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QUESTION:
I
suppose this is an off the wall
question, but it is quite practical,
and I’m not sure where to ask it. A
web search turns up nothing of
interest. I’ve convinced myself that
my coffee cup is interfering with
the 2.4GHz signal between my mouse
and its receiver, which is plugged
into the back of my display. The
spatial orientation of the three
objects is not ideal, but my desk
setup is a bit limited, basically
thus:
2.4GHz
wireless mouse on pull-out
keyboard tray, mouse on the
right.
My ceramic
coffee mug on the surface of the
desk, just above and beyond the
mouse, also on the right.
My display
is just above and beyond that on
the desk, with the mouse
receiver plugged into a USB port
in the middle of the back of the
display.
Ordinarily,
these three things tend
unconsciously toward an unfortunate
syzygy*. I suspect the coffee cup is
interfering with the signal, causing
severe mouse lag. So far
unexceptional. I don’t see why the
ceramic mug or its mostly water
contents might not be reflecting the
signal. But there doesn’t seem to be
any interference when the mug is
full. Only after drinking half or
more do I notice the lag. If I move
the mug way over to the left, that
seems to eliminate the lag. So my
question is, does the water level in
the mug change the overall resonant
frequency of the mug such that it
can interfere with the signal at
some levels but not others? Is it
like filling a jug to different
levels to get different tones when
you blow across the opening? And
considering that microwave ovens
work on a similar frequency by
heating water, would water be an
especially problematic interloper at
2.4GHz?
ANSWER:
Interference and blockage of
electromagnetic waves are tricky to
understand and track down. Like
sometimes when I am driving and come
to a stop there will be a specific
spot where a local radio station
will suddenly almost be totally
blocked. All the things you have
tried are reasonable: I salute you
because physics is, at its root, an
experimental science. If you could
tolerate moving your cup to the left
side, that is likely your best
solution. The 2.4 GHz seems like a
likely culprit here, particularly
since there seems to be negligible
blockage once the water is removed.
Cups of different construction would
be useful to look at, e.g. ,
a metal cup with and without water.
I doubt that it is anything to do
with interference like the blown
over bottle because sound waves are
of the same order-of-magnitude
wavelength as the resonating object
whereas the GHz waves are nowhere
close to the size of the cup. Good
luck!
*If you, are puzzled, as I was, by
the funny word
syzygy, it means when three or more
objects line up as in eclipses for
example.
QUESTION:
Sir,My
doubt is in POTENTIAL ENERGY .I am
really struggling in understanding
this topic .please help me out.Sir
my question is that -"is potential
energy just a term to make sums look
easier?" Sir, taking an example - if
we throw a ball with v velocity
perpendicularly upward then at the
highest point the velocity becomes
zero therefore the kinetic energy
becomes zero but at this point we
say that the energy has been
transferred into potential energy
but acutually where did the energy
go ? And thus can we say that we
defined potential energy so that our
famous rule does not gets violated
i.e. "energy can neither be created
or destroyed it can only be
transferred from one form to another
" is it so ??? What does it mean
that energy is stored???(potential
energy)?????
ANSWER:
My, you
really are struggling. It is clear
that you are just beginning your
physics education, so I will keep my
answer addressed only at that level.
In fact, I will only discuss problem
involving a uniform gravitational
field. First, suppose we know
nothing about potential energy. So,
the example you give is taking a
ball of mass m and throwing
it up vertically with some velocity
v . Then it starts with
energy E 1 =½mv 2
and rises to some height h
where all its energy is gone, E 2 =0.
So energy was not conserved; why
not? Because some force must have
done work on the mass to take its
energy. Of course, that force is
gravity which is mg .
Because the force is constant and
pointing down and the displacement
is positive is positive, the work is
negative, W=-mgh . The work
done is equal to the change in
energy W=E 2 -E 1 =-mgh =0-½mv 2 .
You could use this to find out how
high the ball will go, h=v 2 /(2g );
you have probably done such a
problem before you knew anything
about potential energy.
I next want
to generalize this problem. The ball
is thrown vertically from a height
y 1 with a
velocity v 1 and
at a later time it is at different
height y 2 and
now has a velocity v 2 .
So now the work done is -mg (y 2 -y 1 )
and the change in energy is
½mv 2 2 -½mv 1 2 .
So, rearranging, I can now write (½mv 2 2 +mgy 2 )-(½mv 1 2 +mgy 1 )=0.
If there are no forces other than
gravity which are doing work on the
ball, the quantity ½mv 2 +mgy
never changes! So now I get the
brilliant idea (just one way of
looking at this problem) to call
K =½mv 2 the
energy due to motion (kinetic
energy) and V=mgh the
energy due to position (potential
energy). So you can call
gravitational potential energy a
clever trick which allows you to
never worry about calculating the
work done by gravity. Another great
reason to use potential energy is
that energy is a scalar quantity
whereas force is a vector quantity.
There are
some things you have to be careful
of. If you are using V=mgy
for gravitational potential energy,
you must choose a coordinate system
where +y is up; e.g .,
if the top of a cliff is at y =0
m, the bottom of the cliff cannot be
at y =100 m. You cannot use
potential energy for just any old
force, it has to be a conservative
force. A conservative force is one
which does zero total work over a
closed path (ending where you
started). E.g., gravity
does negative work on the ball and
positive work on the way down. If
you push a book across a table you
must do work against friction; but
now if you push it back it will be
the same amount of work so you
cannot have a potential for
friction. Your "famous rule" is not
very useful. I think you should
think in terms of the energy
equation, E final =E initial +W ext ,
the energy you end up with is the
energy you started with plus what
you added. W ext
is the work done by all external
forces, forces for which you have
not introduced a potential energy;
forces represented by potential
energy functions are not external,
you have internalized them.
QUESTION:
I saw an
older question on your site
regarding how fast Flash could run a
quarter mile. Using similar
assumptions, in Science Fantasy -
verse Warhammer 40000, there are
Space Marines who are biologically
enhanced super warriors with 19-22
implanted extra organs that make
them able to overwhelmingly
outperform a regular human. Space
Marines are roughly 7-8.5t tall
(depending on if it is earlier
variant with 19 organs or later one
with 22 organs then the height is on
the taller end of the spectrum) with
a mass of 750 pounds unarmoured.
Space Marines have been decribed
running 87 km/h taking six meter
strides. Seeing that much shorter
and less massive Flash could run
quarter mile in ten seconds giving a
velocity of 144 km/h, using the
formula, how fast would you put the
maximum velocity a Space Marine
could run given their mass of 750
pounds? Can you give one estimate
using the rubber boot sole, and one
for ceramite boot sole that their
armour is made of? And when a Space
Marine wears armour, his mass is
increased to 500-1000kg, so how does
wearing armour affect the maximum
velocity when affected by laws of
friction, gravity, etc...?
ANSWER:
Oh my,
you missed the whole point of that
caluclation. The mass does not
matter because the frictional force
he can exert without slipping is
proportional to his mass. And his
acceleration is inversely
proportional to his mass; just the
same reason why the acceleration of
a falling mass (in vacuum) is
independent of mass. And strength
doesn't matter either if he can
exert the maximum frictional force
without slipping, μ static mg.
QUESTION:
How an
ellipsoid object, say an American
football...ball is thrown through
the air. It shows projectile motion,
but the other interesting
observation is that the front of the
ball points in the direction of
travel. I have thought that air
resistance possibly applies unequal
torques on different sides of the
ball perhaps, but I read somewhere
that it has to do with gyroscopic
precession - that this axis is
different to the one for angular
frequency, and somehow that explains
it. I'm not sure though, any simpler
way to understand this?
ANSWER:
If we
were to pass the football with
spiral in a vacuum, the spin axis
would remain pointing in the same
direction because of conservation of
angular momentum; this is the case
for the lower trajectory shown in
the figure. In air, though, the
angular momentum mainly remains
pointing in the same direction as
the velocity of the ball. I always
am surprised to find an everyday
phenomenon the physics of which is
unexplained or explained only
recently. This is such a case and
was explained only recently (2020)
in a
paper in American Journal of
Physics . A more accessible
description of the phenomenon can be
found at
this link . Your hunch that the
air has something to do with it was
spot-on.
QUESTION:
so, tell
me if I'm wrong but if an object is
energy it doesn't have mass yet it
is still affected by gravity, but
you need mas to be affected by
gravity 1. why is energy still
affected by gravity 2. is there any
way whatsoever for something without
mass that isn't energy be affected
by gravity.
ANSWER:
No, you
do not "need mass to be affected by
gravity". The
only object we know which has no
mass is the photon (a quantum of
light). And because it moves, it
has kinetic energy which is
well-known; you shouldn't say that a
photon is energy though,
rather it has energy. When
a photon finds itself in a strong
gravitational field it does not go
in a straight line (as you see
things) which you might have
expected, but rather it follows the
curved lines of the space where the
gravity is; in other words, in a
curved space a curved line may be
the shortest distance between two
points. Your second question refers
to something which has
neither mass nor energy; that isn't
really something , is it?
QUESTION:
I hope
this question isnt stupid, also
English isnt my first language, but
here I go: People always say that
you can't travel faster than light
speed, but what is the point from
which you count the speed (0 m/s)?
Because the earth is already moving
through space, so would we need to
accelerate less when accelerating in
the same direction and more if we
accelerate in the direction it comes
from?
Follow up, may
be explained by the answer to the
first question: does the light
emitting from let's say a LED
travelling at 60mph reach a target
100 miles away faster than a
staionary LED (assuming we turn both
on the second they are next to each
other, so when the moving one passes
it)?
ANSWER:
One of
the triumphs of the theory of
relativity is that its success
validates the principle of
relativity: The laws of physics are
the same in all frames of reference.
From this principle we can deduce
that the speed of light in a vacuum
is independent of the motions of the
observer or the source. So that
answers your 'followup': a bystander
would see the two light flashes
arrive at the same time. Your main
question asks which frame of
reference may a moving object not
exceed the speed of light? Since all
observers see the same speed of
light, the answer is any frame.
QUESTION:
Dad I
have a (possibly) physics question:
how come tempered glass just
spontaneously combusts without any
apparent cause? I remember my old
roommate's shower fully falling
apart during the pandemic when no
one was even there
ANSWER:
Hi
Hannah! I don't think you mean
"spontaneously combust" because that
would mean burning/flaming. Glass
does spontaneously shatter or break
in unexpected circumstances,
however. I have answered variations
of this question several times (
e.g. ,
1 ,
2 ,
3 );
1 is probably the most complete.
