8 m/s) with the acceleration g ? how
far will it travel during this interval and what fraction of a light year is
the answer to the previous question?
Reddit .
cheatatmathhomework ! And if you can
read you see that AskThePhysicist.com does not do homework
so I should just trash this question. However, this question
was submitted to AskReddit a year ago so the present
questioner is presumably interested in the answer since it
is marked UNSOLVED . But,
the problem is not unsolved because the one comment points
the reader to a site where the one-dimensional, constant
acceleration kinematics is discussed; this is the
appropriate way to help someone do their homework, not
working it out to the final answer. The solution to the
problem is a simple application of these equations. So here
is my answer:
You have not accurately conveyed the problem. Most important, the final speed is not 3x108 m/s but rather 1% of that number. But that is a small enough fraction of the speed of light that you can do the problem classically, not relativistically
(which would be much more difficult). The problem is not unsolved since the single answer points you to the appropriate equations to use, Specifically,
x ≈½gt 2 and v≈gt . Here,
g =9.8 m/s2 , v =3x106
m/s. Do the algebra to solve for the two unknowns, convert
t from seconds to days and convert x from
meters to light years.
I worked out the answers and
found that the problem, as stated, does not have the correct
answer for the distance traveled. The correct answer is
4.6x1011 m=4.9x10-5 ly.
My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?
c is not
really that big for such a long distance. If you go at that
speed, the distance (as seen by the ship) would be 13.9x109 √(1-0.99992 )=2x108
ly. Since the speed of the ship is 0.9999 ly/yr, the time
necessary would be 2x108 /0.9999≈2x108
yr, not 30. I have seen this kind of estimate before and the
situation being described was probably that the ship has a
constant acceleration (as seen by the ship) equal to the
acceleration of gravity (about 10 m/s2 )
essentially forever; so the speed (as seen from earth) gets
to be much larger than 0.9999c . You can see the
calculation of the speed of accelerating objects as seen
from an inertial frame in an
earlier answer . If you want a really detailed discussion
of getting to the edge of the universe in a few decades, see
this post by
Dave Goldberg ; his number is more like 99.9999999999999999998% of
c !
But your question does
not depend on the details of how far you went and how long
it took you to get there. The radio signals being received
on earth will be Doppler shifted to lower frequency, the
shift getting larger as the speed gets larger. This is the
same idea as the Doppler shift of a siren going away from
you getting changed to a lower pitch. The expression for the
shift in frequency is f observed =f source √[(1-(v /c ))/(1+(v /c ))]=√[(1-0.9999/(1+0.9999)]=0.0007f source ,
a much lower frequency.
earlier answer
as well as the earlier answer it links to. When you have
done that, you should be able to follow my answer to your
question.
First a qualitative discussion. You decide that you want
your spaceship to have a constant acceleration of g
as measured by you, captain of the ship. You therefore
arrange to have your engines always exert a force on the
ship of F=mg where m is the mass of the
ship. So you will always see your ship having the desired
acceleration. But, as I have emphasized in many earlier
answers, acceleration you measure is not the same as
acceleration someone on earth measures. In Newtonian
physics, everyone measures the same acceleration. This is
really a glorified
twin paradox problem and I am not going to calculate how
long the round trip will be for you (but I will calculate
how long it takes in earth time). What I can tell you is that you
will take longer than 4 years to return to earth and the
time elapsed on earth will be longer than your time. Your
desire to accelerate "to
the speed of light" is impossible, as you will see, so
forget that; with your scenario the fastest you will ever go
is about 86% of c .
What I want to focus on is what someone on
earth will measure regarding your motion. This is mostly
worked out in the earlier answers I have referred you to, in
particular v /c =(gt c )/√[1+(gt c 2 ]
and a g =[1+(gt c 2 -3/2 ;
these are plotted in the graph above with the red and blue
curves. Things to notice are that after 1 year you have only
acquired a speed of about 0.7c and your
acceleration has dropped to about 0.35g . What was
not derived earlier was your position as a function of time;
integrating the velocity equation, I find gx c 2 =√[1+(gt c 2 ]-1,
shown by the black curve. Things to notice are that it
starts out looking like a parbola as it would if you had a
constant acceleration of g , but eventually becomes
a straight line as you approach the speed of light which you
never reach. Reading off the graph, the time it takes you to
get halfway, 1 ly, is about 1.7 years (as measured from
earth); your speed will be about 0.86c and your acceleration
(again as measured from earth) will be about 0.13g .
This will be ¼ of your tota
Incidentally, the axes
may be shown to be numerically equal to years and light
years if you approximate g ≈10 m/s2 .
I see that I have misread
the question. For some reason the questioner reverses his thrust at c . Thus the total elapsed time on earth
before your return would be 4.4 years in this scenario.
Light would take 2 years to make this trip. Again, I have
not calculated the time on your clock but it would be
between 2 and 4.4 years.
W /c but
the outside observer would see it take a time (W /c )/ √[1-(v /c )2 ].
If the spaceship were accelerating you would see the light
hit somewhere backward (forward) of where you aimed it
depending if you were speeding up (slowing down).
u across the
ship which moves with speed v , will have a speed u'= √(u 2 +v 2 ),
not u+v as you suggest. However, as you correctly
surmise, this cannot be true for light since, in your
example, you could not have the light with speed √(c 2 +102 )
as seen by an outside observer. The speed must still be
c . So, if the component along the direction of the ship
is cx =v , you must have cy 2 +cx 2 =c 2 =cy 2 +v 2
or cy =c √[1-(v /c )2 ].
Regarding what you would see in the ship, the principle of
relativity states that the laws of physics are the same in
all inertial frames of reference; if your ship is at rest,
we could agree that anything aimed straight acoss will go
straight across. In other words, there is no such thing as
absolute rest —any inertial frame with constant
velocity relative to another inertial frame would have the
same laws of physics.
earlier answer .
(Key to understanding this is the notion that the speed of
light one measures is independent of how you are moving.) It
would also help you to understand if you read about the "twin
paradox ".
My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?
c is not
really that big for such a long distance. If you go at that
speed, the distance (as seen by the ship) would be 13.9x109 √(1-0.99992 )=2x108
ly. Since the speed of the ship is 0.9999 ly/yr, the time
necessary would be 2x108 /0.9999≈2x108
yr, not 30. I have seen this kind of estimate before and the
situation being described was probably that the ship has a
constant acceleration (as seen by the ship) equal to the
acceleration of gravity (about 10 m/s2 )
essentially forever; so the speed (as seen from earth) gets
to be much larger than 0.9999c . You can see the
calculation of the speed of accelerating objects as seen
from an inertial frame in an
earlier answer . If you want a really detailed discussion
of getting to the edge of the universe in a few decades, see
this post by
Dave Goldberg ; his number is more like 99.9999999999999999998% of
c !
But your question does
not depend on the details of how far you went and how long
it took you to get there. The radio signals being received
on earth will be Doppler shifted to lower frequency, the
shift getting larger as the speed gets larger. This is the
same idea as the Doppler shift of a siren going away from
you getting changed to a lower pitch. The expression for the
shift in frequency is f observed =f source √[(1-(v /c ))/(1+(v /c ))]=√[(1-0.9999/(1+0.9999)]=0.0007f source ,
a much lower frequency.
result of the fact that the speed of light is a
universal constant, not vice versa . I would also
point out that these would also be true if the speed of
light were different from what it is; as illustrated by
George Gamow's book
Mr.
Thompkins in Wonderland , they would be much more
evident if the speed of light were very small.
this and all the links in
that answer . [The derivations of
the equations below can be found in the earlier answers.]
From the perspective of your space ship the velocity v
as a function of time t is v =(a 0 t )/√[1+(a 0 t /c )2 ] where
a 0 is the acceleration as measured by the ship (g in your case). In order to achieve this acceleration, you must exert a force
F=ma 0 where m is the mass of the ship.
The velocity is plotted in red in the graph; reading from
the graph, the time when v /c =0.6 is when a 0 t /c ≈0.8=9.8t /3x108 . Solving, t =2.45x107 s=0.78
yr=284 days. The position as a function of t is x =(mc 2 /F )(√[1+(Ft /(mc ))2 ]-1)=(c 2 /g )(√[1+(gt /c )2 ]-1);
for gt /c =0.8, x =2.58x1015
m=0.273 ly. At this point you turn off your engines and
coast at v /c =0.6. The distance you need to
go to get halfway to the star is 16.9 ly and it will take
you 16.9/0.6=28.2 years. The second half of the journey will
take the same time, so the time for the whole trip is T =2(28.2+0.78)=58
years.
Your answers (212
days, 0.17 ly) are different from mine (284 days, 0.273 ly)
because you have assumed that the acceleration, viewed from
the earth, is the same as the acceleration (9.8 m/s2 ) seen from the ship. The
black curve in the graph shows the ratio of the acceleration
seen from the earth compared to the acceleration seen from
the ship; reading from the graph, at 0.6c the ratio
is about 0.5, 4.9 m/s2 . Be sure to read the
discussions of acceleration in the
earlier answers if you want to understand this issue
with acceleration in special relativity.
Since there is no such
thing as an antimatter drive engine, I can make only a rough
assessment of how feasible this is. The force you need to
apply during the 0.78 year accelerations is F=mg. What is m?
The space station has a mass of about 400,000 kg and you
have 40,000 passengers. Let's say each passenger needs one
space station of mass to support his/her needs, so the
required force would be 1.6x109 x9.8=1.57x1010
N. The energy you have to supply to the ship to v /c =0.6
I figure to be about 36x1024 J and the amount of
mass converted with 100% efficiency into energy would be
about 4x108 kg and you would need that much again
to stop it. So your ship of mass 1.6x109 kg would
have to carry 8x108 kg of fuel, one third of the total
weight! Obviously, I have not included this mass change in
my estimates of time. And, of course, there is the issue of
how you are going to store 4x108 kg of
antimatter. The whole scenario certainly does not look
feasible to me!
Finally, note that all
the distances and times are as measured by the earth bound
observer. In other words, the passengers will not be 58
years older when they arrive at their destination. Because
of length contraction, they will have a shorter distance to
travel. Ignoring the accelerating periods, I can estimate
the time of the trip by just assuming they travel 0.6c the
whole trip. In that case, T' ≈T √(1-.62 )=58x0.8=46.4
yr.
ANSWER:
g to get to 0.75c , but any acceleration will do. But,
since you seem interested in a greater than g acceleration, I
will choose a 0 =1.5g. So now a person of
mass m in the ship sees two fictitious forces, mg mg mg g , and make all the "floors" in the ship
perpendicular to the acceleration. If you are interested in how long it
would take to reach 0.75c with a 0 =g ,
you can read off the graph (which I took from an
earlier answer ) that gt /c =1.8,
so t =5.5x107
s=1.7 yr.
ANSWER:
earlier
answer first which has lots of details and discussion of
acceleration. There are two ways to approach this problem:
Suppose
the observer in the inertial frame is doing the pushing and wants to
know how hard to push the particle to achieve a particular
acceleration.
Resistance to acceleration is usually called inertial mass and the
inertial mass m of a particle with rest mass m 0
and speed v is m=m 0 /√[1-(v /c )2 ].
The first figure above plots
m /m 0 as a function of v /c .
So, to achieve an instanteous acceleration of a , a force of
F=ma =m 0 a /√[1-(v /c )2 ];
so for your example of v /c =0.7, you can read off
the graph that you would need to exert a force 1.4 times larger than
you would if the particle were moving slowly. This is probably what
you want if you are interested in electrons accelerating since you
would be accelerating them.
Suppose
the observer was on the particle and the particle was a rocket ship.
You adjust your engines so that the force F which they
exert on the ship causes a constant acceleration of a 0 =F /m
where m is the rest mass of the ship. What
acceleration a does another observer in an inertial frame
(on earth maybe) see when the rocket has a speed v ?
Starting with the velocity derived in the
earlier
answer , v =(a 0 t )/√[1+(a 0 t /c )2 ],
we can calculate the acceleration by differentiating with respect to
t , [dv /dt ]/a 0 =a /a 0 =(1+(a 0 t/c )2 )-3/2 .
Now, reading off the second graph, when v /c =0.7,
a =0.36a 0 ; the stationary observer will
only see 36% of the acceleration seen on the ship.
ADDED
NOTE:
v /c =0.7 and 0.9, will observe the
ship having different accelerations. This is why acceleration, and
therefore also force, in special relativity do not play an important
role as they do in Newtonian physics where all inertial observers see
the same acceleration. This is discussed in the
earlier
answer .
ANSWER: is a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by c 2 .
ANSWER: ≈2x104
m/s and 30x103 km/s=3x104 m/s). So let's just
choose the larger one and see how much time dilation there is. Relative
to the sun, an elapsed time T =4x109 y would
correspond to T'=ΥT wwhere Υ= 1/√(1-(v /c )2 )=1/√(1-(3x104 /3x108 )2 )≈(1+0.5x10-8 ).
Therefore T'≈ (T +20), a difference of 20 years. This
may sound like a pretty long time to you, but relative to 4 billion it
is less than 10-6 %. That is not "significant" to my mind.
You are right that every object has its own clock which, relative to
other clocks, is not necessarily the same; every object also has its own
meter stick, not necessarily the same as other meter sticks in the
universe. The important thing is that you always must talk about
velocity with respect to what.
ANSWER: c . Your statement that the
train rider would see the beams having is speed 0.00001c is
simply wrong. That is not to say that the beams are unchanged —they
would be Doppler-shifted to different wavelengths, but still move with
speed c .
See the faq page to help yourself to accept that this is true.
ANSWER:
there . You might also read an
earlier answer about relativistic velocity addition. You are right,
if light with frequency f is moving in the same direction as
you are, you will see a lower frequency light f' ,
red-shifted. The Doppler equation is f'=f √[(1-β )/(1+β )] where β=v /c.
Suppose we take your case where β= 0.9999999=10-7
so √[(1-β )/(1+β )]=2.24x10-4 and choose λ =550
nm (yellow light). Then f=c /λ =3x108 /5.5x10-7 =5.45x1014
Hz and f' =2.24x10-4 x5.45x1014 =1.23x1011
Hz=123 GHz. This frequency is in the short-wave radio range, nothing he could "…see
with his eyes…"! The Doppler effect has red-shifted the wave
way beyond visible red!
ANSWER: run slowly . Do not mistake that
for saying that your clock appears
to run slowly —it might
appear to run faster or slower
than mine, but is running slowly. Furthermore, it has nothing to do with
mass increasing with speed. But how do you observe things? Your clock
appears to run just the same as if you were at rest; but wait, you are
at rest relative to yourself and you see me moving and therefore find
that my clock runs slowly! Now, the time that I see you getting to the
star is 4.3/0.7=6.14 years. But, you see it as a considerably shorter
time. How can that be if your clock is running normally and you clearly
see your speed as 0.7c ? The answer is length contraction—you
observe the distance to the star to be less than 4.3 light years by the
gamma factor, √(1-0.72 )=0.71, or 3.1 light years so the
time will be 3.1/0.7=4.43 years. If you could get going a speed 0.9c ,
you could get there in less than half a year by your clock but still
6.14 by mine. I have given you a couple of links to the
faq page , but you might want to peruse that page
for more relativity answers which might interest you, particularly the
twin paradox.
QUESTION:
ANSWER:
QUESTION:
ANSWER: …a
very strange thing indeed… x direction while the red-drawn
distances to the mirrors are at rest in the unprimed system. The rest
lengths of the two distances are L and the angle one makes with
the x -axis is θ . The round-trip time along the
x -axis is t 1 =2L /c and
along the other length is the same, t 2 =2L /c .
In the moving system all lengths along the the x' direction are
reduced by a factor √(1-β 2 ) where β=v /c ;
therefore L'=L √(1-β 2 ) so t 1 '=2L' /c =2L √(1-β 2 )/c=t 1 √(1-β 2 ).
For L" only its x' component is contracted, L" cosθ' =L √(1-β 2 )cosθ
while the y" component remains the same, L" sinθ' =L sinθ ;
from these you can easily show that L"=L √(1-β 2 cos2 θ )
and tanθ'= (tanθ )/√(1-β 2 ).
Finally, t 2 ' =2L" /c =[2L √(1-β 2 cos2 θ )]/c =2t 2 √(1-β 2 cos2 θ )≠t 1 ' .
ADDED
COMMENT: I see that I misread your question. I see that
you have put the lengths at rest in the moving frame whereas I have them
in the stationary frame. But in special relativity it makes no
difference which is moving and which is at rest. So, you have found that
the times in the frame not at rest relative to the lengths are not the
same, just as I have; I reiterate that this is expected, not surprising.
FOLLOWUP
QUESTION: Thanks a lot, your competent and prompt
answer is much appreciated!
I would like to reiterate that for two events that occur in the same place
One can think with some justification that this scenario means that the x'O'y' system is somehow more "special" than the xOy system, pretty much in contradiction, if not with Einstein relativity postulate proper, at least with its spirit. In other words, in my opinion, the relativity theory is not devoid of inner contradictions, or at least it leads to conclusions which seem to invalidate the most fundamental notions of cause and effect or of things being what they are irrespective of where we observe them. And to me it is debatable whether a physical theory, however legitimately counter-intuitive it may be understood to be, can lead to this type of conclusions and be nevertheless considered a valid description of the physical world.
ANSWER: not say
that the bomb will not detonate because the device which detonates it is
in the stationary frame and he will agree that the pulses in that frame
are indeed simultaneous. Is the stationary system in some way "special"?
Of course it is —that is where the apparatus resides and all
frames will agree that the bomb will be triggered in that frame.
QUESTION:
ANSWER: E=mc 2 =9x109 J. The melting
point of iron is 1540°C and its specific heat is 449 J/kg·°C,
so the enegy to heat it from 20 to 1540°C is 1x1520x449=6.8x105
J. So, the ball would have increased its mass by 6.8x105 /9x109 =7.6x10-5
kg. You would be hard-pressed to actually measure so small a change in
mass.
QUESTION:
ANSWER: velocity addition .
