauts being
"weightless" when in orbit are
inaccurate—they still have weight but don't
feel it.
QUESTION:
This is a question that goes (I suppose)
to the fundamental nature of things and for
which you may think me the village
simpleton. But I'll give it a go. Let's
assume a single free photon traveling across
the universe at x frequency. What force,
what principle, what property keeps the
photon locked in its cycle of crest to
trough to crest again? What pulls it back to
center? Why can not it simply "break free"
of this cycle and fly off in a static state
to parts unknown. In other words, why? Why
does energy vibrate at all?
ANSWER:
The answers to your questions require
some pretty serious math/physics to answer
rigorously. But I will do my best to try to
give you a more qualitative overview. The
most important thing to keep in mind is that
trying to think of quantum phenomena in
terms of classical phenomena is unreliable
and nothing more than a means of trying to
understand in terms of our everyday
experience. You have some problems with
classical physics as well, so let's start
there.

"Why does energy vibrate?" Energy is not
a thing, it is a property which things
have. One kind of energy is kinetic
energy, the energy of motion. If an
object is moving in a straight line, or
spinning, or vibrating it has kinetic
energy; unless something interacts to
take some or all of that energy away, it
will never go away—that's called
conservation of energy. In fact,
the reason energy is useful in physics
is that physicists have defined it so
that it will obey a conservation law for
any "isolated" system. The energy of the
entire universe never changes.

"…what…keeps the photon locked in its
cycle [?]…" Think of a vibrating guitar
string: It will vibrate forever because
of energy conservation, right? Of course
not, because it is not an isolated
system. It is vibrating in air and as it
moves through the air the pushes on the
string opposite the direction the string
is moving, slowing it down just like a
wind will push on you. Where did that
energy go? The air, if you had a very
sensitive way to measure its
temperature, would warm up just a tiny
bit. Another way the energy gets lost
(from the string) is that the string is
attached to the bridge which is attached
to the body, so the string gives up some
of its energy to the whole guitar as it
starts vibrating too. But that vibration
sends out sound waves which flow away
carrying more of the enrgy with them. If
you just stretched the string across two
nails on a rod it would vibrate much
longer and you would hardly be able to
hear it because because energy is not
"leaking" to the guitar.
Now, back to the
photon. The first thing is that it is not a
particle or a wave, it is a
particle and a wave. This,
particle/wave duality, is hard to swallow
since there is no really good analogy in
classical physics. If you look for a
particle, you will find a particle; if you
look for a wave, you will find a wave. Just
like an ideal guitar string, it needs
nothing to keep it in its state. A photon is
the tiniest amount of a ray of light that
can exist. It has energy which is
proportional to the frequency of the light
which it is part of. You just get confused
if, as you have done, you try to think of it
vibrating with that frequency. If it is in
empty space and nothing interacts with it,
energy conservation just keeps it moving
along with a certain kinetic energy and
linear momentum (also a conserved quantity).
It has no "static state" and cannot "break
out" without violating energy conservation.
However, if it happened to encounter an
electron it would interact with it and some
of its energy might be given to the
electron.
QUESTION:
I have a question about defining types
of heat transfer. Me and a friend got into a
debate about wether the so called "radiant"
heat in the concrete floor that has heated
fluid pipes running through it, is in fact
radiant or convection type. Since the
temperature of the air in the room is heated
by this method i say it was convection
occuring. But its always called "radiant"
heat when we have heated slab floors. Is the
heat emitted from this slab in the form of
radiant waves then, or is the parcel of air
itself in contact with the slab heated and
then convects outward via the air itself?
What type of heat are we talking about here?
ANSWER:
Both play a role in heating the room. If
no air were in the room, the ceiling and
walls would eventually warm up because of
the radiant energy from the floor. But if
there is air in the room, the radiant energy
would heat up the air but not uniformly so
currents of air, convection, would occur.
QUESTION:
I have a 3.7V and 600mah battery, which
has a big bms that can handle a high current
before rendering a short circuit, but now I
have 2 3.7V and 600mah batteries, the
problem is their bms can't hand alot of
current like the last one, and my device
shuts off when i try it on. so can't I
connect these 2 batteries in parallel with
the bms of the first battery, so the device
doesn't turn off, and it gets a better
battery life time?
ANSWER:
I don't know anything about battery
management systems (bls) but I do know DC
circuits, so I will take a shot at your
question. For illustration, let's say that
your first battery can handle a current of 3
A and the other two can handle 1 A; suppose
your device draws 3 A. If they all have the
same the same internal resistance and you
connect them, each will have 1 A flowing
through them and two of the three are right
at the limit before they shut off. If,
however, the two have a resistance smaller
than that of the third, they would need to
have more than 1 A and, I presume, the bms
would shut them down.
QUESTION:
This is a question about sound. If you
were to increase the temperature of a medium
without allowing the volume (of the medium)
to change, how would this affect soundwaves
that travel through it? I've tried
researching this but most of the answers I
get refer to how the change in temperature
changes the density of the medium, which is
not what I am interested in. I want to know
what happens if you prevent the density from
changing. I know such an increase in
temperature would increase the pressure, but
I don't know if that would make the sound
louder or change it at all.
ANSWER:
This question depends a lot on the
situation you have—is it a gas or liquid or
solid? What is it made of? What is its
molecular structure? But you are lucky if
you want to consider a gas which is well
described by the ideal gas law, PV=NRT.
Volume V is constant, N
measures the amount of gas, also constant,
T is absolute temperature (your
variable), and R=8.31 J/mol·K is
the universal gas constant. The sound has a
frequency which will not change but the
speed v will so the wavelength of
the sound will change. It turns out that the
speed depends only on the temperature and
what the gas is. I will not go through the
whole derivation because it requires a lot
of facility with thermodynamic theory, but I
will give you the final result:
v=√[γRT/M]
where
M is the molecular mass of the gas
and γ is the adiabadic index, the
ratio of the specific heat at constant
pressure to the specific heat at constant
volume, C_{P}/C_{V}.
For example, for air, γ=1.4
and M=0.029 kg/mole, so v=20.03√T.
Don't forget that T must be the
absolute temperature, K. I get v=347
m/s at T=300 K.
QUESTION:
The Bohr model correctly interpreted the
atomic spectra of the hydrogen atom and
predicted that the excited atom would emit a
photon whose energy is equal to the energy
difference between two specific levels in
the atom.... After that, quantum mechanics
appeared and the Schrödinger equation for
the hydrogen atom was solved Bohr model
becomes incorrect (or inaccurate, or
incomplete) From the Schrödinger equation,
we learned that the electron does not
revolve in a specific orbit around the
nucleus, but its distance from the nucleus
is a potential, that is, it is not a
specific orbit. The question is here: How is
the energy of the photon emitted from the
atom (a specific energy that is responsible
for the distinctive atomic spectrum) equal
to the energy difference between two orbits,
even though the energy of each orbit is
indefinite?? For example, the energy of the
primary orbital in the hydrogen atom,
according to Bohr = 13.6 ev And the energy
of the second orbit = 3.4 ev, so the
difference = 10.2 ev But in the Schrödinger
equation, the electron in the first level
can take energy greater or less than 13.6.
It is true that the average energy in the
Earth's orbit is equal to this value, but
there are possibilities for it to be greater
than even twice this value or less, so how
is the difference between the two levels
always constant = 10.2. This is proven
practically and experimentally through the
atomic spectrum of the hydrogen atom.
ANSWER:
You have many misconceptions. The Bohr
model is semiclassical and in most respects
just plain wrong. Nevertheless, it played a
very important roll in the development of
modern physics. Just to list a few errors:

Electrons are not simply particles.

The electrons do not move in circular
orbits.

The orbital energies are not determined
by quantizing the angular momentum of
the orbit.

The values of the angular moment are
wrong. For example, the ground state has
zero angular momentum so to think it as
an orbit is dead wrong.

All states are assumed to be stable so
their decay to lower states is not
explained.

It
is empirical and makes sense only in
that it reproduces experimental spectra.
You
have a problem with energies. The zero of
the electric potential of a point charge is
chosen to be infinitely far away. This is
sort of standard of atomic or nuclear
physics. The energy of the ground state is
E=13.6 eV. Negative energy means
the electron is bound, so 13.6 eV is the
ionization energy, the energy you must
expend to free an electron in this state. In
quantum mechanics we find that the electron
is not a particle but simultaneously a
particle and a wave (waveparticle duality).
The solution of the Schrödinger equation for
the hydrogen atom yields not some orbit for
an electron, but a function (wave function)
which is interpreted as a probability
distribution; for example, for a small
volume ΔV=ΔxΔyΔz
at some point (x,y,z) where the
wave function is Ψ(x,y,z),
Ψ^{2} is the probability of
finding the electron in that volume. But the
solution has a definite energy (eigenvalue)
for each state, not an indeterminate energy
as you seem to think. Finally, quantum
mechanics gives us the uncertainty
principle: certain pairs of observable
quantities can't be known to arbitrary
accuracy. The pair of interest to us is
energy and time—the better you know time,
the less well you know the energy. The
ground state of lives forever and so its
energy is exactly 13.6 eV; but the excited
states have definite lifetimes so their
energies are uncertain. In principle, if you
measure the energy of a line in the hydrogen
spectrum you would find that it has a
spread, the shorter the lifetime the larger
the spread. This spread is very small and
also often swamped by other effects: most
spectra come from tubes where the atoms are
moving very rapidly and you get a spread due
to Doppler shifts of atoms moving toward or
away from you. I have no idea what you are
talking about regarding the earth's orbit.
QUESTION:
We have an aboveground pool that
contains roughly 8400 gal of water. It's
dimensions are 25ft long x 16ft wide x 52
inches deep. Our question: Is it possible to
hook a garden hose to the bottom drain
outlet of the pool, and get the water to
rise up to the level of the pool edge (52")
and then drain back down into the pool,
*without the aid of a pump*?
ANSWER:
I was following everything in your
question until I got to "…and then drain
back down into the pool…". I am going to
assume that you want to connect the hose at
the level of the bottom of the pool and fill
the pool that way and that is what I will
answer. The key is that if the pressure of
the hose is less than the pressure at the
bottom of the pool when it is full, it will
not fill. The typical water pressure in a
household is in the range of 4080 psi, so I
wall assume yours is 60 psi as an example.
The pressure at the surface of your pool
when it is full is atmospheric pressure,
14.7 psi. As you go down into the water the
pressure increases at a rate of 0.0371 psi
per inch. So the pressure at the bottom of
your filled pool is 14.7+52x0.0371=19.3 psi.
This is smaller than 60 psi so you can fill
your pool this way. On the other hand, why
would you want to? It would probably be
faster to fill it from the top because the
rate of flow through the hose is bound to
decrease as the back pressure on it gets
larger as the pool fills.
QUESTION:
Suppose a compression spring is placed
between two nearby identical bodies in
space. The stiffness and natural length of
the spring are accurately known. The
gravitational attraction between the bodies
then compresses the spring slightly, and the
compressed length allows the bodies’ rest
mass to be estimated. The spring’s
compressed length should be the same for an
observer moving at high speed at right
angles to the spring, so the mass should be
seen to be the same as before. But using SR
the observer knows this observed mass
exceeds its rest mass. The rest mass would
be less than the mass measured in its rest
frame. So how is this consistent with
invariant rest mass?
ANSWER:
The answer is simple. The spring
constant is not invariant, it depends on
your frame of reference. Related to this is
the realization that force is not really a
useful concept in special relativity.
QUESTION:
Suppose a rocket flies at high speed
toward a stationary laser and carries
equipment to measure the frequency of the
laser’s light. The speed of the rocket is
known and hence so is the nonrelativistic
Doppler shift. After allowing for this shift
the measured frequency of the light should
be slightly blue shifted. This is because
the time rate of the atoms in the rocket
will be slightly reduced due to its high
speed, so in comparison the frequency of the
laser’s atoms will be higher.
ANSWER:
You are making this too difficult. To
calculate the frequency of the radiation
observed on the rocket you only need to know
the frequency in the laser's rest frame;
the way the radiation was created is
irrelevant. And I do not know why you ever
mention the nonrelativistic Doppler shift
because it is incorrect for electromagnetic
radiation. The correct Doppler shift
equation is λ_{rocket}/λ_{laser}=√[(1+β)/(1β)]
where where β=v/c,
v the relative speed and c
the speed of light; this equation has β>0
for the rocket and the laser moving away
from each other. In your case β<0.
For example, if β=0.7, λ_{rocket}/λ_{laser}=0.42,
blue shift.
QUESTION:
The double slit experiment using a
series of particles raises a question: Why
does the electron or photon gun not shoot
straight? Assuming the gun is positioned
between the two slits, it should just hit
the same point on the barrier every time.
CRT’s had accurate electron guns, why not an
accurate gun for this experiment? Can you
describe the nature of the trajectory
variance? Cyclical, random or something
else?
ANSWER:
You miss the whole point of doubleslit
diffraction. Diffraction happens for waves
but a "pure" particle (if there were such a
thing) does not diffract. Any elementary
particle, e.g., electrons and
photons, are both particle and wave—if you
look for a wave you will find a wave, if you
look for a particle you will find a
particle. This is called waveparticle
duality. If you look for wave behavior of
electrons you need to look for diffraction.
The
ruleofthumb for whether diffraction is
important is that the slit or obstruction
not be much larger than the wavelength of
the wave. E.g., the wavelength of
light in the middle of the visible spectrum
is about 5x10^{7} m; the slits in a
good diffraction grating are spaced by about
5x10^{6} m and the diffraction is
easily seen. But what if the slits had been
one meter apart? In that case you would be
hard put to see any diffraction and the
photons could be carefully shot straight
into the gap. There would be singleslit
diffraction from the two slits, but let's
just say they were each a centimeter wide;
if you looked really hard you might find a
little fuzziness of the shadows of the edges
of the slits which would be diffraction.
Now
let's turn to diffraction of electrons. The
wavelength of an electron accelerated by a
voltage of 1 kV (which has a speed of about
1/10 the speed of light) is about 4x10^{11}
m. So your slits would need to be not too
much bigger than that. But you would not be
able to fabricate slits that close together
because the
size on an atom is about 10^{10} m!
So the pictures you see in text books of
electrons and a double slit are a fraud! No
such slits exist. But that slightofhand is
just to get students understand one wave in
terms of some other wave you already
understand without going into all the
technical details of how you can really do
it. There is no doubt that electrons behave
as waves. The first experiment to observe
electron diffraction was the
DavisonGermer experiment; they directed
an electron beam onto an iron crystal, using
the regular spacing of the iron atoms as a
kind of diffraction grating. So, protons,
having much larger mass than electrons, have
much smaller wavelengths. An
earlier answer showed the results for
800 MeV protons scattering from a ^{90}Zr
nucleus. Protons of this energy have momenta
of about 10^{15} m, so to see
diffraction you need a target not too much
larger than that size; the size of nuclei is
about this size. The results are shown in
the graph which shows the intensity of the
proton diffracted from the nucleus. For more
detail, read the answer linked to
here.
This
now answers your question about CRT
electrons not having any problem hitting a
small dot on the screen—the dot is really
huge compared to the electron wavelength.
QUESTION:
How does touch work from a physical
perspective? From a biological perspective
we know that mechanoreceptors carry impulses
along nerves to the central nervous system.
For examples when my fingers touch the
keyboard of my laptop the mechanical
pressure is converted into electrical
signals sent to my brain and I feel the
keyboard. From a physical perspective one
might argue that we never touch anything.
The cloud of electrons buzzing around the
atomic nuclei comprising the atoms of my
fingertips interacts with the the electrons
of the keyboard, virtual photons convey the
information of the electromagnetic field,
but there is no touching involved. The
electrons do not ”touch” each other. Though
there is no touch involved perhaps we might
say that electrons ”feel” forces (and
accumulate energies from them) whenever
there are electromagnetic fields. My
question is: What is going on at the atomic
level when we touch something? How does the
interaction between the electromagnetic
fields of my electrons and the electrons of
the keyboard turn into an electrical current
strong enough for my brain to detect it?
ANSWER:
I have recently
answered a question which touched on the
microscopic origin of contact forces. It
turns out that the contact force is mainly
due to a quantum effect, not to
electromagnetic forces between atoms on the
surfaces.
Wikepedia states that "…in quantum
mechanics the Pauli exclusion principle
(PEP) states that two or more identical
particles with halfinteger spins (i.e.
fermions) cannot occupy the same quantum
state within a quantum system
simultaneously…". In other words, when two
objects get close enough, the wave functions
of atoms on the surfaces begin to overlap
and the electrons in the atoms begin to be
approaching the same state and atoms on both
surfaces stop getting closer to avoid
violating PEP. It isn't really a force but
the effect is the same—the surfaces push
away from each other. The result will be a
compression just below the surfaces which
will then be detected by the nerves but, not
being a biologist, I couldn't give you any
useful explanation how the nerves
work. Regarding whether the objects actually
"touch" requires, at atomic distances, a
definition of what we mean by touch. I would
say that if wave functions of atoms overlap
that they are touching.
QUESTION:
I am confused and need help on
understanding forces acting on a car when it
is moving uniformly on a level road. When a
car is moving uniformly (say forward) on a
level road, what is the direction of
friction acting on it? Some say it is to the
left, because some textbooks would describe
the above situation as a having a driving
force pointing forward, and friction
pointing backward. But then I also have read
some online posts, saying that the socalled
"driving force" is actually friction. So the
friction should point forward. Also, I came
across with something called "rolling
resistance". But I am not sure how I could
include it in my understanding. While this
question may seem simple, it really confused
me a lot because I am not sure how much
textbook situations deviate from real life.
ANSWER:
When you say "…the friction…"
you imply that there is only one frictional
force acting on the car. In the case of a
car going with constant speed on a straight
road, there are three important kinds of
friction acting:

The tires are constantly getting
slightly deformed at the bottom and
deforming them requires a force. This is
called rolling friction and is what is
mainly responsible, at low speeds, why,
if nothing is pushing the car, it will
roll to a stop eventually. The road
exerts a force on the tire and the tire
exerts an equal force on the road.
(Newton's third law.) This force acts
backwards.