QUESTION:
I have
been thinking about a question, like
how we can write the change in unit
vectors over a finite time interval,
now it
may
seem like a homework problem but
trust me it's not it's a genuine
question that I have been thinking
about. I know about the change in
unit vectors in an infinitesimally
small-time interval 'dt' and that
it's given by the magnitude of the
really small angle that the unit
vector has turned about, but what
about a change in the unit vector
over a finite time interval how can
we write that. I tried to find the
answer in many sources, but I was
unable to find any.
ANSWER:
It is not
really clear to me what you are
asking. If you are talking about
Cartesian coordinates, they never
change. But if you have a coordinate
system like spherical polar
coordinates, the directions of the
unit vectors are very dependent on
where a point happens to be. I will
use a two dimnensional polar
coordinate system where the vectors
have both radial and tangential
components to keep the answer from
being too complicated. The figure
shows a red dot moving along some
path (red dotted line) which shows
the path taken by the particle
between two times t 1
and t 2 . The unit
vectors are shown in green. If you
know those two locations you can see
that the unit vectors all rotate
through an angle θ 2 -θ 1
in the clockwise direction. The path
taken by the particle will be
described by some vector function
f which
has components fr
and fθ as shown
in the figure. So the average rate
of change of direction of unit
vectors is ω avg =Δθ /Δt =(θ 2 -θ 1 )/(t 2 -t 1 ).
QUESTION:
A ship
at the speed of light turns on
headlights, understand that a person
on that ship sees light go out at
the speed of light, relativity. What
does a person "see" from outside
that area of relativity? Can they
see that, is that currently
possible, what if, if not current?
Do they see a ship with no lights,
light but no ship, a spectrum of
both that equals the speed of light?
I am, and with more questions, being
as specific as possible: I find it
hard to articulate a concept from
the written word, please accept and
understand my ignorance/lack of
understanding!
ANSWER:
First of
all, no ship can travel "…at the
speed of light…" and I do not answer
questions which stipulate at or
faster than the speed of light. So I
will answer your question assunming
v =0.99c , 99% the
speed of light. As you say, an
observer on the ship sees the light
in front of him having a speed of
c regardless of how fast
the ship is moving. An observer
relative to whom the ship is moving
with speed 0.99c also would
measure the speed of the light to be
c . This is one of the
postulates of the theory of special
relativity: The speed of light in a
vacuum is independent of the motion
of the source or the observer.
However he would not
"see"
the same "color" of the light
because of the doppler shift.
Suppose that the light as seen by
the ship had a wavelength of 450
nm=4.5x10-7 m, blue; for
0.99c speed of the source the
doppler shifted wavelength would be
about 4.5 nm, low-energy x-rays.
Hence the stationary observer would
see nothing but if he could measure
the speed of the x-rays coming out
the front of the ship he would see
them as having a speed of c .
QUESTION:
Lenses/Mirrors
form real / virtual imgs. Let's
consider real images. My query is
how can mere interaction of light
waves create the exact same picture
of the source from which they
originated? Does this mean that
light carries some information in
itself and when it interacts /
superimposes, it creates the
beautiful picture called an "image".
To elaborate, light itself is not
coloured, but why can we see
coloured images (when the light is
reflected from the object) ?
ANSWER:
Imagine
an object to the left of a spherical
lens as seen in the figure. There is
an imaginary line drawn
perpendicular to the lens and
through its center call ed the optic
axis. There is some object
consisting of many colors. I have
chosen to look at just three points
in this object, one red, one blue,
and one green. Each rays from these
points illuminates exactly one point
where the image is formed; although
I have only drawn three rays for
each point, every ray passes through
the the same single point on the
image. Apart from size and whether
the image is inverted or not, the
image is an exact copy of the
object. Your statement that "…light
itself is not coloured…" is not
correct--the color of light is
determined by its wave length.
QUESTION:
Assuming
no engineering challenges like
parasitic losses in materials etc….
Is it theoretically possible to
take: Speaker A. Play a tone. 440hz.
Digitally. Using EXACTLY 100 watts
of power. Speaker B. Play a tone.
440hz. Digitally. Using EXACTLY 100
watts of power. Have speakers A and
B facing each other head on. EXACTLY
180 degrees. The 2 speakers are
playing the same note. Same signal,
timbre, frequency amplitude pitch
etc…. Would these two “sounds”
cancel out perfectly such that “in
between” there would be zero
“sound”. I understand that small
changes in angles and frequencies
will have enormous effects on there
and how the sound will dissipate if
this system is not perfectly
balanced from the get go. Very loud.
But if the experiment were set up
perfectly, is it theoretically
possible to zero out the wave? Full
100% destructive interference?
Chaos? Feigenbaum?
ANSWER:
No, the
result is a standing wave of the
same frequency. In the animation the
original waves from the left (blue)
and right (green) combine to make
the black waves which have the same
frequency and wavelength but zero
velocity. This is due to the
superposition principle which states
that if two or more waves are moving
in the same medium, the net
displacement is equal to the sum of
the individual displacements at each
time and location. Note that the
black curve is sometimes totally
zero, when the two traveling waves
are out of phase with other. If two
identical waves are traveling in the
same direction they could completely
cancel if they were 180° out of
phase. In the second (nonanimated)
gif, imagine the green and blue are
moving in the same direction. The
black would be the resultant wave.
QUESTION:
I am
trying to understand heat, that is,
mostly the idea of hot and hotter.
My understanding is that heat is
basically motion, and heating
something up (cooking, let's say) is
about transferring motion from one
place to another. So that the
changes that take place in cooking
are really caused by the molecules
of stuff rubbing against each other.
So hotter really means MORE motion,
because the actual temperature at
which things burn, like natural gas,
can't be increased just because you
burn more of it. Does that make some
sort of sense to you? Clearly my
mind wanders in the kitchen, and
this has been nagging at me for a
long time.
ANSWER:
I believe
that your question is answered by an
earlier answer . You refer to
motion; in my earlier answer the
kinetic energy referred to is K =½mv 2
where m is the mass of some
atom or molecule in your flame or
pot or food, and v is its
speed (motion).
QUESTION:
I have a
question about fluid
velocity/acceleration and its
effects on tubular friction. Is the
relationship linear? Meaning , if
the fluid velocity is being
increased, is there a moment or a
few moments where the friction
increases more than what it would’ve
been if the relationship was linear?
ANSWER:
From what
I have been able to learn while
researching your question is that
"friction" as in two solids sliding
on their surfaces is not applicable
to fluid inside a pipe. The fluid
exactly on the surface is not
sliding but rather adheres to the
surface and is at rest. The motion
of remainder of the fluid has some
velocity just outside the surface
increases to a maximum at the center
but certainly not linearly.
Everything about this velocity
profile is due to the viscosity of
the fluid, sort of the fluid version
of friction. More detail can be seen
in
this link and links which it
contains. The figure shows a figure
from that link and shows the shape
of the velocity profile, v (r ).
Eventually, at high enough speeds
the fluid becomes turbulent and "all
bets are off".
QUESTION:
How
should I get to calculating the
force that needs to be applied to a
piston to compress a container of
gas? Suppose the temperature is
constant, and by default, as the
volume decreases the pressure
increases. Let's suppose the volume
gets halved.
ANSWER:
The force
depends on the area A of
the piston. And, if the gas cannot
be well approximated as an ideal
gas, you would need detailed
properties about the gas. For an
ideal gas, PV /T
must remain constant and if T
is constant, PV is
constant. Now, it seems easy, like
if you halve the the volume you need
to double the pressure; that would
give you the answer for the force to
hold the piston there once you got
it there, F=P 2 A= 2P 1 A .
Be sure that you understand that
this is what you mean because if you
simply started at the beginning with
this force and continued pushing
with that force, when you got to the
half-volume position you would go
right by it and it would start
slowing down until finally it would
turn around and start coming back.
QUESTION:
For
reference I'm 31. Am I closer to the
age 60 than I am 20? I think I'm
closer to the age 60 because time is
moving forward and we cannot move
backwards to my age at 20. So
wouldn't I be closer to 60 because
time can only move in a linear
motion towards 60 and we are
constantly distancing ourselves from
the age of 20?
ANSWER:
You are
11 years older than 20 and 29 years
younger than 60. You are closer to
20 by almost a factor of 3. If you
had asked which you would be closer
to being, you would be closer to
being 60 than 20; but you didn't.
QUESTION:
Is a
mirror VERY different to how people
perceive me in reality? I am
bothered by the fact that I never
realised before but I guess there's
a reason… How impactful is the
reversal effect? I always thought
that's how I look to others.
ANSWER:
The image
you see is not how others see you,
but a "mirror image". If you look at
the image of yourself in the mirror,
the image of your right hand is the
left hand of the mirror you. If you
want to see an image of yourself
which is what others see, you need
to use a corner mirror. However,
since most of us are pretty much
left-right symmetrical, there is
seldom much difference.
QUESTION:
My eight
year old has a question. Get ready.
If two particles are in entangled,
and one is on the earth and the
other has passed the event horizon
of a black hole. Would the
entanglement be broken or even if
it's unmeasurable still be
entangled. Does the medium "the
ether" transcend the event horizon?
Silas Massicotte (8) and Malachi
Massicotte (6).
ANSWER:
Seriously? Your 8-year old
understands quantum mechanics and
cosmology?! That would be pretty
amazing. I usually don't do, as
stated on the site,
astronomy/astrophysics/cosmology,
but I did a little research and
think I can give you an answer. When
one of the two entangled particles
is totally destroyed, the other one
will not know at all because there
is no communication between the two.
Since the wave function of the
destroyed one simply disappeared,
the wave function of the other will
remain in the mixed state it was in
originally, but it will no longer be
entangled with anything. If a
measurement is made on the survivor,
there will be no change in whatever
remains of the other particle. There
is a pretty good explanation at this
link . There is no such thing as
the ether.
QUESTION:
how can
electromagnetic waves travel in
vacuum? how can something without
any physical body and dimensions
travel in nothing? it should need
something to mediate its propagation
like sound waves require air
molecules. if its pure energy then
what is pure energy?
ANSWER:
This is
exactly the way physicists were
thinking at the end of the 19th
century. Serious searches for the
medium required for light to be able
to propogate were being made. The
best known search was the
Michaelson-Morley experiment which
failed to find it. It was called the
luminferous æther. The punch line is
that it doesn't exist at all. Light
can travel through empty space. One
way to understand this is through
electro-magnetic theory, quite well
understood at the begining of the
20th century. James Clerk Maxwell
unified all that was known about
electromagnetism into four
equations, today known as
Maxwell's equations . One of
those equations says that
time-varying magnetic fields cause
electric fields; another says that
time-varying electric fields cause
magnetic fields. Using all four you
can show that one possible solution
is a wave equation for both electric
and magnetic fields. Since waves are
time-varying, these waves will "keep
each other going" which has nothing
to do with any medium. If you go to
the link above, you can follow other
links telling you more.