In classical physics, v CB =v CA +v AB ;
the notation is "v IJ is the velocity of I relative
to to J". I have written this so that it corresponds to your question —A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
v CB =v CA -v BA
which we can do because v AB =-v BA .
If the trains have speeds v in the same direction, v CB =v-v =0;
in opposite directions, v CB =v- (-v )= 2v— each
sees the other moving with speed 2v . But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer ).
The relativistically correct velocity addition equation is
v CB =(v CA -v BA )/[(1-(v CA v BA /c 2 )]
which reduces to the classical equation for the speeds much less than
c . So, for the trains moving in opposite directions, v CB =2v /(1+v 2 /c 2 );
for example, if v =0.5c , v CB =c /(1+0.25)=0.8c .
Now, you seem
to think that if you double the speed, you double the time difference.
This is not correct— time dilation goes like the gamma
factor, 1/√(1-v 2 /c 2 ).
So, for your situation with v =0.5c , t A =1.33t B ,
t A =1.33t C , t C =1.67t B ,
and t B =1.67t C . These are
confusing, I admit. They are meant to denote, for example, that t B =1.67t C
means that B sees his clock tick out 1.67 s when C's clock ticks
out 1 s; B observes C's clock to be slow.
QUESTION:
ANSWER: twin paradox . Acceleration
just makes everything harder to understand. Basically, just assume
necessary accelerations (departing, turning around, landing) occur in a
vanishingly short time. But, you are not really interested in the twin
paradox, you are interested in how things appear in an accelerated
frame. I am sorry, but nothing in your question after " …clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be c is a
law of physics and the general principle of relativity states the laws
of physics are the same in all (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it appears to run and how it
is actually running are not the same.
I would like
to address how the clocks look from the perspective of the
Doppler effect . The relativistically correct Doppler effect is
usually expressed in terms of the frequency of the light; for our
purposes, it is more convenient to express it in terms of the periods,
T observer =T source √[(1+β )/(1-β )]
where β=v /c and β is positive for
the source and observer moving apart; it makes no difference which is
the observer—each twin will see the other's clock running at the
same relative rate. Let's illustrate with a specific example, β =-0.8,
the traveling twin coming in at 80% the speed of light; T observer =T source √[(1-0.8)/(1+0.8)]=T source /3
so the observer will see the source clock running fast by a factor of
three. But special relativity tells us that moving clocks run slow, not
fast; 1 second on the moving clock will be 1/√(1-0.82 )=1.67
seconds on the observer's clock. How the moving clock looks is thus
demonstrably not a measure of how fast it is actually running. Now, if
the incoming twin puts on the brakes such that he slows to β =-0.6,
T observer =T source √[(1-0.6)/(1+0.6)]=T source /2
so the observer will see the source clock running fast by a factor of
two, apparently slowing down. Now, 1 second on the moving clock
will be 1/√(1-0.62 )=1.25 seconds on the observer's
clock, speeding up compared to when β =-0.8.
QUESTION:
ANSWER: trampoline
model " which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.
QUESTION:
ANSWER:
E=mc 2 , is not lost and the black hole
becomes more massive by the amount of the mass energy. When light is
swallowed by the black hole, the mass increases even though the light
does not have any mass because light does have energy and the energy
shows up as increased mass of the black hole, m=E /c 2 .
QUESTION:
ANSWER:
E=mc 2 =m 0 c 2 /v /c )2 ]
where m 0 is the mass at rest. For example, if v /c
increases from 0.99 to 0.999, that is only a 1% change in speed; but the
energy increases from 7.1m 0 c 2
to 22.4m 0 c 2 , more than triple.
QUESTION:
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?
ANSWER:
f /dq =Aq for some function f (q ),
where A is a positive constant, f is said to be
exponentially increasing with q . For your question, f
is m and q is v . The expression for m (v )
is m=m 0 /v 2 /c 2 ]
where m 0 is the rest mass. Calculating the
derivative, dm /dv =mv /[c 2 -v 2 ].
Thus, although m increases without bound as v
increases, it does not increase exponentially.
QUESTION:
ANSWER: earlier worked out the
velocity of something which would correspond to occupants of your spaceship
experiencing a force equal to their own weight due to the acceleration which
I will adapt to your case later. (See the graph above.) First, though, I
will work out the (incorrect) Newtonian calculation which is presumably what
you want. The appropriate equation would be v=at where v =3x108
m/s, a =1.2g =11.8 m/s2 , and t is the time to
reach v ; the solution is 2.5x107 s=0.79 years. For the
correct calculation, you cannot reach the speed of light; from the graph
above (black curve), though, you can see that you would reach more than 99%
of c when
gt /c =3.
To make this your problem, we simply replace g by 1.2g and
solve for t . I find that t =7.7x107 s=2.4 years.
QUESTION: article
about relativity. The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so?
Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?
ANSWER: To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north. "
The speed in the northward direction will now be v N =65cos450 =46
mph; his velocity in the eastward direction will be v E =65sin450 =46
mph. His total velocity will be √(v N 2 +v E 2 )=65
mph, not (v N +v E x and t for the
stationary observer and x' and t' for the position and
time of the moving observer as seen by the stationary observer. The catch,
though, (which I presume is why the author of your article avoided these
details) is that time and space are slightly different from space (N) and
space (E) in that to get the length of the vector you calculate the square
root of the difference of the squares instead of the sum. So the speed
through spacetime would be √(c2 -v2 ).
So, the stationary observer will see a "spacetime speed" of v spacetime =√(c2- v2 )
and the "space time distance traveled" by the moving observer as seen by the
stationary observer would be d spacetime =v spacetime t'
where t'=t / √(1-(v 2 /c 2 )).
If you do your algebra you will see that d spacetime =ct
indicating that the stationary observer sees the moving observer having
a speed of c through spacetime.
QUESTION:
ANSWER: f . Initially
an energy of hf is removed from the star. But, by the time that
the photon is very far away the star, it has lost some amount of energy Δhf ,
so the net loss to the star and its field is
hf -Δhf . At the instant the photon is created, the
mass of the star is reduced by hf/c 2 . But when the
photon, which is losing its energy to the field, is far away, the energy
of the field will have increased by Δhf /c 2 ;
I would not use the terminology "mass of the field" since the field has
energy density, not mass. But now, it seems to me (not a cosmology/general
relativity expert) that there is a bit of a paradox: we always associate a
given mass with a particular gravitational field, so the field should have
the energy content associated with a mass M'=M -hf/c 2 .
But, in fact, the field would have energy content associated with a mass
M'=M -hf/c 2 +Δhf /c 2 .
I am guessing that the field and the mass somehow "equilibrate" so that the
final mass of the star is consistent with the energy of the final field. (I
could easily be wrong! Perhaps it is only meaningful to look at the total
energy of the star and its field. Whatever the case, I would not talk about
mass of the field.)
QUESTION:
ANSWER:
If you
are interested, I will give here the electric and magnetic fields for one of
the charges moving with velocity v x direction.
E' i Ex + γ (j Ey +k Ez )
and B' =-v E' c 2
where
γ= 1/√[1-(v /c)2 ] and i j k E q is at rest and E' B' q is moving.
QUESTION:
ANSWER: … "
with respect to whom . The
person in the airplane sees time progressing at a normal rate. However, the
person on the ground will see the clock of the person on the airplane run
slowly, so he will perceive the traveling person to be " living
a shorter amount of time", i.e. she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox ,
how clocks run , and the
light clock .
QUESTION:
ANSWER: per
se . When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (E=mc 2 ). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.
QUESTION:
ANSWER: c is a universal constant
is where you should start, not end. When you move with your S' frame with
speed v in the positive x direction, the light beam acquires
an x -component of c x =-v . If the speed of the
light is to remain constant, this must mean that the y-component must
therefore decrease such that c 2 =c x 2 +c y 2 .
So,
c y =c √[1-(v 2 /c 2 )]
which means that the angle θ which the light makes with the y -axis
is θ= tan-1 [β /√(1-β 2 )] where
β=v /c. See an earlier question for
an example of this phenomenon.
QUESTION:
ANSWER: faq page.
QUESTION:
ANSWER: v relative to the earth and the light has a
speed c relative to the train, then surely the speed of that light
relative to the earth will be c+v . This is called Galilean velocity
addition and is only approximately true when all speeds are very small
compared to c . In the theory of relativity the velocity addition
expression is u '=(v +u )/[1+(vu /c 2 )]
where v is the speed of one thing relative to the earth, u is
the speed of another thing relative to the first thing, and u ' is the
speed of the other thing relative to the earth. In your first case, v
is the speed of your train and u is c , the speed of light, so
c '=(v+c )/(1+v /c )=c .
In your second case v '=(0.9c +0.9c )/(1+0.81)=0.99c .
I urge you to look at the FAQ links, why c is the
universal speed limit , and relativistic
velocity addition .
QUESTION:
ANSWER:
First of all, once an object passes the event horizon, you cannot see it at
all. I would say that we really do not know how a clock would behave
inside the event horizon because, as shown below, the field is two
intense for the usual gravitational time dilation expression to be
valid. You might want to look at an earlier answer on
gravitational time dilation . The rate at which the clock runs slower
in a gravitational field (compared to your clock in a very weak
gravitational field) is
√[1-(2 MG /( Rc 2 ))]
where R is the distance from the center of a spherically
symmetric mass M . Note that if R =2MG /c 2
this is zero and the clock stops. That magic R is called the
Schwartzchild radius and so if you are watching this clock, you see it
stop at this R . From the point of view of someone traveling with
the clock, though, time clicks on as usual and you enter inside R .
Your fate would be that as you got closer to the black hole (imagine you
are falling head first), the gravitational force would be much larger on
your head than your feet; this kind of gradient of force is called a
tidal force and it would rip you apart. A nice little animation may be
see
here .
QUESTION:
ANSWER:
This is an example of the relativistic Doppler shift. If the period of the
light waves at the source is T and at the observer is T' ,
the relationship between them is T'=T √[1+β )(1-β )]
where β=v /c. For your question, T'= 14.1T. It
follows, therefore, then that the time it takes for the wave to pass you
is 4x14.1=56.4 min. The light has been extremely "red-shifted".
QUESTION:
ANSWER:
No, not hundreds of times, billions of times more. I calculate that if the
planet were the size of earth, its surface gravity would be such that
the man's weight would be approximately ten billion times greater than
on earth. That man will not be coming back from this planet since his
own weight will crush him. If you had some indestructable clock which
you could put there and pull back when that clock had run for 2 hours,
it would return to you in 20 years as measured by your clocks. Although
I have not done the calculation, it is likely that this would, in fact,
be a black hole and that it would not be possible to return your clock
if it were inside the Schwartzchild radius.
QUESTION:
ANSWER:
You have just encountered what is called the equivalence principle. Since
you have stipulated that the object is in free fall, the answer to your
second question is that, yes, there is a force on the object, its own
weight. The equivalence principle states that if you are inside a box in
empty space with acceleration a , there is no experiment you can
perform which will tell you that you are not in a box at rest in a
gravitational field with gravitational acceleration a . A
corollary is that if you are in free fall in a gravitational field,
there is no experiment you can perform which will tell you that you are
not at rest or moving with constant velocity in empty space. That is why
the accelerometer reads zero in your first question, not "because there
is none"; the object is actually accelerating but cannot detect it.
There is a story about Albert
Einstein, probably apocryphal, but amusing and instructive. As you
probably know, his first job was as a clerk in the Swiss Patent Office
and he could easily complete all his responsibilities quickly and then
spend the rest of the day thinking about physics. One day, while
watching a workman paint a building across the square, the workman fell
from his ladder. Einstein thought to himself, "there is no experiment
which he can do which could distinguish whether he was in free fall in a
gravitational field or was in an inertial frame in empty space".
QUESTION:
ANSWER:
3000 years sounds pretty crazy to me. And, don't you think it would be
pretty boring? Anyhow, we can estimate the time (on the ship) required
pretty easily. The gamma factor would be γ =1/√(1-0.82 )=1/0.6
and the distance is about d =1.6x105 ly. So, the
length-contracted distance would be d '=d /γ= 96,000
ly so the time to get there would be t=d '/v =120,000 years.
3000 years ain't gonna get it!
QUESTION:
ANSWER:
The higher the temperature, the more energy the metal contains. So,
technically, a higher temperature means a higher mass. The effect,
though, is so small that you would be hard-pressed to measure it.
QUESTION:
ANSWER:
Good question. Relativity tells us that the mass of an object which is m 0
when at rest and is traveling with a speed v becomes m=m 0 /√[1-(v /c )2 ]
where c is the speed of light. In the LHC the speed of the
protons is
0.999999991 c ,
so their mass is m=m 0 /√[1-(0.999999991 )2 ]=7456m 0 ,
plenty of mass available!
QUESTION:
ANSWER:
As light falls into a black hole, it gains energy like a mass does, but it
does not do that by increasing its velocity. The energy of a photon is
hf where h is Planck's constant and f is the
frequency of the light. So, the energy of the light increases by the
frequency increasing (or equivalently, by the wavelength decreasing).
QUESTION:
ANSWER:
It is my understanding that general relativity predicts than any region in
space which has energy density creates a gravitational field, it need
not be static, cold mass. Since you must add energy to heat something
up, it would have more mass but very little more mass.
QUESTION:
ANSWER:
First of all, it is not due to E=mc 2 , it is due to the
redefinition of linear momentum in special relativity. There are lots of
links on the faq which address the question of relativistic mass (1 ,
2 ,
3 ).
Basically, think of mass as inertial mass, the resistance to
acceleration. It is well known that objects cannot exceed the speed of
light, no matter how hard you push them. Therefore, inertia must be
increase as you go faster. In general relativity, Einstein was able to
show that gravitational mass is identical. So you have more mass (and
therefore weight) when you are running but by an immeasurably small
amount (unless you can run at speeds near the speed of light). All your
rambling about chemical energy, kinetic energy, etc . just
confuses the issue�you are either moving or you're not, don't worry
about how you got there.
QUESTION:
ANSWER:
mv 2
and mc 2 mean. K ≈�mv 2 is the
kinetic energy of in classical mechanics of a particle of mass m and
speed v ; it is only (approximately) true for v<<c . E =mc 2
is the energy which a mass m has, by virtue of its mass, when it it
at rest. The correct kinetic energy of a mass m with speed v
is K=mc2 [1/√(1-v 2 /c 2 )-1].
It is fairly easy to show that, if v 2 /c 2 <<1,
K ≈�mv 2 . Letting β=v /c ,
QUESTION:
ANSWER:
at the same time and the difference is the length. I think you
would agree that this is a perfectly reasonable definition of length. It is
not that things appear shorter , it is that things are shorter .
You might be interested in an
earlier answer .
FOLLOWUP QUESTION:
ANSWER:
3 m and the 0.86 c
is about 2.6x108 m/s. Special relativity says that the diameter,
as measured by the moving frame is 12.7x103 √(1-0.862 )=6.5x103
m. Now, the time which the moving frame observes that it takes for the earth
to pass is 2.5x10-5 s. If the diameter of the earth was 12.7x103
m, the time would have been 4.9x10-5 . That is how long an
observer on the earth observes the moving frame to pass the diameter of the
earth. Both length and time depend on your reference frame. Experiments
equivalent to this have been done and everything I have written in this
paragraph has been experimentally verified. (If you had read my original
answer, you would have seen that I did answer your question about apparent
vs. physical contraction.)
QUESTION:
ANSWER:
E is given by E=hf and its momentum
p is p=E /c . Therefore, the photon gains momentum by
increasing its frequency f .
QUESTION:
ANSWER:
F only only has meaning in
terms of the acceleration a it causes some mass m to have, F=ma .
This can be more generally stated as the force is the time rate of change of
linear momentum, F =dp /dt . Using the relativistic
expression for p , this defines force in special relativity. Usually,
though, it is much more useful, both in classical mechanics and relativistic
mechanics, to solve problems using the potential energy V (r V (r F∙ r
QUESTION:
If you were able to experience a time dilation field and you were to place
a living organism half way in to the field, what would be the hypothetical
result?
QUERY:
What is a time dilation field?
REPLY:
It's a field where inside it, time is either going by super slow or super fast relative to outside the field. It is used in scifi quite a bit so I'm not sure if it's just science fiction or if it is theoretical
ANSWER:
QUESTION: MV 2 on them because they are slow in terms of relativity. So my question is at around what velocity should the kinetic energy of an object be calculated with the relativistic kinetic energy calculation? Like I have heard that relativistic effects become noticeable at around 15% lightspeed, so around there? And at what velocity would you say that an object becomes noticeably relativistic as a rule of thumb?
ANSWER:
γ =1/√[1-(v /c )2 ].
At v/c=0.15, γ =1.01; a nonrelativistic calculation would only have
about a 1% error. The graph to the right shows that relativistic effects do
not really start kicking in until around v /c =0.5.
QUESTION:
ANSWER:
QUESTION:
ANSWER: is and how time appears and involves only special relativity. See an
earlier answer . If the moving clock is moving in a circular orbit of
radius R , it is more complicated and involves both general and
special relativity. In that case there will be time dilation due to its
speed, v , but also due to its acceleration a=v 2 /R
. To calculate the time dilation due to the acceleration you would use
the
equivalence principle and say that it would be the same as the
gravitational time dilation in a gravitational
field with acceleration g =a=v 2 /R . For the
orbit, there would be no difference between how time is and appears to be
because the distance from observer to clock is constant. Questions like
yours must be considered if GPS systems are to be accurate.
FOLLOWUP QUESTION:
ANSWER:
q&a about the father and son, it
would have told you how sound would be sped up or slowed down and the picture of the clock would behave the same way. I guess I could have added for the circular orbit part that both the special and general time dilations are to slow down, so the
video of the clock would be be slower than the local clock. However, to have a significant slowing would require an enormously large orbit to have a large enough speed for special time dilation
to be significant and a small enough acceleration that it did not crush you;
the reason that the transmitted clock video is slower is simply because the
clock being recorded is running slower in the frame of the "stationary"
ship. Of course there is the time delay due to the time of transmission in
all cases, so if you synchronized all clocks at the start you would see a
delay when comparing clocks, so it would be important to look at the rates
of the clocks, not their absolute readings.
