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How does SR cause the speed of light to be constant?

by kmarinas86
Tags: constant, light, speed
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kmarinas86
#1
Jul1-10, 03:32 PM
P: 1,011
Let's say you are receiving photons from all over. The velocity of each photon coming to you is c. If you moved at velocity v compared to your initial inertial frame, without SR you would say that photons going the same direction as you are moving at c-v, while the photons going the opposite direction do so at c+v. How does time dilation bring both values back to c? It makes no sense to me. Length contraction seems to reverse this perception. So how does the speed of light become constant?
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LostConjugate
#2
Jul1-10, 04:55 PM
P: 842
If you dilate time, then your measurement of speed (distance over time) will be dilated. Time is the denominator which means it denominates speed.
lenfromkits
#3
Jul1-10, 11:39 PM
P: 108
According to Hendrick Lorentz (and adopted by Einstein):

The light is travelling at a very high speed, and for anything travelling at such a high speed, distances are smaller.

Now, if you take the photon that is travelling at c+v, it will essentially have increased in speed by an amount of 'v'. But since it would then be travelling even faster, the distance in front of it shrinks even more. When you look at it, what 'would' have been a v+c velocity is now over a distance that is kind of squished more. Relative to you, for say one second, it is not travelling as far (because it's squished). Less distance per second means it would appear to you as though it was just plain travelling slower....and appear to be travelling at c.

For the other photon, at c-v, remember the distance across which it is travelling is already squished to a large degree, since it is travelling so fast. When it 'slows down' by an amount of 'v', at the same time, the distance in front of it kind of un-squishes (opposite of the v+c). This causes it to appear to be travelling faster - which again works out to exactly 'c'.

The real brain-bender to keep you awake at night is to consider just how big that 'distance' is when a photon is travelling at c. Just think, distances shrink for faster travelling objects. Then continue to shrink towards zero as an object continues to approach c. What is the distance between two points then for a photon when travelling at c? And therefore how much time passes as the photon travels between any two points? So given that, how much longer would it take for the photon to travel twice as far? Or 1000 times as far? And what does that say about its 'state' and speed relative to itself? Don't take this too seriously though, Einstein didn't and it will contradict what you are trying to learn. It's just not a theory based on perfect philosophical reasoning, so when you come across things that seem like holes, they probably are.

Austin0
#4
Jul2-10, 12:17 AM
P: 1,162
How does SR cause the speed of light to be constant?

Quote Quote by kmarinas86 View Post
Let's say you are receiving photons from all over. The velocity of each photon coming to you is c. If you moved at velocity v compared to your initial inertial frame, without SR you would say that photons going the same direction as you are moving at c-v, while the photons going the opposite direction do so at c+v. How does time dilation bring both values back to c? It makes no sense to me. Length contraction seems to reverse this perception. So how does the speed of light become constant?
I agree completely. Length contraction is equal in both directions relative to the direction of travel and time dilation is the same for all clocks in the system.
My conclusion is that the answer lies in a shift in simultaneity or clock desynchronization.
I.e. the clocks at the back relative to travel are running ahead of the clocks at the front.

This is of course the expected case regarding our clocks as measured by another inertial frame. This observation is of course relative but if you consider that simultaneity or desynchronization might be both actual and relative, then this would explain the equal measurements of c in our frame at whatever velocity we would accelerate to.
SO desynchronization would be a real intrinsic effect of those changes in velocity but at the same time would be relative in that they would be undetectable by any means within the system and any quantitative measures of the degree of desynchronization as measured by clocks in other inertial frames would be relative , differing from frame to frame and so without real meaning. Also if this is the case other frames measurements would be effected by some unknowable degree of intrinsic desynchronization in those frames themselves.
If this is the case then it is a catch 22. The desynchronization results in the constant measurement of c and that constant measurement of c makes it impossible to detect the desynchronization of the clocks.

As for the desynchronization itself there are three interpretations:
1) It is purely relative and all clocks are actually synched within their frames.
2) It is a result of the light synchronization convention. That after going inertial after a period of acceleration it would be neccessary to resynch the clocks.
3) That the clock desynchronization is an inevitable result of an actual temporal shift in simultaneity, either from the acceleration required to change inertial velocity or directly from the instantaneous velocity.