QUESTION:
I have a question
about a calculation in the USPSA (United States
Practical Shooting Association) rulebook. The
power factor of ammunition is measured by the
weight of the bullet in grains times the
velocity in feet per second, divided by 1000. A
165 grain bullet at 1000 fps would be 165 power
factor. This is intended to keep recoil the same
between competitors. Felt recoil is subject to a
lot of variables that have more to do with the
firearm than anything else. But the rule is to
ensure that everyone has a minimum level of
recoil to manage. Does a 110 grain bullet at
1500 fps, or 165 power factor exert the same
force backwards as a 250 grain bullet at 660
fps, or 165 power factor? All else being equal,
is power factor an accurate measure of recoil?
The muzzle energy is much higher for the lighter
bullet.
ANSWER:
The quantity power
factor is nothing more than linear momentum
(mass times velocity) in physics but in unusual
units, milligrain feet per second: mgr·ft/s);
physicists usually express it in kg·m/s. But the
units you use don't affect the physics so I will
use yours. The thing about linear momentum is
that if there are no external forces on a
system, it never changes its magnitude; this is
called conservation of linear momentum. So, if
the rifle has a mass of M =3 lb=21,000
gr=21,000,000 mgr and the two bullets both have
equal power factors of 165 you can calculate the
speed V of the rifle after the gun is
fired: Before the gun is fired momentum is zero,
so it must be zero afterwards: 0=165-21,000·V
or V =0.00786 ft/s=0.094 in/s. Now, you
asked for the force on the shooter. There is not
enough information to calculate this because
that depends on how quickly the rifle takes to
stop—if you have a bony shoulder it would hurt a
lot less if you stopped it more slowly by having
a squishy pad between you and the rifle butt.
Suppose it takes 1 ms to stop; then the average
force on you over that time would be [3 lb x
0.00786 ft/s / 0.001 s] = 24 lb. But you said
"all else being equal" so I presume that the
recoil force would be the same for both
scenarios because the masses of the guns and the
recoil velocities are the same.
QUESTION:
When text book gets
to quanta, it would say E=hf. It would say f = n
* fundamental frequency. Ok, I can understand
there are 1Hz and 2Hz and so on. But couldn’t
one build a circuit that makes 1.5Hz? At the
end, it is a voltage completing a cycle within
some time. And if f can be decimal, then it is
continuous. So E is continuous. I must be
missing some concept here.
ANSWER:
Not everything is
quantized, but even if everything were in some
way quantized, you could only detect with your
senses in some situations. Usually we think of
quantum physics as what happens in very small
systems, systems on the general order of atomic
sizes. The thing which is bothering you, I
think, is that something which is macroscopic
which shows no signs of being quantized means
that an a microscopic version of that system
would not be quantized either. Take, for
instance, a mass m hanging on a spring. So if
you pull it down to some amplitude A ,
say 1 cm and let it go, it will oscillate up and
down with some frequency f determined
by the stiffenss of the spring k and
its energy will be E =½kA 2
and its frequency will be
f =2 π √(k /m ).
You could quadruple the energy by doubling the
amplitude. But, of course, you would be able
choose any amplitude you liked? Like 1.2379 cm?
Of course, assuming you could actually measure
that accurately. So quantum mechanics is
nonsense, right? No, I could contend that the
macroscopic oscillator is also quantized
(theoretically at least). What is the spacing of
the energy levels of this quantum system? hf ,
of course. what is hf if f =1
cycle/second? ΔE =6.6x10-34
J! Suppose that the spring in this oscillator
had a spring constant k =1 N/m; then,
since E =½kA 2 , going
to the next allowed level would require a change
of amplitude of ΔA =√[2x6.6x10-34 /1]=3.6x10-17
m. That is about 1/10 the size of an atomic
nucleus! I don't think you would notice this
constraint!
Your intuition tells you that
quantum mechanics can't be right because you
have never noticed its effects. What is your
intuition based on? Your experience! Do you have
any experience regarding such tiny sized things?
QUESTION:
I am working on a
piece of science-fiction centred around the mid
– career physicist, upon whose desk objects
appear for significantly timed short periods, a
significant pattern. At some point in the peace,
the working assumption becomes That these
objects are being transmitted from "Parallel
universe"/other reality in the Multiverse. This
may seem to be an odd question if such objects
were assumed to be capable of being transmitted
between parallel universes or other reality
planes in a Multiverse, do you think they would
be capable of being photographed and or scanned
in some other forms such as x-ray?
ANSWER:
Of
course, this is all speculative, but I would say
that if your eye can see them a camera could see
them because the eye is, essentially a camera.
But you could also say that maybe the eye is not
really seeing them but rather only your brain
was seeing them, like a halucination. I remind
you that I do not usually do cosmology, the
realm in which multiverse is included.
QUESTION:
I have a theory (or
just an idea) concerning faster than light
travel that I would like you to answer. My idea
is that if FTL travel is possible, then the
energy required to make a warp drive, wormhole,
etc. work is exactly the same amount of energy
as would be required based on Newtonian physics
if there were no law of relativity. For example,
suppose you have a 1000 metric ton FTL spaceship
that can travel four light years in four days.
Please calculate how much energy it would take
to accomplish this if only Newtonian physics
were involved. Then look up some FTL scheme.
Here’s one you can use
https://newatlas.com/physics/ftl-warp-drive-no-negative-energy/
Or just pick another than you prefer. Then
calculate how much energy it would take to move
the 1000 metric ton spaceship four light years
in four days. My idea is that perhaps the
universe is designed in such a way that the
energy required is the same. In other words even
if FTL travel is possible, you don’t save any
energy by using it. In other words, there is no
“free lunch.” I donated $20 for your answer,
since I suspect this will take a lot of work.
REPLY:
(Other readers note: I am only answering
this question because the questioner has sent a
donation.) It would been good
if you had read the site ground rules before you
had sent me a donation since I have no way to
refund it. No faster than light questions is one
of the ground rules. You can also be sure that
anybody designing a "warp drive" is liable to be
a crackpot and the amount of work I would have
to do to evaluate any such design would be
unreasonably large. I am willing to compute the
energy using Newtonian mechanics and also
calulate the energy to do your four-day trip
relativistically. I will assume negiligable time
to get up to speed.
ANSWER:
One light year is 365 light days. Using
Newtonian mechanics, the speed you would have to
go 4 ly in 1 day is 365 times the speed of
light, 365x3x108 =1.1x1011
m/s. The energy which a your spaceship would
have at that speed is
E =½mv 2 =½x106 x(1.1x1011 )2 =6.2x1027
J=1.7x1021 kW·h.
Relativistically, the
distance 1 light year (ly) would have to be
length-contracted to 1 light day (ld) would be
1=365√(1-(v 2 /c 2 ))
or v =0.999992494c or β 2 =0.999984988.
Now, total energy is E =γmc 2
where γ =1/√(1-β 2 )=258.
But we don't want the total energy, we want the
kinetic which is K=E-mc 2 =(γ- 1)mc 2 =257x1.x106 x(3x108 )2 =2.31x1025
J=6.42x1018 kW·h.
So you see, contrary to
your idea, it takes less energy to go
relativistically than classically.
QUESTION:
I am a new and
upcoming math and physics teacher at a local
high school. A more senior physics teacher posed
this question to me before he retired, and I am
rather embarrassed that I cannot solve it.
Apparently, he said this system is accelerating
and that acceleration varies with the angle.
[COMMENT:
The questioner shared a copy of a
textbook problem with m =50 kg, M =100
kg, θ =13°, μ s =0.41,
and μ k =0.35. I chose to do
it in general and we can stick those numbers in
later.]
ANSWER:
I would first choose the knot (massless) as the
body; this is a good idea because we don't know
whether the system is accelerating or not yet,
but it doesn't matter because all the forces on
the knot must equal zero because it is massless.
Because of the symmetry of the system, the
tensions in the strings attached to m
and m must have the same magnitudes,
t 1 =t 2 =t ,
so we have
ΣFy =0=-T +2t sinθ .
The x -equation
just tells us what we already know, that the
x -components cancel each other out. We now
have one equation with two unknowns, T
and t , so we need to choose another
body. I choose m on the left.
ΣF y =0=N-mg-t sinθ
ΣFx =ma =t cosθ -f
f=-μ k N
if m is moving or f=-t cosθ if t cosθ <μ s N
Now
we have five unknowns, T, t, N, a, and
f but only 4 equations. The other m
is not going to give us any new information
because of the symmetry, so we are left with
only M . This gives another equation,
ΣF y =MA=Mg-T .
Note that I have chosen +y
to be down so that m and M are
moving in their positive directions. The sixth
equation will be the relationship between a
and A,
a=A ·tanθ.
I also calculated that
the net force on m goes to zero at 55°
so the static coeffecient of friction would kick
in and the system would stop.
t=Mg /(2sinθ )
f=μ k (mg+Mg /2)
t cosθ-f =0
and therefore,
θ =tan-1 (500/350)=55°.
I
leave it to the reader to do the algebra above.
When I did the calculation using
WolframAlpha it gave me confidence that
the equations of motion are all correct since
A =0 at 55°.
So I now had 4 equations
with 4 unknowns (I incorporated the a-A
and the f-N relations to reduce the
number of equations and unknowns to 4.) I put in
the value of θ =13° and the solution is
shown below, calculated in
WolframAlpha . I should note that I
checked that the system is actually accelerating
and not stuck by static friction at θ =13°.
The
results of calculations over the whole range
0-55 are shown in the graph below.
ADDED
THOUGHTS: Just a few comments. I
have to admit that I have not rigorously shown
that a=A tanθ although I did
enough qualitative thought to the relation that
I feel pretty confident. Also, the behaviors of
all the variables are just what I expect them to
be using that relation. It is interesting that
at θ =0, because when we usually think
if a weight hanging on a horizontal string, that
the string would have to have infinite tension
to not deviate from the horizontal; that
situation, though, requires the (massless)
string to be tied between two immovable walls.
The equations here have a kinetic frictional
force acting on the "walls" so they are not
immovable. Since the string at 0° has no
vertical component, T must be 0 N, as
the calculation shows; M must therefore
be in free fall for the instant that θ =0°.
Since the frictional force is 0.35x490=-171.5 N,
the solution finds that t =171.5 N also;
I guess technically t could be anything
and still have a= 0 because of the
tangent factor, but small angle nonzero θ
will have whatever t is required.