QUESTION:
ANSWER:
predict waves which propagate with a speed
of 1/√(μ 0 ε 0 ) which just happens to be
exactly the measured speed of light. The two constants, μ 0
and ε 0 , are the permeability and permittivity of free
space and essentially quantify the strengths of electric and magnetic
forces. You might also want to read earlier related answers linked to on my
faq page.
QUESTION:
ANSWER:
17
kg/m3 and the radius is about 12 km, so the mass is about M =5x1017 x4xπ (12x103 )3 /3=3.6x1030
kg. To find the acceleration due to gravity, g=GM /R 2 =6.67x10-11 x3.6x1030 /(12x103 )2 =1.7x1012
m/s2 . So your weight on the surface would be about a trillion
times what it is on earth. You would be crushed totally flat. We can
estimate how far light would "fall" in such a gravitational field. The time
it would take to go 100 m would be about t ≈100/(3x108 )=(1/3)x10-6
s. The distance it would fall would be y =�gt 2 ≈10 m.
On earth the speed of something which dropped 10 m in 100 m would be about
71 m/s≈160 mph, quite a bit faster than a garden hose. But qualitatively,
yes the light would fall quite a lot.
QUESTION:
ANSWER:
All clocks run slower because time itself runs
slower as the gravitational field gets stronger; this includes biological
clocks. You would never notice it for biological clocks in the earth's
gravitational field because the effect is very tiny and biological clocks
are not accurate enough to detect the slowing. You could detect it if you
were close to a black hole where gravity was much stronger.
QUESTION:
ANSWER:
appear
shorter but they are shorter. See an
earlier answer .)
QUESTION:
ANSWER:
a=F/m and so any force will cause an infinite
acceleration. In fact the theory of special relativity requires that any
object with zero mass can move only at the speed of light in vacuum. So a
massless object at rest cannot exist.
QUESTION:
ANSWER:
a and being at rest in a
gravitational field with local gravitational acceleration a .
Therefore, it is as if the two were at rest in a uniform gravitational
field. Then it is a consequence of general relativity that the clock
"higher" in the gravitational field will run faster. Now, you suggest having
accelerometers at each end and that they will record the same; that is true
because it is no different from being at rest in a uniform field and
measuring the field at both ends. Now, if you are at rest in a field, there
is no problem because there is no acceleration and therefore no velocity
change. So the results are required (by the equivalence principle) to be the
same if you are in a zero field with an instantaneous acceleration. I think
the problem is that since acceleration is not really a useful quantity in
relativity, you should not conclude that equal accelerations in unequal
times necessarily implies a relative velocity between two observers in the
same frame. In fact, it clearly cannot because both are at rest in the same
frame.
FOLLOWUP QUESTION:
ANSWER:
tidal
force . You may have read somewhere that when you fall into a black hole,
you get "all stretched out" as you fall; this is due to the extreme tidal
force from one end of your body to the other.
QUESTION:
ANSWER:
γ =1/√(1-(v /c )2 ),
T' =γT (T is the lifetime for a particle at rest) but in
the particle frame, a distance toward the earth is shortened by the factor
1/γ , L'=L /γ (L is the distance for the earth at
rest). To check that this makes sense, the earth sees the particle
move with speed L /T' =γL' /γT=L' /T and the
particle sees the earth move with a speed L' /T =(L /γ )/(T' /γ )=L /T' ,
both the same.
QUESTION: 2 for half the trip and -1g for the other half. Newtonian kinematics states xf =xi +vt+at2 /2 yet over the course of time my velocity will become relativistic and the equation no longer applies.
What is the correction needed for me, a stationary observer, to solve how long it will take the rocket to reach its
destination?
ANSWER:
this problem ;
I have found a solution for the distance as a function of time for a
constant force F rather than the time as a function of distance, but
that can easily be inverted. I showed that x =(mc 2 /F )(√[1+(Ft /(mc ))2 ]-1);
in your case, F=mg , so x =(c 2 /g )(√[1+(gt /(c ))2 ]-1).
Solving for t , t =√[(x /c )2 +(2x /g )].
For the first half of the trip, x =2.175 ly and I calculated that g =9.8
m/s2 =0.11 ly/yr2 , so t =6.58 yr. Given the
symmetry of the situation, the total time for the trip, as observed by you,
is 13.2 yr. Using the result from the
earlier answer , v =(gt )/√[1+(gt /c )2 ],
I find that the speed at the midpoint of the trip is 0.89c . The
average speed for the whole trip was 4.35 ly/13.2 yr=0.33c .
FOLLOWUP QUESTION:
ANSWER:
relativistic rocket . For a rocket with acceleration g halfway to
a distance D in light years and then with deceleration g the rest of the way, the
elapsed time T in years on the rocket is T =1.94∙arccosh[(D /1.94)+1)].
So, for D =4.35 ly, T =3.58 yr.
QUESTION:
ANSWER:
c
and the light hit the target. Your experiment would be different, however,
if you were accelerating forward; the light would miss the target for the
same reason that if you were to aim a rifle directly at a target on earth it
would fall some on the way there. You might be interested in the
light clock . Your space ship could be thought of as a light clock which
ticks once when the light hits the target.
QUESTION:
ANSWER:
c and a piece of foam insulation (much softer than iron) had
catastrophic results. You might be interested in another
recent answer .
ADDED ANSWER:
17
J, about the same as 2,000 Nagasaki atomic bombs. Curtains for this space
ship!
QUESTION:
ANSWER:
QUESTION:
ANSWER:
2 )=0.14 light years
away as you observed it if going 99% the speed of light; that gives you a
lot less time to maneuver. But there is a much more important barrier to
being able to navigate at very high speeds. The upper picture to the left
shows the space craft as seen by an outside observer; light from a distant
star comes with velocity c θ
relative to your velocity v θ'
because of the velocity transformation to the new light ray c' c'=c , but the directions are
different, θ'≠θ. It can be shown that θ'= tan-1 [sinθ √(1-(v /c )2 )/(cosθ+ (v /c ))].
I have plotted this for several values of v to the right. The effect is
quite dramatic. For example, at 99.9% the speed of light everything in your
forward hemisphere and much from behind you will appear inside a 50
cone in front of you! Just having optics which could give good enough
resolution to see anything but an extremely bright light in front of you
would be a challenge in itself. But wait, it even gets worse! There will be
significant Doppler shifts and much of the light at very high speeds will be
shifted out of the visible spectrum. A nice little animated gif below (which
I found on
Wikepedia ) demonstrates this but only up to 89% of c .
QUESTION:
ANSWER:
K =�mv 2 where
m is mass and v is speed. Relativistically, K=E-m 0 c 2
where c is the speed of light, m 0 is the mass of
the object when not moving, the total energy E =√(p 2 c 2 +m 0 2 c 4 ),
and the linear momentum p =m 0 v /√(1-v 2 /c 2 ).
So, you see, mass appears all over the place. The momentum of a massless
particle, the photon, is p=E /c ; its energy is all kinetic, and
K=E=pc
since a photon has momentum even though it has no mass.
QUESTION:
ANSWER:
d'
the traveling friend travels in 3 weeks (as measured by him). d'=vt' =0.99999c ∙3=2.99997c
or d' =2.99997 lw each way (1 lw is a light week, the distance light
travels in 1 week). The earth-bound friend, however, sees a much longer
distance that his friend has traveled because the friend sees the distance
length-contracted by a factor √(1-.999992 ); so the distance which
the earth-bound friend sees is d =2.99997/√(1-.999992 )=670.8
lw each way; the time someone traveling 0.99999c takes to go this distance
out and back is 2x670.8/0.99999≈1342 weeks. The earth-bound friend has been
healed for about 1336 weeks, more than 25� years, when his friend returns
with a just-healed arm. You still cannot really accept that moving clocks
really do run slowly, can you?!
QUESTION:
ANSWER:
QUESTION:
ANSWER:
FOLLOWUP QUESTION:
ANSWER:
etc .
trying to find a spot with small gravitational field is totally unnecessary;
all that is required is that you need to be far away from any mass and
nearly the whole universe satisfies that criterion. We generically call such
places intergalactic space . So, choose some place, maybe halfway
between here and the Andromeda galaxy, our nearest galactic neighbor. For
all intents and purposes the field there is zero and a clock would run at
some rate. Now, take that clock and put it on the surface of a sphere of
radius R and mass M . The
gravitational time dilation formula you need to compute how much slower
the clock would tick at its new location is √[1-(2MG /(Rc 2 ))] where G =6.67x10-11
N∙m2 /kg2 is the universal constant of gravitation and
c =3x108 m/s is the speed of light. Putting in the mass and
radius for the earth, I find that the space clock runs about 7x10-8
% faster than the earth clock, certainly not "extremely fast"! The
bottom line here is that, except very close to a black hole, gravitational
time dilation is a very small effect. An interesting fact, though, is that
corrections for this effect must made in GPS devices because they require
extremely accurate time measurements to get accurate distance measurements.
QUESTION:
ANSWER:
2 )=1.06 ly and so the time it takes to reach
you is 1.06 yr. An observer on the ship says it takes 1 yr.
QUESTION:
Two people are on a high speed, constant velocity train on opposite ends. One person standing on the platform. The two people aim a laser beam at the center of the train at the same time in their reference frame. In the center of the train, there is a special gun that will shoot and kill whoever's laser beam reaches it first; but if the two beams reach it at the same time it will be disarmed. From the person on the platform's standpoint, one of them dies due to the relativity of simultaneity. From the two people on the train's standpoint, they both live. Later, the two people get off their high speed train and meet with the person on the platform. Is this not a paradox? The platform person clearly saw one of them die.
ANSWER:
QUESTION:
I am inside a train in which all the sides/windows are sealed i.e., I cannot see anything from inside. If the train moves in a friction-less rail, i.e., without jerking, how can I prove that there is motion taking place or there is displacement?
ANSWER:
QUESTION:
(This question refers to an
earlier
answer .)
With respect to your
ANSWER above, I have read elsewhere that one of the observations from particle accelerator experiments is that the mass of a particle increases the more it is accelerated AND that there is a concomitant release of energy (I recall reading �release� vs expenditure or other terminology, but the ol� memory could be faulty, too). I read in your
ANSWER above about inertia being what increases versus size/"amount of stuff" as a given mass accelerates. Per my outside reading I am attempting to conceive the source of the energy released, not in mathematical terms, but with creative logic, if there is such a thing. Maybe not. Math might be the only way to explain the phenomenon. My attempt with English in question form: Is the measurable release of energy due to the fact that so much more energy is required to continue accelerating the particle? Is the source of the "released energy" I read about actually the instrument itself, the particle accelerator? For some reason, prior to reading your
ANSWER , I conceived that as mass/inertia increases [as acceleration takes place] there was some inherent energy in the particle itself that with the added force of acceleration, inevitably had to be expressed / released. Creative huh? What IS the source of that energy "released" as the particle accelerates and its inertia increases? Or am I so way off you�re having a cleansing, releasing, hearty laugh? And, can this concept be explained in language that an extremely uneducated but by no means unintelligent person can understand? Are you that good? I�m giving you lots of credit, much deserved for your dedication to askthephysicist. If answering requires math, my reading said answer is not likely to be helpful so save yourself the time and move on with your day.
ANSWER:
E=mc 2 which I hope you can work with�this simply says that any particle
has an inherent energy due to is its mass m which is the mass (inertia) times the speed of light
c
squared. If a particle is at rest, it has energy because it has mass. If a
particle is moving, the classical view is that it also has what we call kinetic energy,
energy by virtue of its motion. There are two
ways you can look at it. One is that we observe that the mass m
increases as the particle speeds up, eventually approaching infinite mass as
the speed approaches the speed of light; in this view, kinetic energy is
simply folded into the quantity mc 2 . The other way is to
say that the particle has an inherent mass energy m 0 c 2
where m 0 means the value that m has if the particle is at
rest; in this view we interpret the energy of the particle to be rest mass
energy plus the kinetic energy K it has by virtue of its motion, m 0 c 2
+K .
These are just two ways of looking at the same thing; therefore, if the
particle retains its identity, if it now interacts with its environment it
can release the energy it previously gained by accelerating and if it ends
up at rest it will have released K Joules of energy. Any energy this
particle "releases" can only come from energy which it previously acquired.
ADDED
THOUGHT: An accelerated particle
does radiate electromagnetic energy (light, x-rays, etc .) called
bremsstrahlung ; this is how an antenna works. So, when a particle is
being accelerated, it does "release" a bit of energy in this way. The rate
of this radiation is usually quite small compared to the rate at which
energy is being added and is often ignored when considering accelerators.
That which you "read elsewhere" might have been referring to
bremsstrahlung.
QUESTION: ?
ANSWER:
QUESTION: W hat is the formula to calculate the total energy of an object that is moving at relativistic speeds? (Ignoring any potential energy)
ANSWER:
m 2 c 4 +p 2 c 2 )
where p=mv /√(1-v 2 /c 2 ) is the
linear momentum, m is the rest mass, v is the speed, and c
is the speed of light. This can also be written as E=mc 2 +K
where K is
the kinetic energy (which is not �mv 2 ). Comparing these two
expressions and doing some algebra, you could find an expression for K .
Also, it may be written E=mc 2 /√(1-v 2 /c 2 ).
QUESTION:
ANSWER:
k =2,000 N/m and you compress it by x =1 cm=0.01 m. The work
you did is E =�kx 2 =0.1 J. This energy will end up as added
mass m=E /c 2 =10-18 kg.
QUESTION:
ANSWER:
QUESTION:
I'm in a space ship moving at the speed of light. I put my space suit on, go out on the deck of the ship and tee up a golf ball. Assuming I have the skill necessary to hit the ball, will I be able to hit the ball forward? It seems to me that in the vacuum of space, the energy required to hit the ball forward would not change with my vehicle's speed, but it's also my understanding that the speed of light is an absolute maximum speed. One of these is apparently wrong. I think you are going to tell me I can't hit the ball forward. Can you explain why?
ANSWER:
faq page if you
have a problem with this. However, we can address your question by saying
your ship has a speed 99.999% the speed of light u =0.99999c .
Suppose that you can launch a golf ball at a speed of 10% the speed of
light, v =0.1c . (Of course, it is impossible to make it go that
fast because you are not strong enough, but it makes this example easier to
quantify.) If you were able to do this, there would be absolutely no reason
why you could not do it on the moving ship: you would simply see the ball
move away from you with speed of 0.1c just as if you were here on
earth. Now, classical physics (and your intuition) would have the speed seen
by someone watching this to be V =(0.99999+0.1)c =1.09999c ,
almost 10% faster than the speed of light. However, classical physics does
not work here; again, see the faq
page. Instead, V =(0.99999+0.1)c /(1+0.99999x0.1)=0.99999182c .
QUESTION: y question deals with the "potential energy"
of a gravitational field, in relativistic terms. As an object falls in a
gravitational field, a tiny bit of its mass gets converted to kinetic
energy, a small reduction in mass. However, according to Special Relativity,
as an object accelerates, gets closer to the speed of light its mass (how it
interacts with spacetime) increases. So, does an object accelerating
(increasing kinetic energy) in a gravitational field lose mass, as a result
of falling in a gravitational field, or gain mass as a result of approaching
the speed of light? OR is there some breakeven point were the reduction of
mass from falling stops, and then mass starts increasing again as a result
of approaching the speed of light. If the mass starts increasing because of
approaching the speed of light, wouldn't the gravitational potential
approach infinite?
ANSWER:
earlier similar questions . The reason you do
not want to simply say that there is "a
small reduction in mass" if it is falling is that it is not at rest when you
look at it later. As I explained in the earlier answers, you need to
calculate things relativistically to find out how the mass changes. That is
essentially what the
last link I gave you in my earlier answers does, calculates
β= v /c
as a function of time for a constant force. To know how the mass varies (if
you interpret relativistic momentum that way), just calculate m 0 / √(1- β 2 )
where m 0 is the rest mass. It occurs to me, though, that
it might be of interest to redo that calculation for a force which is not
constant but which has a value mg =m 0 g /√(1- β 2 )
where g is the acceleration due to gravity in whatever strength field
you wish to examine. I will not give all the details here, just the results.
We start by integrating the relativistically correct form of Newton's second
law,
m 0 g /√(1- β 2 )=dp /dt =(d/dt )[m 0 v /√(1- β 2 )]
integrated givesgt /c =�ln[(1+β )/(1-β )]
solved for β gives
β= (1-exp(-2gt /c ))/(1+exp(-2gt /c ));
put this into the mass and get m /m 0 =1/√(1-β 2 ).
These are plotted to the right above.
Note that the
result for a constant force m 0 g
is plotted in red for comparison. Potential energy is not a useful concept here. Note that the time to get to
near c , t ≈3c /g , is about 280 years for
g =9.8 m/s2 ! At that time the moving mass is about 10 times
greater. A more lengthy discussion of a mass in a uniform gravitational
field (including general relativity) can be seen
here .
QUESTION:
Around 1905 when Lorentz came up with his tranformation equation, it was assumed that c was constant, but was it also assumed that c is the speed limit of the universe?
Why didn't he assume that v could be greater than c in some cases? Why didn't he put a corrective constant next to c in the equation to allow for the case that time might stop BEFORE or After v reaches c?
The reason I'm asking this question is because the Lorentz transformation has time stop exactly when v=c. How did he know that should be the case?
ANSWER:
c is a constant and why it has the value it does, go
to my FAQ page; there you can also see earlier questions about why c
is the speed limit. Your question about a "corrective constant" makes no
sense since universal constants by definition do not need correcting. Since
no clock can ever reach c , time never "stops". I doubt that Lorentz
ever thought about time stopping in any context.
QUESTION:
I have points A, B, and C which are moving in 3 dimensional space. If I use point A as a frame of reference, I know the direction and velocity of points B and C. If my points are moving arbitrarily close to the speed of light, what equations would I use to calculate the relative velocity of C from B's frame of reference, and the difference in clock speeds between those two points?