A
second force is static friction which
causes your "driving force" to impart a
forward force on the tire. The mechanics
of your car causes there to be a torque
on your wheels which tries to cause them
to rotate counterclockwise (as seen from
the side). In fact, if the road were
frictionless (icy, for example) the tire
would spin and not cause the car to move
forward. This static friction also keeps
your car from slipping if the road
curves; in that case, the static
friction will have a component in the
direction you are going and a component
toward the center of the curve you are
negotiating.

Finally there is air drag which exerts a
force on the car opposite its velocity.
It becomes more important as the speed
increases, increasing quadratically (e.g.
3 times more speed gives 9 times more
drag.)
You
might wonder why static friction could drive
the car forward—the car is certainly not
static. But where the "rubber meets the
road" is certainly static.
I hope
this clears it up for you.
QUESTION:
I am in high school and I conducted an
experiment that involved a parallel spring
systems, mass and vertical displacement. The
system had an equivalent spring constant of
around 80N/m. I noticed that a 200g mass
produces close to 0 displacement, but the
difference in displacement in between 1000g
and 1200g is close around 23cm. So why does
the increase from 0200g so insignificant
compared to the increase from 1000g2000g.
According to hooke's law shouldnt this
similar increase in force result in the same
displacement?
ANSWER:
The problem here is that Hooke's law is
not really a law, it is an approximation
which works very well in some situations and
not for all situations. For example, suppose
you hung a 1 metric ton mass (1000 kg), a
9800 N weight, from your spring; would you
expect it to stretch 122.5 m? Springs in the
real world are approximately linear for a
limited range of applied forces. If I were
you, I would make more measurements by
adding masses in 200 g (m=0.2 kg) increments
making careful measurements of the stretch
and then plot the stretch as a function of
the weight (W=mg); this should
yield for small masses the slope of the line
is smaller than 80 N/m for small W
but becomes 80 N/m. It might also be
interesting to then take weights off to see
if it traces back on the stretching data; if
it doesn't, that is called hysteresis.
QUESTION:
A motorcycle has to move with a constant
speed on an overbridge which is in the form
of a circular arc of radius R and has a
total length L. Suppose the motorcycle
starts from the highest point. What maximum
uniform speed can it maintain on the bridge
if it does not lose contact anywhere on the
bridge? My Trouble: I found the answer in
internet saying the Normal force acting on
motorcycle will be zero. But I am not
getting this part that Why and How will the
Normal force be zero? Isn't Normal force a
repulsive (electrostatic) force between
atoms of two bodies in contact, then why
will it be zero!?
ANSWER:
First of all, it is not at all helpful
to think of the microscopic origins of
normal forces in classical mechanics
problems like this one. Also, almost every
physics book you will read will say that it
is the result of electromagnetic forces, but
this is incorrect; the origin of this
force is mainly due to quantum effects—the
Pauli exclusion principle—as first shown by
Freeman Dyson. All we need to say is
that when two objects are in contact with
each other there are forces that each exerts
on the other which, by Newton's third law,
are equal and opposite. As is often done
with vectors, it is convenient to resolve
this force into two components, the force
tangent to the surface of contact (usually
called the frictional force F)
and the normal component (usually called the
normal force N).
See my figure. So the motorcycle in this
problem has two forces acting on it, its
weight W and the
contact force which has components
F and N.
There is zero acceleration in the xdirection
so
FWsinθ=0,
F=mgsinθ.
There
is a centripetal acceleration in the
negative ydirection, a_{y}=v^{2}/R
mv^{2}/R=NWcosθ,
N=m(gcosθv^{2}/R).
Some
things to note:

Both N and F are
functions of θ.

F is positive on the way up but
negative (braking) on the way down. It
is zero at the top.

N is always positive.

If
gcosθv^{2}/R≤0,
N becomes zero or negative; for
a motorcycle, N can't be
negative because this particular contact
force can only push, not pull. This does
not mean that the normal force can never
pull down. What if the road were made of
iron and the motorcycle carried a very
powerful magnet? Also, roller coasters
are not just rolling on wheels but have
to have a mechanism such that if you
were to stop at the top of an inside
looptheloop you would not crash to the
ground. What does it mean if N=0?
It means that the motorcycle is no
longer on the road; once N=0
the only force acting on the motorcycle
is its own weight and so it becomes a
projectile; F will disappear
also so its speed will no longer remain
constant; this happens all the time in a
motocross race.