At θ= 55° everything has stopped and
must be in equilibrium: both N and
T are at 980 N and t =980x0.35/cos(55)=598
N. Because all these accurately describe what we
know they should actually be, I am pretty
confident that a=A tanθ is
correct.
QUESTION:
My understanding is
that atomic clocks circling the earth run 1)
slower because they are moving fast but 2)
faster because they 'feel' less of earth's
gravity than clocks on the surface of the earth.
Which wins out, speed or gravity? How much is
the time dilation/contraction due to gravity and
how much due to speed? Could there be a planet
of some mass that you could put a satellite in
orbit around at a certain speed so that the two
would balance out so that the orbiting clock
would measure time at the SAME rate as one on
the planet's surface? And, how much math would I
need to figure out the answer to that last
question?
ANSWER:
I started trying to use the expressions for the
two and doing the algebra demanding that their
magnitudes be equal. But then I stumbled on this
figure which addresses exactly what you were
asking. The red curve is the kinematic time
dilation (slowdown) for circular orbits. The
green curve is the speedup due to gravitational
time dilation. Also shown, in blue, is the sum
of the two, the function which you wanted to be
zero somewhere. Lo and behold! There is a spot,
looks like it is about one third of an earth
radius above the surface. Note that where GPS
satellites are, the gravitational time dilation
is much more important. If your GPS didn't have
general relativity corrections built into it, it
would not work. The math would not be very
demanding, but the physics might be. You need to
have an analytic expression for the blue curve
and set it equal to zero and solve for r .
Everything below in
this answer is incorrect. I was misled by an AI
error, several incorrect internet posts, undue
confidence in myself to be able to do the GR
calculation! It is an excellent example of how
incorrect an internet post can be and how AI
probably learns from its memorizing everything
on the web. I did more research because it
seemed extremely odd that the geosynchronous
orbit and the zero dilation orbit would be the
same and the fact that my incorrect answer was
about 10% smaller than the actual geosynchronous
orbit.
I am leaving it posted
as an example of how wrong the internet can be!
I
really liked this question because it was one of
those questions which gave me the opportunity to
learn something new. I wanted to understand
everything which went on in making the graph
above. Well, I worked on it for a number of
hours and finally got where I understood what
should be the functions plotted. I got this
graph from the
Wikepedia article on gravitational time
dilation; I would recommend it because it gives
the form of the gravitational dilation (I leave
it to the reader to understand the notation
here. t 0 is at the surface
of the earth and t f is at
some distance r from the center of the
earth):
t 0 =t f √[1-(2GM /(rc 2 ))]
What needs to be plotted
is (t 0 -t f )/t f
which must be zero on the surface. So I
conclude that
(t 0 -t f )/t f =√[1-(2GM /(rc 2 ))]-√[1-(2GM /(Rc 2 ))]
The kinematical time
dilation turned out to be easier for a circular
orbit of radius r. Without any details except to
note that GM /r=v 2
(t 0 -t f )/t f =-1/√[1-(GM /(rc 2 ))]+1,
ensuring that an orbit at r=∞ have zero speed.
So, here's my
calculation. The kinetic calculation agrees very
well with the Wikepedia graph but the
gravitational one dies not. The result is that
the zero net dilation is at about 6 earth radii,
but the Wikepedia calculation was at about 1.3
earth radii, a really big discrepancy. I stared
and stared and could find an error. So I decided
that I would ask AI about this; here is my
interaction:
Six earth radii is
approximately where the geosynchronous orbit,
where all the communications satellites are. It
would appear that my calculations are right. I
did check by googling and the usual answer was
the same as ChatGPT's.
I think AI got it
backward, though: kinematic dilation is slower,
gravitational dilation is faster.
QUESTION:
Picture an attic
that covers a house and its driveway. The area
of the driveway and the house are the same, so
they are each covered by 50% of the attic. In
the center of both the house and the driveway
are 2 identical circular holes into the attic
above. The temperature in the house is 68°F,
attic is 86°F and driveway 104°F. Since warmer
air flows upwards, will the driveway air get
into the attic, or will the colder air in the
house work as a barrier to keep the attic air
from coming down?
ANSWER:
There are a few things we need to understand
before we get to the answer:
The zeroth law of
thermodynamics is that heat always flows
from a higher temperature to a lower
temperature.
The zeroth law is
true for an isolated system. If there is
some other agent which acts to add,
subtract, or redistribute the heat energy,
it may not be true.
Particularly for
gases, the notion of "warm air rises" is a
result of the fact that if you increase the
temperature the volume increases and so the
gas becomes less dense; it rises because of
gravity so the warm "floats" on the cold
just like wood floats on water.
There are three ways
that heat is transferred—radiation,
conduction, and convection.
First, lets consider
these systems are totally isolated—no gravity,
perfect insulation all over the outsides, no
furnaces, ACs, etc . Looking at all
three at once would muddy the water, so I will
imagine there is a perfect insulating wall in
the middle of the attic. Heat would flow from
the hot to the cold by convection through the
holes and conduction through the ceilings.
Eventually all the air would be at the same
temperature, somewhere between the two starting
temperatures, which would depend on the amount
of air in each of the two volumes. The same
thing would happen for the garage/attic except
heat would flow up, not down. Next, turn on
gravity; then there would be a continuous
increasing temperature from bottom to top. The
ceiling would have the same temperature top and
bottom. Similarly for the garage/attic half. You
might want to imagine that the ceilings are
perfectly insulating; the net result would be
the same but only convection through the holes
would transfer the heat.
Finally, your scenario,
all three volumes connected through the holes
and no wall in the middle of the attic. With no
gravity everything would end up with the same
temperature. With gravity added, there would
then be a continuous increase from floors to the
roof of the attic. Don't forget that this is the
real world and any sources or sinks of heat will
change the details.
QUESTION:
I have searched the
internet and not found an answer. If there are 2
identical blocks of ice. One is at -1 Degrees C
One is at -50 Degrees C Do they melt at
different rates? Will the colder piece take
longer to melt? I’m assuming the colder piece
will take longer, but I don’t actually know.
ANSWER:
I am assuming that each piece is in the same
environment, say a room at constant temperature
T somewhere above the freezing point or
below the boiling point of water. Heat flows
from a higher temperature environment to a lower
temperature environment. When the heat flows
into the lower tempeature environment (the ices
in your case) it is absorbed and can do two
things: it increases the temperature or it can
cause a phase change (melting, freezing,
condensing, or evaporating) without a change of
temperature. (You might want to read a
recent answer to a
question similar to yours.) At the beginning of
the experiment each piece of ice will heat by
increasing its temperature. The 1°C piece will
get to 0°C in some short time, t 1 ,
then energy coming in is used to melt the ice
until it is fully melted in a time t 2 .
As the ice is melting, the water from the
melting will start warming while the remaining
ice will continue to melt without warming.
Eventually all the water will be heated to the
temperature of the room; call the time from when
all the ice was melted until all the water was
at T to be t 3 . It
will take the first piece t 1 +t 2 +t 3
to get to the end of the experiment. The only
difference between the two pieces is that it
will take the colder piece fifty times longer to
get to 0°C. So it gets to the end of the
experiment in a time 50t 1 +t 2 +t 3 .
The times for the two pieces to completely melt
will be the same, but it will take the colder
one longer before it starts melting.
QUESTION:
In the known
universe, is there anything else in existence
besides matter, waves, forces, movement and
change? In other words what are the very basic
essential physical things and the effects,
excluding any measures that we seem to give a
reality to quite often? Maybe waves could
exclude matter or maybe waves are just matter in
motion? Space itself seems not to be truly part
of existence? Time to me is a measure of forced
motion and change and the same with energy. What
are the bare essentials of the known universe
that I can detect through my senses, not just
concepts alone? Would you include fields? What
are the indisputables?
ANSWER:
This is a strange a very strange question. There
is only one way I know to encompus everything
which exists. Mass is energy, light is energy,
electric and magnetic fields are energy, motion
of a mass is energy, etc .. The answer
is that the universe is ENERGY.
REVISED
ANSWER: It occurs to me that time
also exists. Time is not energy, so the revised
answer is ENERGY and TIME.
FOLLOWUP
QUESTION:
What is energy and where has it come from? Does
the ability to do work come from energy or the
forces or both? It seems to me that of all the
forms, shapes, colours, sizes, amounts of energy
that can be stored or transferred but not
created or destroyed.. energy is a quantitative
measure of the movement that forces create on an
object or particle. Could energy be dispersing
waves of just motion through fields since the
big bang. Anyway, if not a measure, what is it?
Please not the ability to do work.
ANSWER:
You seem to have a talent for submitting
questions which, contrary to
site ground rules , are not "…single,
concise, well-focused questions…" You are
expecting me to give you a complete tutorial on
energy, something which takes a couple of years
of classes to fully understand. Essentially we
invent a quantity which, for any isolated
system, is conserved, i.e . remains
constant no matter what happens inside it. By
being isolated we mean that there are no
external forces doing work on the system.
[Sorry, there is no way to talk about energy
without mentioning work.] Conceptually,
isolated means any system to which energy
is neither being added to nor taken from the
system. Assuming that the entire universe is all
that there is (or not interacting with other
systems outside the universe), the total energy
of the universe never changes.
QUESTION:
While water is
freezing, energy is being released in the form
of heat- how does this jive with a temperature
plateau during the freezing phase change?
ANSWER:
Suppose that we have a glass of water which has
been put into a freezer which is at -10°C. Heat
will flow from the relatively warmer glass of
water to the cold freezer and will and will be
removed from the freezer into the room so as to
keep the freezer at the set temperature. After a
couple of hours all the water is frozen at a
temperature of -10°C. Now there is a power
faliure and, since no freezer is perfectly
insulated, heat will leak in from the room so
the temperature will will start warming. When
the temperature has risen to 0°C heat will
continue to leak in and the air will continue
heating and the ice will start melting. The air
is using the heat to increase its temperature
but the ice will use the heat to melt the water,
not to increase its temperature. When all the
ice has melted, it water will start increasing
in temperature.
Well, I see that my
little story has not answered your question, it
is the reversal. So my story continues: The
water and the air are now at some temperature,
say +10°C. Now the power comes back on and the
water and air start cooling, the freezer
removing heat from both. When the water gets to
0°C it starts freezing and the heat being
removed results in freezing, not cooling; but
the air continues to cool. Finally when all the
water is frozen the ice now cools because heat
is being removed by the freezer. Eventually
everything is at -10°C again.