I understand how I would do this if the speeds were nowhere near the speed of light, but things get a tad complicated after accounting for relativity.
ANSWER:
velocity addition if the velocities of B and C are colinear: in your
notation, V CB =(V CA +V AB )/(1+(V CA V AB /c 2 ));
read V XY as velocity of X relative to Y. This may be
generalized
to noncolinear motion as V CB =(V CA +V ‖ AB +γ CA -1 V ┴ AB )/(1+(V CA ∙V AB /c 2 ))
where V ‖ AB
and V ┴ AB
are the components of V AB
parallel and perpendicular, respectively, to V CA
and γ AB -1 =√[1-V CA 2 /c 2 ].
Note that if the speeds are small compared to c , V CA ∙V AB /c 2 ≈0γ CA -1 ≈1
so V CB ≈(V CA +V ‖ AB +V ┴ AB )=(V CA +V AB ).
QUESTION:
How can the speed of light be the fastest thing in the universe if it is a relative measurement? I mean that if you had two light particles going in opposite directions, wouldn't one go twice the speed of light relative to the other?
ANSWER:
FAQ page. It is
a relative measurement, but your idea of the relative velocities is
incorrect at large speeds. You want to use
v' =( u + v ) ,
but the correct expression is
v' =( u + v )/[1+( uv / c 2 )]
where
c
is the speed of light; notice that if
u
and
v
are both small compared to
c ,
your expectation is very nearly correct. For your particular example,
u=v=c
and so
v' =( c + c )/[1+( c ∙ c / c 2 )]= c.
You can also find other answers about light speed, the maximum velocity
possible, etc . on the FAQ page.
QUESTION:
Suppose you have radiation detectors fixed on the ground on Earth. Will they detect radiation coming from a charged particle in free fall near them?
The first answer that comes to mind is: Yes, they will detect radiation because the particle is accelerated, and electrodynamics predicts that accelerated charges must radiate in this situation.
According to the Equivalence Principle, this situation is equivalent to detectors fixed on an accelerated rocket with acceleration g moving in the outer space and far away from the influence of other bodies. If the answer to the previous question is yes, then the detectors on the rocket should also detect radiation coming from a charge in free fall as observed by the reference frame of the rocket. But a charge in free fall in this reference frame is at rest in the inertial reference frame fixed with respect to the distant stars, and a charge at rest in an inertial frame should not radiate.
Is it possible that detectors fixed on the rocket detect radiation but detectors at rest in the inertial frame do not? Is radiation something not absolute, but relative to the reference frame?
ANSWER:
Almeida and Saa ,
has evidently laid the paradox to rest. They demonstrate in this article
that observers for whom the charge is not accelerating "�will not detect any radiation because the radiation field is confined to a spacetime region
beyond a horizon that they cannot access�" and "�the electromagnetic field generated by a uniformly
accelerated charge is observed by a comoving observer as a purely electrostatic field."
Like all "paradoxes" in relativity, there is not really a paradox; rather a
radiation field in one frame may be a static field in another. Basically,
you nailed it when you said "radiation
[is] something not absolute, but relative to the reference frame."
QUESTION:
How fast a person would have to move to be completely invisible to our eyes? Like Goku from Dragon Ball.
ANSWER:
λ '/λ =√[(1-β )/(1+β )] where λ ' is the observed wavelength, λ
is the laser wavelength, and β is the ratio of the velocity
v of Goku relative to the speed of light c , β=v /c ;
β> 0 is for Goku moving toward you. When he is coming
toward you, β >0 and so he must be going fast enough that 700 nm
light is shifted to 390 nm light (blue shift). I find then that β toward =0.53.
Similarly, when he is moving away from you, β <0 and so he must
be going fast enough that 390 nm light is shifted to 700 nm light (red
shift). I find then that β away =-0.53. So, he needs to
be moving with a speed of about half the speed of light.
If he is not moving directly in a line to or from
you, the situation is a little more complicated. The more general
expression for the Doppler shift is λ '/λ =(1+β cosθ )/√(1-β 2 )
where θ is the angle between his path and your line of sight.
(Note that when θ= 00 , this reduces to the equation
above.) When θ= 900 he is directly in front of you and
so λ '/λ =1/√(1-β 2 ),
always a red shift because 1/√(1-β 2 ) is
always greater than 1 so λ '>λ . Solving, β transverse =0.83.
One more thing: he will be emitting other
frequencies beside visible. For example, he will be emitting infrared
radiation which will be doppler-shifted into the visible when he is moving
toward you. I have ignored this possibility and assumed that he only emits
visible light. After all, he is just a made up character anyway, right?!
QUESTION:
If you see a laser
beam as blue light of 420 nm from a spacecraft. How would you be able
to tell how fast the spacecraft is going and in what direction? I believe the wavelength of the laser pointer is 620nm.
ANSWER:
λ '/λ =√[(1-β )/(1+β )]
where λ ' is the observed wavelength (420 nm), λ
is the laser wavelength (620 nm), and β is the ratio of the velocity
v of the spacecraft relative to the speed of light c , β=v /c ;
β> 0 is for the spacecraft moving toward the observer. Solving
this equation for β , β= -0.371. The negative sign means that
the spacecraft is moving away from you. This is called a redshift.
QUESTION: his question should be fairly quick. Are there any know situations were momentum isn't conserved? I would say no, as momentum is always conserved if you make your system big enough. The only time momentum appears to be not conserved is when you put restrictions on the size of your system, and don't account for the momentum transferred outside your system. When you include the system " outside" your system, momentum is in fact conserved.
ANSWER:
isolated system; an isolated system is
one for which there are no net external forces acting. This is actually a
result of Newton's third law which essentially states that the sum of all
internal forces in a system must equal zero. This works really well until
you get to electricity and magnetism where it is easy to find examples of
moving charges exerting electric and magnetic forces on each other which are
not equal and opposite (you can see an example in an
earlier answer ). In that case you would say that such an isolated
system obeys neither Newton's third law nor momentum conservation. However,
looking deeper, we find that an electromagnetic field has energy, linear
momentum, and angular momentum content and, in the end, momentum
conservation, because of the momentum contained in the field, is still
conserved for an isolated system. Thus, Newton's third law is saved, but not
always in the simplistic "equal and opposite forces" language. Finally, if
linear momentum is p =mv p =mv p ≡γmv =mv v /c )2 ];
this then unifies energy E and linear momentum p into a single
entity which is conserved, the energy momentum 4-vector: E 2 -p 2 c 2 =m 2 c 4 .
Not as quick as you expected!
QUESTION:
How fast must one move away from a typical light bulb so that it Doppler shifts into microwave radiation and cooks him/her?
ANSWER:
δ≡f observed /f source =√[(1+β )/(1-β )]
where β is the ratio of
the relative speed of the source and observer to the speed of light. Taking
f source ≈600 THz=6x1014 Hz and
f observed ≈2.5 GHz=2.5x109 Hz, I find
δ ≈4.2x10-6 . Now, the
Doppler equation may be rewritten as
β =-[(1- δ 2 )/(1+δ 2 )]≈-(1-δ 2 )2 ≈-(1-2δ 2 )=-0.999999999966,
pretty fast! I used binomial expansion, (1+x )n nx +�
if x <<1, a couple of times to approximate.
QUESTION:
My question is in regards to relativity and quantum mechanics. I�ve heard physicists say on t.v programs and in books that relativity and quantum mechanics are incompatible and describe two different phenomena. Do these two theories contradict one another? Is the universe contradictory or is there anything in the universe that can definitively be defined or described as a contradiction?
ANSWER:
E=mc 2 ,
etc . is perfectly compatible with quantum mechanics; there is
something called relativistic quantum mechanics. What is probably being
referred to by your sources is the theory of general relativity which is the
best current theory of gravity. No one has been successful in developing a
theory of quantum gravity. I would not so much describe them as
"incompatible" as I would call them "nonunified".
QUESTION: hen a subatomic particle is listed as having a voltage of, say 1.8 Gev does that mean approx or exactly 1.8 and are all such particles exact.
ANSWER:
-19 Coulombs (C)) and has been
accelerated across a potential difference of 1 Volt (V). Since 1 C∙V=1 Joule
(J), 1 eV=1.6x10-19 J. Also, 1 GeV=109 eV=1.6x10-10
J. The energy of a particle can mean several different things, depending on
context. Most likely, since you used the word "listed", E =1.8 GeV
refers to the rest mass energy of the particle, E=mc 2 ; so,
you see, specifying this energy is equivalent to specifying the mass of the
particle at rest. Sometimes the mass is specified as, in your example, m =1.8
GeV/c2 . For example, the mass of the proton is 0.9383 GeV/c2 ;
every proton has exactly the same mass as every other proton because, as I
will explain below, a proton is a stable particle, it lives forever. If a
particle is listed with a mass of 1.8
GeV/c2 , however, not every one of those particles will have
exactly the same mass because I know for a fact that there is no particle
with that mass which is stable�it will decay with some lifetime. The reason
the lifetime makes a difference is the Heisenberg uncertainty principle
which states that the uncertainty of the energy of a particle can only
be zero if you measure that energy over an infinite time, or ΔE Δt >ћ
where ћ≈ 6.6x10-25 GeV∙s is the rationalized Planck
constant, ΔE is the uncertainty of energy (mass), and Δt
is, essentially, its lifetime. So, the shorter the lifetime of a
particle, the greater will be the spread of possible masses for that
particle. Suppose that your particle has a lifetime of 1 ps=10-12
s; then ΔE ~6.6x10-25 /10-12 ≈6.6x10-13
GeV. So there would not be much of a spread of measured masses but it would
not be zero.
Other things which
could be meant by specifying the energy of a particle are its total energy
E total =mc 2 /√[1-(v /c )2 ],
where v is its velocity, or its kinetic energy, K =E total -mc 2 .
QUESTION:
Let's say we have planet A and B. Planet A suddenly fly away from planet B at 1/2 the speed of light. Planet B also suddenly fly away from Planet A at a little bit more than 1/2 the speed of light. If we observe from Planet A, we can see Planet B because light travelled from Planet B to Planet A. However, after all the flyings, Planet A and Planet B will be 'separating' from each other at a little bit faster than the speed of light. Does it mean that after a while, observers from Planet A can no longer see Planet B? How does this workkkkk???????? :(
ANSWER:
v A =0.5c
and the speed of B relative to D is v B =0.6c let's
say. Your assumption is that the speed of A relative to B is 1.1c . (c
is the speed of light); that is wrong. In an
earlier answer I discussed this kind of problem. The speed v with
which A and B each see the other going away is given by
v =( v A + v B )/[1+( v A v B / c 2 )]=1.1 c /[1+( 0.5x0.6c 2 / c 2 )]=0.85 c .
Not intuitive, but true!
QUESTION:
How can a stationary radio receiver set to a particular frequency receive transmissions from a moving transmitter, say, from an aircraft?
Why doesn't the movement of the aircraft cause its radio transmission to be off frequency relative to the static receiver?
ANSWER:
f of
electromagnetic waves relative to the source frequency f 0
is given by f=f 0 √[(1+β )/(1-β )] where β=v /c
and v is the speed of the source and c= 3x108
m/s is the speed of light; β >0 is when velocity is toward you.
If β is much smaller than
1, it may be shown that the fractional change in frequency is |f-f 0 |/f 0 ≈β .
For example, suppose the source is an airplane with speed 300 m/s (about 700
mph) and the frequency is 90 MHz. Then |f-f 0 |/f 0 ≈10-6
or the Doppler shift is |f-f 0 |≈90 Hz. A tuner does not
have nearly the sensitivity to detect a 10-4 % change in
frequency.
QUESTION:
How can a fast moving physical ship with a slow ticking clock get to Andromeda 89 thousand times faster than light, as you claim?... 28 years on the ship's slow clock but 2.5 million Earth orbits around the Sun, just for a "reality check."
ANSWER:
REPLY:
Ask *a* Physicist , where I read his claim awhile back on his
"We Can Make It to the Edge of the Observable Universe in a Few Decades" article/blog. I take it that you do not agree with him.
ANSWER:
similar answer on my site
which we can use for comparison. The speed which Goldberg uses is 99.9999999999999999998% of
c . In my calculation I used a very fast speed
but much slower than his, 99.999% of c . When I calculated the
elapsed time on the ship I got about 11,000 years, he got only 28 years. My
ship was going the same speed the whole time but his started from rest and was constantly
accelerating, so he would have gotten a smaller time yet if his ship had
been going with a constant speed the whole time; if the Goldberg space ship
had been going 0.999999999999999999998 c
the whole way, it would get to Andromeda in about an hour, its time! Your
problem, I think, is how either his or mine could be right since it takes
2.5 million years for light to travel that distance, right? But, it depends
on who is measuring the time. An observer on earth will see the light take
2.5 million years to get there and the space ships arrive just a
little (compared to 2.5 million years) later. But an observer on the space ships will see a greatly
shortened distance to Andromeda, so he will observe both himself and light
to take a much shorter time to get there, the light always just beating
the space ship. You are right, nothing can travel faster than light, but
everybody sees the same light speed in their own frame; there is no
contradiction in saying one observer sees light taking 2.5 million years to
travel between two points in space while another observer sees light as
taking 10 years to travel between the same two points.
QUESTION:
In the TV-series Stargate SG-1 a Goa'Uld Ha'Tak is capable of going 50% lightspeed with sublight engines. Wouldn't these kinds of speeds already be relativistic and what happens excactly to the crew in the ship when the relativistic effects start? And what is the minimum relativistic speed in which relativistic effects become noticeable? A friend of mine said that at 8% lightspeed relativity kicks in making space travel above 8% lightspeed really challenging.
ANSWER:
γ= √(1-β 2 )
where β is the ratio of the speed to the speed of light. For example
if β= 0.1, 10% the speed of light, then γ= 0.995 which would
mean only about 0.5% error using nonrelativistic calculations. For β= 0.5,
50% the speed of light, γ= 0.867,
still only about 13% error. But, there are no "relativistic effects" for
somebody on the ship; everything appears normal on the ship. They would,
however, see distances to objects toward which they are traveling shrink. I
do not know what it means that "space travel�[is] really challenging". Of
course, it is challenging to get to relativistic speeds in the first place,
but I do not see any hazards "kicking in" at
β= 0.08.
QUESTION:
Is it possible, and if so how to calculate the magnitude of a 4-dimensional vector?
ANSWER:
A x , A y ,
A z ) and a time-like part A t ,
then the magnitude of the vector is A =√[A t 2 -(A x 2 +A y 2 +A z 2 )].
For example, the space-time vector has a magnitude of R =√[c 2 t 2 -(x 2 +y 2 +z 2 )]=√[c 2 t 2 -r 2 ]
and the energy-momentum vector has a magnitude of √[E 2 -(p x 2 +p y 2 +p z 2 )c 2 ]=
√[E 2 -p 2 c 2 ]=m 0 c 2 .
If you want a bunch more detail, see the Wikipedia article on
4-vectors .
QUESTION:
Is there a Doppler shift between the centre and outside of a wheel.
Say for example a standard merry go round you find at a sideshow, if I stand in the middle and look at something on the outer does my image Doppler shift.
The reason it is confusing to me is because your angular rotation is the same for both people but not your linear speed.
ANSWER:
v and light comes at an angle θ as shown.
In this case, f'=f /[γ (1+(v cosθ /c ))] where
γ =1/√[1-(v 2 /c 2 )] and f and
f' are the frequencies at the source and observer, respectively. If
θ= 900 ,
f'=f /γ ; this is called the transverse relativistic Doppler
effect. But, in your scenario, each will see the other to be at rest. But
both are actually accelerating relative to an inertial frame, so the
situation is not so simple. You can see a discussion of this scenario on
Wikepedia . Suppose that the source is at a distance r 1
from the center, the observer is at a distance r 2 , and the
angular velocity is ω. Then assuming that the
distance traveled by the observer in the time it takes the light to travel
between the two is very small, I find that f'≈f √[(1-(r 1 ω/c )2 )/(1-(r 2 ω/c )2 )].
I just realized that you actually asked an easier question, namely that r 2 =0
and r 1 =R where R is the radius of the
merry-go-round, so, f'=f √(1-(Rω/c )2 ). This is
exactly the same as the transverse Doppler effect above, f'=f /γ.
[Sorry that the answer is much longer than needed. Whenever I misread a
question and do a lot of interesting and more general work, I am loathe to
delete it! The concise answer for you is that f'=f /γ. ]
QUESTION:
I regularly contribute to another website that I will not mention here, but we had a discussion on a particular subject matter and one topic which came up was kinetic energy of an object moving close to the speed of light.
Specifically an assertion was made that an object traveling 85% of c would have kinetic energy equal to a matter / antimatter reaction of the same mass as the traveling object.
0.5mv^2 did not produce compatible numbers unless one assumed quite low efficiency of the M/A reaction. It was then stated to me that formula is not valid to the situation, as it fails to account for the increased mass of an object traveling at such speeds.
I'm no physics professor, but that seemed a little fishy to me. I attempted to divine an answer to this with the use of some Google-Fu, however it seems my Google-Fu is not strong enough, as I found multiple conflicting answers (and even multiple conflicting formulae to determine the kinetic energy in such a situation.)
So since my Google-Fu has failed I figured I would ask you. Do you need a different formula to calculate the kinetic energy of an object traveling at such speeds, and if so what is that formula?
ANSWER:
right place ! From your
discussion, that unnamed website is the wrong place to get physics right.
The expression �mv 2 for kinetic energy is true
(approximately) only for speeds much less than the speed of light. There is
no mystery about what kinetic energy is in special relativity, but I will
not go through the derivation. The total energy E of a particle with
rest mass m is E = E=mc 2 /√(1-v 2 /c 2 ).
If the particle is at rest, v =0 and E=mc 2 . If the
particle is not at rest, it still has its mc 2 energy and
everything else must be kinetic energy K : K=E-mc 2 .
That is all there is to it.
[It is not at all
clear what is meant by " �a
matter/antimatter reaction of the same mass�" in your question; I will just
briefly discuss particle/antiparticle pair creation and required kinetic
energy.]