For a given angle where N=0,
what is the speed v? v=√(Rgcosθ).
That speed is greatest when when
θ=0°. So we finally come up
with the answer to your question, v_{max}=√(Rg).
QUESTION:
I'm finding there are 2 seemingly
contradictory concepts when it comes to air
pressure and temperature. The first is that
when temperature rises, it increase the
pressure of the air because the molecules
move more quickly causing them to hit the
walls of their container more forcefully and
more often. The second is that colder air
has more pressure because the molecules come
closer together, making it more dense which
results in it having higher pressure.
ANSWER:
The problem is that you are looking too
narrowly. There are other variables besides
pressure P and temperature T.
We should start with the simplest behavior
of a gas, the ideal gas law, PV=NT
where V is the volume of the gas
and N is some measure of the amount
of that gas. Usually, if we want to find the
relation between two of these four
variables, we hold the other two constant.
If we do that for P and T
we obviously find that pressure increases
linearly with temperature, P∝T,
your first "concept," but it should have
been modified by "…if N and V
are held constant…". It is now clear that
your second "concept" cannot also be true
under the same proviso. So let's hold just
N constant (don't add any new air
to V); so now P∝T/V.
So your second "concept" is true only if the
rate which T decreases is smaller
than the rate at which V decreases;
in other words, the pressure will increase
in a cooling gas if the volume is
sufficiently compressed.
QUESTION:
We are working on a gravity energy
system. How much electrical output (kWs) can
we expect from a 50,000 lbs weight, falling
at a rate of 6 inches per second, and what
the kWs would be if the same 50.000 lbs was
falling at a velocity of 1 foot per second?
ANSWER:
First, let me be a bit of a scold. A
Watt is a unit of power, energy per unit
time, whereas a Wattsecond is a unit of
energy, force times distance. You give me
enough information to tell you the power but
not the energy where I would have to know
the time the weight fell. Also, a Watt is a
Joule per second (J/s) or a Newtonmeter per
second (N·m/s); but you use Imperial units
(inches, pounds) so I will have to first
convert these to SI units. End of scold! The
mass of 50,000 lb is m=22,680 kg,
so the weight is 22,680x9.8=2.22x10^{5}
N; 1 ft=0.305 m. So, since it falls 0.305 m
in 1 s, the energy per second is 2.22x10^{5}x0.305=67.8
kW. So, you could get this much power for
however long as it fell. If it falls for 100
ft, you would deliver this power for 100
seconds, a little less than 2 minutes. Of
course if you correct for friction losses it
would be less than that. But more
importantly, I hope you have a cheap and
easy source to pull it back to the top.
QUESTION:
I was asked this question many years ago
and passed it on without results: A30 foot
wooden freestanding pole on a concrete
platform is leaning 10 degrees from vertical
and held in place by a wire. The wire
breaks, how many seconds does it take for
the pole to fall to the horizontal ground?
Don’t loose sleep over it. Greetings from
Klaus a 87 yr old forgetful physicist
ANSWER:
This is basically the simple (physical)
pendulum problem where Newton's second law
results in d^{2}θ/dt^{2}
∝ sinθ. But you will recall that
this differential equation can only be
approximately solved for θ<<1 and
the solution is the simple harmonic
oscillator. When you and I were students
solutions of the largeangle simple pendulum
were very tedious employing power series or
sometimes obscure special functions. Today
one could write a simple computer program to
solve the equation numerically pretty
easily.
QUESTION:
If Entangled particles separated over
great distances act simultaneously, then the
"force" binding them must act faster than
light speed? If a particle was on earth and
entangled with a particle in the sun (for
example) then interacting with the earth
based particle would generate an immediate
effect on the sun based particle. But if the
force / energy that binds the two moved at
the speed of light, then it would take 8
minutes for one to impact another, so does
this force move faster than light speed? And
how does this reconcile with Einstein?
ANSWER:
You at least partially understand why
Einstein referred to entanglement as "spooky
action at a distance". The particles need
not be bound and there is no "force" which
is somehow keeping track of the other
particle. The entangled particles have been
prepared so that they are, in essence, a
single quantum system. Although we would
feel that there was something
"communicating" between the two particles,
it has been proven that whatever it is
cannot be used to send information.
QUESTION:
Why do 2 spheres hung by a light string
with the same mass but different charges
(+ve) make the same vertical inclination
with the string?
ANSWER:
Because of Newton's third law: If object
A exerts a force on object B, then object B
exerts an equal and opposite force on object
A.
QUESTION:
In a rigid body like a coin, if an
impulse is given along its centre of mass
then we calculate the change in momentum by
multiplying mass and velocity change of the
body.then energy is it kinetic energy due to
velocity of its centre of mass. but if
impulse of same magnitude is applied not
along the centre of mass then we get both
linear and angular velocity. But linear
velocity is same as in the first case. But
if we observe the energy then we get both
linear kinetic energy and angular kinetic
energy. How is this possible? How can we get
extra energy without giving more impulse to
the rigid body?
ANSWER:
This is a little tricky because, as you
state, an impulse of J
causes an increase in kinetic energy, so if
you apply J
anywhere on the body you should get the same
increase in kinetic energy. (I am assuming
no external forces here so we need not worry
about potential inergy.) So, if I apply a
force, the direction of the vector passing
through the center of mass of a rigid body
of mass m, I can write
J=mv_{cm
}where v_{cm}
is the velocity of the center of mass. So
the energy resulting from the impulse is
E=½mv_{cm}^{2}.
In this situation, all energy goes into
translational energy. But if we exert
J elsewhere, it
now also exerts a torque about the center of
mass which causes the object to also acquire
a rotational kinetic energy. Now,
J=mv'_{cm}+Iω/d
where ω is the
angular velocity about the center of mass,
I is the moment of inertial about
the center of mass, and d is the
moment arm of the torque about the center of
mass. I believe this answers your question
because clearly v_{cm}>v'_{cm}
so the translational energy gets smaller. A
very useful theorem of classical mechanics
is that the kinetic energy of a rigid body
is equal to the kinetic energy of
the center of mass plus the kinetic energy
about the center of mass, so
in this case, ½mv_{cm}^{2}=½mv'_{cm}^{2}+½Iω^{2}.
QUESTION:
is there an equation that could define
the oscillatory motion of a screw as it
rolls down a ramp
ANSWER:
See an earlier
answer.
QUESTION:
Hi, I was in my house the other day at
night when I noticed a bizarre light effect
coming through a window sun blind. Looking
across from the hill I live on are street
lamps. When the light of these shined
through the netting I was perceiving what I
could only determine to be an interference
pattern. A point of light was split into 3
definite points. It didn't matter how far
from the netting I was and if I moved my
head side to side the pattern did not
change, the spacing remained specific. I
will be happy to send a video of the
phenomena.
ANSWER:
The questioner, at my request, sent a
video and a picture illustrating grid size.
I have taken a screen shot of the pattern.
The pattern certainly looks like a
diffraction pattern. But, on closer
inspection, there are some issues. The first
thing that comes to mind is that a
streetlight is normally white light and
seeing a diffraction pattern from white
light is not the familiar pattern of bright
and dark spots because the pattern for a
particular grating depends on the wavelength
of the light so all the wavelengths from
that white light would overlap. On the other
hand, some street lights are sodiumvapor
lamps and sodium has two very close very
bright lines (the sodium Dlines) with
wavelengths a little less than 600 nm=0.6 μ
(microns); this wavelength has a color which
is yelloworange, and so is the pattern we
are examining. Now, looking at the the
gridsize picture, I would estimate that the
holes are on the order of a quarter of a
millimeter, 0.25 mm=250 μ, with a spacing of
about 0.5 mm=500 μ. There is a general rule
of thumb that for diffraction to be
significant the sizes of the diffracting
object should not be big compared to the
wavelength of the light, and this grid has
features almost three orders of magnitude
bigger than the wavelength of the light if
it is from a sodium lamp. If it is not a
sodium lamp, I have no idea why it is the
color it is. So I have no definitive answer
to the question. My best guess is that the
pattern is simply the pattern of light
passing through the mesh, essentially just
the projection of the mesh pattern and that
the source is a sodium lamp.
ADDED
THOUGHT:
I was just proofing
this answer before I posted it. The question
included "…A point of light was split into 3
definite points. It didn't matter how far
from the netting I was…". If this was the
case, it is definitely not diffraction which
would spread as you moved farther away. It
was worth doing all the work, though,
because it illustrates how scientists try to see all the angles of a problem,
not just jump to a quick conclusion.FOLLOWUP
QUESTION:
Thanks for taking
the time to do a thought experiment on this
puzzling phenomena. I would like to
challenge the concept that it is the grid
mesh projecting through. If this was the
case, then depending on the angle the grid
should move, my hand is not steady in the
video, but the grid pattern does not move.
As the camera pans surely the pattern should
move respectively and the central bright
spot would get split as a fibres moved
across it? The same pattern is present on
both white and yellow sources. Could it
possibly be a diffraction through the actual
strands if they are translucent nylon,
similar to a rainbow diffraction through a
water droplet? I have attached another image
I noticed a couple of days ago, this shows a
window from a long distance away reflecting
sunlight through the mesh. We can see a
splitting of the light into its constituent
colours. Hopefully this adds some useful
information? ANSWER:
First, a little correction:
The "rainbow" phenomenon is refraction,
not diffraction; that, in fact, may
be the best clue yet to the final answer to
your question. Let me argue one more time
against diffraction. I think I have done
about as exhaustive an analysis as I can,
given what I have. Maybe if I had a piece of
the material and could do experiments on it
I could learn more. One question: The
pattern—is it ON the mesh itself?
Or can the light be projected on a screen of
some sort? If it is something just on the
mesh, it is not diffraction. If you have
adequate intensity and could get it to
project onto a screen, the pattern would
spread more as the screen got farther from
the mesh. Just to get a better grasp of how
the diffraction can be expected to look we
can compare the spacings W of the
between maxima or minima for a double slit
diffraction pattern, W=λL/d
(see my figure), where λ is
the wavelength of the light, ~0.40.7 μ for
visible light. This is just a rough
approximation to spacing we would expect,
but one which an excellent
orderofmagnitude estimate of spacings in
any multiple appatures array. First, it
depends on how far away you put a screen; If
L is 2 m away, the spacing is twice
what it is when L=1 m. Taking a
typical visible wavelength to be 0.5 μ,
W=0.5x1/500=1 mm at L=1 m.
Somewhat to surprise, this is not at odds
with your observation. But it is not
definitive because the spots must get
farther apart as you move away from the
mesh. If you find that the spacing remains
the same, relative to the grid, regardless
of the distance you are from the grid, it is
not diffraction. I do like your idea that
perhaps refractions of the light passing
through translucent fibers is responsible.
So I guess I still stick with my original
conclusion about what we are seeing.
QUESTION:
Let we have the result of a physical
measurment like this: "Experiments have
Dirac’s number at 1.00115965221 with an
uncertainty of about 4 in the last digit" Is
this gives any interval that we could say
certainly Dirac's number is within it? I
mean exact certainty like 3<π<4. If
so what is smallest such an interval?
ANSWER:
The answer to your question hangs on
what you mean by "about 4". Normally we
would say what your question says but
without the 'about' which would be denoted
by 1.00115965221(4) or
1.00115965221±0.00000000004. But the "about"
indicates that there is uncertainty in the
uncertainty, and I don't really know how to
notate that.
QUESTION:
Does it require more nuclei to fuse the
heavier elements and is it the reason for
why the heavier elements are more rare?. For
example it takes 4 hydrogen nuclei to make
one helium nuclei so it takes more helium
nuclei to create the next element.
ANSWER:
It is not fruitful to think of fusion as
how many nuclei are fused to make a
particular nucleus because there are so many
ways any element could be made by fusion
reactions. For example, you could make ^{
12}C by fusing two ^{6}Li or
three^{ 4}He. What needs to be
considered is the energetics of fusion, can
you make a new nucleus by fusing lighter
nuclei? At the beginning of the existence of
the universe there was almost nothing but
hydrogen, a little helium, maybe a bit of
lithium. Now, if enough of these primordial
elements become bound by gravity, they will
eventually collapse to the point where
fusion ignites and you have a star. Why do
stars shine and get very hot? Because fusion
results in huge amounts of energy being
released as it happens. As the star gets
older and older the average nucleus is
getting heavier and heavier. But now there
is a "barrier" for this process to continue
forever until you have one giant nucleus: it
turns out that when you reach iron, fusion
happens only if you add energy, that is it
is no longer a source of energy when heavier
nuclei fuse. So the earliest stars had
almost no elements in them heavier than
iron. Then why are there elements heavier
than iron in the universe? It is because
when the a star is near the end of its life
it ends with a terrific bang called a nova
which creates an enormous energy which
forces fusion to continue as the star is
exploding creating many new elements. After
the nova, all the debris along with hydrogen
clouds coalesce and make new stars and those
stars have heavier stuff like lead and gold
and silver etc. in them. This is
why Carl Sagan liked to say "We are all made
of star dust".
QUESTION:
Why is Chernobyl still a toxic wasteland
and Heroshima and Nagasaki populated and
safe today?
ANSWER:
An atomic bomb and a reactor meltdown
are very different events. A bomb carries a
relatively small amount of fissile
materials. For example, Fat Man, the
Nagasaki bomb had about 14 lb. of
plutonium239; about 2 lb. actually
fissioned in the explosion. Since longlived
radioactive isotopes from the bomb come from
those two pounds, there is not really all
that much. Furthermore, that bomb was
detonated at 1650 feet altitude, the
radioactive fission products would have been
spread over a pretty large area. Most of the
casualties were due to burns, pressure
burst, flying debris, collapsing walls and
roofs, etc.; only those in a relatively
small area directly under the explosion were
injured of killed because of direct
radiation (gamma rays, x rays, neutrons).
A
large nuclear reactor like at Chernobyl has
a core of about 100 metric tons of uranium
when first installed and as that uranium is
used up it is converted to many radioactive
isotopes, both shortlived and longlived,
which are still confined in the cylinders
the core is composed of and highly
radioactive. At Chernobyl a combination of
human error, design faults, and mechanical
failures caused the core to overheat, the
resulting steam at too high a pressure to be
contained, and the steam containment
exploded sending out huge amounts of
radioactive steam to escape. Since the water
also cools the core, the core overheated,
burned, and melted and the explosions spread
all the nuclear waste over the adjacent
land. A whole different orderofmagnitude
from a bomb.
QUESTION:
Does it require more nuclei to fuse the
heavier elements and is it the reason for
why the heavier elements are more rare?. For
example it takes 4 hydrogen nuclei to make
one helium nuclei so it takes more helium
nuclei to create the next element.
ANSWER:
It is not fruitful to think of fusion as
how many nuclei are fused to make a
particular nucleus because there are so many
ways any element could be made by fusion
reactions. For example, you could make ^{
12}C by fusing two ^{6}Li or
three^{ 4}He. What needs to be
considered is the energetics of fusion, can
you make a new nucleus by fusing lighter
nuclei? At the beginning of the existence of
the universe there was almost nothing but
hydrogen, a little helium, maybe a bit of
lithium. Now, if enough of these primordial
elements become bound by gravity, they will
eventually collapse to the point where
fusion ignites and you have a star. Why do
stars shine and get very hot? Because fusion
results in huge amounts of energy being
released as it happens. As the star gets
older and older the average nucleus is
getting heavier and heavier. But now there
is a "barrier" for this process to continue
forever until you have one giant nucleus: it
turns out that when you reach iron, fusion
happens only if you add energy, that is it
is no longer a source of energy when heavier
nuclei fuse. So the earliest stars had
almost no elements in them heavier than
iron. Then why are there elements heavier
than iron in the universe? It is because
when the a star is near the end of its life
it ends with a terrific bang called a nova
which creates an enormous energy which
forces fusion to continue as the star is
exploding creating many new elements. After
the nova, all the debris along with hydrogen
clouds coalesce and make new stars and those
stars have heavier stuff like lead and gold
and silver etc. in them. This why
Carl Sagan liked to say "We are all made of
star dust".
QUESTION:
Why is Chernobyl still a toxic wasteland
and Heroshima and Nagasaki populated and
safe today?
ANSWER:
An atomic bomb and a reactor meltdown
are very different events. A bomb carries a
relatively small amount of fissile
materials. For example, Fat Man, the
Nagasaki bomb had about 14 lb. of
plutonium239; about 2 lb. actually
fissioned in the explosion. Since longlived
radioactive isotopes from the bomb come from
those two pounds, there is not really all
that much. Furthermore, that bomb was
detonated at 1650 feet altitude, the
radioactive fission products would have been
spread over a pretty large area. Most of the
casualties were due to burns, pressure
burst, flying debris, collapsing walls and
roofs, etc.; only those in a relatively
small area directly under the explosion were
injured of killed because of direct
radiation (gamma rays, x rays, neutrons).
A
large nuclear reactor like at Chernobyl has
a core of about 100 metric tons of uranium
when first installed and as that uranium is
used up it is converted to many radioactive
isotopes, both shortlived and longlived,
which are still confined in the cylinders
the core is composed of and highly
radioactive. At Chernobyl a combination of
human error, design faults, and mechanical
failures caused the core to overheat, the
resulting steam at too high a pressure to be
contained, and the steam containment
exploded sending out huge amounts of
radioactive steam to escape. Since the water
also cools the core, the core overheated,
burnt, and melted and the explosions spread
all the nuclear waste over the adjacent
land. A whole different orderofmagnitude
from a bomb.
QUESTION:
I have a fairly simple question (for
you) regarding tangental and cebtripital
force acceleration. Assuming a pendulum with
a rigid shaft 44" long weighing 60gms with a
mass of 300gms on the end (44" from pivot)
how would I calculate the effect of adding
[say] 2" on the opposite end (weighting
100gms) on speed of the swinging pendulum.
What would that weight or force be called?
Radial accelerator? I'm not even sure I am
asking this correctly.
ANSWER:It
is not as simple as you might think, but
just a little messy algebrawise, not hard
conceptionally if you know basic classical
mechanics including rotational motion which
I will assume that you do. I will assume
that the shaft is uniform and thin, the
other two masses are point masses, and the
friction is negligible, so energy must be
conserved. (See my notation on my figure.)
The pendulum is lifted to some angle θ and
released from rest. I will choose the level
of zero potential energy to be the bottom
end of the stick when it is vertical. When
released there is no kinetic energy T,
only potential energy V_{1}:
E_{1}=V_{1}=[m_{2}L_{2}+m_{1}(L_{2}L_{1})+m_{3}(L_{2}/2)](1cosθ)g.
When it has rotated down to the vertical position
where the angular velocity ω will
be largest, it has both kinetic energy T
and potential energy V_{2}:
V_{2}=[m_{1}(L_{2}L_{1})+ ½m_{3}L_{2})]g
T=½Iω^{2}
where
ω is the angular velocity and I
is the moment of inertia:
I=[m_{2}L_{2}^{2}+m_{1}(L_{2}L_{1})^{2}+m_{3}L_{2}^{2}/3].
So we
may now solve for ω=√{2(V_{1}V_{2})/I}.
I will
now look at your particular case.
Unfortunately you mixed your units and I
always prefer to work in SI units, so I will
choose θ=60° so cosθ=0.5
and use SI units:
m_{1}=0.1
kg (or m_{1}=0), L_{1}=2"=0.051
m, m_{2}=0.3 kg, L_{2}=44"=1.12
m, and m_{3}=0.06
kg, and g=9.8 m/s^{2}.
First
the pendulum without m_{1}:
V_{1}=1.81
J
V_{2}=0.33
J
I=0.40
kg·m^{2}
ω=2.72
s^{1} (radians/second)
Since
the velocity u a distance R
from the axis of rotation is u=Rω,
u of m_{2} is
u_{2}=1.12x2.72=3.05 m/s=6.82
mph.
Second, add m_{1}:
V'_{1}=2.33
J
V'_{2}=1.05
J
I'=0.44
kg·m^{2}
ω'=2.08
s^{1} (radians/second)
u_{2}'=1.12x2.41=2.70
m/s=6.04 mph.
So the
effect of adding m_{1} has
been to decrease the speed of the pendulum.
Increasing the moment of inertia decreases
the speed but the increased potential energy
difference (V_{1}V_{2})
increases it more. Regarding your question
"What would that weight or force be called?
Radial accelerator?", the force of the 100
gram weight isn't called anything—it is an
added weight to the system, an added force,
an added torque, an addition to the moment
of inertia of the system. It would be wrong
to call it an accelerator because its effect
here is to slow it down.
QUESTION:
I have a question about and relativity.
I have recently discovered a planet that is
exactly the same as Earth but it orbits its
star at twice the rate that Earth does.
There is an advanced rate there that I want
to talk to. I know that there will be a
delay as the message travels but will there
also be problems owing to differing
experiences of time because of relativity?
What if instead of "orbits its star" I meant
"rotates on its axis"?
ANSWER:
Your question is a little strange. What
does "exactly the same" mean? Twice the rate
could occur in many ways determined by the
mass of the star and the distance of the
planet from the star; furthermore, it the
earth had a mass twice as big as it does, it
would orbit the sun with the same rate as it
does now. But none of that matters since you
are asking me to assess the relativity
effects on time. In order to for relativity
to have any significant effects (except for
the most demandingly accurate time
measurements, such as required for GPS), the
quanty v/c must be not
much smaller than 1; here v is the
speed the planet is moving (relative to the
star, let's say) and c is the speed
of light. v=3x10^{4} m/s
and c=3x10^{8} m/s, so
v/c=10^{4.}
Relativity has a negiligible effect
for almost all situations involving
planetary motions.
QUESTION:
I was wondering in a typical graph that
shows the x,y,z electric and magnetic
components of an oscillating electromagnetic
wave where the y and z axes are used to
describe the behavior of the electric and
magnetic fields, is it representing the
strength of the fields at any given
distance/time along the wave (x,t) or is it
showing how far the fields extend into space
along their axis? or both?
ANSWER:
The former, i.e. the relative
magnitudes of electric fields and magnetic
fields at the instant when the graph was
drawn; the wave, as it has been drawn, is
moving in the +x direction.
However, the relative magnitudes of the
electric and magnetic fields is meaningless
because they are measured in different SI
units (electric field in Volts/meter or
Newtons/Coulomb, magnetic field in Teslas).
Using SI units, the electric field is much
larger than the magnetic field. Regarding
the information the graph gives about
spacial properties of the fields, only
directional information is meaningful;
e.g., the electric field is
perpendicular to the magnetic field and lies
in the (y,x) plane.
QUESTION:
Today my wife placed two identical jars
in our microwave. Both were open and filled
with the same amount of water at the same
temperature. The jars were placed directly
across from one another, approximately
halfway to the edge of the rotating glass
plate. One jar began to actively boil, while
the other jar did not. About 30 seconds
later the water in the nonboiling jar burst
explosively into a full boil. She shut off
the microwave and had to wipe down the
entire inside. What could possibly explain
this unexpected result?
ANSWER:
Water at atmospheric pressure normally
boils at 100°C. But to boil, bubbles of
water vapor must form; these bubbles
normally form in existing tiny air bubbles
or around microscopically small impurities
in the water and are called nucleation
points. Water with few nucleation points can
get hotter than 100°C and is called
superheated water. Superheated water is
quite unstable so when one bubble of vapor
starts to form it expands very rapidly and
the result looks like an explosion like your
wife witnessed.
QUESTION:
What kind of equipment would I
theoretically need to build a time machine?
I'm writing a novel based on the subject.
ANSWER:
A
DeLorean, a
flux capacitor, and a
Mr. Fusion. I hope you get the joke
here; if not, google Back to the Future,
a trilogy of movies. I hope that you
don't think I am ridiculing you, but there
is no good physics answer: the current laws
of physics forbid backward time travel but
you can travel forward in time by taking an
extremely long, very fast trip returning to
the place you started from (the
twin paradox). But you are writing a
novel and can let your imagination run wild!
QUESTION:
I once saw in a science museum,
experiment. They had a piece of thin metal
that connected to some kind of permanent
magnet, and the metal became red hot, and
fell awa, when it became red hot. The metal
was touching the magnet. I was wondering if
you could have two of those magnets,real
big, attract to each other almost all the
way. Then have two pieces of thin metal
become red hot, and you could then separate
the magnets. The exhibit did this over and
over, all day.
ANSWER:
The magnet holding up the metal is not a
permanent magnet but an electromagnet which
creates not a constant magnetic field but a
constantly changing (AC) field. Varying
magnetic fields induce varying electric
fields. The alternating electric fields
causes an electric current in the thin metal
and that current causes resistive heating in
the metal, just like AC currents in your
toaster heats up the wires. The reason that
the cool metal is held up is that it is
probably a ferromagnetic material which
becomes magnetized in an external magnetic
field and therefore is attracted to the
magnet. But this magnetization will not
happen at a temperature above some
temperature called the Curie temperature.
For iron this temperature is about 770 °C so
the hot metal will drop.
QUESTION:
I was wondering if you went deep in the
ocean, where there is a lot of pressure due
to the water, if you took a open container
down with you and closed it (air tight)
would it still have the same pressure
inside? And if you took it to the surface
would it still contain that pressure?
ANSWER:
I never thought about it, but I cannot
think why not. For example, suppose you take
the open container (full of water) to the
bottom of the Mariana trench where the
pressure is more than 1000 times atmospheric
pressure; the pressure inside and outside
the container is now 1000 atmospheres. Now
close the container and bring it up to put
the surface. There would be a gauge pressure
of about 999 atmosphers pushing out. It had
better a pretty strong container!
QUESTION:
How to calculate the downward force at
the weight 2, in grams, not inch grams and
(not torque).
ANSWER:
Your question has lots of problems. Is the
system in equilibrium? If so there needs to
be some other force present to keep it in
equilibrium, so where would that force be
applied? You ask "what is the force from the
weight at this angle" What weight? What
angle? Force on what by what? The pivot is
not at the center. Is the line connecting
the two masses to be considered weightless?
REPLY:
Do not want equalibriam, I want to know
what the downward force is under the 78
grams as if I had a scale under it. What
I calculate is not what the value I
physically measured. The angle is 90
degrees, parallel to earths gravity.
Consider the pivot point is at the
centre of the 2 weights.
(Not the way
you drew it. I will use the numbers on
your figure.) The Force is from
the heavier weight 78 gms.
(No it
isn't. Yes, the 78 gm weight does exert
a downward force on the line equal to
its weight, but that force is not what
your scale will read.) The line
connected to the 2 weights is
neglectable; I have calculated the line
and know has very little affect on the
system.
ANSWER:
You can't have it both ways. If you put
a scale under it, the scale holds it in
equilibrium and is therefore another force
on the seesaw. If you simply have it as you
have drawn it at rest and let it go it will
have an angular acceleration (even though it
is at rest at that instant); it is therefore
a little trickier to calculate.I will show
each way:

Equilibrium (with scale)
The net torque must be zero:
9.7x24+10.7x(F78)=0, so F=(834.6232.8)/10.7=56.2
grams

Nonequlibrium (free to rotate)
This is a little trickier so I will
switch to SI units (kg, m, s) shown in
green; N is Newtons (kg·m/s^{2}).
There is a net torque about your pivot
which is τ=0.76x0.270.24x0.25=0.145
N·m. Note that I have chosen clockwise
to be positive. A torque causes an
angular acceleration α=τ/I
where I is the moment of
inertia about the axis about which τ
is calculated. Since the moment of
inertia of a point mass m a
distance r from the axis is
mr^{2}, the moment of
inertia is I=0.0245x0.25^{2}+0.0775x0.27^{2}=0.0072
kg·m^{2} so α=20.2 s^{2}.
The linear acceleration is a=rα=0.27x20.2=5.5
m/s2, about half gravitational
acceleration. Now, you are interested in
the forces on your 78 gram mass. There
are two forces, the weight of 0.76 N
pointing down and the force F
which the bar exerts up; Using Newton's
second law, 0.76 F=0.0775x5.5=0.33
N=33.7 gram (force). If there were a
massless scale attached to the bat, this
is what it would read at the instant you
released it.
QUESTION:
If photons of light are massless
(without mass) why are they constrained from
escaping a black hole's intense gravity?
ANSWER:
In classical physics an object which is
in a gravitational field is affected only if
it has mass. But, general relativity tells
us that massless particles do indeed see
gravitational fields. Very massive stars and
galaxies bend the paths of passing photons;
this is called gravitational lensing
and was first observed more than 100 years
ago. Now, think of throwing a ball straight
up giving it an initial energy of ½mv^{2};
because of gravity it loses its energy as it
goes up and, at some height it will have
lost all its kinetic energy. Now, consider a
photon in a very strong gravitational field.
As the photon moves away from the object
causing this field (black hole, for
example), it loses energy just like the ball
did. But, unlike the ball, it cannot lose
energy by slowing down because the speed of
light must always be a constant c.
The energy of a photon is given by E=hf
where f is the frequency of the
light and h is Planck's constant.
Since we can write f=c/λ where
λ is the wavelength, so the photon
loses energy by increasing its wavelength.
(This is called the gravitational red
shift.) So now, if the field is strong
enough the photon will lose all its energy
and simply be gone. But its energy is not
gone and contributes to the growth of the
mass of the black hole.
QUESTION:
Consider sitting in an airliner
traveling between cities. If a parakeet is
free inside the plane and lets go of, lets
say, the seat, and wants to simply fly
inside the plane…will the parakeet have to
begin flying faster and faster to remain
inside the aircraft unattached? What
happens?
ANSWER:
The bird flies relative to the air
inside the airplane. The inside air is at
rest relative to the airplane. The airplane
is not accelerating. Therefore it maies no
difference whether the airplane is going 600
milea per hour or sitting at rest at the
airport; the way that the bird flies is
identical in both situations.
QUESTION:
I'm thinking electromagnetic waves and
their propagation, and I'm having trouble
seeing how energy is transferred. In the
analogy with a string, I know that there is
a tension on the string enabled by the bonds
between the atoms of the string. I know how
energy can be stored and transferred through
these bond interactions. Back in the
electromagnetic field, what is it about the
field in one point of space that allows it
to transfer energy to the next point in
space? Referring to the analogy, what is the
analog for tension or atomic bonds in the
electromagnetic field? I feel certain that
what I'm asking for is either a facet of the
speed of light, or perhaps vacuum
permittivity, but I'm having a difficult
time fitting those concepts into my
understanding.
ANSWER:
Think about the electric field at some
point where the light is passing by. It is
capable of exerting a force on a charge
which might be there. The charge will
acquire kinetic energy as it is accelerated
by the field. Therefore the light must carry
energy. Another way to look at it is that
light may be thought of as photons which
have energy which they can transfer to
something else.
QUESTION:
I am a tennis pro, and have been a
tennis stringer for over 40yrs. I always
have had the idea that Sound (Musical Notes)
maybe more accurate then the pull weight
measured by the pound or kilo, on a
stringing machine, when stringing a racquet.
Is this feasible? I have strung a racquet by
using a guitar tuner to insure every string
plucked when pulled, hits the same note and
then clamped. Am I way off base and the
physics/math of weight vs sound, behind this
is wrong?
ANSWER:
The frequency f ("musicsl
note") of a vibrating string depends on
three things: the tension T in the
string, its length L, aad its mass
density μ. Mass density might be
measured as ounces per inch which depends on
the string you chose. The equation is f=[√(T/μ)]/(2L).
From all that I read, the strings should all
have the same tension. But the strings are
not all the same length, so they would not
have the same frequency (note). So yes, you
are "way off base"!
QUESTION:
Can increase in pressure inside a hollow
semi circular sphere or a paraboloid or
other half cut closed figure cause it to
produce lift?
ANSWER:
No. Why would you think that? There is
one way it could happen, but only with a
decrease of the pressure of the air inside
the container: If the weight of the
container and the air in it were smaller
than an equal volume of the air outside
(normal atmospheric pressure); this is just
Archimedes' principle. The reason a helium
balloon rises is because its weight is less
than the weight of air which it displaces.
QUESTION:
I have a question about time. Time on
earth moves at one rate. Satellites in orbit
operate in a slightly different time and
have to be updated to maintain accuracy. If
I travelled to Mars to live what would be
the time difference between the people on
earth and the people on Mars? Would people
age differently on the different planet.
ANSWER:
The difference between rates of clocks
on Mars and clocks on earth would be
negligibly different. Satellites,
particularly those used by GPS, need
extraordinarly accurate time synchronism
because the sighals (EM waves) it deals with
are moving so fast that a tiny error in time
could give rise to a large error in
distance.
QUESTION:
THE MOTION OF THE STARS AND PLANETS IS A
PUSH AND PULL OF ENERGIES. THE STARS GREAT
GRAVITY KEEPS THE PLANETS IN ORBIT AROUND
THEM. MY QUESTION IS ABOUT THE TWO FORCES
THAT EITHER KEEP THE PLANETS TRAPPED IN THE
STARS GRAVITY WELL OR ALLOW THEM AS IN AN
EXPANDING UNIVERSE TO OOPOSE THAT GRAVITY BY
THEIR "CENTRIFICAL" FORCE AND EITHER ESCAPE
THE STARS GRAVITY OR SUCCOMB TO IT AND
EVENTUALLY HAVE THEIR ORBITS DECAY GIVEN
THAT THE LIFE OF THE STAR IS LONG ENOUGH?
HYPOTHETICALLY DO NOT ALL PLANETS ORBITS
DECAY AND SUCCOMB TO THIS? IF NOT, WHAT
FORCE KEEPS THE PLANETS IN THEIR ORBITS
OTHER THAN THEIR ELLIPSES, THE CONSTANT
INWARD THEN OUTWARD DISTANCE FROM THE STAR,
WHAT FORCE ACTS ON THE PLANET KEEPING THEM
FROM EVENTUALLY GOING TO THE SUN?
ANSWER:
It seems as if you have not studied
physics much; that's ok, I can give you an
overview and point you in the right
direction for further study on your part. We
must first ask what forces act on planets.
There is no such thing as a centrifugal
force, it is what is called a fictitious
force because it is simply something we make
up in certain circumstances where Newton's
laws do not work. Perhaps you are thinking
of the accelerating universe expansion
caused by "dark energy" when you refer
"EXPANDING UNIVERSE…'CENTRIFICAL' FORCE."
That is unmeasurably small. Therefore, there
is no "PUSH AND PULL" tug of war going on,
it is only pull, i.e. gravity. The
problem you are wrestling with,
understanding planetary morion and generally
called the
Kepler problem, was solved about
350 years ago by Isaac Newton. Essentially
what he found is that any body moving around
a much heavier body moves in a stable
periodic orbit (does not decay and does not
change its shape), not "EVENTUALLY GOING TO
THE SUN". I urge you do learn about
Kepler's laws and the solution of the
Kepler problem.
QUESTION:
I am a medical student currently
studying the cardiovascular system and I am
having difficulty in understanding how a
pressure drop is created inside blood
vessels. My question is: in a laminar
constant flow where the pipe contains a
viscous liquid what creates the pressure
drop along the pipe between point A and
point B assuming they are on the same level?
In other words, how does friction lower the
pressure? Does it decrease the rate of
collisions between the liquid molecules? If
yes, how does it do that? If no, is there
another parameter that contributes to
pressure apart from the rate of collisions?
ANSWER:
I presume that you are familiar with
Poiseuille’s law,
R=ΔP·4πr^{4}/(8ηL),
where
a fluid with viscosity η is flowing
through a tube (artery?) of length L
and radius r, R is the
rate of fluid flow, ΔP is the
pressure drop over L. Now, I see a
real problem with your question: you state
that there is a pressure drop "created" and
you try to understand this microscopically.
Since we are talking about cardio, let's
just say that, if the blood if moving,
something has to be pushing it. Let's go
back to the heart; the heart is a pump and
muscles in this amazing pump are
exerting
a force on the artery we are looking at. The
force is distributed over the whole area of
the artery so there is a pressure at that
end. After the blood has moved a certain
distance L we will find a lower
pressure; why? Because the viscosity is just
friction and exerts a backward force over
the whole volume. If it were an ideal fluid
with no viscosity, the pressure would be the
same on either end. Now I have sketched a
cartoon of what is happening: There is some
force (left red vector) which is pushing the
fluid ahead of it and, some distance down
the pipe, there would necessarily be an
analogous force (right red vector) pushing
the fluid in front of it (if viscosity were
zero). But, because there is viscosity,
there is a drag force which lessens the
pushing force as you move along the fluid
flow; I have represented this by the center
red vector as the net effect on the pushing
force across this length of pipe. Adding
this net drag force to the right red vector
results in the rightpointing green vector
which means the pressure on the right end
must be smaller than on the left. We
normally think of friction as slowing
something down, but because we assume this
to be an incompressible fluid, this is not
possible for a fluid. When I am presented
with an equation, in this case Poiseuille’s
law, I like to ask if it makes sense.
Assuming that I change one variable while
holding the others constant:

If
you increase viscosity, flow rate
decreases—honey flows more slowly than
water;

if
you increase the length the rate
decrease or the pressure drop increases;