QUESTION:
I have a question
about time. I am a nursing student, taking
anatomy courses and learning the atomic
physiology of bodily processes. For example, how
ions flow in and out of heart cells to make it
contract. The whole process takes so much time
and thought to track it all out, when in
reality, it takes .8 seconds to actually
complete a cycle. My question is, how fast are
these atoms moving in order to move at the speed
to complete our bodily processes in the small
fraction of a time we experience as humans? Or,
is it that atoms are so small, they experience
time different than we do? or is that not an
accurate thought? thanks for the help!
ANSWER:
do not know much about electric currents in the
heart, but I do understand electric currents in
a conducting wire like copper. In a wire, when
you have a voltage between two ends of a wire,
electrons will flow in the wire with some
average speed v. It is a common exercise in an
introductory physics class to have the students
estimate the speed with which those electrons
move in the wire; a typical answer for household
wiring is about ½ inches per minute! So, if the
switch for a lamp is 10 feet from the lamp, it
would take an electron starting at the switch
(10 ft)x[(12 in)/(1 ft)]/(0.5 minutes)=240
minutes=4 hours to get to the light. This is, of
course, nonsense. What actually happens is that
when the switch is turned on there is a voltage
across the ends of the wire which establishes an
electric field* in the wire; the other end
receives the information that the field is there
at the speed of light, almost instantly at 10
feet. So all the electrons in the whole wire are
moving, however slowly, almost immediately.
Hearts are not conducting
wires but they can conduct charge carriers
because flesh is an electrolyte which plays the
roll of conductor and the charge carriers are
ions (atoms or molecules either missing an
electron or carrying an extra electron). So when
some action in the heart requires a message to
be sent from point A to point B, a voltage
between A and B is turned on at A resulting
almost immediately in the pulse arriving at B;
an ion created at A does not get sent
to B. I believe that calcium ions are important
charge carriers in the heart; I take a drug
which is called a Ca channel blocker but I
couldn't tell you in any detail how it works.
So the answer to your
question is that the ions move very slowly but
the information which they carry arrives very,
very quickly.
*An electric field is the
thing which causes there to be a force on the
charge carriers which drives the electric
current.
QUESTION:
When light goes
from, say, air to glass or glass to air at a
non-90° angle, it's refracted. Does that use
energy? Are the air and the glass heated? And if
so, is that what's happening when a microwave
oven heats things the most at the interface
between different things (like the ceramic plate
and the ham).
ANSWER:
Read the Q&A right after yours. It should be
clear that the photon emerging from the first
medium has the same energy as when it entered
that medium. Therefore no net energy has been
lost at boundaries. However, real materials also
absorb some of the photons so a net loss of
energy occurs when an ensemble of photons
traverse the medium and it shows up as thermal
energy.
QUESTION:
Concerns the speed
of light in a medium. I understand the speed of
light varies slightly with hot and cold earth
atmospheres, primarily (apparently) with
density, as it does in other mediums.
Intergalactic space is a medium, however
tenuous, and not a true vacuum. It would
therefore seem that the speed of light would
vary slightly even in it. The standard speed of
light is calculated as in a true vacuum. Is this
an answerable question?
ANSWER:
To understand refraction in classical E&M is
difficult; essentially the alternating electric
fields induce vibrations in the atoms in the
medium and the result of complicated
calculations is that the speed of the wave
changes. But it is easier to understand, at
least qualitatively, in terms of photons; the
photons are absorbed by atoms and promptly
re-emitted. This takes a tiny amount of time and
is repeated enough times to cause a change in
the average speed of the photons. In
intergalactic space the density of hydrogen is
about one atom per m3 . The likelihood
of there being an interaction which would
measurably change the average speed of a single
photon are about as close to infinitesimal as we
could imagine.
QUESTION:
When light goes
from, say, air to glass or glass to air at a
non-90° angle, it's refracted. Does that use
energy? Are the air and the glass heated? And if
so, is that what's happening when a microwave
oven heats things the most at the interface
between different things (like the ceramic plate
and the ham).
ANSWER:
Read the Q&A right after yours. It should be
clear that the photon emerging from the first
medium has the same energy as when it entered
that medium. Therefore no net energy has been
lost at boundaries. However, real materials also
absorb some of the photons so a net loss of
energy occurs when an ensemble of photons
traverse the medium and it shows up as thermal
energy.
QUESTION:
Concerns the speed
of light in a medium. I understand the speed of
light varies slightly with hot and cold earth
atmospheres, primarily (apparently) with
density, as it does in other mediums.
Intergalactic space is a medium, however
tenuous, and not a true vacuum. It would
therefore seem that the speed of light would
vary slightly even in it. The standard speed of
light is calculated as in a true vacuum. Is this
an answerable question?
ANSWER:
To understand refraction in classical E&M is
difficult; essentially the alternating electric
fields induce vibrations in the atoms in the
medium and the result of complicated
calculations is that the speed of the wave
changes. But it is easier to understand, at
least qualitatively, in terms of photons; the
photons are absorbed by atoms and promptly
re-emitted. This takes a tiny amount of time and
is repeated enough times to cause a change in
the average speed of the photons. In
intergalactic space the density of hydrogen is
about one atom per m3 . The likelihood
of there being an interaction which would
measurably change the average speed of a single
photon are about as close to infinitesimal as we
could imagine.
QUESTION:
Air is moved
downwards, how can momentum be conserved if the
helicopter remains stationary? Should there not
be an equal and opposite momentum upwards?
ANSWER:
Momentum isn't always conserved, only if there
are no external forces acting on system (or if
all external forces add to zero). The rotating
blades are pushing down on the air which causes
the air to accelerate downward. But if the
blades push down on the air the air must exert
an equal but opposite force on the helicopter
(Newton's third law). So, if we ask if there are
any external forces forces on the helicopter,
the answer is yes— the earth exerts a downward
force W called the
weight and the air is exerting an upward force
L often called the
lift; we can therefore conclude that momentum is
not conserved for the helicopter. In the special
case L =-W ,
and the helicopter is either at rest or moving
with constant speed down or up, the momentum is
conserved. For the air it really makes no sense
to talk about momentum unless you look at it
microscopically; fluid dynamics is very
complicated and performed by computers. There is
no doubt, though, that the helicopter and air
exert forces on each other.
QUESTION:
My question is with
regard to an apparent mathematical disagreement
with inverse square law. Let's say you have
point source in 3D space positioned at (x_0,
y_0, z_0). It emits photons in uniform 3D
directions. --- I just want to exclaim that
properly picking uniform 3D directions is not
obvious! See Wolfram's Sphere Point Picking page
---. Anyway, you have an infinite length/height
vertical y/z plane detector at x=1. One can
derive the governing distribution of hits on
this detector and it turns out to be the
multi-variate Cauchy distribution given by:
p(y,z) = 1/(2*pi) * (1-x_0)/sqrt((1-x_0)^2 +
(y-y_0)^2 + (z-z_0)^2)^3 This density does not
seem to follow the inverse square law. It has a
3/2 power in the denominator! It would need to
be 2/2. I have derived densities for other 3D
geometries (e.g. cylinder) that are not exactly
inverse square law either. I am not a physicist
but I am struggling with a physicist who claims
it must be inverse square law. It is not. It is
similar, but agrees less than the Cauchy density
when compared to Monte Carlo simulation. What am
I missing?
ANSWER:
Why would an infinitely large detector give you
information on the r dependance of what
ever the "source" is emitting? By definition, a
point source of a vector force field creates a
field pointing radially and having a magnitude
which decreases like 1/r 2 . I
don't know what a multi-variate Cauchy
distribution is, or if I ever did I don't
remember. You seem to be talking mathematics
here and not physics.
REPLY:
The Cauchy
distribution (maybe you know it as Lorentz?) is
the distribution of where rays hit a vertical or
horizontal line when emitted in uniform
directions from a 2D point source (x_0, y_0).
The bivariate one I wrote is the distribution of
where rays hit a y/z plane (at x=1 in this case)
when rays are emitted in uniform directions from
a 3D point source (x_0, y_0, z_0).
ANSWER:
I have absolutely no idea what you are talking
about! But whatever it is, it is not a means of
observing or proving an inverse square law. If
you want to find out what the field is, since
whatever is coming from the source (I will call
it "stuff") has to be isotropic, your detector
should be either a sphere with the source at the
center or a segment of that sphere. Since the
area of a sphere increases like r 2
and the total amount of stuff hitting the
surface of the sphere does not change (because
there are no other sources or sinks), it follows
that the density of the stuff must decrease like
1/r 2 .
REPLY:
I think I have an
answer for you! Though it has stumped me for
more than a month now. If I look at the behavior
of probabilities on a small area of the y/z
detector plane and I allow the plane to move...
say x=1,2,3,etc... Then the ratio of
probabilities of detection in this small area
will approximately follow the inverse square
law. A PDF is not a probability itself. You have
to integrate over an area of the detector plane
to calculate probability of detection within
that area. So if I calculate the probability of
detection in, say, [-0.1,0.1]^2 for the y/z
detector plane positioned at x=2, then it's
approximately 1/4th the probability of detection
in [-0.1,0.1]^2 for x=1. If I do the same for
the y/z plane positioned at x=3, then it's
approximately 1/9th the probability of detection
relative to the x=1 scenario. And so forth and
so forth. I think this other physicist assumed
you could just arbitrarily define a density
function with inverse square law. This seems to
only be true for histograms! The bins of a
histogram can be thought of as the integral of
some PDF over a small area. Still, that
Cauchy/Lorentz density is an inverse cube where
the calculated probabilities behave like inverse
square law. That's pretty weird and unintuitive
to me!
OK. I'm a poor communicator at this
stuff, I'm sorry. So the way this physicist
thinks is that if a point source (x_0,y_0,z_0)
emits a number of photons in 3D uniform
directions, then the number of photons that
should hit a specific location on a surface that
is a distance r away from the source should be
proportional to 1/r^2. So he asserts that you
can just define a probability density of photons
hitting a detector surface directly using the
inverse square law. For example, if it's a y/z
plane at x=1, then r=sqrt((1-x_0)^2 + (y-y_0)^2
+ (z-z_0)^2) and you get your inverse square law
probability density function of p(y,z) = C*1/r^2
where C is a normalizing constant. That seems
reasonable to me! And it's quite elegant, right?