For your case (v =0.85c ):
E =mc 2 /√(1-0.852 )=1.90mc 2 ,
so K =0.90mc 2 .
There is therefore not enough kinetic energy to create another particle of
the same mass (which I am assuming the remark in your question about
"antimatter" refers to). The way to create an antiparticle, for example an
antiproton, is to collide a fast proton with a proton at rest; the energy
required to create one particle of the same mass takes quite a bit more than
mc 2 of kinetic energy because momentum must also be
conserved, so the particles cannot be at rest after a collision. Also, you
cannot simply create a mass out of kinetic energy, you have to create a
particle-antiparticle pair so in the proton experiment you end up with three
protons and an antiproton. It turns out that the incident proton kinetic
energy energy needs to be 6mc 2 . The advantage of a
collider is that the momentum is zero so you only need to have each proton
with kinetic energy mc 2 to create a proton-antiproton
pair.
Here is an added detail since I am sure that the
expression for K above looks unfamiliar to you. If v <<c ,
1/√(1-v 2 /c 2 )≈1+�v 2 /c 2 +�;
this is just a binomial expansion and, if you are not familiar with
that, try it yourself: e.g . if v /c =0.05, 1/√(1-v 2 /c 2 )=1.001252
and 1+�v 2 /c 2 =1.001250. So now we can
write K ≈mc 2 (1+�v 2 /c 2 )-mc 2 =�mv 2 .
Tada!
FOLLOWUP QUESTION:
Just to clarify a bit the discussion which led me to asking you arose out of one concerning hypothetical weaponry. One individual asserted that a projectile traveling 0.85c would have kinetic energy equal to a matter-antimatter bomb with a similar warhead yield. [Thus a 500kg projectile would have energy equal to the energy released by the reaction of 250kg of particles annihilating with 250kg of antiparticles] I figured being a real physicist you would not want to waste too much time drawn into a discussion on the purely hypothetical topic. Thus my question came to you, since my original calculation for kinetic energy using 0.5mv^2 with m=500kg provided significantly less energy than E=mc^2 with m being 500kg.
The website in question is TVtropes.com which analyzes elements common in fictional works literary, TV, Film, and others.
ANSWER:
mc 2 of
kinetic energy. The speed to get exactly mc 2 is 0.866c .
But, you have no comprehension of how much energy is involved here. I think
you would be interested in an earlier answer along
these lines based on the movie Eraser .
QUESTION:
how much kinetic energy does a beam of hydrogen atoms moving at 5% the speed of light have.
ANSWER:
K of one hydrogen atom of mass M and you
can figure out how many atoms are in your beam. K=Mc 2 (1/[√(1-0.052 )]-1)=0.00125235Mc 2 =1.88228x10-10 J where c is the speed of light. Since the speed is quite small
compared to the speed of light, you could get a good approximation using the
classical value of K , K =�Mv 2 =1.87875x10-10
J.
QUESTION:
Just read Ender's Game, and it got me thinking of relativistic travel. What would be the optimum speed (relative to Earth) that you could send a probe (say, 1 light-year), and return to Earth? Go too slow, and it takes a long time. Faster is better, but if you go too fast, then relativity starts to work against you.
ANSWER:
relativity
starts to work against you." And, what are the criteria which determine the
"optimal speed"? If you are on earth and want to get the probe back to earth
as quickly as possible, the faster the better. If you are on the probe and
want to get there and back as quickly as possible, the faster the better.
How does relativity work against you?
FOLLOWUP QUESTION:
So, say there is an object 1 light-year away from the Earth. I want to send a robotic probe there, spend 1 day gathering a sample, and then return it to the Earth. If I send it at 50% c, (if I did my math correct) I calculate it would take (from my point of view on Earth) ~ 4.6 years (plus one day) to get the sample back (with only 4 years + 1 day elapsing relative to the probe). But, if I set the probe speed to .999 c, it would take me ~ 44.8 years ( + 1 day ) to get my sample, which is almost 10 times longer than .5c. This is what I mean by relativity working against you.
ANSWER:
c to take two years to go one light year and
2 years to get back: 4 yr + 1 day. If the speed is 0.999c , the time
out (and back) is 1/0.999=1.001: 2.002 yr + 1 day. From your point of view,
relativity is irrelevant if all you are interested in is measured by your
clocks and your meter sticks.
You might ask how much
time elapses for a clock attached to the probe. (See my discussion of the
twin paradox .) Each probe will
see the distance you see (1 ly) Lorentz contracted, the slower by √(1-0.52 )=0.866
and the speedier by √(1-0.9992 )=0.045 and so the elapsed times
will be 1.73 yr + 1 day and 0.09 yr + 1 day.
QUESTION:
It's well established that a rocket propelled spaceship accelerating forward will gain mass as it goes faster and faster. Appreciably so as it approaches light speed. My question is, how does this mass gain manifest itself? That is to say, does it occur by the atomic particles themselves gaining mass (protons, electrons etc.) or do new particles simply pop into existance due to the energies involved? Also, by what mechanism does this occur? How does energy being expelled from the rear result in the addition of mass to the ship and it's contents?
ANSWER:
QUESTION:
If I person could build wings for themselves like Icarus and fly at 99.999 percent the speed of light, all the way to the Andromeda galaxy, two and a half million light years distant, how much time would pass from their perspective? Would they experience said trip as taking two and a half million years, or would time pass more quickly for them? Would they even be alive upon reaching said galaxy?
Do we even possess the mathematics to work a problem like that out?
ANSWER:
γ , is γ = 1/√[1-(v /c )2 ]
which, for your case, is γ = 1/√[1-(0.99999)2 ]=223.6. For
Icarus, time would seem to pass as normal, but he would see the distance he
has to travel shrink by the Lorentz factor: D' =D /γ =2.5x106 /223.6=11,180.7
light years. So, the time that would elapse on his clock would be
11,180.7/0.99999=11,180.8 years. I do not think he would live that long.
Note that only arithmetic was used here, so, yes we certainly "possess
the mathematics to work a problem like that out"!
QUESTION:
I am trying to estmate the energy of a subatomic particle like a proton,by making the v=c or extremely close to c,but i am having a problem that the momentum is becoming Zero when calculating momentum at relativistc speed ,I my completely making a mistake,or am i missng somethin,
ANSWER:
2 =(pc )2 +m 2 c 4
where m is the rest mass. Also, p may be written as p =mv /√[1-(v /c )2 ].
As v gets closer and closer to c , p gets bigger and
bigger until eventually the m 2 c 4 term becomes negligibly small
and E≈pc . So, I guess I do not see where your problem is.
QUESTION:
I know Einstein said E=MC2 and basically all mater can be equated to some quantity of energy; then why do we go to the gas station to fill our cars? Why can't we use garbage, which is mass and has energy, to power our cars? How can we convert matter to energy? I know we can burn gasoline to use perhaps 1/4 the heat content in the form of expanding gas to apply pressure to the piston in the engine. Has anyone invented a converter that changes matter to energy yet? We eat food and basically run on sugar which fuels a chemical based process. Any other matter converters?
ANSWER:
fission ),
the mass of the fragments is smaller than the mass of the initial nucleus by
an amount much bigger than with chemistry, something like 0.1% which is a
huge improvement over chemistry; this is how nuclear reactors work. Also
from nuclear physics, you can take very light nuclei and make them combine (fusion )
and get like 1% of the mass converted into energy; this is how stars work
and so, you see, solar energy comes from "matter converters" too and so does
wind energy since the sun is the energy which causes winds to blow. If you
want to get 100% efficient you have to go to particle-antiparticle
interactions in particle physics. When an electron and its antiparticle the
positron meet, their mass completely disappears and all the energy comes out
as photons. Did you ever see the Back to the Future movies? Doc came
back from the future where they had invented a small appliance called "Mr.
Fusion" to do what you want, to convert garbage into the huge amount of
energy needed to power the time machine.
QUESTION:
My Question is, if you were in a spaceship in space with no windows, what experiment could you perform in order to prove whether you were moving or not moving?
ANSWER:
QUESTION:
Hey, so muons falling down towards the Earths surface survive decaying before they reach the surface because of length contraction as a result of Special Relativity. But I read SR doesn't apply when the object is accelerating. Aren't the muons accelerating due to Earths gravity? What have I understood wrong?
ANSWER:
-6 s
and has a relativistic speed, say 2x108 m/s; with an
acceleration of about 10 m/s2 , its velocity changes by about
10-5 m/s over its lifetime. Hardly accelerating, is it?
QUESTION:
I'm hoping you can settle a bet between my father and myself. We are
both movie buffs, and work together. While working we discussed the
Ah-nuld movie "Eraser". I'm going to make the assumption you have never
seen the movie which revolves heavily around the manufacture of
man-portable super-railguns that fire aluminum rounds at 'close to the
speed of light'. Being firearms enthusists as well, we discussed the
flaws in their idea (beyond the fact no technology in the near future
would allow such a weapon as the one depicted to be made, let alone one
man-portable.) Where we came to a disagreement was why for other reasons
it would be useless. While we agree that the mooks Ah-nuld tears through
as he does in every movie would be out of luck. In practical situations
the weapon would be useless but we disagree why. My father believes that
a 5.56mm / 0.00401753242 kg bullet (our assumptions) traveling 80% of c
would immediately flatten out and quickly lose all of its kinetic energy
due to its low mass and aluminin's relatively soft nature upon striking
anything solid such as a door, or brick. I say a round with the same
dimensions would explode on contact with something solid (again like a
brick) as the work-heating would vaporize a piece of aluminimum with
that mass. If it could even get there without being melted by
atmospheric friction. As I've submitted questions to you before we
decided we would --drum roll--- ASK THE PHYSICIST!!! Are either of us
correct? The loser has to buy the next Pizza we order.
ANSWER:
14
Joules. To put that in perspective, if you took that energy and delivered it
to the power grid over a period of a day, this would be the equivalent of a
6 gigawatt (6x109 watts=6 billion Joules per second) power
station. The largest power station in the US has a power output of about 4
gigawatts. Or, the energy of the Nagasaki atomic bomb was about 1014
Joules, 1/5 of the energy of your 4 gram bullet. So to argue whether the
bullet's flattening or exploding is the reason it would not do much damage
sort of misses the point, don't you think?! If the bullet takes 10 seconds
to deliver its energy to whatever can take it, you are still talking about
the energy of 5 WWII-era atomic bombs being delivered. I would not want to
be within 50 miles of that. Finally, it means that the "gun" has to deliver
all that energy in an unimaginably short amount of time (I presume that
since, if it is "man-portable", the gun must be no more than a few feet
long). I figure that a force on the bullet of more than 2 million pounds
would be required during the time it was in the gun; do you think an
aluminum bullet could withstand such a force? And, Newton's third law says
that, if the gun exerts that force on the bullet, the bullet would exert
that force on the gun. Imagine the recoil! This is such a ridiculous
scenario, I don't think we even need to beat a dead horse and talk about
what air friction would cause to happen as it flew to the target.
QUESTION:
ANSWER:
QUESTION:
Why exactly does the speed of
light (c) come in to play in Special Relativity? It seems to be rather
arbitrary in the sense that the Lorentz Transformation would (from what
I can gather in my admittedly novice research) yield the same results
using any speed that is accepted as being constant, whether it be the
speed of light, sound, or the 4:17 to Charleston. Is light specifically
used for any reason other than it's bearing on optics?.
ANSWER: c is a universal
constant seems very ad hoc . In fact, that postulate is not needed.
Here is the reasoning: the cornerstone of special relativity is that the
laws of physics must be the same in all inertial frames of reference.
Maxwell's equations, the laws of electromagnetism, predict the speed of
light to be exactly what it is and to depend only on two fundamental
constants of nature (constants which quantify the strengths of electric and
magnetic forces). Since Maxwell's equations are laws of physics, they must
be the same in all frames of reference and hence the speed of light must be
the same in all frames of reference.
QUESTION:
Can the total mass of a closed system fluctuate at any given
time, while the total energy remains of course constant, considering the
existence of mutual conversions of matter and energy ?
ANSWER:
There is an
electron and a positron (an antielectron). They encounter each other and
annihilate resulting in two photons. The original mass was twice the
electron mass and the final mass was zero.
A uranium nucleus
undergoes fission. Afterwards there is less mass but more kinetic energy of
the fission products.
QUESTION:
I would like to understand why time slows down at high speeds. Any information i find on this topic seems to complicated to get my head around, even my physics tutor said its too complicated and i should just assume 'it just does'. Could you explain it to me in a way i may be able to understand?
ANSWER: �the
light clock. Imagine a stick with a flashbulb and a light detector at one
end and a mirror at the other end. A flash of light goes up to the mirror,
bounces back and hits the detector. The detector is connected to a little
electrical circuit and a speaker which says "tick" whenever it detects a
flash of light and also tells the flashbulb to send out another flash of
light. So, the clock goes "tick, tick, tick, ..." Now, this is a perfectly
good clock, right? What you now do is make this clock go past you very fast
(velocity being perpendicular to the stick). Now the light has to travel
farther than when it was at rest, right? But, since the special theory of
relativity asserts that the speed of light is the same as seen by all
observers, the clicks are less frequent as seen by you because the flashes
have farther to travel. So, the moving clock runs slow. You can play with a
nice
simulation of a light clock.
QUESTION:
what is the real meaning of momentum & how we can relate it with photons which has no mass at rest and not defined mass while moving with speed of light?
ANSWER: p=mv.
One of the most important features of momentum is that the total
momentum of an isolated system never changes; this is called conservation of
momentum. However, in the theory of special relativity, if you choose
momentum to have the same definition, you find that momentum conservation is
lost. Conservation is such a powerful concept that we choose to redefine
momentum. The details of all this are given in an
earlier answer . The
result is that if a particle of mass m has an energy E , then
its momentum p is p=√{(E /c )2 -m 2 c 2 }
where c is the speed of light. So, you see, it is not necessary for a
particle to have mass to have momentum�it need only have energy.
QUESTION:
Hello I used to have a Feynman book that had this scenario and I
forgot how he explained it I have since lost the book and was wondering
if you could explain it. I have a spaceship moving at 180,000 kms inside
of that spaceship I have another spaceship moving at 180,000kms. To the
observer on the ground the second spaceship is moving at 360,000kms.
That exceeds the speed of light please explain what would happen.
MY FIRST
RESPONSE:
FOLLOWUP:
ANSWER: v' =(u +v )/[1+(uv /c 2 )]
where u is the speed of the first ship, v the speed of the
second ship, c the speed of light (300,000 km/s), and v' is
the speed of the second ship seen by the outside observer. Note that if u
and v are both very small compared to the speed of light, then
the quantity (uv /c 2 ) is very close to zero so that
v' =(u +v ), which is what you expect to be correct, is
approximately true. However, in the example you cited the speeds are not
small compared to c (they are 60% of c ). If you do the
arithmetic you will find that v' =265,000 km/s.
QUESTION:
What is the formula for displacement as a function of time for a particle undergoing constant acceleration? It's d=1/2 a*t2 at a Newtonian scale, but relativity kicks in as it gets closer to the speed of light. By "constant acceleration" I mean constant from an observer at "rest" relative to the accelerating particle (as would be the case of a particle falling towards a gravitational well) as opposed to "constant in the accelerating particle's frame of reference" (which would be the case for a rocket burning fuel).
Here's the function for the case I DON'T need:
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html .
ANSWER: redefining momentum .
Acceleration is essentially a useless quantity in relativity. In fact, it is
not possible to have uniform acceleration in the sense that you define it
(that is why it is defined differently in Baez's discussion). Think about
it: Suppose that the "uniform acceleration" were 1 m/s2 and your
speed right now was 0.5 m/s below the speed of light; if acceleration stayed
constant, your speed in one second from now would be 0.5 m/s above the speed
of light, a forbidden situation.
FOLLOWUP QUESTION:
If you could drop a rock down an infinitely deep well with a constant gravitational "pull," what formula would describe its velocity ("v") in terms of time falling ("t")? I know it starts as v = at (where "a" is the gravitational pull) and approaches v = c (where c is the speed of light), but what does it do in between?
ANSWER: m is
constant. I know that an object with zero net force has dp /dt =0
where p =mv /√[1-(v /c )2 ]
(again, see the link to
relativistic momentum ). Now, I am going to define a constant
force F to be one for which the rate of change of momentum is
constant, that is dp /dt =F where F is the constant. Now, we
know that v≈ Ft /m
for small time and v ≈c
for large time. It is very easy to integrate p /dt =F to get momentum as a function of time,
p=Ft . Putting in what p is in terms of v and solving for
v, I find v /c =(Ft /(mc ))/√[1+(Ft /(mc ))2 ].
This function has the correct properties at small and large t and is
shown in the graph at the right. This is the correct v (t ) for
a constant rate of change of momentum F. (For purists, I am talking
about 3-momentum.)
NOTE ADDED
LATER: dx=vdt.
Doing this I find x =(mc 2 /F )(√[1+(Ft /(mc ))2 ]-1).
Note that this has the expected properties that for small time, x ≈�(F /m )t 2 ,
and for large time, x≈ct . (I assumed x =0 at t= 0.)
QUESTION:
Much is made of the fact that the speed of light is independent of its source speed (a speeding rocket, etc.). However, isn't the same true for other waves such as sound? No matter how fast a train is moving, its sound waves still travel at sound speed. I realize there is a
Doppler shift, but this does not change the wave speed. So why is the big deal made about light speed being constant?
ANSWER: any relative velocity between the source and the
observer. No matter how those two move, the wave speed is the same. Sound is
not the same since it moves relative to a medium (the air, normally). If the
source moves toward you, as you note, you will still measure the same speed
of the waves. However, if you move toward the source, you will measure the
waves moving with a velocity of your speed plus the speed of the sound in
still air. The key is that there is no medium with respect to which light
waves move; they may move through completely empty space, unlike sound which
requires a medium.
QUESTION:
According to Google the mass of an electron is 9.10938188 � 10 (power of) 31 kilograms. Does that mean that a charged capacitor has more mass than the same capacitor without a charge?