if
you increase the radius of the pipe, the
flow rate increases or the pressure drop
decreases.
This
all makes sense.
QUESTION:
If a rigid rod is suspended at its ends
from 2 strings of equal length but hung at a
slope (approx. 15degrees from horizontal),
will the strings hang vertically (plumb) or
at an angle?
ANSWER:
I assume that you will have the strings
attached to the slope a distance L
apart (see the diagram); if they weren't, it
would not be possible for
them
to be plumb. Now, we want to find when the
rod will be in equilibrium. That means that
the sum of all forces on the rod must add to
zero (Newton's first law). So I have drawn
the diagram showing the strings making an
angle θ with the vertical (y),
and the angle the slope makes with the
horizontal (x) is φ. There
are just three forces, the weight
W, the tensions of the
strings on the rod T_{1}
and T_{2};
also I have shown the horizontal and
vertical components of the tensions (black).
I assume that the rod has uniform mass
distribution so that the weight acts at the
center of the rod. If the tensions were big
enough, the rod could be in equilibrium in
the vertical direction because the weight is
down and the vertical tension components are
up. But in the horizontal direction the only
forces are the x components of the
tensions and they both point in the negative
x direction; therefore it cannot be
in equilibrium. So we conclude that the
strings must be plumb (θ=0). We can
also conclude that the tensions must be
equal be equal because there must be no net
torque on the rod. Also, T=W/2 to
satisfy translational equilibrium.
QUESTION:
When making a large roll of paper, does
the surface temperature of the roll change
as its wound up. Does it change as it is
unwound (rate of paper coming off roll is
constant, so core of roll spins faster as
roll gets smaller. I always wanted to know
the physics behind this when printing
newspapers for 25 years! ~Shawn (The Boston
Globe)
ANSWER:
The paper should certainly heat up. The
reason is that friction is present. Imagine
taking a sheet of paper and hanging it from
one edge (#0 in the figure) You now pull it
out to position #1 and release it. Gravity
will pull it back and, perhaps it will swing
down like a pendulum. It will either swing
back to #0 or even keep moving until it gets
to #2; but it certainly will not go as far
to the right as it was dropped from.
Similarly when it swings back to the left
again to #3 it will not go as far as it did
to #2. Eventually it will stop at #0 again.
To start this pendulum you needed to do
work, so you endowed the paper with some
energy. But, after all was said and done,
the energy you gave it is gone. Imagine this
happening inside a large sealed box so we
can think of our paper pendulum as an
isolated system; then one of the most
important laws of physics is conservation of
energy, the total energy of an isolated
system never changes, should apply. So where
did the energy you added go? Heat! The paper
and the air in the box will become slightly
heated. There are two main ways that
friction works. Probably the most important
one is the air drag slowing it down as it
moves through the air; but this does not
seem like it would be important for paper
going onto or off of the roll since in that
case the motion is not perpendicular to the
surface, rather the motion slices through
the air. The other way is that the paper is
not perfectly elastic, some of the energy
you put in is lost just by bending the paper
which would heat it; think of a thin sheet
of a soft metal which, if you bend it back
and forth a few times, you can feel that it
is getting hotter. So, if you were pump all
the air out of the box, it would still stop
swinging eventually but probably quite a bit
more slowly.
QUESTION:
If two objects are moving away from each
other at half the speed of light, then are
they each travelling at the speed of light
relative to each other? And is that
possible?
ANSWER:
No that is not possible because
velocities do not add the way you expect
them to if the speeds are large. In your
world, if two cars are moving with speeds
u and v in opposite
directions, you would say that v'=u+v
where v' is the speed which
v is going relative to u;
this is called Galilean relativity.
But for objects moving with speeds
comparable with c, the speed of
light, the correct expression is
v'=(u+v)/[1+(uv/c^{2})].
In your question, u=v=0.5c
and you will find that v'=0.8c.
See an
earlier answer. Note that, if the speeds
of the objects are much smaller than c,
uv/c^{2}≈0 and you
get the familiar answer.
QUESTION:
How are the mass of the moon and earth
related to the moons orbit?
ANSWER:
I will only give the results for
circular orbits. In general, the period
T of the orbit is T=2π√[R^{3}/(G(M+m)]
where M and m are the
masses of the earth and moon, respectively,
G=6.67x10^{11}·m^{3}·kg^{−1}·s^{−2}
is the gravitational constant, and R
is the radius of the orbit. If m<<M,
this may be approximated as T=2π√[R^{3}/(GM)];
this is Kepler's third law, the square of
the period of a circular orbit is
proportional to the cube of its radius T^{2}∝R^{3}.
QUESTION:
This one has been bordering me for over
2 decades : What would be the reading on a
spring balance if 2 equal in magnitude but
opposite forces are applying on it ?
ANSWER:
If we are going to use a spring balance,
let's first calibrate it. Take a spring, lay
it on a table, and attach it to the wall as
shown on the figure. Next, exert many forces
on the end to stretch it and mark on a scale
the forces corresponding to what you have
pulled with. Since we assume that this is an
ideal spring the distance it is pulled (S)
is proportional to the force exerted The
proportionality constant, usually denoted as
k, is called the spring constant:
F=ks. For example, if S=10
cm in my figure, k=1/0.1=10 N/m. So
I have taylored my example to your question
and the scale at this stretch tells us that
you are pulling to the right with a force of
1 N. Note that the spring is in equilibrium
because it is at rest, so Newton's first law
must apply (the sum of all the forces must
equal zero). Vertical forces are the
spring's weight down and the force upward of
the table which must be equal to each other
in magnitude. Wait a minute! Is there
anything other than you which might be
exerting a horizontal force on the spring?
Of course, since the spring is attached to
the wall, it must exert a force of magnitude
1 N to the left. That is what you asked
about.
QUESTION:
Is there a similar dopplerlike shift in
sound frequency if the sound source and
receiver are stationary relative to each
other but the air is moving at high speed
across both. A rocket team I am mentoring is
doing this experiment on their rocket with a
buzzer and microphones above and below the
buzzer, then recording the frequency of the
buzzer during flight.
ANSWER:
This is no different from the source and
observer both moving with the same velocity
through still air for which there is no
Doppler shift.
QUESTION:
I am writing a scifi novel and have a
question related to the physics of
astronomical bodies. I know quite a bit of
physics but wanted to double check
something. I wanted to know if a "cracked"
moon would retain a gravitational pull? Now
by "cracked", imagine the planet (moon in
this case, orbiting a gas giant)in its past
was "destroyed" by a super weapon that
fractured its hemispheres apart. I figure if
the hhemispheres were still close enough
they would rejoin into a sphere... but I
like the aesthetic of a ruined world. Also
this is a moon orbiting a gas giant so I
figure some parts would be flung out due to
the forces. So could this body still hold a
gravitational pull (even if weak) in the
broken state?
ANSWER:
Strange that you are talking about a
"moon", since that usually means something
orbiting a planet; smaller things orbiting
the star are ususally called asteroids or
dwarf planets; I will assume that this
halved object is not orbiting some other
larger object, just the gas giant. When the
"super weapon" broke up your moon it was
probably a very violent event where each
half would have gone flying away with likely
enough speed to escape each other. In that
case they would just end up in a new orbit
around the star and interact only slightly
with each other. Another possibility is that
the two would end up in orbit around their
center of mass and the center of mass would
orbit around the star the same way the whole
moon did before its destruction. I find that
the second scenario would be unlikely
because they would have flown directly away
from each other and, if they are not going
fast enough to escape from each other, they
will eventually come back and crash into
each other. The crash would not be gentle
and would not result in the original sphere
simply returning. Now, to your question.
There is nothing magic about a sphere and if
you have two hemispheres each of them would
have a gravitational field half the strength
of the two of them. Any object, regardless
of its shape, has a gravitational field.
Near a nonspherical object the strength of
the field depends on where, precisely, where
you measure it; but if you look at the field
far away from it the field will be very
close to that of the whole sphere. The force F on a point (or spherical) mass
m due to another (point or
spherical) mass M a distance R
away is given by F=MmG/R^{2}.
QUESTION:
A common golf practice putting aid
consists of a 9foot green mat that rises 5
inches in the final 1 foot, to a standard
sized golf hole. Balls putted to the hole
from any distance away that reach the lip of
the hole but do not go in, will roll back
down the small hill and return to a point
well beyond the starting point of the putt,
often rolling beyond the 9foot length of
the mat. Thinking of the conservation of
mass and energy, can you explain how the
ball can return to a point beyond its point
of origin? It seems like it would need more
energy on the return trip than was imparted
upon the ball initially in order to do this.
ANSWER:
You are forgetting that the ball begins
its journey with considerable kinetic
energy. So it would have to lose quite a lot
of energy to friction to not get beyond the
starting point. If it lost no energy to
friction it would be moving with the same
speed you hit it with when it got back to
the same place it started.
I'm a
little disturbed by your reference to "…the
conservation of mass and energy…" because
there is no such thing as conservation of
mass, mass being a form of energy. You
should just think of conservation of energy.
Many times in physics the mass of a system
increases or decreases or disappears
altogether.
QUESTION:
What does it mean that a black hole rips
time? Does time stop or not work or changes
in a black hole?
ANSWER:
Because of time dilation, the closer
something gets to the black hole, the more
slowly time runs. At the event horizon,
inside of which nothing, even light, can
escape, time stops. So, from the perspective
of someone outside the black hole watching
something falling into the black hole, it
simply stops. However, we certainly know
that it does continue into the black hole.
Nobody, I believe, knows what the laws of
physics inside the the event horizon and
what time does. Of course, since nothing can
get out of a black hole, we can never do a
direct measurement of what is going on
inside. So hypotheses can be tested using
only indirect evidence.
I will
remind you that, as stated on the site, I
normally do not do
astronomy/astrophysics/cosmology. I would
urge you to find an astronomy site which
will answer your question (again a couple of
recommendations on the site).
QUESTION:
I was watching videos on some strange
properties of mercury, and its ability to
generate electromagnetic fields(EMF) when
having electricity applied to it. I was
wondering if EMF's have any effect, or
reaction with gravity? For example, if a car
was hypothetically inside a EMF stronger
than the one it is in on earth, then would
that cause the car to be less effected by
the earth's gravity? My hypothesis is if an
object is inside a considerably stronger EMF
than its surrounding environment, then that
object behaves as if gravity is like water
to a HydroFin. If you could clarify the
relation between gravity and EMF's, and tell
me if my hypothesis holds any water I would
greatly appreciate it
ANSWER:
Your question is very confusing.

First, it is usual notation in physics
that EMF is an acronym for
electromagnetic force, not field; this
itself is an anachronism because it is a
voltage (potential difference), not a
force!

Secondly, "…having electricity applied
to it…" is ambiguous; I will take that
to mean that a current is caused to flow
through it. Interestingly, it does not
appear that you ever talk about this
again

Third, electromagnetic field is broad
and not fully specified, so I shall
assume you mean an electric field.

Fourth, your entire sentence "My
hypothesis is…HyfroFin." makes no sense,
so I will ignore it.
The
crux of what your question is, as I see it,
the sentence "For example…gravity." What you
want to know is, essentially, if you put a
car into a strong electric field, will it be
"less effected by…gravity?" I will use a
conducting sphere to represent your car
because that will still convey the same
ideas but allow me to draw an easy to
understand figure. First, ignore the blue
lines; there is no electric field yet. Your
car sits on the road (black line) and is not
moving. The earth exerts a force (green
arrow) W down on
your car and is called the weight. The car
is in equilibrium because the road exerts an
upward force (yellow arrow) N
on the car which is of the
same magnitude as the weight; N
is what you would measure if the car were
sitting on a scale. You can't change the
weight but you could exert another upward
force on the car which would make it
seem to have less weight because
N would get
smaller. So you want to exert an additional
upward force by putting it in a strong
electric field. I will put it in a uniform
electric field. (A uniform field is one
which has the same strength everywhere and
points in the same direction everywhere; if
your car weren't there all the blue arrows
would be parallel and equally spaced.) Now,
what happens to the car when you turn on the
field? Because it is made of metal which is
a conductor, electrons in the car experience
a force opposite the field direction and
positive charges are left behind where the
electrons migrated from . This charge
distribution on the surface of the car now
creates an electric field of its own and the
blue arrows in the figure are the vector sum
of the original uniform field and the field
created by the car. The positive charges on
the top of the car experience a force upward
but the negative charges on the bottom
experience a force downward and so the net
result of adding the electric field is zero.
So your hypothesis that the field will
counter the gravity is not correct.
QUESTION:
Hi, i am an out of school adult,
electronics engineer, good understanding of
phisics, however my question: We know like
charged objects repel, we can determine how
strong the repelingness is, however HOW IS
THE MOTIVE FORCE IMPARTED TO MOVE THE
OBJECTS APPART? THERE IS A COMMUNICATION A
'FIELD' WHAT IS THE REPELINGNESS CONSTRUCTED
OF AND HOW DOES IT WORK?
ANSWER:
Classical electricity and magnetism
(E&M) is an empirical field theory, based on
experiments. A simpler field theory is
Newtonian gravity; here we find that any two
masses will attract each other. We imagine
that masses are the source of gravitational
fields but also masses feel gravitational
fields. If one mass m is near
another mass M, it feels a force at
all places around it, a vector, so the force
field is essentially a vector specification
for every point in space, a "continuous
table" which can be expressed as a function;
for example, if the masses are point masses,
the force M exerts on m is
F_{Mm}=u_{0}Mmk/r^{2
}where u_{0}
is a dimensionless vector of magnitude u_{0}=1
which points from m to M,
and r is the distance between the
masses, and k is just some
constant; similarly, the force m
exerts on M is F_{mM}=u_{0}Mmk/r^{2};
you probably know this Newton' universal law
of gravitation. But we want something which
is more general than specific to a
particular situation. So, if we divide out
m for example,
F_{Mm}/m=u_{0}Mmk/r^{2}≡G_{M};
this is called the gravitational field due
to M and may be thought of as the
way that M alters the space around
it and also lets you know the force any
point mass nearby feels due to its presence.
So now
you have had your tutorial in fields and we
can move on to talk about E&M. The electric
field is pretty much the same as the
gravitational field except the masses are
replaced by electric charges and k
is obviously not the same (I will call it
k'), E_{Q}=u_{0}Qk'/r^{2}
for point charges. A big difference from
gravitational theory is that while there is
only one kind of mass, there are two kinds
of charge, so the possibility of repulsive
as well as attractive forces exists. As you
no doubt know, opposites attract, sames
repel.
So far
I haven't probably told you anything you
didn't already know. I surmise, from the
capitalization you have used, that you don't
want to know about the field but what is
actually happening that you can get your
teeth into. It turns out that force fields
can be quantized, which in the case of E&M,
means that electric waves (light, radio,
etc.) may be thought of as streams of
tiny "pieces" of the wave called photons;
for a particular frequency wave all photons
have the same energy. If you now fully flesh
out E&M into the quantized field theory
(quantum electrodynamics, QED), photons may
be interpreted as the "messengers" of the
field. Think of an electric charge as being
surrounded by photons, the density of which
gets smaller as you get farther away. A
nearby charge, with its own "cloud" of
photons, interacts with the other charge via
these photons; there is your mechanism you
seek! The complete theory of E&M, including
QED, is generally considered the most
complete theory we have in physics. Also,
you can't directly observe this photon cloud
because the photons are called "virtual",
popping into and out of existence.
Incidentally, nobody has been successful in
quantizing the gravitational field, but if
it were the "field messenger" analogous to
the photon for E&M would be called the
graviton.
QUESTION:
Equivalence principle. I understand that
if I'm in an elevator in deep space and it
starts accelerating upwards at 9.8 m/s^2, if
I drop a ball I would see it move to the
floor the same as it would on Earth. If the
elevator continues to accelerate and I drop
the ball again I would see the ball move to
the floor but in a much shorter period of
time and so this wouldn't match what happens
on Earth. Does this mean the equivalence
principle only applies to that initial
period when the elevator starts
accelerating?
ANSWER:
The equivalence principle states that
you cannot tell whether you are in a frame
accelerating relative to an inertial frame
or in an inertial frame with a gravitational
field. The principle does not depend at all
on what the current speed of the
accelerating frame is, only its
acceleration. So, for example, if you have
an acceleration of 29 m/s^{2}, no
experiment you perform will be different
than if you were in a gravitational field
with g=29 m/s^{2}; even if
you have a speed of a million miles per
hour, your acceleration is not changing.
QUESTION:
The Wright brothers invented controlled
flight. In the years since, we have gone
from a fabric covered frame to synthetics
and carbon composites and the exploration of
space. Einstein and others came up with the
physics that developed atomic energy, both
bad and good. Since the mid 1940's what has
physics come up with, a material, process,
better mouse trap, that has made life
better? Are not all these creations in
technology the results of garage experiment,
engineers, both mechanical and chemical, and
inventors looking for a better way? What
have physicists been doing for the last 80
years, other than smashing things together
in accelerators to make pretty pictures?
ANSWER:
Wow, you seem to have quite an axe to
grind! First, let me address the
"philosophical" issue. One of the things
which makes us human is that we are curious,
always seeking better understanding of our
world, in fact the whole universe. Even
though new discoveries, to people like you,
seem pointless because they seem to have no
application you can use in your everyday
life, they often pan out in the end. So I
will enumerate just a few examples

So, one of the most important components
of modern technology was created by
physicists. Can you imagine modern life
without transisters? Sure, engineers
collaborated after the discovery by
miniturizing but without physics your
mobile phone would be absolutely
impossible.

How about lasers? Guess who you can
thank for lasers—that's right
physicists. They are everywhere,
scanners in grocery stores, the little
cliponyourfinger blood oxygen
sensors, surveyers instruments, etc.

Had a CT scan or MRI lately? Physicists
invented them.