If you know how to calculate the distance to the
detector surface, you can trivially describe the
probability density of photons hitting any
location of the detector surface. Well... but
there's this Cauchy distribution that defines a
distribution of photons hitting a planar surface
when emitted from the point source. That one is
proportional to 1/r^3... and it lines up almost
perfectly with Monte Carlo simulation. So how
can that be 1/r^3? The physicist asserts inverse
square law must be true. It's well-tested and
known to be true. I understand that. But the
Cauchy distribution describes the simulation
better than the inverse square law-based
density. From a math point-of-view, it's so
straightforward to derive the Cauchy
distribution that it would be baffling if it
wasn't the correct probability density. But the
darn thing really has 1/r^3 in it!
ANSWER:
Since I can not understand your Cauchy approach
or what it tells you about about the point
"source", I thought I would start at the
beginning with a point "source", in particular a
point charge, and ask what can be learned by
asking about that field in the vicinity of an
infinite plane. I believe I can square the
confusion between my physics and your
mathematics; as you will see, your 1/r 3
function appears quite naturally. I will look at
an infinite sheet occupying the xy
plane and a point charge Q , a distance
d from the sheet. The electric field
has a magnitude E=kQ /r 2
and points radially away from Q .
Although your calculations were in Cartesian
coordinates, cylindrical coordinates are much
more natural. At the surface of the sheet the
electric field has no azimuthal component and so
the area element to consider is dA =2πρdρ .
My aim here is to get the total flux through the
infinite sheet, but I will look at other things
as I go along. A few things we need are:
r =√ (ρ 2 +d 2 )
E=kQ /r 2 =kQ/ (ρ2 +d2 )
cosθ=d/√ (ρ2 +d2 )
Ez =E cosθ=kQd /(ρ 2 +d 2 )3/2
sinθ=ρ /√(ρ 2 +d 2 )
Eρ =E sinθ=kQρ /(ρ 2 +d 2 )
dA =2πρ dρ
The first thing to note
is that E z is where your 1/r 3
is; just because you
happen
to get a 1/r 3 behavior on
the x,y plane does not contradict what the
behavior of the field is. Ez * is
what is important here if we are interested in
something spread over the entire sheet because
for every Eρ on the area
element, there is an oppositely-pointing Eρ
180° away which cancels it. dA·E cosθ is
what physicists call electric flux and, taken
over the area dA , the flux is dΦ =2πkQρ dρ/ (ρ 2 +d 2 )3/2 .
Finally we can calculate the total flux through
the sheet:
Φ =2π
0 ∫∞ Ez ρ dρ= 2πkQ
* The graph shows Ez
normalized to Ez (ρ= 0,
z= 0) as a function of ρ in units
of d.
QUESTION:
I recently learned
that at CERN they produced streams of protons by
applying large magnetic fields to hydrogen gas.
This removes the electrons and these only the
nucleus specifically protons. This made me
wonder as alpha particles are merely the nucleus
of the helium gas, were you to apply a large
magnetic field to helium gas would you merely be
left with alpha particles?
ANSWER:
I don't where you "learned" this, but it is
almost 100% wrong. In a proton accelerator you
need to do three things:
Ionize hydrogen gas
to make a plasma. This is done with a radio
frequency electric field.
You then extract the
protons from this plasma using electric
fields and inject those into an accelerator
which speeds them up.
Then you repeatedly
speed them up, again using electric fields
and magnetic fields to steer and focus the
protons being accelerated.
Most accelerators,
including for alpha particles, work the same
way. In fact, magnetic fields cannot ever exert
the forces necessary to accelerate charged
particles. The reason is that the force which a
magnetic field B
exerts on a charged particle with a velocity
v is perpendicular to
v .
QUESTION:
So I just read an
article about virtual particles and Hawkins
radiation, and how a black hole can take virtual
particles, separate them, and take one of them
while the other Hawkins radiation is ejected out
and that's how a black hole can evaporate, which
I still find hard to believe. If that is the
case, and a virtual particle pops existence and
one is taken away and the other is ejected what
happens with that energy? So, if there's energy
between the virtual particles and that energy
can't be destroyed only transferred could that
energy end up as dark energy? I'll explain it
again in the hopes of clarifying, near a black
hole, virtual particle, can pop into existence,
be separated, with one particle, being devoured
by the black hole, and the other one being
ejected out into space, if there is energy
between the two particles at the moment of them
being separated could that energy simply turn
into dark energy or would it stay with the
Hawkins radiation being pushed into outer space?
ANSWER:
If a particle/antiparticle pair is created near
the Schwarzschild radius, and one of them is
inside and one outside, the one inside cannot
escape but the one outside can. If this had
happened either inside or outside the black
hole, they would quickly have recombined to
conserve energy consistent with the uncertainty
principle. But, now, because the particle
outside has energy, energy would appear to be
not conserved. But that does happen, so that
energy excess has to be provided by the black
hole which means that it must lose a bit of
mass. The details doesn't matter of how this
happen, it has to happen to conserve total
energy.
QUESTION:
When a golf putt
hits the edge of the hole, sometimes the ball
will "lip out," whipping around the edge and
heading off in a new direction. Some lip outs
can look quite violent and travel surprisingly
far from the hole. My question: can hitting the
edge actually increase the distance to the hole?
Formally, if point H the center of the hole, is
it possible for a lip out to finish further from
H than an identical putt would if there were no
hole punched in the ground? Or does conservation
of momentum make this impossible, and this is
all just an optical illusion caused by the brain
misinterpreting the abrupt change in direction?
ANSWER:
Two golf questions in a row! In an
earlier question I discussed the motion of a
putted ball in great detail. If you want to
understand my answer to your question, you must
read this
Q&A first to see how I use the frictional
force which is what I used there. I will also
assume that the parameter
μ R =0.093 and the mass of a
golf ball is m= 0.046 kg. So the
frictional force of a rolling golf ball f =-mgμ R =-0.045x9.8x0.093=0.042
N if the green is level which I will assume to
be the case. The acceleration of the ball along
a straight-line path will be a=f/m =-0.91
m/s2 . The figure shows the two paths
you stipulate, one rolling a distance D 1 ;
the second, rolling a distance D 2 +D 3 +D 4 .
D 3 , not labeled in the
figure, is the path during which the ball is
"lipping out", moving on the rim around
an arc of angle θ ; so D 4 =Rθ
where θ is in radians and R= 0.054
m is the radius of the hole. I will assume*
that the acceleration along the curved path is
the same as along the straight paths. Now, we
can find D 1 because, from
the earlier answer, D 1 =0.55v 0 2 =where
v 0 is the velocity which the
ball has as it leaves the putter. I will choose
v 0 =1.5 m/s which yields
D 1 =2.05 m. Since I have assumed
all paths have the same acceleration along all
paths, each of the two putts must travel the
same distance, so
D 2 +D 3 +D 4 =D 1 =2.05
m
D 4 =2.05-D 2 -0.054θ
Since we are trying to
find out whether or not it is likely that the
ball which lipped out is farther from the hole
or not, knowing D 4 will
likely tell us. D 2 is just
the distance from the hole center where the lie
was; θ will probably not be bigger than
90°=1.6 radians and, besides, it is very small
in the scheme of things. If we choose D 2 =1
m and θ= 90°, D 4 =2.05-1-0.085=0.965
m. It makes more sense to look at a straight
shot which just missed getting lipped with the
one that lipped by 1.6 radians. The second
figure shows how far each are from the center
after stopping. The lipped one is closer by
about 8%.
*The question about what
the rolling friction should be in the arc was
what worried me most when I started analyzing
this problem. But the normal force on the ball
by the ground has to be bigger than it is when
going straight because of a centrifugal force
acting outward. But the conclusion would be the
same that the lipped ball will end closer to the
center of the hole would would not be changed;
if the normal force is bigger, the friction is
bigger, so it loses more speed and would be even
closer.
QUESTION:
I was trying to
measure the bounce of a golf ball to determine
whether its lifespan was up. In order to control
the ball's bounce as to not get away I thought
bouncing it in a tube with viewing hole would be
a good environment. Unfortunately I noticed the
balls bounce approx 50% less in the tube than
outside of it. I would like to know why the golf
ball bounces higher outside than inside a 2 inch
PVC tube from 5 feet off the ground? Is there
anything I can do to alter the tube to get the
same results (ie drill holes in it, cut out
sections in the bottom to release air)?
ANSWER:
This is so far from how the freely falling ball
falls, that I won't even discuss air drag which
is the main source of friction for the ball
which is not in the pipe. But something else is
going on here. Imagine that the diameter of the
ball and the inner diameter of the pipe were
exactly the same and that there was no friction
between the two; also imagine the bottom is
sealed. You release the ball and all the forces
on it are its weight down, the force of the
atmospheric pressure on the top which is also
down, and the force of the pressure of the air
(which starts as atmospheric pressure) which is
up. So the net force on the ball is initially
only its weight down; so it starts out just like
the ball without the pipe. But, as soon as its
starts falling the pressure of the air below it
begins increasing. The farther it falls, the
higher the pressure gets, so eventually the net
force from the pressures is up and has a value
exactly equal to the weight. But, when it gets
to that point the ball which has been
accelerating downward begins to be more up than
down so the ball starts slowing down eventually
coming to a momentary stop. Now the ball has a
net force up so it starts accelerating upward
and eventually goes back to where it started. It
keeps oscillating up and down just as if it were
on a spring.
Now your situation is not
like this but it is very similar except it is
leaky; the air will leak out as the ball falls,
but the pressure in the air below the ball will
still increase. All the air which the ball is
pushing through has to squeeze through a much
smaller space than if the pipe weren't there.
That is going to considerably impede the fall of
the ball. If you used a pipe much larger to keep
from having to chase the ball it would probably
better match the freely falling ball.
QUESTION:
This is the 90 year
old man in a nursing home and would like to know
if it is possible to have a total vacuum in a
closed space.
ANSWER:
In a very excellent vacuum system there are
about 20,000,000 molecules/cm3 . In
intergalactic space there is typically 1
molecule per m3 . We now understand
that even if we had a volume with absolutely no
molecules in it, it would still not be truly
empty because there are constantly
particle-antiparticle pairs which pop into
existence and pop back out, called virtual
particles; this is called vacuum polarization.
QUESTION:
I am in a state-run
nursing home that has toilet facilities in a
small enclosure with two doors measuring 5ft by
8ft and height 10ft. The temp rises to 6 degrees
above the adjacent rooms. Is this due to the
heating of the particles in the atoms creating
energy in the form of heat?
ANSWER:
If the toilet room has its own heat register or
radiator and this room is much smaller than the
adjacent rooms, the heat will increase the
temperature more compared to larger rooms; and
if the doors are closed for most of the time or
even less, it will naturally be hotter than a
larger room on the same thermostat. This could
be corrected by partially closing the register
or partially reducing the water/steam flow to
the radiator.