ANSWER: E=mc 2 , the whole mass of
the capacitor must have increased. However, since the energy stored is very
small, this increase of mass would be impossible to observe.
QUESTION: his is something which came
to my mind. A book is on a table at rest, so the weight
of the book or downward force must be equal to reaction force or upward
force. The weight of the book=force given by the table on the book. Now if
we put another book on this book the equation goes: weight of first
book+weight of second book=force given by the table on the books. since
from above equations: force given by the table on the book=force given
by the table on the book or weight of the first book=weight of first
book+weight of second book then weight of the second book is zero(0) how
come its possible. plz plz make me clear.
ANSWER: earlier answer . The weight of the
first book is the force which the earth exerts on the book; the reaction
force is the force which the book exerts on the earth. The reason the force
which the table exerts on the book is equal and opposite to the book's
weight is that the book, being in equilibrium, must have the sum of all
forces on it equal to zero; this is Newton's first law (N1). I will try to
go through each of your two questions individually:
The book has two forces on it, its own weight and a force from the
table. Those are the only two forces. Because of N1 , the force
from the table must be up and equal in magnitude to the weight.
I will call the top book T and it has a weight W T .
I will call the bottom book B and it has a weight W B .
The table exerts a force on the bottom book and I will call that F .
Book B exerts a force up on book T and I will call that F TB .
Book T exerts a force down on book B and I will call that F BT .
Focus your attention on book T. There are only two forces on it, its own
weight and the force from book B. It looks exactly like the one book
problem (it is the one book problem!) and so the force F TB
must be equal in magnitude to W T and pointing
upward because of N1 . Now, focus your attention on book B. There
are three forces on book B, its weight W B , the force
from the table F , and the force from book T F BT .
These three must add up to zero because of N1 . But we do not know
F BT . But wait, aha! We know that F BT
is the force which T exerts on B and F TB is the force
which B exerts on T, so N3 tells us that these must be equal and
opposite. Therefore the magnitude of F BT must be equal
in magnitude to W T and point down. Putting it all
together and using N1 , the magnitude of F must be W T +W B
and F , of course points up.
A final word of warning: never say something like "the weight of the top
book is a force down on the bottom book" because the weight of anything is a
force on that thing, not something else. Just because it works out here that
the magnitude of the force the top book exerts on the bottom book is equal
in magnitude to the weight of the top book does not mean the "forbidden
statement" above is true. If you had a book on the floor of an upward
accelerating elevator, the magnitude of the force of the book on the floor
would not equal the magnitude of the weight of the book.
QUESTION:
Can a massless particle have or carry energy?
ANSWER: E =√[m 2 c 4 +p 2 c 2 ]
where p is the linear momentum. If m =0 then E=pc.
Massless particles have momentum. The only massless particle we know is the
photon which has an energy E=hf where h is Planck's constant
and f is the frequency. So the momentum of a photon is hf /c .
QUESTION:
Einstein's famous E=Mc2 doesn't seem to hold for a photon which is massless but has energy. It doesn't even hold for the creation of energy since the photon is not created by annilating mass, but rather by an electron shifting orbit. Further, the photon doesn't convert into mass when absorbed. What am I missing?
ANSWER: E=mc 2 is the
energy of a particle of mass m at rest; a photon is never at rest and
therefore this equation is not applicable to it.
The energy of any particle is E =√[m 2 c 4 +p 2 c 2 ]
where p is the linear momentum. Note that if p =0, the particle
is at rest and indeed
E=mc 2 . If m =0 then E=pc.
Massless particles have momentum. The only massless particle we know is the
photon which has an energy E=hf where h is Planck's constant
and f is the frequency. So the momentum of a photon is hf /c.
Regarding the rest of your question, when an electron drops to a lower orbit
the mass of the atom decreases by exactly the right amount to supply the
energy to the photon. When a photon is absorbed by an atom, conversely, the
atom becomes excited and therefore more massive.
FOLLOWUP QUESTION:
In a previous answer you stated the mass of an electron decreases as it shifts to a lower orbit and releases a quantum (hf) of energy. How does this mass change occur? Does the electron divide itself and become smaller in order to shed the mass required for energy release? What's the mechanism at work here?
ANSWER: m
when alone. Their bound mass is less than 2m . You can say this because it
takes work to pull them apart. Where does this work go? Into mass. In
atoms, it is almost impossible to measure the mass difference because it is so
small. But in nuclei,
where the binding is much stronger, it is measurable. The mass of a 4 He
nucleus is measurably smaller than the masses of two protons and two neurtons.
QUESTION:
A vertical narrow tube has a photon emitter at one end, so that photons travel through the tube and are emitted along the verical axis of the tube (all in a frame moving left to right near the speed of light). Do the photons emitted from the end of the tube leave at an angle to the vertical, as seen by a non relativistic frame? If so, do the photons hit or move toward the wall of the tube on their way inside the tube?
ANSWER: v along the direction of the
motion of the tube but their total velocity is still c . The angle
they make with the vertical is therefore
θ= sin-1 (v /c ). What you are describing is
very similar to the light
clock.
FOLLOWUP QUESTION:
you said that both the moving observer and the stationary observer must agree on whether the photon hits the wall of the tube or not. You said the moving observer see the photon traveling straight up the tube.
but you also said the stationary observer sees the photon traveling at an angle to the tube wall. you didn't say whether that angled path resulted in the photon hitting the wall and how this discrepancy is resolved.
ANSWER:
�the
tube has the same horizontal velocity as the photon does, v . So,
every time t that the photon moves a distance vt , the tube
also moves a distance vt and so the photon stays in the middle of the
tube
QUESTION:
During a discussion about basic physics I was having with my daughter, she asked an interesting question. She saw a story about helicopters that mentioned the helicopter blade tips could exceed the speed of sound creating the characteristic �whop, whop� sound, or small sonic booms. She wondered if it would be possible for the blade tips to ever exceed the speed of light. I was intrigued by this so I worked for a few hours to come up with an equation to answer her question. I am not a math expert, but using simple algebra I came up with the formula: r=(60V) / (2 π R) where r = radius of the blades in meters, V = velocity in meters per second, and R = blade RPM. From this I calculated that the blade tips would exceed the speed of light if the blades had a radius of 286.28 kilometers and were spun at 10,000 RPM. This of course assumes the materials used could withstand the forces involved. My question: is this formula accurate and given the radius and RPM mentioned, what is the formula to calculate the forces along the blades starting from the center moving to the tips (centrifugal, centripetal?)?
ANSWER: r from the end
would be v=r ω =9.55Rr
where R is rpm as per your notation and ω is the angular
velocity in radians/s; it looks like your equation got 60/(2π )
where you should have had
( 2π ) /60.
So, I get the length would be just 3.14 km if it could work. Suppose the
blade had a linear mass density λ kg/m and a length L . So, the
mass of the whole blade would be Lλ. If you calculate the force
classically , ignoring relativity, you would get F (x )=(L-x )λxω 2
where x is the distance out to the point where you are calculating
the force necessary to keep the mass beyond from flying off. But,
relativistically, λ depends on x , λ =λ 0 /√[1-(ω 2 x 2 /c 2 )]
where λ 0 is what the mass density is if the blade is at
rest. So, you see, if ωx=v=c , the mass of the whole blade becomes
infinite meaning the force needed anywhere is infinite.
QUESTION:
Have there been any theories presented by rational and peer-accepted scientists, trying to pair the combining force of Gravity with the separating force of Dark Energy into a single equation or system of equations?
ANSWER:
QUESTION:
The formula I learned in school for kinetic energy is E = mv^2 / 2. But the famous atomic energy formula is E = mc^2... why doesn't the 1/2 apply?
ANSWER: m=m 0 /√(1-v 2 /c 2 )
for a particle with velocity v ; here m 0 is the
rest mass, the mass something has when it is at rest. I have shown it
before and will not do it again, but it can be shown that if the speed
is much smaller than c, the energy is approximately given by E ≈m 0 c 2 +�mv 2 ,
the energy something has because of its mass plus the energy it has
because of its motion (kinetic). So relativity reduces to Newtonian
physics at everyday speeds.
QUESTION:
This may be a stupid question but i was just curious about this. I know with modern technologies and advanced medicine people live longer. My question is even in the slightest amount do you think that the way people get around today; seeing as our modes of transportation are a lot faster than they were say 60 years ago, are people living longer because of it? Or does the twin paradox just relate to actual travel in space and not on earth and if so what factors keep that from being true on earth?
ANSWER:
γ =√[1-(v /c )2 ] where v is the
speed of the clock and c is the speed of light. Consider the
speed of a near-earth satellite, about 18,000 mph≈8000 m/s. Then γ =√[1-7x10-10 ]≈1-3.5x10-10
which means that a clock on the space station runs about 0.000000035%
slower than your clock.
QUESTION:
Doesn't the fact that a black hole can bend light prove that something can travel faster than the speed of light? As light is pulled toward the black hole it would accelerate, since it is already traveling at the speed of light the moment it started moving toward the black hole it would be going faster than the speed of light would it not? Just curious.
ANSWER:
�mv 2 , but this obviously cannot be true for light since
it has no mass. The energy of a photon is hf where h is
Planck's constant and f is the corresponding frequency of the
electromagnetic wave. So, what happens when a photon gains energy is that
the frequency increases; this is known as a gravitational blue shift (the
color of the light moves to shorter wavelengths) and happens when a photon
approaches any massive body, not just a black hole.
QUESTION:
My daughter has a stumper for me. It comes from a first-year survey university course on time. What if B and A were to take off in opposite directions to planets both 8 light years away, and travelling at 80% of the speed of light. They can see each other�s ship. What would A�s clock read when she reached her destination? At the same time (for A), what would B�s clock show? Now assume that they both turn around immediately to return to Earth, and at the same speed. At what location and date would A see B arrive at her destination?
ANSWER:
earlier answser which you should read so that you understand what the
picture below is showing. At the midpoint each observer's clock reads six
years to himself and he sees the other clock reading less than one year (I
will let your daughter work out exactly what since this is her assignment).
When they return to earth, each sees both their own and the other's clock to
read twelve years. It is important to note that how the other's clock is
seen to be and what it is are different and this is not due to
relativity, it is because of the time it takes for the light you see to
reach you. The similarity of the question to my answer makes me think the
instuctor must have used AskThePhysicist.com as the source of the question.
QUESTION: i know that nothing can travel at or faster than
the speed of light. but, just simply why? what equations or whatever
says no...
ANSWER: Because the mass of an
object, that is its inertia, increases as the velocity increases.
Therefore it gets harder and harder to accelerate it as it goes faster
and faster. The expression for the mass of an object m as a
function of its velocity v is m =m 0 /√(1-(v 2 /c 2 ))
where c is the speed of light and m 0 is the
mass when it is at rest. Note that as v approaches c , m
approaches ∞ so it is impossible to push beyond c .
Another way to look at it is from the perspective of energy. The energy
of a particle is E=mc 2 =m 0 c 2 /√(1-(v 2 /c 2 )),
so the energy required to accelerate the mass to the speed of light is
infinite and there is not an inifinite amount of energy in the universe.
QUESTION: we were discussing time travel in class and learned that it is theoretically possible. We were thinking about this and wondered if it is possible to actually perform an experiment to prove this. Have there been any experiments that demonstrate this phenomena. Will traveling at great speeds actually cause a temporal shift? Also, does going back in time and "changing the past" affect present events or is it impossible to affect the present. We are aware of the "paradox of time travel" (going back and killing Mozart...would his music cease to exist?....or....going back and preventing your parents from uniting, would you cease to exist?)
ANSWER: twin paradox . Moving clocks
running slow (called time dilation) has been
verified experimentally in many ways. Since, as far as we know from
physics, backward time travel is not possible, we do not need to worry
about the kinds of paradoxes you ask about. QUESTION:
I was wondering if you could explain how length contraction works. I've already done some background research and I understand the mathematical reasons my text book gives me, I was just wondering if you could give some kind of analogy that would enable me to picture the effects of length contraction, and better yet allow me to explain it to my friends in a way they can understand.
ANSWER: earlier answer about
the light clock and be sure you understand that and have a good
intuitive feeling for why moving clocks run slow. From that point,
length contraction can be understood as a natural consequence of time
dilation. Here is how it goes. Imagine a bomb which has a fuse of 1 s.
That is a clock. If that bomb is moving by us with a speed of 99% the
speed of light, the elapsed time before detonation will be 1/√[1-(0.99)2 ]=7.09
s. So, the distance it will travel is 0.99x3x108 x7.09=2.1x109
m. But, the bomb will measure in his own frame that he should last 1 s
and go 0.99x3x108 x1=2.97x108 m, only about 1/7 the
distance we measure. If you think about the distance we measure as a
long stick of length 2.1x109 m, then the bomb sees this stick
moving by him with speed
99% the speed of
light, so to reconcile his results with ours he must measure that length
to be
2.1x109 x√[1-(0.99)2 ]=2.97x108
m.
QUESTION:
I'm reading "Relativity" by Albert Einstein and in it he says that simulteneity is relative and uses the example of two lightning strikes at points "A" and "B" and an observer at the mid point between the two "M". Einstein states that an observer on a train moving from "A" to "B" observing the lightning strikes just as the observer reaches "M" will see that strike "B" occurs before "A" (as the train it moving toward the light from "B" and away from the light coming from "A").
Does this mean that an event only becomes real when the light from it reaches the observer? Or could one legitimately says "These two events happened simultaneously, but I saw the flash from "B" before I saw the flash from "A"?
ANSWER: The relativity of simultaneaty is not "how
it looks", it is how it is. In relativity you have to be careful how you
define a time interval so that you correct for such things as the time
it takes light to reach you. The converse is also true, a distant star
is seen right now but as it was long ago. I always find the following
example more convincing than the lightning at the ends of the train
example. In the center of one of the cars there is a flashbulb which
flashes. Light reaches the front and rear walls of the train at the same
times. However, an observer on the side of the track watching the car go
by sees the rear wall come forward to meet the oncoming light and the
front wall running away from the light coming at it. Hence the event
corresponding to the flash at the back wall occurs earlier for the
observer at rest. (It is crucial to realize that both observers see all
light moving with the same speed.)
QUESTION:
We have two clocks starting out in an the same inertial framework, and
then accelerate one of them (Clock A) to some constant speed V, but the
two clocks are also tethered by a rope. At some point, the accelerated
Clock A will tug on the rope and begin accelerating the clock that is
still at rest in the original framework (Clock B). Clock A will begin to
decelerate and clock B will begin to accelerate until they are both
traveling at the same speed V2. The question is, will they both show the
same time (from the point of view of someone still at the initial
framework)?
ANSWER: I have been puzzzling over how to best explain this one! I believe that
the problem, as stated, is equivalent to clock A moving away at constant
speed for a while and then stopping. In both cases, clock A starts in the
same inertial frame as B, then jumps out of that frame and stays there for a
while, and then jumps back into B's frame. I recommend first reading my
earlier explanation of the twin
paradox to see what my paradigm is for the explanation illustrated by
the graph at the right. The graph represents clock A traveling at 80% the
speed of light to a point 8 light years from the earth; then clock A stops
moving. The graph shows clock A's ticks (once a year) with black crosses,
and clock B's ticks (once a year) with red crosses. Clock B measures 10
years for the trip [(8 light years)/(0.8 light years/year)]. But clock A
measures only 6 years because of length contraction (the distance is only
4.8 light years for clock A). Thereafter, the clocks run at the same rate
but clock A is 4 years behind clock B.
QUESTION:
Say there is a father and son. The father is the first astronaut to attempt to "time travel" into the future by flying his rocket around a black hole at extremely high velocities, so that for a given period of time (say an hour) it is a lot longer on earth (say 10 hours). My true question is, if the father could somehow communicate to his son in real time, what would this conversation sound like? Would the son hear his dad talking extremely fast? or would the distance between them make up the distance in "time travel". If it was the distance that is the factor, what if the dad was on a super train traveling at similar speeds on Earth?
ANSWER: Below I have copied a figure from my
earlier answer where I talk about
the twin paradox, the situation you allude to in your question. Here there
are specific numbers (given in the caption) to make it concrete. As
explained in the earlier answer, the traveling twin (the father in your
question) takes 6 years to go each way while the stationary twin (son) has
ten years elapse. Each (father and son in your case) sends out one light
pulse each year and by looking at the spacing of those pulses you can deduce
how a conversation would sound. Here is a summary of how each sounds to the
other:
On the trip out, the father hears the son slowed down by a factor of
3 (2 yearly signals from home in 6 years).
On the trip home, the father hears the son speeded up by a factor of
3 (18 yearly signals from home in 6 years).
For the first 18 years, the son hears the father slowed down by a
factor of 3 (6 yearly signals from dad in 18 years).
For the last 2 years, the son hears the father speeded up by a
factor of 3 (6 yearly signals from dad in 2 years).
Of course, it cannot really be a conversation because of the long transit
times of the signals; rather each is just speaking, reciting poetry or
something. Higher speeds would lead to more extreme numbers but similar conclusions.
Overall, note that the father has aged 12 years while the son has aged 20
years. I always like to emphasize that how time appears to elapse on
a moving clock is not the same as the time which actually does
elapse; for example, during the last two years for the son, the father's
clock looks like it is running faster than the son's whereas it
actually is running slower.
QUESTION:
Due to the length contraction, you notice that a passing train appears to be shorter than when it is stationary. What do the people in the train observe about you?
If you are on a train that is going really fast, do the people on the ground look shorter, longer, or the same?