Computers? When I was in college I had a
summer job in the company where research
and development for making components
for nuclear submarines was done. They
had a computer in a room with an area of
probably about 10 yards by 30 yards made
using vacuum tubes; it had far less
computing power than the iPhone I have
in my pocket. Who first built computers?
Need I really ask?
I hope
you get the idea.
ADDED
THOUGHT:
By the way,
fundamental "discoveries" are never found by
inventors' garage experiments any more.
These "discoveries" are extensions of what
is already known—a better mouse trap but not
the fundamental idea of a mouse trap.
QUESTION:
In nuclear physics we talk about
bombardment of a nucleus with neutrons. Can
the same be done with protons? A neutron can
decay into an proton (after in average 9
minutes). At the moment of this change, a
"travelling neutron" might change into a
proton before it hits the nucleus. What
happens then?
ANSWER:
Protons are much easier to use to study
nuclear structure than neutrons. Because
protons have electric charge they can be
accelerated and steered using magnitic and
electric fields. Neutrons have no charge and
therefore you need to get them from nuclear
decay or from reactions of other particles
with nuclei, for example bombarding a
nucleus with protons and look for neutrons
coming out. Protons are much easier than
neutrons to detect making protons (or other
charged particles) more useful for
experiments. Regarding your second question,
if a neutron decays before interacting with
the target, it makes no difference since it
is neutrons which the experiment is designed
to observe and unwanted protons would be
very small compared to the flux of neutrons.
QUESTION:
If time slows as we approach the speed
of light, and stops at the speed of light.
And photons are particles of light, then
surely from the photon’s perspective, they
are all still inside the Big Bang. They have
not started expanding with the universe,
which is travelling slower than the speed of
light. Because photons only travel at the
speed of light. Which implies that all
photons we interact with, are still
connected together inside the Big Bang. Does
this not explain some of the mysteries of
quantum physics we are trying to solve?
ANSWER:
Photons do not have a "perspective". It
is pointless to talk about how things look
from a photon because, being a stable
particle, it has nothing "aboard". You can
read about this issue in an
earlier answer.
QUESTION:
I have a pot I bought from Goodwill. One
of those gorgeous well balanced copper
bottom pots. When I got it home and I put it
on my electric stove top and boil water, for
example, that the pot shakes and shakes and
shakes while it’s preparing to boil and
boiling water. Turns out the pot has kind of
a point in the center of it so It’s not a
flat bottom pot. Heres my question. I’ve
always wanted to know is why the heat from
my electric stove top makes my pot shaken,
shake shake. If I did not watch the pot, it
would possibly fall off the stove top
because the hotter the element is the more
it shakes and jiggles. Why does it do this
to my pot of water?
ANSWER:
You must have one of those glass surface
ranges. So the pot never has its whole
bottom on the surface. If it were sitting
with its center of gravity exactly above the
"point" it would be called unstable
equilibrium. The tiniest bit off and it
is not any longer in equlibrium and it
topples over until something stops it, in
your case the bottom edge of your pan hits
the stove top. (Think about trying to
balance a pencil on its point.) But it is
still very close to equilibrium so it takes
a very small force to cause it to rock back
toward equilibrium but, again, it won't stay
there and will drop to some other point on
the bottom edge, etc., etc.
Heating up something very hot causes all
sorts of things to happen—the metal expands
and warps a tiny bit more in one place or
another causing the position of the center
of gravity to move around; as the water
heats it convects around or starts bubbling
up in one place and not another; as the pot
rocks, the water sloshes around a little.
All these things keep changing and causing
the pot to shake around
QUESTION:
I have a question about gravitational
waves. What happens to a gravitational wave
if it encounters a massive object? If a
gravitational wave, created, perhaps, by two
black holes colliding with each other,
reaches some massive object and passes
through it, does this cause the object to
move? And if it does move, does it, in turn,
create further gravitational waves? Does it
strengthen the original wave such that it
lasts longer? I guess another way of putting
it is, would a gravitational wave from two
colliding black holes be more or less likely
to reach the earth if it passed through
other planets on the way? I assume the
effect would be very small, but would there
be any effect at all?
ANSWER:
I normally do not answer questions on
astronomy/astrophysics/cosmology as stated
on the site. But, I will take a stab at this
one. I think we can assume that the
superposition principle applies to
gravitational waves, so any gravitational
waves passing the same point in space would
simply add. It has taken many decades of
many physicists trying for gravitational
waves to be actually observed. And, if any
mass accelerates it will create
gravitational waves. So let's consider
Jupiter, a quite large mass which is
accelerating constantly because of its
orbital motion. The gravitational waves
predicted are undetectably small. It takes a
catastrophic event like the merging of two
black holes to make detectable waves. Even
though they are large distances from us, we
still can detect them although their
intensity decreases quadratically with the
distance they have traveled. These waves are
enormously bigger than any they might
encounter, so the addition of others makes
no detectable difference.
QUESTION:
What would happen if fermions had no
spin? Would this violate the Pauli exclusion
principle?
ANSWER:
They would not be fermions would they?
Then the Pauli exclusion principle would not
apply to them. And, atoms could not exist,
nuclei could not exist, stars could not
exist—our universe as we know it could not
exist.
QUESTION:
If I look at the polar plot of the
KleinNishina differential cross section
with respect to solid angle for Compton
scattering, it appears that for highenergy
photons, forward scattering dominates. That
is, the scattering angles are near zero, and
the energy transfer to the electron is low.
But then I look at the differential cross
section for recoil electrons with respect to
their energy vs. recoil energy and I see
that the most highest probability for
scattering an electron into a particular
energy occurs at the Compton edge (i.e. a
photonscattering angle of ~ 180 deg). These
two things seem contrary to one another.
Clearly I am missing some basic concept.
ANSWER:
You
should read the details of this issue in the
discussion on
Physics Forums. I urge you to download
the book (Evans) referred to in the Physics
Forums discussion. I will use results from
the book and discussion to give you my
"quick and dirty" explanation of the energy
dependence of dσ/dT where
T is the electron kinetic energy.
Suppose that the angle which the electron
recoils is φ then the differential
cross section is dσ/dΩ'
where dΩ'=2πsinφdφ.
(Note that Evans notates the cross section
as _{e}σ.) In the book it
is shown (eq. 4.5) that (dσ/dφ)=2πsinφ(dσ/dΩ')
which is the angular distribution of the
differential cross section. Now, cleverly,
the chain rule is applied (eq. 5.1): dσ/dφ=(dσ/dT)(dT/dφ).
Finally one can show (eq 5.2) that
dσ/dT=(dσ/dΩ)[2π/(α^{2}m_{0}c^{2}]{[(1+α)^{2}α^{2}cos^{2}φ]/[(1+α)^{2}α)(2+α))cos^{2}φ]}^{2}.
Here
α=hν_{0}/m_{0}c^{2}
where hν_{0} is the energy
of the incident photon and m_{0}c^{2}
is the rest mass energy of the electron. Be
sure to note that (dσ/dΩ)
is a function of θ; I have looked
up reference D12 of the Evans book and
gotten the
equation for (dσ/dΩ)
and plotted it for hν_{0}=10
MeV; your intuition was right: the cross
section for electrons scattering is largest
when the photon scatters to 180°. There will
be, at each angle φ, an energy
spectrum and that is what [dσ/dT](α,θ,φ)
gives you. Now the spectrum you are
interested in is the spectrum of all
electrons so you integrate over θ
and φ. The resulting spectrum you
will see cannot be be larger than hν_{0}
because that is all the energy the photons
brings in and, since the photon will not be
totally absorbed, the highest energy
electrons will be a little less than that. I
guess that wasn't so "quick"!
QUESTION:
I am an Indian,I came across a section
from my religious scripture that States.
"When atoms move to cover space, this is
called time, and time is measured according
to the amount of space covered by atoms
[Srimad Bhagavatam 3.11. 34]. It is
generally accepted that when time stands
still (although impossible), there is no
movement of anything. This also confirms
that the movement of atoms is time" can you
answer the validity of this?
ANSWER:
We know, from the successful theories of
special and general relativity, that time
and space are not independent things, that
they are coupled to each other such that
different observers have different
observations regarding the values of space
and time in their own frames. Your has
statement some validity in that it
recognizes that time and space are in some
way related. The statement that "…the
movement of atoms is [my
emphasis] time…", however, is vague and its
meaning not operationally defined. I have
heard some of our best cosmologists state
that if only we really understood what time
is we would understand the
universe much better than we do.
QUESTION:
Recently I've been thinking about
alternative explanations to the doubleslit
experiment and there is one idea that I
can't debunk. Accelerating and decelerating
electrons creates an electromagnetic field
that can later affect the movement of
electrons. So is it possible that the double
slit experiment doesn't prove electrons are
waves but was rather caused by an
electromagnetic field that affected the
movement of electrons?
ANSWER:
What electrons are you thinking about?
So, we arrange the situation so that only
one electron per second goes through the
slit(s) and there are no other electrons to
be affected by the presence of the incident
electron (.e.g., make the slits out
of an insulating material). You will still
find the same double slit diffraction
pattern. Further, you will observe only one
electron at a time striking the detector
screen. I think you should not waste your
time trying to disprove waveparticle
duality; the number of measurements and
experiments which prove it are beyond
number.
QUESTION:
As a massage therapist, I often see
images of a head hanging lower and lower,
labeled with the increased weight at each
degree of tilt. Now the head only weighs so
much and it's hard to believe that a couple
vertabrae add all the kilograms. So what is
really being said?
ANSWER:
When I first started to research this I
was appalled at how badly the analyses were
from a physics standpoint. First of all,
you, unlike almost anybody, are correct in
saying that the head has a weight, period.
The weight of something is a force which
points toward the center of the earth and is
the force which the earth exerts on it. So
the fact that 90% of articles say the weight
of the head increases as we lean forward is
nonsense. I have lifted a generic diagram of
what happens when you lean forward. A
simplified model of what the forces on your
head are its weight W,
the force F_{M}
due to a muscle attached to the vertebra
behind your head, and a force F_{V}
due to the top vertebra pushing up. Now, the
head is in equilibrium so the sum of these
three forces must equal zero; another way to
say this is that the sum F_{M}+W_{
}must equal the negative
of F_{V}
which is why I have labeled the sum on the
diagram to the left as F_{V.}
F_{V
}is also the force which
the head exerts on the vertebra . This force
is what is usually called the weight of the
head and is clearly larger than the weight.
When you are standing straight up
F_{M}=0 so
F_{V} =W.
Note also that all forces have been assumed
to pass throught the center of gravity so
that we do not need to think about torques.
QUESTION:
Would it be possible for someone to
construct a funnel or cornucopia shaped
object with a polished mirror surface that
directed all light incident to any surface
to reflect down to a point like a fiber
optic cable?
ANSWER:
Probably the closest you could come to
doing this is a parabolic mirror which
focuses all light from distances much
greater than its focal length to a single
point. Reflection telescopes are parabolic
and have have the advantage over lens
telescopes because they have neither
chromatic nor spherical aberations.
QUESTION:
If you take a flat piece of black metal,
say iron, it is not emitting light/photons
at that moment. If anything it could said
that it is absorbing them. If you then pass
a powerful electrical current through the
piece of metal it can then become so bright
that you cannot look at it. Is the
brightness being seen truly being caused by
the sudden release of photons? If so how is
the current passing through the iron
generating photons?
ANSWER:
Every object is constantly emitting
photons. If that object is in thermal
equilibrium with its environment, the rate
at which it emits energy is equal to the
rate at which it absorbs energy. Even if the
object is cool, say room temperature or a
little above, it emits photons. That is how
nightvision goggles work, "seeing" infrared
radiation. If you now pass a current through
this bar it heats up because of resistive
heating (just like your toaster) and it is
now hotter than its environment and it emits
more photons than it absorbs and, because it
is much hotter, it emits photons which are
in the visible spectrum, red.
QUESTION:
Hi, is there any work being done on
wavelengths/wave forms being helical in
structure? The more I look at wavelengths,
the more it seems counterintuitive to me
that they exist in on a flat plane, and
that, surely they must have at least some
level of 'spin' or some level of helical
movement? I am just wondering if any work
has been done in this area, or if it has
been 100% proven that they move within a
flat plane.
ANSWER:
I commend you for thinking, great
intuitive leap! You are absolutely correct,
the waves which are first taught in a
physics class are what are linearly
polarized, or said to have linear
polarization. For an electromagnetic
wave, a linearly polarized wave always has
the electric field perpendicular to the
direction the wave is moving and always in a
single plane in which the line of travel is
entirely contained. In situations where that
plane is rotating about the line of travel
is when you get the "helical" behavior you
have been speculating about; this is called
circular polarization. The little animation
shows how the electric field (little red
arrow) varies; note that the electric field
in this polarization has two components
which are 90° (¼ of a wavelength) out of
phase with each other. If you want to learn
more about polarization, see the
Wikepedia article. You ask whether there
is "…any work being done…". No, all this has
been well understood for about 200 years!
QUESTION:
I have an assignment for my year 11
physics but I accidently made the experiment
I conducted for it much harder to understand
then it should be. The original prac was
simply attatching different coloured paper
around 4 beakers and recording the change in
temperature in a heat box using heat lamps.
This was simple to explain and I did a
shorter practise assignment on it. In the
experiment we have though we instead decided
to change the colour of the water, and heat
it up with the sun. We can't retake the prac
but I cannot find a formula that works for
the expected relationship, neither do I
understand what any of it even means.
I was
able to get an AI to create a fomula:
ΔT =
(P × t) / (M × C × AF) where:
P =
solar radiation (1361 W/m2)
t = time
(600 seconds or 10 minutes)
m = mass of
dyed water (300 g)
C = specific heat
capacity of water (4.184 J/g°C)
AF =
absorption factor (let's assume it is 0.5)
But
this formula doesn't take into consideration
the outside temperature, and I don't
understand what is actually happening.
ANSWER:
The first thing I want to emphasize is
that physics is not formulas. Yes, equations
are the way we approach physics problems,
but if you think physics is getting a
"formula" and plugging numbers into it to
get an answer, then you will never really
know how to understand nature. Let's take a
look at the formula which the AI gave you.
Does it make sense? What causes the
temperature to change? Some of the light
entering the beakers is absorbed which means
that the energy heats up the water. So the
more light which is absorbed (large
absorption factor) the more the temperature
will change; but the AnotsobrilliantI
has put AF in the denominator which
means large AF results in small ΔT,
blatantly wrong! Also an exercise worth
doing when determining whether an equation
is possibly correct is to do a dimensional
analysis—are the dimensions the same on each
side of the equation? I will do this
analysis in terms of units rather than
dimensions which makes it easier here. Left
side of the equation has units of
temperature (K); K is for Kelvin, the unit
for temperature in SI units. Look at the
right side: P has W/m^{2}=(J/s)/m^{2},
t has units of seconds (s), M
is kg, C is J/(kg·K), AF
is m^{1}. Putting them all
together, [(J/s)(s)/(m^{2})]/[(kg)(J/(kg·K)(m^{1})]=K/m;
so the equation is almost dimensionally
correct. The culprit here is the AF;
perhaps your AI bot did not use the usual
definition of absorption fraction, maybe
simply the fraction not getting through (i.e.
50% if AF=0.5), but to put it in
the denominator still is wrong because this
would say that twice as much energy is
absorbed as comes in! But now I think I see
the problem. If we take AF to be a
fraction or percentage, the dimensions of
the right side are now K/m^{2},
still different from the left (K). I think
that AF should be changed to A
times F where A is the
area the light is shining on and F
is the fraction of the light energy absorbed
and both A and F should be
moved to the numerator:
ΔT=P·t·F·A/(M·C)
and
this will be dimensionally correct. The
outside temperature does not enter because
we assume that the water is at that
temperature to start and all we care about
is the change in temperature. Also, I have
used Kelvin temperatures but all we want is
change in T and ΔT is the
same for °C and K. Let me try to see if this
gives anything at all like we would expect
just guessing A and F:
P=1361
W/m^{2}
t=300
s
F=0.5
M=0.3
kg
A=100
cm^{2}x(1 m/100 cm)^{2}=10^{2}
m^{2}
C=(4.184
J/g°C)x(1000 g·K/1 kg)=4.184x10^{3}
J/kg·K
I find
ΔT=1.63 °C which seems reasonable.
But,
what does your assignment ask for? It asks
only what the temperature is after 10
minutes, it does not ask you to have an
equation. But I hope that you have learned
something about how you should approach
understanding the meaning of an equation. I
had to spend quite a bit of time on your
question because you started me with an
equation where there were some clear errors
and also a serious misunderstanding of what
you had (AF rather than A·F).
I hope that you now "…understand what is
actually happening…".
QUESTION:
I just went bowling and was able to roll
a 14 Pound ball at 23.36MPH. I was just
wondering how many pounds of force was
needed to reach that speed? I threw out my
back and arm in the process but it was
totally worth it.
ANSWER:
If you had bothered to look in the faq
page you would not have to ask this
question. And, how in the world did you
ascertain the speed so accurately? The
answer is that it is impossible to answer
this question because there is not enough
information. As one quick example, suppose
we view you pushing on the ball with a force F_{1} for 0.1 seconds; it
would require a force of approximately F_{1}=148
lb to achieve the ball's speed of 23.36 mph.
Now, if you instead push for 0.2 seconds, it
would require a force of approximately F_{2}=74
lb to achieve the ball's speed. Two
different forces would result in the same
speed! If you want to have more detail, go
to the
faq page and follow the link there.
QUESTION:
Hello, I am a motorcycle rider and an
argument that comes up repeatedly in the
community is regarding the effects of sheer
winds pushing a motorcycle to the left or
right. Obviously, wind pushes a
motorcycle,but the issue in questions is
whether solid wheels like that on a Harley
Fatboy "catch" more wind than that of a
wheel with spokes. I've always been of the
understanding that a spoked wheel spinning
at the same rpm's would essentially be the
same as a solid wheel and there is no real
difference to the rider on the motorcycle.
Are you able to explain this and which is
correct?
ANSWER:
I feel quite certain that the force of a
cross wind on a solid wheel would be larger
than on a spoked wheel, spinning or not; for
a wind to exert a force on something the
wind must push on it and, whether the wheel
is spinning or not, it has a much smaller
area for the wind to actually hit. There is
a fair amount of research which has been