QUESTION:
I just really
confused myself with a strange hypothetical
situation during the impulse/momentum unit at my
school. If a kid is riding a sled on completely
frictionless ice with negligible air resistance,
and then separates himself from the side of the
sled in a way such that both the sled and the
kid are moving in the same direction, would the
sled still speed up? I know the sled's mass
would seem to decrease from the sled's
perspective, so it should speed up to conserve
momentum, but if the kid is still moving at the
same velocity as before, wouldn't any increase
in speed for either thing not conserve momentum
for the system? I tried to explain this to my
teacher, and we clearly didn't understand each
other.
ANSWER:
This depends on how the kid separates from the
sled. There are two things you need to be aware
in questions like this: Is linear momentum
conserved and is kinetic energy conserved. If
there are no external forces acting on
a system the linear momentum will be conserved;
most problems you are likely to encounter have
momentum conservation. If there are no external
forces doing work on a system the kinetic energy
would be conserved; if it is (which it usually
isn't), it is called an elastic event. I shall
look at two scenarios:
What is the relation
between the two final velocities if we
demand that linear momentum is conserved? If
the subscripts labeled 1 and 2 are for the
boy and the sled, then
(m 1 +m 2 )u =m 1 v 1 +m 2 v 2
or v 1 =u [1+(m 1 /m 2 )]-(m 1 /m 2 )v 2
where
u is the speed before separation and
v i are the velocities
after. For example, suppose
that v 2 =u /2,
then v 1 =u [1+(½m 1 /m 2 )].
As you can see, there are an infinite number
of possible velocities after the collisions
depending just on how the kid got off the
sled. Of course, that is not surprising
because we have only one equation for two
unknowns.
Now, suppose that we
impose the condition that energy must also
be conserved. The derivation of the
solutions for a perfectly elastic event can
be found in any introductory physics
textbook or go to the
Wikepedia article on elastic scattering
in one-dimension. Now we have two equations
for two unknowns which means of course there
will only be one solution. If both objects
have the same speed
u before
the collisions, then these equations become
v 1 =u and v 2 =u .
So, you see, the boy and the sled have
the same speed afterwards as they did before
the boy got off but only if he got off in
such a way that no energy was added nor
taken away from the system.
It looks to me that you
did not give me all the information about the
problem. If it had been something like "Suppose
that after the separation the sled had a speed
of 3u /2, how fast is the boy moving?"
you could find out that the boy was moving more
slowly, v 1 =u [1-(½m 1 /m 2 )].
QUESTION:
I am a new adjunct
instructor for a introductory physical science
online class. My background is in chemistry.
Anyway, I ran across a conceptual problem in
OpenStax Physics 2e, that I am confused about.
In the chapter reading, Coulomb’s law is
introduced followed by electric fields which are
derived from Coulomb’s law. Neither of which I
find confusing. Here is the problem I am
confused about: Compare and contrast the Coulomb
field and the electric field.
ANSWER:
Electric
field is a general term, Coulomb field is
presumably referring to a field which falls off
like 1/r 2 . The prototypical
Coulomb field originates in an isolated point
charge q : E =r o kq /r 2
where r o is
a dimensionless vector of magnitude 1 which
points directly away from the point charge,
called a unit vector; the constant k is
sometimes written as 1/(4πε o ).
This Coulomb field plays a crucial role in the
determination of the field of any charge
distribution which is not a point charge. If you
have a chemistry background, you surely know
calculus, so I will give you the whole general
picture so you can understand what goes on. If
the course you are teaching is "physical
science" you surely will not teach it at this
level, probably the electric field due to 2 or
more point charges. In the figure a tiny
(infinetesimal, really) piece of the blob of
charge has been focused on. The field at the
point labeled P due to the tiny (point)
charge labeled qi is a
Coulomb field; now you have to integrate over
the whole volume for every point in 3-D space.
Except for simple shapes like a sphere, this is
an extremely difficult calculation to do
analytically. That is why your course will
probably not talk about anything much beyond
point charges. You might also touch on the field
of a large uniformly charged plane which is
pretty easy to conceptionalize and introduces a
uniform electric field.
QUESTION:
Einstein, in his
theory of General Relativity, stated that
Gravity is not a force. So, why the phisicists
are still trying to develop a theory to explain
Gravity as a force next to the other three
fundamental forces of nature?
ANSWER:
I doubt very much that he did said that gravity
is not a force. Although thinking of space-time
warping is the most popular way to visualize the
results of his theory, it is not a unique way.
General relativity (GR) is also a field theory
and interactions in field theory are
manifestations of forces. Einstein was aware of
this alternate view, in fact he embraced
it—because GR is deterministic, he believed in
predestination. Also, the search for grand
unification including gravity is rooted in the
hoped-for quantization of GR; field theories are
quantizable. See
earlier Q&As on this topic; be sure to see
links in that Q&A.
QUESTION:
I am writing you to
seek your expertise in the field of physics. As
you know, Newton's law of universal gravitation
demonstrates that the gravitational force
between two objects is proportional to the
product of their masses and inversely
proportional to the square of the distance
between them. The use of r2 instead
of r arouses my curiosity. Any further
explanations would be highly appreciated.
ANSWER:
At the Newton was proposing his three laws there
was a lot known about the solar system. Most
importantly, Tycho Brahe had made thousands of
precise measurements of planetary orbits, their
shapes and periods. His assistant Johannes
Kepler continued after Brahe's death but he also
used those data to formulate the properties, the
three Kepler's laws. Keep in mind that Kepler's
laws were purely empirical, nothing we would
call a theory. It was now known that the orbit
of a planet was an elipse one focus being at the
sun, that the quantity T 2 /a 3
was the same for all the known planets (T
is the period and a is the semimajor
axis of the elipse); also, the line from the
center of the sun to the orbit swept out equal
areas
in
equal times. Newton's laws introduced the notion
of force and its relation to motion. He realized
that the planets moving around the sun implies
that some force was acting between the sun and
the planets. And surely he realized that the
force would not be constant nor would it get
bigger as the objects got farther apart. Maybe
he did try a force which went like 1/r ,
but if he did it would not explain the
data—Kepler's three laws were the data. But it
would not work. Nor would 1/√r , 1/r 3 ,
etc., but a 1/r 2 force did
work.
There is another way you
can look at forces. A force usually may be
thought of as a field where we think of
lines of force directions in the direction
a mass would be attracted; the closer the lines
are together, the stronger the field; and total
number of lines is proportional to the mass. The
red lines in the figure show the earth's
gravitational field but only in one plane. The
blue lines show spheres again only in two
dimensions. You should imagine the lines coming
in from all directions and many imaginary
spheres. The strength of the field would be the
number N of lines per unit area A.
But every sphere is punctured by exactly the
same number of lines as all the others. The area
of a sphere is 4πr 2 so the
number of lines per unit area, strength of
field, is N /(4πr 2 ).
Eureka! There's your 1/r 2 .
QUESTION:
I am a student in
Belgium. Near my school, there is a wind turbine
that has caught my interest. I have been
wondering if it's possible to determine the wind
speed by measuring the time it takes for the
turbine blades to complete one full rotation,
given the radius of the wind turbine. I am
assuming a constant wind speed for this
calculation. I believe that the wind speed (Vw )
can be expressed as Vw = (3.6 * 2 * π
* Rturbine ) / T (in kilometers per
hour), where Rturbine is the radius
of the wind turbine, and T is the time it takes
for the blades to complete one rotation.
However, this formula does not seem to yield
realistic results. Despite conducting some
research on this matter, I haven't been able to
find a satisfactory answer.
ANSWER:
You would have been right that the rotational
speed of the blades depends on the speed of the
wind if you were looking at a simple pinwheel.
There are no other significant torques on this
machine than that caused by the wind on the
blades so, to a pretty good approximation, the
only thing affecting the rotational speed is the
wind. The wind turbine, however, is a very
complicated machine. Here are some things which
are interesting:
The blades must move
slowly, usually around 10-20 rpm, because of
structural limits and noise polution.
So there are
mechanisms which keep the rotational speed
within those limits.
Because this is too
slow for the generator to work there is a
gear box to increase rotation to the
generator by a factor of about 100.
The blades do not
work like paddles or sails, rather like the
wings of an airplane or rotor of a
helicopter.
The speed of the tips
of the blades is more important than the
rotational speed of the blades themselves
for maximizing efficiency.
Don't forget that,
unlike the pinwheel, energy is being
constantly taken away from the blades and
converted into electrical energy which will
be sent out to the world; so there must some
energy balance where the energy taken out is
equal to the energy being supplied by the
wind minus energy lost to friction,
inefficiency, etc .
There is a very good
webpage going into the details of wind
turbines; be sure to look at the video too.
There is another
video which is also pretty detailed.
QUESTION:
if pi measures a
circle and and it never repeats it self, but its
used to measure a circle doesn't that mean the
numbers would eventually go around the circle
and repeat.
ANSWER:
This isn't physics. π does not "go
around the circle", whatever that means; it is a
measure of the ratio of two lengths, the radius
R and circumference C of a
circle, π =C /(2R ).
QUESTION:
For matter that is
falling into a black hole, even before it gets
to the event horizon, it would be impacted by
both gravitational and velocity time dilation in
accordance with standard and generally
relativity. This time dilation can be very
extreme at these extremes. So would this mean
that through the perspective of the matter
falling in, it's still hasn't reached the center
or even the event horizon? Because from our
perspective outside the black hole it is going
to take them billions or even trillions of years
to get there. So while for them it's a quick
decent, for us it's a very long time. And since
we are here now could that possibly mean that
all that matter in black holes has yet to reach
the center because they are nearly frozen in
time dilation? If I'm wrong what aspect is not
working the way I think?
ANSWER:
Your question is not really a "single, concise,
well-focused" question as required by site
ground rules. I suggest this
link to get a tutorial about falling into a
black hole.
QUESTION:
Is infinite energy
possible in a frictionless surface
ANSWER:
There is not an infinite amount of energy in the
whole universe. The only way it could get
infinite energy is if it goes with a velocity
equal to the speed of light and that is
impossible exactly because you cannot get an
infinite amount to give to it.
QUESTION:
I was learning about
a STEAM activity for my children. This activity
involved cutting up pool noodles and putting
toothpicks in them to stick the noodle pieces
together. I was wondering, why doesn't the
toothpick fall out? What force(s) are acting on
it to keep the two together? It seems strange to
me, because the pool noodle is porous. I want to
know so I can teach my children the physics
behind the activity.