ANSWER: Length contraction causes the lengths parallel to the direction of
motion to be shortened. So, a fat man standing at the station would
become a skinny man (side to side, but not front to back) but no shorter
as measured by someone on the train. Similarly, as measured by someone
on the platform, a fat lady standing on the train would become a skinny
worman but no shorter. You will note that I did not say that these folks
"look skinnier" because physicists normally do not care how something
looks , they care about how something is . This is a very
important distinction and one which even authors of physics books often
fail to make. How something looks may be very different from how
something is. Hence, your question is incorrectly stated ("� appears
to be shorter� ")
although I believe I know what you meant.
the same time ; the
difference of those positions is the length. When you look at a stick,
you are not observing the stick ends where they were at one time but you
are seeing the farther end as it was sometime earlier than when you see
the closer end. Of course, this does not matter if the stick is at rest,
but if it is moving it does matter. For everyday moving sticks, there is
no perceptable change in the
looks or appears , we better be sure we
understand how long the moving stick is . The result from special
relativity, using the definition of length I gave above, is that the
moving stick is shorter by a factor of
√(1-(v 2 /c 2 )), so if v =0.8 c
(i.e . 80% the speed of light), the length of the moving stick is
only 60% its length when at rest. (This effect is called length
contraction.) So now, the first picture shows the situation if the stick
is coming toward you (you are on the right). Light (red arrow) leaves
the far end of the stick and does not catch up with the near end of the
stick until the stick has gone a long way (four stick lengths) and now
light from the far (red arrow) and near (green arrow) ends move forward
to your eye. So the stick looks to be 5 times longer than it actually is
and 3 times longer than if it were at rest! Now, if the stick is moving
away from you, the situation is very different. The moving stick is
still 60% its at-rest length, but now the near end moves away to "meet"
the light from the far end; the result is that the stick, as shown in
the second figure, looks much shorter than it is. It now appears to be only 1/3 the
rest length or 5/9 the actual (moving) length. Note that in neither case
does the stick appear to be its actual length. (The scales of the two
figures are different; note the different rest lengths. I had to do this
so the "much-shorter" and "much-longer" figures would be about the same
size.) So, maybe you can now understand why I often make a big deal
about relativity being about how things are, not how things appear .
The same kind of arguments may be made about
time dilation : moving clocks run
slower, they may or may not appear to run slower.
QUESTION:
I understand that nothing with mass can travel the speed of light because an infinite amount of energy would be required to accelerate the mass and there is not an infinite amount of energy in the universe. As a statement, it makes perfect sense. But I don't understand the math:
". . .
The expression for the mass of an object m as a
function of its velocity v is m =m 0 /√(1-(v 2 /c 2 ))
where c is the speed of light and m 0 is the
mass when it is at rest. Note that as v approaches c , m
approaches ∞ so it is impossible to push beyond c .
Another way to look at it is from the perspective of energy. The energy
of a particle is E=mc 2 =m 0 c 2 /√(1-(v 2 /c 2 )),
so the energy required to accelerate the mass to the speed of light is
infinite and there is not an inifinite amount of energy in the universe. "
Can you explain without math why an infinite amount of energy would be needed?
ANSWER: No, you really cannot do it totally without math, but maybe I can make
it more explicit. Let's calculate the quantity 1/√(1-b 2 )
for various values of b. If b is the ratio of speed to light speed, then
this factor is what determines how big the energy of the particle is.
For example, if the speed is 50% the speed of light, b =0.5, b 2 =0.25,
and 1/√(1-b 2 )= 1/√(1-.25)=1.15.
Go faster, say 80% the speed of light, then b =0.8, b 2 =0.64,
and 1/√(1-b 2 )= 1/√(1-.64)=1.67.
Go faster, say 99% the speed of light, then b =0.99, b 2 =0.98,
and 1/√(1-b 2 )= 1/√(1-.98)=7.07.
Go faster, say 99.999% the speed of light, then b =0.99999, b 2 =0.99998,
and 1/√(1-b 2 )= 1/√(1-.99998)=223.6.
Go faster, say 99.99999% the speed of light, then b =0.9999999, b 2 =0.9999998,
and 1/√(1-b 2 )= 1/√(1-.99998)=2236.
Can you see where this is going? When (never) we go 100% the speed of
light, then b =1, b 2 =1, and
1/√(1-b 2 )= 1/0=infinity.
QUESTION:
If a plasma rocket were able to maintain an acceleration that mimicked 1
G to humans on board, how soon would the rocket reach the closest star?
ANSWER: x =(mc 2 /F )(√[1+(Ft /(mc ))2 ]-1).
In your case, F=mg . Solving for t , I find t= √[(x /c )2 +(2x /g )].
I chose to calculate t for 4 light years, approximately the
distance to the closest star, and found t ≈5 years. This is the
time elapsed on earth for the trip. For a passenger on the rocket the
time will be much shorter because the velocity for much of the trip is
apparently near the speed of light (it takes only about 5 years to go 4
light years).
QUESTION:
I have seen at MIT site that "A ping-pong ball moving near the speed of light still looks spherical." I thought that the ball would contract in the direction of movement. Why would it still look spherical?
ANSWER: are , not how they look . But
it is, nonetheless, interesting to ask how things look. I tried to
illustrate that things look different than they are in a simple example
of a stick moving at high speed directly at or away from you in an
earlier answer .
The key is that when you look at the object, you simultaneously see
light from different parts of the object which were emitted at different
times; hence, if it is moving at a speed comparable to the speed of
light you are not seeing the object as it is at some time but how
different parts of it were at different times. I found a
wonderful movie on the web which illustrates this for a moving
object which would be a sphere when at rest (like your ping pong ball).
QUESTION:
I have been told that as an object accelerates to the speed of light that it becomes more 'massive'. I have also checked some other answers on the site and they use the term 'increases in mass'
My question is how can something increase in mass? it can't right, there are only some many atoms in the object? Perhaps they are confussing weight with mass?
Also Force = M*A so M=F/A therefore if anything as A increase M decreases?
ANSWER: F=ma argument, you are
confusing acceleration with velocity�you
can have a large acceleration without having a large velocity. Anyway,
Newton's second law, in the form
F=ma , turns out to be wrong
at high speeds. If you want to read an answer which gives the big
picture leading to why inertia increases with speed, my best answer is
probably
this one .
QUESTION:
Two trains 200 km apart are moving at a relative speed of 50 km/h. A fly takes off from one train, flying straight to the other one at the speed of 75 km/h, touches and flies back to the first train. How far will the fly have gone when the trains collide?
ANSWER:
QUESTION:
I understand that matter gains in mass as you accelerate it towards the speed of light, and that as you approach the sped of light, mass tends toward infinity, meaning you would need an infinite force to "push" it beyond the speed of light, and that's one of the reasons why nothing can exceed the speed of light. However, I also understand that with increased mass comes increased gravity, so... When they accelerate those sub-atomic particles to velocities approaching the speed of light at CERN, why don't they suddenly start exerting lots of gravity and sucking the entire CERN campus toward them?
ANSWER: m =1.7x10-27 /√(1-0.99999999992 )=1.2x10-22
kg. The proton has increased mass by a factor of about 70,000 and that is
still far from a baseball mass.
QUESTION:
How fast do you have to travel before you experience significant relativistic effects. For the purposes of this question, let's say 1 hour of travel at "X" velocity equals 1 hour plus 10 seconds.
ANSWER:
√[1-(v /c )2 ] where v is the speed and
c is the speed of light. I am glad you gave me a concrete example (10
seconds for an hour) since "significant" depends on how accurate you need to
be. For example, although satellites used for GPS navigation are nowhere
near the speed of light, not correcting for relativity leads to
significant errors in the computation of where you are. Your situation is
about 0.3% time difference, so √[1-(v /c )2 ]≈0.997≈1-�(v /c )2 +�
where I have used a binomial expansion of the square root because I assume
that v /c is small compared to 1. Solving, v /c ≈0.055,
so the speed would about 1/20 the speed of light.
QUESTION:
ANSWER: E=mc 2
is that it is often used when not appropriate. If you write E=mc 2 ,
then this is the energy of a mass which is at rest; or else it means that
the mass has a different meaning from what you usually think of when it is
moving with speed v , namely m =m 0 /√(1-(v 2 /c 2 ))
where m 0 is the mass of the particle at rest, what you
usually think of as inertial mass. As I have said in many
earlier answer s, I
prefer to not think of mass as increasing with velocity, m to me just means
rest mass. The correct equation for energy is E= √ (p 2 c 2 +m 2 c 4 )
where p is the linear momentum. So, if a particle is at rest, momentum is
zero and E=mc 2 ; if the mass is zero (as is the case for a
photon), E=pc. So a photon has momentum even though it has no mass.
One thing to be careful of, as explained in my
earlier answer s, is
that momentum is no longer mv but rather mv /√(1-(v 2 /c 2 )) .
QUESTION:
ANSWER:
QUESTION:
I have a question regarding special relativity. I
was pondering over a thought experiment that I came up with using a
modified version of a light clock. This modified light clock, with every
tick, sends a signal to a space ship which is moving towards it at a
uniform speed. This is where I get a little confused. It would seem that
since the spaceship is moving towards the light clock that the frequency
of the signals being received would increase. Intuition would tell me
that the rate of the ticks has therefore increased. And yet I find it
hard to reconcile with the fact that, since the light clock is in motion
relative to the spaceship, that the ticks ought to be slowing down. I
realize that my argument is flawed somewhere as special relativity has
already been proven through countless experiments. I ,however, need help
understanding where exactly that flaw is. Do you have any thoughts on
this?
ANSWER: are
is not necessarily how things appear to be . To answer your question,
read my explanation of the twin
paradox where I emphasize that the rates of clocks as you see them is
not the same as the rate at which clocks are ticking. See also
another answer . Time
is defined as the time interval between two events at the same point in
space and the clock coming toward you is not at the same places when it
sends you sequential ticks, so the ticking you observe is not the time
interval between ticks on that moving ship. If the space ship is going away
from you the observed frequency of ticks would be less than the real
tick rate. The same kind of phenomena are observed for length contraction.
Length is defined as the distance between two points in space measured at
the same time. But if a stick is moving toward you, it appears longer
than its rest length even though it is actually shorter and the stick moving
away from you looks shorter than its actual contracted length. Again,
see an earlier answer and links
there.
QUESTION:
I am not getting the idea of length contraction.Does length of an object actually shrink when in motion or is it just apparent to the observer which is at rest.Is moving rods shrinking a measurement problem?
ANSWER: looks is not the
same thing as how long the stick is . For much more detail, see an
earlier answer .
QUESTION:
When we heat something, the particles in it move faster and faster till they reach the speed of light. But what would happen if we keep heating? The particles can move faster than the light, or there's a maximum value for the temperature (like a minimum: 0 Kelvin), or what?
ANSWER: answer ed this
question before and the answer is that there is no upper limit. I have
actually been corrected since there is an upper limit imposed by the total
amount of energy in the universe, but that begs the question. If you have
something, there is no limit to the amount of energy you can put into it,
assuming you can get that energy somewhere. There is a minimum temperature
possible, as you note, but that cannot actually ever be reached because of
the laws of quantum mechanics. So the speed of particles in an object are
constrained by 0<v<c corresponding to temperatures
of 0<T <
∞ and the extremes are
unreachable.
QUESTION:
Why are only mass, length, and time changed as an object approaches the speed of light? Why not other "fundamental" properties of matter e.g., charge? Does this imply that charge is not fundamental or is somehow very different from mass, length, and time?
ANSWER:
QUESTION: What is the medium through
which electromagnetic waves propagate? It's easy to see physical waves
go through water. And I can understand sound waves traveling through
air. But what is the stuff that carries electromagnetic waves?
ANSWER:
QUESTION: My 11 year old has asked me,
"Does gravity bend light?" I did highschool physics about 20 years ago,
so am very rusty and not up to date with current thought. I have looked
at this discussion topic on the archives of many forums, but have ended
up very confused by the differing opinions/explanations. I would really
like an answer which is easy to explain to a child, but yet not so
simplistic that it is inaccurate. Am I asking the impossible? She has
read about the nature of light and also about gravity, and can't
understand how light can be affected by gravity, when it has no mass.
Is it because photons are energy and so can be used instead of mass? Or
is it that the gravitational pull around massive stars affects the
"space" around it and light just follows the stretched paths? If this
is true, how can some authors say that the light is still moving in a
straight line even when it is following a curved path?
ANSWER: Here
is one explanation, probably the easiest for your daughter to
understand: Light being affected by gravity is a result of the
principle of equivalence in general relativity. This states that there
is no experiment which you can perform to distinguish between your
being in a gravitational field or in an accelerated frame of reference.
Thus, for example, imagine that you are in an elevator which
accelerates upward; if light enters through a hole in the side of the
elevator it will clearly appear to fall like a projectile because of
the acceleration of the elevator. So, the same thing will appear to
happen in a gravitational field the acceleration due to which is
exactly the same as the acceleration of the elevator. Hence, light
will "fall" in the earth's gravitational field with an acceleration of
9.8 m/s2 . You might be interested in the answer to an earlier question .
Here is
another: If we look at the world as having a Euclidean "flat" geometry
and watch a ray of light pass a very massive object, we see the light
bend. But, the way that general relativity describes the world
says that, if we are in the vicinity of a massive object the space
itself is not Euclidean but is curved; in this space the light follows
a "straight line" in that non-Euclidean geometry.
QUESTION: Taking the light as a
guiding agent (its invariant speed) in several thought experiments,the
lorentz transformation of coordinates can possibly be completely
derived,and time dilation,
lenght contraction,non synchronisation of simultaneous events,and such
things can be understood therefrom.But mass increase seems to be
detached from this continuity of explanations.
The explanation of "Mass" gets carried away by referring to Sir J.J
thompsons electron experiments.
Hence the question arises-
Is it possible to deduce the effect of mass increase
("Mass increases by gamma factor as velocity increases"{where
gamma=((1-(v/c)2 )0.5 )})
by purely referring to lorentz transformations?without referring to
explanations and definitions from electromagnetism.
The relativistic addition of velocities seems to provide some clue.
Also an explanation as to how the time part of four momentum can be
treated as energy is needed.
ANSWER: All the quantities which you can derive from
the Lorentz transformation are what we call kinematic quantities. Mass,
force, and more importantly linear momentum and energy are what we call
dynamic properties. So, just like in an introductory physics course
where, after we learn how to describe motion (kinematics), we next want
to understand how motion can be changed; in classical physics this
leads us to Newton's laws. What happens in relativistic physics is that
we quickly find that Newton's second law, in the form F =ma m they will get different
answers for a F =dp /dt p =mu u dp /dt= 0,
that is the momentum is conserved. If mu is the
definition of momentum, where m is what classical physicists
call the inertial mass, we find that momentum conservation is no longer
a true thing for isolated systems. So, what we do is to redefine what
we mean by momentum such that momentum is conserved and the new
definition becomes the old definition for small speeds. If we define
momentum as p = g mu we find that momentum is
conserved in an isolated system and p @ mu
for small u m is the inertial mass of an
object at rest and that p = g mu . Your
question about energy being the fourth component of four momentum is
too involved for this site.
QUESTION: if two people are moving
past each other at a constant speed in an infinite empty space, to
either one the universe would be static and the other person would be
in motion (relativisticly).
if two people were at a constant unmoving distance from each other, but
one of them was rotating at a constant speed, to either one the
universe would be static and the other person would be in motion,
correct?
so if they were far enough apart, wouldn't the rotating person see
their friend travelling faster than the speed of light (tengentially)?
ANSWER: The two situations are not equivalent.
Assuming that in your first scenario each was an inertial frame (one in
which Newton's first law is true), then both are inertial frames and
each can rightly claim that he is at rest. In the second scenario, if
one is an inertial frame then the other (revolving around it) is not,
so the rotating frame cannot claim to be at rest.
QUESTION: I was recently watching a
programme about the Hubble Space Telescope and the pictures it had
captured. It mentioned that the pictures it captured were of galaxies
forming millions of years ago-If this is so- Could it then be possible
for man, should he a find a way to manipulate space travel, be able to
position himself at a certain area of space, at a certain distance and
witness the birth of the milkyway itself? or indeed the birth of
Earth?..........perhaps even the birth of man?
ANSWER: Suppose that something happened here on
earth 500 years ago and you were located at a distance of 500 light
years from the earth with a very good telescope. Then you would be able
to witness that event right now. But, here is the rub. You cannot get
there from here before the light from the event reached there because
it is physically impossible to travel faster than the speed of light;
in other words, you could get there in just under 500 years and you
would be able to witness what happened here tomorrow.
QUESTION: I’m a PhD student in
philosophy interested (but not working in) the philosophical
foundations of relativity. I have a couple of questions dealing with
the speed of light and special relativity. The main question is number
3, the other two are auxiliary. Your response would be greatly
appreciated. I will also greatly appreciate it if you could let me know
whether there is some bibliography available dealing with this kind of
issues.
Why is the relativistic
length contraction – as it appears in the Lorentz transformation – a
function of the speed of light? Conceptually, it is possible
that the speed of light had a different value than it actually has. How
would that have affected special relativity (in particular length
contraction for fast-moving bodies)? I recommend your looking into the
Mr. Tompkins in Wonderland books by George Gamow. Suppose we will discover a
"form of light" (call it light*, symbolized c*) whose speed is double
the speed of light, and it is also invariant. This would enable us to
send signals at this speed and to use these signals to synchronize
clocks so that we may provide an empirically meaningful definition of
simultaneity. Do we need to revise special relativity and base the
Lorentz transformation for length on this new value (c*), instead of c?
Which of the following possible responses should be the case and why:
a) we can do that; b) is it impossible to do that; c) we must do that.
ANSWER:
The answer is that the fact
that light has the same speed in all frames of reference leads to this
result. It is simple algebra and, if you are seriously interested, you
should take the time to learn the basics of special relativity. Special relativity would have
the same form it does. If the speed of light were 100 miles/hr, effects
like time dilation and length contraction would be everyday phenomena
which we would not find puzzling at all. I recommend your looking into
the Mr.
Tompkins in Wonderland books by George Gamow. This seems extremely unlikely
since light is just a manifestation of electromagnetism and its speed
is predicted by Maxwell's equations. For a new kind of invariant speed
to exist, there would have to be another fundamental force in nature
the theory of which predicts radiation of that speed. Since there is no
good quantum theory of gravity, that is a remote possibility; gravity
waves, however, have never been directly observed although their
existence has been inferred from energy loss of binary systems. It is
usually assumed that gravity waves move with the speed of light, but
that is unverified. Regarding your simultaneity question, there is
never any problem defining when two events are simultaneous. The
problem is that events simultaneous in one frame of reference are not
simultaneous in other frames; having some other universal speed is not
going to change that.