done on bicycles rather than motorcycles,
but the emphasis seems to be mainly on the
effect on drag forces along the direction of
the velocity rather than laterally. I think
that strong opinions are based on perception
of the rider who has ridden cycles with both
kinds of wheels and perceives no difference.
For example, a rider is likely to think that
the two wheel contributions are the same
because the force of a cross wind on the
rider and the rest of the cycle (red arrow
in figure) is probably much larger than the
difference between spoked and solid wheels,
making any difference difficult to feel.
Similarly, the force tending to tip the
cycle over (as opposed to the tendency to
push it laterally) is relatively smaller for
the wheels (green arrows) than for the rider
and the rest of the cycle (red arrow)
because, the red arrow is much higher off
the ground and much larger than the green
arrows and therefore exerts a much larger
tipping torque.; again this lessens the
ability to percieve a difference in the two
types of wheels.
QUESTION:
I cannot find the answer to this
question anywhere. Consider 1kg rock
traveling through space at a constant
velocity. It has a constant amount of
kinetic energy. If it comes near a large
planet and gets caught in its gravitational
field, and eventually hits the surface of
the planet, its velocity will increase as it
approaches the planet, thereby increasing
its kinetic energy. How can we reconcile
this with the first law of thermodynamics,
namely energy can neither be created or
destroyed but only transferred from one
place to another. Where does this new energy
come from? Gravity is a field, not energy.
ANSWER:
You misunderstand conservation of
energy. Only the energy of an
isolated system is conserved. In your
example, the rock was originally isolated
and so its energy was conserved, i.e.
it moved with constant velocity in a
straight line. However, when it encountered
the planet, it was no longer an isolated
system. It experienced a force (the gravity
of the planet) which did work on it and
increased its kinetic energy. Incidentally,
a gravitational field is energy, it
has an energy density.
QUESTION:
I am new student of diagnostic imaging 
ultrasound and have a question regarding the
acoustic variables. As per the information
that I collected I came to an understanding
that the particle motion is the distance
covered or displacement of the particle.
However my Professor was disagreeing with
this and says that particle motion is not
the distance. Could you please help me to
clarify this?
ANSWER:
There is no way I can know the jargon
used in your specialized field. "Particle
motion" could be one or more
quantitiesfrequency and amplitude for
example. But what particles are being
described?
FOLLOWUP
QUESTION:
So basically, we are
learning about sound waves and its
properties and its appplication in
ultrasound imaging. The lecture was
regarding the acoustic variables and were
having discussion on "particle motion" as
the sound travels through the medium. As per
what I read, the particle motion is because
of the oscillations in the particles of the
medium as sound travels through it and it is
distance and can be measured in cms, etc.
However our instructor said that particle
motion is not particle displacement and
hence distance is not an acoustic variable.
I am not really convinced by his explanation
and was seeking some advice on this.
ANSWER:
Again I emphasize that I am not an
expert in ultrasound symantics and you would
be much better advised to talk with your
instructor to find out what he/she means.
Generally, in acoustics, we do not pay much
attention to what the particles in the
medium are doing. We know they are moving
around in some way but we are more
interested in the wave itself.
What
are the properties of the wave which you are
using for imaging, for example.
 The three most important variables
are its velocity v in the
medium (made up of molecules which
presumably are your "particles"), the
wavelength λ of the wave, and
the frequency f; these three
are related, v=λf.
 Also important in understanding the
wave is its amplitude which is a measure
of how "loud" the sound is.
 A fifth important variable is the
phase of the wave, important if
interference of more than one wave is
considered.
What is the motion of the particles which
underlie the sound wave?
 Sound waves are known to be
longitudinal, i.e. the particle
motion is oscillatory parallel to the
direction of the velocity of the wave.
 Being oscillatory, the each particle
moves about fixed point in space, it
does not move with the wave; so
"displacement" and "velocity" of the
particle depends on the time and their
timeaveraged values are zero. This may
be the reason your instructor does not
like your choice of displacement to be
part of particle motion.
 The individual particles vibrate
with a frequency which is the same as
the frequency of the wave and with an
amplitude which is proportional to
whatever you observe the amplitude of
the wave to be.
The animation illustrates the particle
motion. The little red arrow tracks the
longitudinal oscillatory motion of a single
particle. Watching the open circles shows
them getting more and then less densly
populated which indicates that sound may be
more macroscopically viewed as regions of
high and low pressure which propogate in the
direction the waves are traveling.
I hope that this
helps you. Anyway, it is the best I am going
to be able to do!
QUESTION:
in a scene from Apollo 13, astronaut Ken
Mattingly and engineer John Aarons are
seeking to help the crew of the Apollo 13
transfer the remaining power from a battery
on the LEM to a battery on the Command
Module to which the 3 astronauts had
returned after the had used the LEM as a
"lifeboat"...how can that be done?
Especially when the LEM battery has less
"juice" than the battery on the Command
Module? No doubt an umbilicus would be
attached from one battery to the otherwould
the power simply flow spontaneously from one
battery the other? And however they were
caused this to occur, wouldn't the current
flow from the battery with MORE power to the
one with less? How can the power be "coaxed"
from the weaker battery to the stronger?I
ANSWER:
There is a
nice article on the web which addresses
your question.
QUESTION:
I am a freelance writer, and I am trying
to build a scenario, only if it is
theoretically possible. Here it goes. In the
movie Interstellar, which I know is likely
more film than proven science, there is a
planet in which time passes slower than time
on earth because of its proximity to a
massive black hole. So every hour on Planet
A equals ten or so years on Earth. I have
found some theories and answers online that
all suggest this is possible, time dilation,
relatively stuff. What I am trying to
understand is can the opposite be true? Is
it possible with specific conditions that a
planet could exist in which time moves
faster than it does on Earth. For instance,
on Earth one year is 365 days. Could there
be a scenario in which for every year on
Earth, 20 or 50 years passes on Planet B? I
have tried to find this answer online, and
by asking people who have a better
understanding of this subject matter than I
do, but I haven't gotten close to sniffing
this.
ANSWER:
First, Interstellar has
received many accolades for the accuracy of
the physics. At the beginning of the
production they hired Kip Thorne, a renoun
theoretical astrophysicist and Nobel
laureate as an advisor; he kept a very tight
rein on maintaining good science,
forbidding, for example,
fasterthanlightspeed of space ships. A
picture is worth 1000 words! Look at the
graph I have included. This graph
illustrates the rate at which clocks in
circular orbits run faster than a clock on
the surface. Both time dilation due to
velocity (red curve) and gravitational time
dilation due to earth's gravity (green curve)
as well as the net time rate change (blue
curve). The clock considering only special
relativity (speed) runs slower but the
effect is zero infinitely far away. The
clock considering only general relativity
(gravity) runs faster and is fast as it can
get infinitely far away. Infinitely far away
the clock will run faster at a rate of about
700x10^{12} second for each second
on earth's surface; that is the fastest you
can get a clock to go compared to earth. One
year on earth would be shorter by 0.022
seconds than the distant clock. What you
want is not possible.
QUESTION:
I would appreciate it if you could
explain how a screw/conical shaped object
moves down an inclined surface. I understand
that it may have to do with circular motion
of some sort, but in real life, I have
noticed that it seems to oscillate, with its
oscillations increasing as it rolls down in
some scenarios, I'm just curious as to what
concepts are involved in this!
ANSWER:
A cone on an inclined plane can be very
easy to describe, pretty easy to describe
but just qualitatively, or really hard to
describe. If there is no friction between
the cone and the plane and it
starts
from rest, it will slide straight down the
incline just like a cube or ball would with
no friction; thats the easy one. If the cone
can roll only without slipping (infinite
friction), it will not go down the incline
at all but oscillate like a pendulum as
illustrated in the figure; this is the easy
situation to understand qualitatively. It
could be described analytically because
there are only two variables, the angle of
swing of this pendulum, and the rotational
angle about the axis of the cylinder, but it
would not be easy; also, it would likely
require require small angle approximations
as in the simple pendulum. If the cone can
slip and roll, the problem is much too
difficult for the purposes of this site. I
can imagine it slipping and rolling, as you
have observed, and having an oscillating
behavior. For more detail, try this
link.
QUESTION:
let's assume that a race is taking place
between the speed of light and a person with
a stick. The stick is as long as the
earthsun distance, and it needs a
millimeter to touch the earth. My position
is in the sun. The goal of the race is for
one of these contestants to touch the earth.
So the contestant with the stick only needs
to touch the stick for the stick to touch
the earth. Which of the contestants will
win?
ANSWER:
Your question is very much like a
similar
earlier question. The answer is that the
light would touch the earth before the
stick, long before. When you push your end
of the stick to move it forward 1 mm, the
other end does not instantaneously move. The
information that you have moved your end
moves down the stick atom by atom with the
speed of sound in whatever the material you
have made the stick of. The speed of sound
is far slower than the speed of light.
Besides, the mass of such a stick would be
so enormous that it would take an enormous
force to provide the necessary acceleration
to move it forward in a relatively short
time.
QUESTION:
Why can't energy just be seen as a
measure of change? It seems to me that if
something changes then there must be a
compensatory change somewhere else. Consider
the case of a mass at a elevation of 3m
above the earth. What we call its potential
energy is both a measure of the amount of
change that occurred to raise the mass and a
measure of how much chage will occur if it
is released. Its kinetic energy is at each
moment as it falls is a measure of how much
change has occurred. More mass moved more
change; moving faster more change happening.
Thus, kinetic energy as a measure of change
would involve both mass and velocity as it
does. When it hits the ground and stops that
a change that needs to be compensated for.
The earth is deformed, i.e., other things
moved, i.e., had an increase in their
kinetic energy although this would probably
be measured as an increase in heat. The
object would also be hotter, as its own
atoms would have be jolted into motion, so
actually just kinetic energy, i.e., change
in motion. There would probably me some
sound energy also just kinetic energy of air
molecules, etc. Isn't this all just
different ways of measuring different
changes in different parts of a system? I
have seen answers to this question. All in
the negative, but they are unconvincing.
Mostly they say something like, "No, energy
is ... " or something really unhelpful like,
"Everything is energy"
ANSWER:
Your question is qualitative. Science is
primarily quantitative. I think you hit on
wellthought out ideas but they are
qualitative ideas and that may be useful for
understanding things conceptually, but not
for putting you in the position to study
nature
quantitatively. One of physics' most
important laws is that the total energy of
an isolated system never changes; indeed,
when energy is not conserved, i.e.
changing, that is important to measure and
understand why. But you are not really going
to appreciate that if you think that energy
is "...just...a
measure of change" are you? Science requires
clear quantitative definitions of the
variables which you can measure to test
whether your laws and hypotheses are true or
not. And I might ask what is the mass energy
of a particle, you know, E=mc^{2},
got to do with change? Or, how can the
energy density of a constant electric field
be understood "just" in terms of change?
QUESTION:
If I am trying to find what length is
needed for a pendulum for it to complete one
back and one forward motion to the beats per
second of a song, how would I do that?
ANSWER:
A simple pendulum is defined to be a
point mass M attached to a massless
string of length L. The period
T of such a pendulum is approximately
T=2π√(L/g)
where g is the acceleration due to
gravity which has a value of 9.8 m/s^{2}
(if L is in meters) or 32 ft/s^{2}
(if L is in feet). All times are in
seconds (s). Therefore L=T^{2}g/(4π^{2});
putting in the values for g and pi,
L=0.248T^{2} m or
L=0.811T^{2} ft. For
example, suppose that T=2 s; then
L=3.24 ft=38.9 inches.
All
this comes with a couple of provisos though.
First, there are no such things as point
masses or massless strings; this should be
pretty easy to ignore as long as the size of
the mass is much smaller than the length of
the string. Second, you will note that I
used the word approximately when describing
the equation for T. It turns out
that when deriving this equation you find
that the differential equation you need to
solve does not have a closed form solution,
you have to make the approximation that the
maximum angle of swing is small, for your
purposes you should probably keep the angle
of swing (from vertical) to be less than 30°
for the formulae to work well. By the way,
note that the equation for T does
not depend on either the mass or the
amplitude; for larger angles this is not the
case.
QUESTION:
In the twins paradox, twin A travels
away at nearly the speed of light and, some
time later, returns at nearly the speed of
light. Twin A has aged less than twin B.
However, since velocity is relative, twin A
sees twin B travel away and then return, so
twin B should have aged less than twin A.
Obviously both can't be right. The
difference is that twin A accelerated. Did
the rate of aging actually only reduce
during the periods of acceleration, i.e.
while twin A was subject to a force? If so,
is this related to the slowing of time while
subject to a gravitational force?
ANSWER:
The twin paradox is not a paradox at
all; it is easily understood. It has nothing
to do with the acceleration or gravity. What
you are missing is length contraction. The
trip is to some distant star and back. The
distance between this star, as measured by
the earthbound twin, is D. But the
traveling twin sees that distance to be
D√[1(v/c)^{2}]
because of length contraction. I urge you to
go to my
faq
page where you can find a link to my
discussion of the twin paradox.
QUESTION:
My question in regard to the
photoelectric effect. All electromagnetic
waves travel at the speed of light, yet
their frequencies can change. There's
something here I'm not getting.
ANSWER:
I don't see how this relates to the
photoelectric effect. Perhaps you are
thinking about the Compton effect where a
photon scatters from an electron and its
frequency after the collision is smaller
than before the collision? If the light has
some frequency f then the energy of
one photon is E=hf where h
is Planck's constant. And, although a photon
has no mass, its linear momentum is
p=hf/c where c is the speed of
light. Now, let's think classically for a
moment. Suppose one ball collides with
another which is at rest; after the
collision the incoming ball will be going
more slowly than before the collision, that
is, both its linear momentum and its energy
are smaller. The same is true for a photon
colliding with an electron and its energy
and momentum are both smaller and the only
way that can happen is if it has a lower
frequency. It is hard to swallow, at first,
that something can change its momentum
without changing its speed, but if it has no
mass, it may be hard to imagine it has any
momentum at all! But, that's the way it is
in special relativity.
QUESTION:
My question in regard to the
photoelectric effect. All electromagnetic
waves travel at the speed of light, yet
their frequencies can change. There's
something here I'm not getting.
ANSWER:
I don't see how this relates to the
photoelectric effect. Perhaps you are
thinking about the Compton effect where a
photon scatters from an electron and its
frequency after the collision is smaller
than before the collision? If the light has
some frequency f then the energy of
one photon is E=hf where h
is Planck's constant. And, although a photon
has no mass, its linear momentum is
p=hf/c where c is the speed of
light. Now, let's think classically for a
moment. Suppose one ball collides with
another which is at rest; after the
collision the incoming ball will be going
more slowly than before the collision, that
is, both its linear momentum and its energy
are smaller. The same is true for a photon
colliding with an electron and its energy
and momentum are both smaller and the only
way that can happen is if it has a lower
frequency. It is hard to swallow, at first,
that something can change its momentum
without changing its speed, but if it has no
mass, it may be hard to imagine it has any
momentum at all! But, that's the way it is
in special relativity.
QUESTION:
I had a little fun question. I was
watching an old Shrek
commercial and he threw donkey into
space. Donkey weighs 400 lbs mind you. How
strong would Shrek have to be and the force
needed to do that?
ANSWER:
So, I attached a picture from the video
showing Donkey just as he stopped and began
to fall back down. Let's make a guess that
he is at a distance 10 earth radii (R_{0})
from the center of the earth (i.e.,
at an altitude of 9R_{0}).
The potential energy U of mass
m at a distance r from the
center of the earth is U=MmG/r
where M is the mass of the earth
and G is the universal
gravitational constant. The donkey's total
energy at the instant that Shrek releases
him is his potential energy plus kinetic
energy, E_{1}=MmG/R_{0}+½mv^{2},
where v is the speed of Donkey; at
the peak of his trajectory his energy is
E_{2}=MmG/(10R_{0}).
If we neglect friction as he passes through
the atmosphere, the energies must eqal each
other; so if we equate them and put in all
the data we would need, M, m,
G, and R_{0}, we
can solve for the velocity. You are probably
not interested in all the arithmetic, so I
will skip it and give you the answer, v=7500
m/s=17,000 mph. Now you can find the kinetic
energy to be 5.1x10^{9} J=1400
kilowatthours. If Shrek exerts an average
force over, let's estimate, about 6 ft, the
force would be about 2.55x10^{9}
N=4.6x10^{11} lb. Even if Donkey
were made from the strongest material known
to mankind, there is no way he could survive
this force!.
QUESTION:
How much wind force would be required to
move an air cargo container measuring
6'x6'x6', with an empty weight of 138 pounds
sitting on a bed of rollers, ball bearings
or casters? Is it less than would be
required if sitting on the ground?
ANSWER:
So, it depends on the material it is
made from, what the "ground" is like, and
how good the the rollers are. I will assume
that the ground is very hard and horizontal.
If you were doing the pushing other than the
wind, there would be two things you could
ask: how hard must you push to get it moving
and then how hard to keep it moving at a
constant speed. In a situation like this
(horizontal) the force of friction (f)
is proportional to the weight (W),
f=μW where μ is called the
coefficient of
friction;
if μ is a small number it will be
easy to move, if large it will be hard.
There are two kinds of μ, static
and kinetic, which give you the force for
starting or keeping at a constant speed,
respectively; μ_{static}>μ_{kinetic}.
That takes care of the sliding part; now we
have to address the speed of the wind and
how much force it exerts on something. There
is a very useful way to estimate this force
¼Av^{2} where A is
the area presented to the wind in m^{2},
F is in Newtons (N), and v
is the wind speed in m/s. Since this is in
SI units, I will have to convert to your
Imperial units to SI units, W=138
lb=614 N, A=36 ft^{2} =3.34
m^{2}. So now, the push must be
equal to the frictional force, 614μ=3.34v^{2}
or v=√(184μ) m/s=30.4√μ
mph. We could now try some examples:

wood sliding on concrete, μ=0.62,
v=24 mph

steel sliding on steel, μ=0.8,
v=27 mph

nylon casters on concrete, μ=0.04,
v=6.1 mph
I have
included a graph of speeds as a function of
coefficient of friction.
QUESTION:
What are the physics principles of
throwing boiling water in extreme cold and
instantly becomes cloud (no snow i guess).
ANSWER:
See this
article and this
video.
QUESTION:
OK, maybe a stupid question, but it
doesn't leave me alone: If you are in a room
where all walls are mirrors and the only
light source inside the room is a flashlight
you are holding, then shouldn't the room
stay perfectly lit, when you turn the
flashlight off? ...at least for a decent
amount of time, as all the photons keep
bouncing around and only slowly get absorbed
(by the mirrors' imperfections and your own
body, including your eyes)? Intuitively, I
assume the room goes pitch black
immediately, as you turn off the light, but
I do not know why.
ANSWER:
The reason is that the light goes so
fast and loses a tiny bit of light with each
reflections but there are so many that the
light decays very fast. To see some
numerical details worked out, go to an
earlier answer.
QUESTION:
Can we see further around the Earth due
to spacetime curvature from gravity? Because
the equations for how far our visible
horizon is and my own personal experience,
don't quite seem to mesh.
ANSWER:
The curvature would be incredibly small,
far to observe using your "…own personal
experience…. I don't know what your
experience is but it is probably due to the
assumption that the earth is a perfect
sphere which is which it certainly is not,
particularly over small distances.
QUESTION:
Our class was discussing the observer
effect in the tripleslit experiment. I
understand it pretty well up until the
observer effect comes into play. If no one
is observing the outcome, how do you know
it's not doing the same thing as when you're
observing? This reminds me of the 'if a tree
falls in the woods, and no one is there to
hear it' riddle. It's not really a riddle as
it surely does make a sound be no one was
there to hear it.
ANSWER:
The "observer" is not just someone
standing around watching the experiment.
What it means is that you set up some kind
of measurement apparatus to determine
through which slit the particle passed. That
measurement will destroy the interference
pattern.
QUESTION:
Is it possible to entangle more than two
particles? Is there a limit?
ANSWER:
Not only is it possible, it has
been done. In principle, there is no
limit.
QUESTION:
Quick physics question! If the sun
starts to die could it be restarted with a
trillion nuclear bombs? I know it’s
impossible to make a trillion nuclear bombs,
but if it were possible and the sun was
dying.. what would happen if we tossed said
bombs into the sun? Would the sun explode or
will it reignite
ANSWER:
The reason a star dies is that it runs
out of fuel, mainly hydrogen and some
helium; so unless you can add enough
hydrogen to make a new sun, you cannot
kickstart it somehow because there is
nothing left to restart. One H bomb contains
about 300 kg of hydrogen. One trillion is 10^{12},
so a trillion bombs would add about 3x10^{14}
kg of hydrogen. The mass of the sun, mostly
hydrogen, is about 2x10^{30} kg. The
bombs have 16 orders of magnitude less mass
than the sun; so tiny are the bombs that the
sun would not even notice.
QUESTION:
I have an odd question I can't stop
thinking about; I'd love some help, and I'm
doing everything I can to find an answer to
this, including sending messages to multiple
people that may carry the ability to help
me. No this is not a botted message, I'm
writing this in person I assure you. taking
c as the speed of light. Now, considering an
Object moving away from me to my right at
60% of c, if it emits a light towards me
that light will travel towards me at 100% of
c itself. Now the light being emitted
towards me has a relative velocity of 160%
of c to the object it's emitting the light
from. Now let us name the previous object in
question object D, and let us take another
object; object B and say it's moving away
from me to my left at 60% of c. Now the
light being emitted from object D is
traveling at c relative to me, and 60% of c
relative to object B, and the light is also
traveling towards object B, and we can
safely assume it's bound to reach object B
at some point. This is from my perspective
as an observer (which we shall state as an
observer on object A) at rest relative to
objects B and D. Now as Special Relativity
states, as long as an object is moving at a
constant velocity it's safe to assume in
relation to another object that it very well
may be at rest and the other object may be
moving. Knowing Special Relativity doesn't
constrain relative speed if we shift our
frame of reference to object B, we observe
object A moving away from the speed of light
from object B but we do not observe object D
or any light emitted from it as it is moving
away from us faster than the speed of light
(I'm going to assume it's possible because
of relative speeds), however considering out
frame of reference from object A, we clearly
saw light from object D reach object B, here
lies my confusion. What is actually
happening? Do I lack a good understanding of
special relativity hence I give rise to
faulty and stupid questions? I'm eagerly
waiting for a response, please let me know.
ANSWER:
This question should really be thrown
out because it is neither concise nor
wellfocused as stipulated in the site
ground rules. However, I can address it
concisely so I will answer it. Yes, you do
"…lack a good understanding of special
relativity…"; the speed of light in vacuum
is the same for all observers. You should
see an
earlier answer discussing velocity
addition in special relativity. Using
v'=(u+v)/[1+(uv/c^{2})]
for your first situation, v=c and
u=0.6c, v'=1.6c/(1+0.6)=c;
both observers see the same speed of the
light.
QUESTION:
I am attempting to figure out the
pressure or impact the following would
exert. Picture a square plate (filter
plate), 48" X 48", 2" Thick, 125lb... These
plates are, hanging, supported by two
topmounted angle iron bars. The plate
supports have rollers at each end and the
plates while hanging can be slid for
cleaning. The Question: If one of that
support (hanging the plate) would fail, and
the plate would swing (one holder still
intact) approximately 2 ft. and strike an
object; what impact pressure and strike
force would be applied? Summary: Square
plate, appx. 125lbs. swinging appx. 4 ft.,
what would the impact be when the plate
strikes an object?
ANSWER:
I am not sure that I get the picture
from your description. Seems like the plate
rotates about a horizontal axis along one
edge and swings until it is vertical as
shown edgewise in my sketch. Is this right?
I will proceed assuming that this is right.
What I can do, assuming that there is
negligible friction of the pivot and the
plate, is calculate the angular velocity of
the plate when it is vertical, how fast it
is rotating. However, I should warn you that
it is impossible to answer your question
because the answer depends on the nature of
the collision between the plate and whatever
it is colliding with. Let me give you a
couple of examples to demonstrate that your
question cannot be answered without a lot
more information. Pressure is the force
divided by area. So if a small object is
struck and experiences some force F,
it will experience a larger pressure than a
large object experiencing the same force
would. Also, note that the speed of the
plate close to the pivot but gets larger
farther away from the pivot; therefore, the
pressure during colliosion increases as you
get farther from the pivot. Now, how do you
compute the force? When the plate hits some
object, the object experiences some average
force which is equal to the change in the
linear momentum Δmv (its mass m
times its speed v) divided by the
time t the collision acted, F=Δmv/t.
To illustrate, imagine two balls, each
having a mass m, one hard like a
billiard ball and another which is very
squishy. When the plate strikes the billiard
ball, the ball will take off almost
immediately, in a very short time, so the
average force over that time will very
large. When the plate strikes the squishy
ball, they will remain in contact far longer
so the average force will be much smaller.
This also is why if you fall onto concrete
you get hurt much more than if you fall onto
a mattress. Given the fact that to answer
your question, even approximately, is
impossible without more information, I will
stop now.
FOLLOWUP
QUESTION:
Let me give you just
a little more information if I may; Here's
the problem I have a worker standing on a
work platform, cleaning, and spraying
"Filter Plates", these are plates on a large
Water Treatment Filter Press. The plates are
"hung" by two topmounted hangers. They are
in line with each other, and cascade,
pressed to squeeze out water and
contaminants; about 20 plates on the unit.
Concern: If (and it has happened) one of the
plate "Hanging Brackets" breaks, get's
loosen the plate may lose its hanging
capability, not completely fall since the
other hanger is intact, but Swing Out and
possibly strike the worker while they are on
the work platform.
REPLY:
So, we have a uniform 4x4 square hanging
down vertically from supports approximately
at the two corners at the top. One corner
fails and the square rotates about a
horizontal axis through one corner and
normal to the plate? Keep in mind that, as I
explained in the original answer, it is not
possible to simply associate a force with
some moving object, you must have
information about what it hits and how.
Therefore you can only expect an
orderofmagnitude approximate calculation
making assumptions about the collision time
and the angle the square has rotated through
before hitting the worker. Your picture
shows the square to have not yet rotated
through 45^{°} and the corner
hitting the worker approximately in his
trunk. Are those approximately true? Would
it matter if I did my approximate
calculation assuming 45^{°} where
the square would have its greatest speed?
FOLLOWUP
QUESTION:
Subsequent exchanges
with the questioner brought up the following
important point: The rotating stops on its
own by the time it reaches the bottom where,
without friction, it would be moving the
fastest. This means that friction plays a
very important role in this problem. So I
have concluded that the best approach is
more qualitative.
ANSWER:
The figure shows the plate swinging from
one support only. L=4 ft and W=125
lb are the size and weight of the square.
The angle θ denotes how far the
plate plate has rotated from its initial
position, 45°; when θ=0° the center
of the plate is directly below the
suspension point. The length of a diagonal
of the square is 2L/√2 and the
weight acts at the center of the square
(assuming uniform mass distribution). I am
interested in what horizontal force (shown
as F in the
figure) must be applied to hold the plate
from rotating. I am ignoring friction for
now. Summing torques to zero about the pivot
point I find that F=W[sin(θ)/sin(θ+45°)/√2].
The graph shows F plotted as a
function of θ; the required force
is modest, 62.5 lb at the largest, and less
than 40 lb at smaller angles where striking
a worker is more likely. Furthermore, if it
were moving at smaller angles it would be
going slowly because of the frictional
force; also the frictional torque would be
opposing the torque due to the weight and
reduce the necessary force F needed
to bring the system into equilibrium. If it
is going slowly and the torque necessary to
stop it is rather small, I would conclude
that a worker could rather easly just reach
up and grab it to stop it. My only
reservation would be that the suspension
point at the pivot is not designed to be a
pivot and might not always behave the same;
if the friction were such that it would pass
the equilbrium position with a significant
velocity, it could be more dangerous.
QUESTION:
Holding a car FOB against your head will
increase its effective range. For example,
if you are trying to lock/ unlock your car
from a distance and the FOB battery isn't
strong enough to make a strong enough signal
to reach your car, you can hold your car key
against your head... which extends the
effective distance of the FOB by several/
dozens of yards, and may allow you to then
lock unlock your car if you are just out of
your FOB's range. There are many websites
which document this. I know it sounds crazy,
but it's real. What is your opinion
regarding how this trick amplifies the FOB
signal?
ANSWER:
It is not my "opinion", it definitely
works and the reason is very well
understood. The explanation is fairly long
and technical and involves the wavelength of
the beam from the FOB, the size of your
body, the size of your head, resonance, and
the dipolar property of water molecules.
There is a pretty clear yet entertaining
video which you can watch at this
llink.
QUESTION:
Hello, this is to settle an argument
between my brother and I and we can't find
an answer, so I hope you can help us. There
are two soda cans cold out of the fridge,
both at the same temperature and sitting in
the same environment (say on a coffee
table). One can has been opened and the
other remains sealed. Does the can that has
not been opened warm faster because its
still sealed and under pressure.
ANSWER:
Let me first address how heat is
transmitted to and through the interior of
the unopened can. We generally think of heat
transfer as occuring via three ways:
radiation, conduction, and convection.

The can has two surfaces, the bottom and
side, where heat may be conducted
through the aluminum directly to the
soda; the third surface, the top, has
aluminum in contact with the gas at the
top through which which heat may be
conducted to the surface of the soda.
The rate (R) of heat transfer
through a medium of thickness t
with a temperature difference ΔT
is proportional to ΔT
and inversely proportional to t,
R=σΔT/t,
where σ is the conductivity;
clearly, since the conductivity of the
gas is much smaller than for the
aluminum and its thickness much larger,
much less heat is conducted through the
top.
 Now, the soda near the inner
surface is warmer than it is inside, so
now the heat could conduct into the
cooler volume. But, in a fluid, heat is
much more efficiently distributed via
convection, currents in the fluid.
Similarly, heat through the gas at the
top of the can will occur much more
efficiently by convection.
 I don't believe that radiation plays
a particularly important role in the
heating of the soda.
So, now, what changes when the top is
opened? As you noted, the pressure in the
gas will decrease meaning a smaller density
and therefore a smaller conductivity. But,
convection will play the major role in the
gas and I believe that the gas will quickly
equilibrate with the outside air; then, the
temperature at the top surface of the soda
will be the same as outside air. Also, the
soda in the open can will be losing its
carbonation, but since there is only about
0.04 oz of CO_{2} in a 12 oz soda, I
would expect no significant changes in the
properties of the soda as it loses the CO_{2}.
So, I expect the
open can to warm up faster. But, I am not
absolutely positive and suspect that there
will not be a big difference. Here is what I
would do: get an instant read thermometer
and put it in the open can; then wait until
the soda gets to about ¾ of the room
temperature and quickly open the closed can
and measure the temperature. If you do this
experiment, please let me know the result!
QUESTION:
My question is: assuming space was a
perfect vacuum, (for simplification
purposes) as light travels through it, and
then slows down (however minutely)by
entering a planet's atmosphere, once it
passes through the amosphere and reenters a
vacuum, does it resume (accelerate to) its
original speed or does it remain at the
slower speed caused by the atmosphere?
ANSWER:
In any medium light has a speed v,
a frequency f, and a wavelength
λ. The energy of any single photon is
E=hf where h is Planck's
constant; also, v, λ, and
f are related by v=λf. The
energy of any photon cannot change, so f
remains constant regardless of v.
Therefore when the light moves from the
atmosphere to a vacuum, its wavelength
increases so that v=c where c
is the speed of light in a vacuum.
QUESTION:
Calculate the kinetic energy of a 5,000
kg meteoroid traveling at 11.2 km s1. If it
were to impact the surface of Earth at this
velocity, what would be the equivalent
strength of the explosion using the units of
Tons of TNT?
ANSWER:
The speed is very small compared to the
speed of light (c=3x108
m/s), so classical mechanics can be used.
The kinetic energy is K=½mv^{2}=3.14x10^{11}
J. 1 Joule (J) is 2.39x10^{10 }tons
of tnt, so the energy released is 3.14x10^{11}x2.39x10^{10}=75
tons of tnt.
QUESTION:
I watched the "trip to infinity" doc on
netflix and there was a part that said if
you put an apple in a box, that apple will
eventually decompose but the trapped energy
will eventually create something else. Can
the same concept be applied to humans and
rebirth? Is our body just energy that takes
another shape eventually after we die?
ANSWER:
Assuming you have a box where nothing
can go in or out, you would have an isolated
system consisting of the apple and whatever
air is inside. If there are no bacteria
present there would be no mechanism for
decomposition. Even so, the water in the
apple would evaporate out, eventually drying
the apple. One apparent energy transfer
would be in the form of heat and it would
warm up inside. But, the heat can't escape.
When the water evaporates, some chemistry
might happen inside the apple releasing heat
energy and changing the composition of what
is left of the apple. Etc., etc.,
etc. But you can be sure that the
apple will not eventually become a mouse or
a peach or even a new apple. Putting you in
the box would eventually result in the same
fate as the apple, but you can be sure that
rebirth is not in the cards. Of course,
there is no such thing as a perfect
isolating box and heat, the main result of
the apple's demise, would eventually leak
out and radiate away until the inside and
outside of the box were in thermal
equilibrium.
QUESTION:
In general science class years ago I
learned that when a balloon is inflated and
the stem released, the pressure of the air
in the balloon pressing against the front
side opposite the stem is greater than the
ambient air pressure in front of the balloon
and since the pressure at the stem is being
released, the balloon moves toward the lower
pressure area ahead of it. This is not hard
to understand for a balloon or rocket.
However, I don't understand how this
principle works in a jet engine. Since the
forward end of a jet (particularly a ram
jet) is open also, what is the interior
pressure pushing against to move the jet
forward?