ANSWER:
I have raised 4 children, and I never heard of
this—must be pretty new. It is friction between
the toothpick and the noodle piece that keeps it
in. The frictional force can be felt because if
it takes a force to push it into the noodle (or
pull it out). An interesting thing is that the
frictional force between two surfaces depends on
how hard the surfaces are pushed together—when
you push the toothpick in it pushes the noodle
over to make room for itself. Since the noodle
stuff is elastic, (if you squeeze it, it springs
back), the two surfaces are pushed together
pretty hard which gives you a lot of friction.
If you were to drill a hole exactly the same
size as the toothpick is thick, there would be
much less friction. If you think about it, that
makes sense because you know it would much
easier to push the toothpick into that hole than
to push it into the noodle without a hole.
QUESTION:
If sound like light
travels on earth, does the sound like light just
continue on continuously.
ANSWER:
Sound does not travel anything like light does.
The only thing they have in common is both are
waves. The most important difference is that
sound needs a medium through which to travel,
air in the case you are thinking of I presume.
Light does not need a medium, it can travel
through a perfect vacuum. So light keeps right
on going when the atmosphere disappears but
sound does not because there is no air to travel
in.
QUESTION:
As a solid object
(ex a box) travels in a circular path, the side
of the box closest to the inside of the curve
moves a shorter distance than the side of the
box towards to the outside of the curve. Yet,
the sides of the box remain in the same relative
position to each other............ How can one
side of the box travel a greater distance than
the other and the box still remain intact?
ANSWER:
That's just the way it is in Euclidian
geometry—every part of a rigid body which is
rotating about some axis remains locked to the
where it is in the rigid body. In fact, that is
what a rigid body is defined to be. Think of a
CD rotating about its central axis. Any point a
distance r from the axis has a speed
v=rω where ω is the angular
velocity in radians per second. So the farther
out any point is the faster its speed is. But if
the spinning object is not a rigid body it will
not remain intact. If you start with a spinning
pancake made of soft putty it will spread out.
If the box you were thinking about was made of
rubber it would not retain the same shape if you
started to spin it.
FOLLOWUP
QUESTION:
Thank you so much for replying to my question,
but.... I'm sorry, even tho your answer makes
perfect sense, it did not answer my question .
How can one side of the box travel a greater
distance than the other and the box still remain
intact? Imagine the box is going around a
circular race track (like a race car). The track
has a diameter of 1000' at the center of the
track - the circumference of the center of the
track is 3141.59' . Say the box is 4' long and
2' wide. In one trip around the track, the
center of the box travels 3141.59'. The outside
of the box (at a diam of 1001') travels
3144.73'. The inside of the box (at a diam of
999') travels 3138.44'. So, the outside of the
box has travelled 6.29' farther than the inside
of the box.... How can one side of the box
travel a greater distance than the other side
and the box still remain intact?. I have
wondered about this for many years of my life
and have never been able to find someone to
explain it to me. (straight-line physics is so
much easier than curved-line physics....) Thank
you so much for your attention to this question.
ANSWER:
The answer is actually simple, if one part of
the box is farther from the center it moves with
a larger speed than another part of the box
closer to the center. What I think the problem
is that you do not know the basics of rotational
motion. Central to rotational kinematics is what
the linear velocity v of any
object rotating about some axis with angular
velocity ω and at some point a distance
d from the axis: v=ωd. The
angular velocity must be measured in
radians/second (s-1 ); since there are
2π radians in a circle, an angular
velocity of 1 rotation per second has ω =2πd.
I have drawn a diagram of a stick rotating
around one end with angular velocity ω
showing one end having a speed v but
the middle having a speed of v /2 and
the end it is rotating around is at rest;
v=Lω . Your instinct that if it wasn't
strong enough to have those differing speeds,
remain a rigid body, it will break. A good
example of that is when a very tall chimney
falls, it breaks before hits the ground.
QUESTION:
I suspect that Elon
is not producing trucks because of the way
electric vehicles brake.. Am I correct in saying
that there is a transfer or energy to the road
surface on EVs, not on normal vehicles that rely
on pads.a big rig, heavy.. those wheels, the
weight, the method of slowing, I believe that e
trucks will not suit our roads. Do you get me?
ANSWER:
Yes, you are wrong that there is "…a transfer of energy to
the road surface…" when EVs brake. Let's discuss
what braking does. When a vehicle is moving it
has energy by virtue of its motion called
kinetic energy. In order to stop the vehicle
from moving, you must get rid of that kinetic
energy. The traditional way to do that is to use
friction of brake pads (or shoes) rubbing on a
metal disc (or drum). Where does the kinetic
energy of the vehicle go? Into heat energy
because the brakes get very hot. Wouldn't it be
nice if you could capture some of that energy
and store it to use to change back to kinetic
energy later rather than get the energy from the
burning fuel. There is a way other than friction
that could be used; that method is called
regenerative braking (regen) and involves using
electromagnetism to create an electric current
which could be stored by sending the current to
a battery (not to the road surface!) Before
recent times there were no batteries to store
all this energy except the usual 12 V battery
which gets recharged but not by a braking system
but by the alternator. But with hybrids or EVs,
there are big batteries which are hungry to grab
any energy they can to lengthen the time before
they next need recharging. But no vehicle relies
solely on regen because regen might not slow the
vehicle down fast enough or it might fail
altogether; all hybrids and EVs have both
conventional brakes and regen brakes
and usually you can adjust what fraction of the
braking is done by each. Furthermore, if the
vehicle is not braking fast enough with just
regen, conventional brakes would jump in.
Therefore, as you can see, braking is not an
issue at all in whether the vehicle is feasible.
But adding regen to any vehicle is to your
advantage if you have the ability to store the
electric energy.
QUESTION:
I read that quantum
spin is intrinsic angular momentum. I do not
understand what intrinsic angular momentum is.
Could you please tell me? In ignorance, I guess
that it is some number which quantum physicists
derive from the positions, movements and
interactions of electrons, other particles and
photons, and which predicts their positions,
movements and interactions, but does not refer
to classical rotation because the number is so
high that under the equation E = MC2 the
electron, other particle or photon would have
infinite mass, which is wrong.
ANSWER:
I believe that it is always good to have a clear
picture first of what a quantity is in classical
before trying to understand corresponding
quantities in quantum physics. We often break
angular momentum (AM) into two pieces, orbital
angular momentum (OAM) and intrinsic angular
momentum (IAM). IAM corresponds to the AM an
object has by virtue of spinning about an axis
in the object itself. OAM corresponds to AM
which an object has by virtue of its moving
relative to some other point. An example is the
earth which has OAM by virtue of its motion
around the sun and IAM by virtue by virtue of
its rotation about its axis. A rough, but often
successful model of the atoms can be understood
by thinking of electrons being in orbits around
the nucleus (OAM) but also you must assume that
the electrons have IOM to understand what is
going on in detail. Electrons have IOM which
never changes. In general, however, in quantum
mechanics the angular momentum of a system
cannot be just anything you like (for example,
if you spin a ball there is no limit on how fast
it spins, 1, 23, 32.5, 0, etc . RPM);
instead they can only have certain discrete
values which is called quantization of AM.
Elementary particles have one fixed AM but the
total angular momentum of a quantum system in
some particular state which has quantum numbers
either integers n=0, 1, 2, 3… or half-odd half
integers, n=1/2, 3/2, 5/2… The first set
are called bosons and the second set are called
fermions. The angular momentum J of a
system with quantum number n is J is
given by J =ℏ√(n(n+1)) where ℏ
is the rationalized Planck's constant. IOM is
often referred to as spin. But if you try to
make a semiclassical model of an electron as a
little spinning ball you get impossible results,
like the surface of the ball traveling faster
than the speed of light. So we can think of it
like a spinning ball but have to keep in mind
that it is not a classical object and just has a
property that behaves sort of like classical
objects do.
QUESTION:
I am trying to
understand something about gravity. To it seems
that the gravitational pull equations I was
taught is flawed. It is stated that it is
directly proportional to the mass of the object
and inverse to the distance from the object. The
distance makes sense. The mass does not because
gravity itself is needed to make the mass.
Without it, mass would not exist. It is like
trying to describe the color red using crimson.
ANSWER:
You have it backwards, gravity is the result of
the presence of mass (or any energy density for
that matter), not vice versa . Also,
gravitational force is inversely proportional to
the square of the distance.
QUESTION:
Light is able to be
bent by both gravity and it easy bent when
traveling through clear objects like a prism.
Could it be that the secrets to gravity and
anti-gravity might be found in light?
ANSWER:
These two bendings are for two entirely
different reasons. So the answers are no and no.
QUESTION:
I'm not sure if this
counts as astrophysics, but if an object was
falling towards something, like the earth (no
air resistance), and then the earth disappeared,
would it keep the inertia from the fall? Because
if considering gravity as just space-time
curvature, it disappearing seems like it would
leave the object motionless, as it always was
motionless in freefall, because standing on the
earth is the same as accelerating up?
ANSWER:
I seem to always be saying...if you ask about
velocity you have to specify velocity relative
to what. If you are watching this scenario in a
frame at rest relative to the earth, the object
will move with the velocity it had at the
instant earth disappeared.
QUESTION:
My question is does
it matter if we have two Dynamos which produce
the same voltage but different frequency does
the frequency matter and how?
ANSWER:
It depends on what you are powering with the
dynamos. But most devices are designed to
operate at a particular frequency.
QUESTION:
Is it theoretically
possible for a black hole to contain other
smaller black holes?
ANSWER:
If you mean by contain inside the Schwarzschild
radius, the answer is yes, although it wouldn't
last for very long. If you mean inside the black
hole itself, what does it mean for a point to
"contain" anything? Keep in mind that I, as
stated on the site, do not normally do
astronomy/astrophysics/cosmology.
QUESTION:
How fast would a
missile be flying if it was fired from an SR-71
blackbird flying mach 3.2?
ANSWER:
As always with questions about velocity, you
must specify velocity with respect to what. If
the blackbird has a
velocity v with respect to the ground
and the speed of this particular missile has a
velocity u with respect to the ground
if fired from the
ground , and the velocities are in the
same direction, then the missile has the speed
u+v with respect to the ground
if fired from the
blackbird .
QUESTION:
So let’s say my
buddy and I are driving along the highway in our
Pontiac Trans Am’s at 99% the speed of light
maybe a couple hundred yards apart. I’d like to
chat with him on the CB radio. Are we traveling
faster than the radio waves can be received?
ANSWER:
It sounds like you and your
buddy are traveling in the same direction with
the same speeds. In that case, according to the
principle of relativity, there is no difference
from your both sitting at rest separated by "a
couple of hundred yards apart."