QUESTION: Why is the speed of light given by 1/sqrt(permittivity
*permeabillity)? What is the great mistery behind such a simple
relation?
How these two parameters combine to give the speed of light?
Why does the vacuum (nothing...) has physical properties such as
permittivity and permeability?
ANSWER: This is the great triumph of Maxwell's work
in the 19th century. There are laws of electromagnetism
which can be summarized in four equations, now known collectively as
Maxwell's equations. The quantity e 0
(permittivity of free space) is just a proportionality constant which
tells you how strong the electric force is and, of course, it appears
in the equations. Similarly, the quantity m 0
(permeability of free space) is just a proportionality constant which
tells you how strong the magnetic force is and, of course, it appears
in the equations. (In this context, there is nothing wrong with empty
space having permittivity and permeability because one certainly does
not need matter between charges or currents for them to exert forces on
each other.) When Maxwell messed around with the equations he
discovered that they could be rewritten as wave equations and that the
speed of these waves had to be 1/[e 0 m 0 ]1/2 . That this happened
to be the speed of light was the point in the history of physics that
we understood what was doing the waving in light waves--electric and
magnetic fields.
QUESTION: i was once concidering the
effect of propelling an object backward off a moving vehicle with the
exact velocity that you are travelling. From my point of view the
object is moving away from me, but to an observer standing at the side
of the road at the instant i throw the object it merely drops to the
ground. This made me wonder if it is possible for light to behaive in
the similar manner. In this hypothetical situation an object must be
travelling the speed of light. Would the light from the object
travelling in the opposite direction of travel merely stand still in
space becoming "dead light" or would it accelaterate to it's regular
pace. In this situation we would also have to assume that the object is
generating it's own light since light itself would not be able to
reflect off it due to it's speed.
ANSWER: It seems like I answer this question (in
various forms) at least weekly! There are two problems with your
hypothesis, both violations of the theory of special relativity. First,
no material object may have a speed equal to or greather than the speed
of light; the reason, simply, is that to accelerate it to the speed of
light you would have to supply an infinite amount of energy, obviously
not available. Second, one of the basic postulates of the theory of
theory of special relativity is that the speed of light in a vacuum is
the same for all observers. Thus, even if your object were moving with
99.99% the speed of light (which is possible), an observer at the
roadside would see the light from its back end moving with the speed of
light, not 0.01% the speed of light. The theory of special relativity
is completely verified experimentally and no serious scientist doubts
its correctness.
QUESTION: I have read
that the speed of light cannot be broken. I have also read that all
motion is relative. Suppose point A is a stationary reference point.
Point's B and C are start near point A and travel away from
each other at 51% the speed of light relative to point A. Does this not
mean that relative to each other, points B and C are traveling apart
from each other at 102%? Do you not add their speeds together as if two
automobiles are each traveling away from each other at 60 mph are
traveling at 120 mph relative to each other?
ANSWER: No, your intuitive way of adding velocities does not work in
the theory of special relativity. Because of the fundamental
postulate of special relativity, that the speed of light is the same in
all frames of reference, the correct formula for velocity addition is v= (v 1 +v 2 )/(1+v 1 v 2 /c 2 )
where c is the speed of light. So, for your example, v 1 =v 2 =0.51c
so v =0.81c , 81% the speed of light. Note that the
formula reduces, to an excellent approximation, to your v= (v 1 +v 2 )
if the speeds involved are very small compared to the speed of light
because the term v 1 v 2 /c 2
is extremely small compared to 1.
QUESTION:
Is all
that right? - if so, then what about if the Car begins Rolling West,
and again the Light on the East End of the Car is Switched On - this
time it would take Longer for the Light to reach the West End, so by
the same Reasoning, it seems like the Special Theory would say that now
the Car's Length has Increased!
Is that a
Flaw in the Special Theory? - maybe the Wording should be something
like "Sometimes an Object in Motion becomes Shorter, but Sometimes it
becomes Longer"
Do I
deserve a Nobel Prize? - or am I Overlooking Something?
ANSWER I will answer your last questions first--no, you do not
deserve a Nobel prize! And it is not so much that you have
overlooked something, but that you have computed the length of the car
incorrectly. In order to measure the length of something, you
need to measure the positions of each end at the same time, not
different times. So first you must use special relativity to get
length contraction so that you know what the length of the moving car
is and then you can analyze your experiment.
You
cannot deduce the length of something by simply measuring the time it
takes light to travel between its ends (which is what you have
discovered here!). Let's do your problem but do it
correctly. Suppose we call the length of the car, in its own rest
frame L 0 . Then, if it has a speed v relative
to some observer, its length as measured by that observer is L=L 0 [1-v 2 /c2 ]1/2 .
For your first experiment (car at rest) the time is t =L 0 /c .
For your second experiment, (car moving east) let us call the time t 1 .
Looking at the first picture (dashed lines show where the car has moved
to when the light, red, strikes the other end), it is clear that L=L 0 [1-v 2 /c2 ]1/2 =vt 1 +ct 1 ,
so t 1 =t [1-v /c]1/2 (you will
have to do a little algebra to get here). t 1
is, as you state, indeed less than t but not because L
is shorter than L 0 , although it is. In your
third experiment, refer to the second picture and call the time t 2 .
Now we have that L=L 0 [1-v 2 /c2 ]1/2 =ct 2 -vt 2 ,
so t 2 =t [1+v /c]1/2 . So
now, t 2 >t but not because the car is any
different length from experiment #2 (it has exactly the same length);
rather because the light had to travel farther than the length of the
car.
QUESTION:
ANSWER: x,y ) are not the same
as in the rotated (red) coordinate system (x',y' x',y' x,y )
are:
x'=x cosq +y cos q y'=-x sin q +y sin q .
The
rotation "mixes up" the spatial dimensions. It turns out that in
special relativity, the transformation between space and time if one
system is moving with speed v with respect to the other looks
very much like a rotation in "four-dimensional space" because x
(the direction of the relative velocity) and t get mixed up:
x'= g (x-vt )t'= g (t-vx/c 2 ) g =(1-(v/c )2 )-1/2
The fact
that this transformation mixes up these two dimensions leads us to
recognize that space and time are no longer separate unrelated
concepts. Instead, we should think of "space-time" which includes
both time and the three dimensions of space, thereby suggestive of a
four-dimensional world.
Regarding
your second question, I found an interesting
example on the web which might be instructive for you.
QUESTION: This question is purely theorical. If you somehow got a
ship moving at 4/5 the speed of light in one particular direction. Then
the same ship is accelerated to the same speed in a direction
perpendicular to the original direction. So given that, wouldn't the
resultant speed be greater than the speed of light? And if so wouldn't
that go against the theory that faster than light travel is impossible?
ANSWER: To answer your question you must know the Lorentz
transformation for velocities. I shall call the direction of the
velocity of the ship relative to the ground, before the acceleration in
the perpendicular direction, the x -axis; so the velocity of the
ship in its own rest frame is, of course, u'x = 0 and
the velocity of the ship in the ground frame is ux =0.8c
(where c is the speed of light). Now the ship
is given a velocity in the primed frame (the one moving in the +x
direction with speed 0.8c ) of u'y = 0.8c .
(It may be easier to visualize the ship firing a projectile in the y'
direction with speed 0.8c and asking what the speed of the
projectile seen from the ground is; it is equivalent to your
question.) The inverse Lorentz transformation for components of
the velocity perpendicular to the direction of relative motion (i.e.
the transformation which tells you uy if you know u'y )
is
uy =u'y /[ g (1+(u'x v /c 2 ))]
where g =1/[1-(v /c )2 ]1/2
and v is the speed of the moving frame (in the x
direction), 0.8c in your question. It is now a simple
matter to compute uy =0.48c . Therefore
the speed of the ship in the ground frame is u =[ux 2 +uy 2 ]1/2 =0.933c
which is, as it must be, less than the speed of light.
Let me
speculate where you may have gone astray in your reasoning here.
You were probably assuming that uy =u'y
because the Lorentz transformation for lengths in the y direction
is y=y'. As shown above, however, this is not the case
because of the fact that the two observers use clocks which run at
different rates to measure the velocities in their own frames.
QUESTION: In 1998, researchers at Stanford University's Linear
Accelerator Center successfully converted energy into matter. This feat
was accomplished by using lasers and incredibly strong electromagnetic
fields to change ordinary light into matter. The results of this
experiment may allow for the development of variety of technological
gadgets. One such development could be matter/energy transporters or
food replicators that are commonly seen in some of our favorite science
fiction programs.
My
question is, if we were to build a transporter that worked by
converting people into energy, would they be able to theorically
survive the procedure once their energy has been reconverted back into
matter, since both are interconvertible?
ANSWER: Changing electromagnetic energy into mass has been seen for
decades. It is called pair production and happens when a very
high energy photon spontaneously creates an electron-positron pair.
Conversely, any time a particle meets its antiparticle the mass usually
totally disappears and electromagnetic energy appears. Regarding your
question, it has no short answer. But, there is enormously more
to the transporter idea than simply destroying and creating mass--you
have to put it back together the way it was. I should also note
that it has never been possible to take some macroscopic thing like a
person and totally annihilate all the mass. Anyway, if you want
to read a good discussion of the physics of why it is extremely
unlikely that a transporter will ever work, I would recommend a book
called The Physics of Star Trek by Lawrence M. Krauss.
QUESTION: 1. Can you tell me how and why did the special theory of
relativity lead to the atomic bomb and revealed the secret of stars
?
ANSWER: The answer to your first question is that one of the
consequences of the theory of special relativity is that there is a
connection, an equivalence, between mass and energy. That is,
mass may be converted to energy and vice versa. As a
result, if you cause a very heavy atomic nucleus to split (nuclear
fission) energy is released because the mass after the fission is less
than that before; this is the principle behind a nuclear reactor or a
conventional atomic bomb. But if you take two very light nuclei
and fuse them together (nuclear fusion) you also get energy released
because the mass after the fusion is less than that before; this is the
principle behind how stars generate energy and also behind the hydrogen
bomb.
The
answer to your second question is that general relativity is the theory
of how gravity works. In essence, in order to satisfy the
postulates of the theory, the structure of space and time themselves
are altered by the presence of gravitational mass; this alteration is
often described as a warping of the space around massive objects which
then explains why massive objects attract each other (they simply fall
along "straight-line" paths in this warped space). I don't think
that I would say that general relativity "gave us" the big bang or
black holes. They just happened. Of course, to properly
describe a black hole, you need to use general relativity. But
the reason a black hole forms is that when a star is all "burnt out"
the force of gravity (which is always trying to compress any large
object) has no force opposing it which is strong enough to keep the
object from collapsing.
QUESTION: what is "energy" really? heat, motion, capacity to do
work.....all of these?! i am very confused about the concept of
energy....any info where it is looked into philosophically?
ANSWER: Yes, all of these and more! When I teach the concepts
of energy to students in introductory physics, I always emphasize that,
in many respects, energy is just a clever trick to either make problem
solving easier or to allow us to solve harder problems. But it is
not a new fundamental thing; it is just casting Newton's laws (which
are the fundamental things in classical mechanics) into a new
form. In more advanced courses, energy arises naturally from
mathematical manipulation of the differential equation which is
Newton's second law. The "elemental" form of energy, the one most
easily grasped, is kinetic energy, the energy something has by virtue
of its motion. The way that kinetic energy can be changed is to
do work on the system. Kinetic energy is useful in the
kinetic theory of gasses and solids--temperature is nothing more than a
measure of the average kinetic energy per molecule. The next
thing we introduce in mechanics is the potential energy which
attributes energy to something by virtue of its position; I always tell
my introductory students that potential energy is basically a
bookkeeping device--it automatically keeps track of work done by some
force which is always present. What you gain with energy
compared to forces is that energy is a scalar which leads to easier
problem solving; the price you pay is loss of information about the
time development of the system. An important thing we learn from
studying energy in classical mechanics is the beauty and power of
conservation principles--some things, like the total energy of an
isolated system, are constant with respect to time.
Physics
at more advanced levels attributes more importance to energy than a
cute trick or a bookkeeping device. In the theory of special
relativity, the energy associated with the mass of an object emerges as
an unexpected important property of material objects (E=mc 2 ).
In electromagnetism we are often interested in the energy contained in
electromagnetic fields. In quantum mechanics, energy is one of
those variables in nature which you can not know to arbitrary precision
(that is, a Heisenberg uncertainty principle applies to energy).
Even empty space contains energy (called vacuum fluctuations) whereby
the vacuum is viewed as particle/antiparticle pairs popping into and
out of existence (since the uncertainty principle allows conservation
of energy to be violated if for a short enough time).
I do not
know much about philosophy of science (although I do a lot of
philosophizing on my own!), so I cannot point you to a philosophical
reference on energy. Surely there must be good ones, though.
My answer to an earlier question might also be of
interest to you.
QUESTION: What is the speed of gravity? If you don't have a exact
speed calculated, is it faster than the speed of light?
ANSWER: No one has ever measured the speed at which the gravitational
force propagates. In Newtonian physics we calculate orbits by
assuming that the forces are instantaneously transmitted, but nobody
believes this--it just works pretty well because realistic propagation
times are bound to be very short over the distances of interest.
A better theory of gravity is general relativity. Here the theory
predicts that gravity is due to distortions of spacetime caused by the
presence of mass and these distortions are predicted to propagate with
the speed of light. In general, all the physics we know forbids
any information propagating faster than light speed. Finally, we
believe that a theory of quantum gravity will someday be found in which
the quanta which transmit the force will be massless particles called
gravitons (similar to photons which transmit the electromagnetic
force). Massless gravitons, like all massless particles, would
have a speed equal to the speed of light so the force which they
transmit would move with that speed. There is a very enlightening
essay at http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html
QUESTION: I have heard through books that I read and other sources that
objects cannot excede the speed of light. Why is this? If you are
travelling at any speed and you turn on a flashlight, how fast is the
light moving? Is it moving at C + the speed you are moving? Or is it
just moving at C? Please send I response, I would be thankful.
ANSWER: The cornerstone postulate of the theory of special relativity
is that the speed of light is a universal constant. Therefore, no
matter who measures the speed of light coming out of the flashlight you
allude to, the measured speed will be the same, 3x108
m/s. This is hard to swallow because it conflicts so seriously
with our "Newtonian-Galilean" intuition, but it has been shown with
great accuracy to be true. With this postulate under our belts,
we can reformulate mechanics. One of the consequences of this
reformulation is that the amount of energy it takes to accelerate a
mass m to a speed v is given by
E=mc 2 / (1-v 2 /c 2 )1/2
- mc 2
where c
is the speed of light. Notice that if we want to accelerate the
object up to v=c, an infinite amount of energy needs to be
supplied. Sorry, but nobody has an infinite amount of
energy! You need to find a popularized book on special relativity
(there are many); it does not take very sophisticated mathematical
ability to understand the theory.
QUESTION: Contrary to the usual description of the twin paradox, where
one of them stays on earth and the other is travelling at relativistic
speeds, consider the situation where both of them ( say A & B ) are
travelling at 'identical' speeds but away from the same point ( some
arbitrary point in space, away from the influence of any gravitational
field so as to neglect the effects of General theory of relativity ) in
diametrically opposite directions and then returning to the same point.
Then, according to a third 'neutral' observer at rest with respect to
the 'point' from where A & B started off, both of them would have
aged equally. But in A's Frame Of Reference (FOR in short ) B has been
constantly travelling ( except at the moment where they stop and start
retreating ) and hence B has aged according to A; the same applies even
to A from B's FOR. Here, though we ( the 'neutral observers' ) know
that they are of the same age how is one going to resolve the paradox
keeping in mind the views of A & B ?
appear
to run. The earthbound brother sees his twin age only 6
years in his first 18 years and then age 6 more years in only 2
years. The traveling brother sees his twin age only 2 years in
his first 6 years and then 18 more years in his remaining six years of
travel.
So now we
come to your question. How would we analyze the situation if one
twin went on the original trip and the other twin went to a star 8
light years in the opposite direction and back? The graph below
illustrates this scenario. It is clear that each twin sends and
receives 12 pulses so each sees both himself and his twin age 12 years
over the duration of the trips. Again, note the interesting
disparities between actual aging and the appearance of aging.
Neither twin receives any pulses until the return trip.
QUESTION: I want to know (among other things i consider wrong) why you
believe light is massless? Do you believe these to be true: light bends
around stars, light can't escape from black holes, light is able to
push light-craft such as futuristic light-sailships and spherical craft
that WAS able to be propelled 200 feet into the air by a high-powered
laser, or that we can see light at all? Well MY point is that anything
that acts on anything else must have mass, otherwise there is 0
momentum to transfer onto the next item. If light was to be attracted
by gravity, wouldn't it have to have some sort of mass that the gravity
acts on since the amount of attraction between two objects by gravity
is related to the mass of BOTH objects?
ANSWER: None of the things you cite require light to have mass. The
problem is that you are thinking classically. In fact, even in
classical electromagnetic theory, light has momentum and energy but no
mass. In modern physics, the electromagnetic field is quantized, i.e.
the light behaves like photons in certain circumstances which are
massless particles with velocity c =3x108 m/s, energy
E=hf (f is frequency, h is Planck's
constant) and momentum p=E/c. Light being affected by gravity
is a result of the principle of equivalence in general relativity. This
states that there is no experiment which you can perform to distinguish
between your being in a gravitational field or in an accelerated frame
of reference. Thus, for example, imagine that you are in an elevator
which accelerates upward; if light enters through a hole in the side of
the elevator it will clearly appear to fall like a projectile because
of the acceleration of the elevator. So, the same thing will appear to
happen in a gravitational field the acceleration due to which is
exactly the same as the acceleration of the elevator. Being able
to see something has nothing to do with mass; the detection only
requires that we have a detector sensitive to the electric or magnetic
fields associated with the light.