As far as I can see this concept is consistant with the explicit fundamental postulates of SR although it may conflict with the interpretated implicit postualtes or informally derived theorems of many individuals.
In any case it does not change the basic reality that we are left with purely relative measurements . Its only possible value is that it provides a logical hypothesis that explains the measured invariance of c.
lenfromkits
#5
Jul2-10, 12:24 AM
P: 108
Quote Quote by kmarinas86 View Post
Let's say you are receiving photons from all over. The velocity of each photon coming to you is c. If you moved at velocity v compared to your initial inertial frame, without SR you would say that photons going the same direction as you are moving at c-v, while the photons going the opposite direction do so at c+v. How does time dilation bring both values back to c? It makes no sense to me. Length contraction seems to reverse this perception. So how does the speed of light become constant?
Oh, in addition to that, time dilation is essentially where an object arrives at its destination in less time that you measured on your own clock. If something arrives at its destination sooner than expected, the trip must have then been shorter. So if your twin brother flew away on a 1000 kilometer trip and returned and his clock was 5 minutes behind yours, you'd have to assume the trip was shorter for him than what you observe. In other words, the length of his trip must have contracted because the time it took him to do it was shortened (dilated).
my_wan
#6
Jul2-10, 09:00 AM
P: 863
Technically, SR is because the speed of light is constant, the speed of light is not constant because of SR.
Rasalhague
#7
Jul2-10, 09:15 AM
P: 1,402
Quote Quote by my_wan View Post
Technically, SR is because the speed of light is constant, the speed of light is not constant because of SR.
bcrowell has a good discussion of this issue in the FAQ he posted in a recent thread:

http://www.lightandmatter.com/cgi-bin/meki?physics/faq

(And if anyone's wondering which came first, the chicken or the egg, it was the egg, obviously!)
Fredrik
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Jul2-10, 09:36 AM
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Quote Quote by kmarinas86 View Post
How does SR cause the speed of light to be constant?
It doesn't. I would define SR as the set of all theories of particles and fields in Minkowski spacetime (a mathematical spacetime that by definition has an invariant speed built into it). So the invariant speed is simply part of the definition of SR.

It still makes sense to ask for the cause, but it's important to realize that the only thing that can give you an answer is another theory, and that such a theory would have to be based on some other set of assumptions.

Quote Quote by kmarinas86 View Post
Let's say you are receiving photons from all over. The velocity of each photon coming to you is c. If you moved at velocity v compared to your initial inertial frame, without SR you would say that photons going the same direction as you are moving at c-v, while the photons going the opposite direction do so at c+v.
This is wrong. Velocities in SR don't add up that way. If an object has velocity v in a frame that has velocity u, the velocity of the object is (u+v)/(1+uv/cē). If you use that formula you don't get c-v and c+v. You get (c-v)/(1+c(-v)/cē)=c and (c+v)/(1+cv/cē)=c.
Aaron_Shaw
#9
Jul2-10, 10:20 AM
P: 54
Quote Quote by Fredrik View Post
It doesn't. I would define SR as the set of all theories of particles and fields in Minkowski spacetime (a mathematical spacetime that by definition has an invariant speed built into it). So the invariant speed is simply part of the definition of SR.

It still makes sense to ask for the cause, but it's important to realize that the only thing that can give you an answer is another theory, and that such a theory would have to be based on some other set of assumptions.


This is wrong. Velocities in SR don't add up that way. If an object has velocity v in a frame that has velocity u, the velocity of the object is (u+v)/(1+uv/cē). If you use that formula you don't get c-v and c+v. You get (c-v)/(1+c(-v)/cē)=c and (c+v)/(1+cv/cē)=c.
Perhaps the op is asking whether, and how, it is that time dilation that causes the addition of velocities to be counter intuitive.
HallsofIvy
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Jul2-10, 10:30 AM
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Quote Quote by kmarinas86 View Post
Let's say you are receiving photons from all over. The velocity of each photon coming to you is c. If you moved at velocity v compared to your initial inertial frame, without SR you would say that photons going the same direction as you are moving at c-v, while the photons going the opposite direction do so at c+v.
No. That's the whole point of "relativity". The correct formula is, rather, "if object b is moving toward you at speed v relative to you and object a is moving toward you at speed u, relative to object b, then object a is moving toward you at speed
[tex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/tex]
relative to you". NOT "u+v".

If u= v= c, that reduces to
[tex]\frac{c+ c}{1+ \frac{c^2}{c^2}}= \frac{2c}{2}= c[/itex]

How does time dilation bring both values back to c? It makes no sense to me. Length contraction seems to reverse this perception. So how does the speed of light become constant?
The fact that c is a constant, to all observers, is a result of experimental results. The rest of relativity is derived from that.
yossell
#11
Jul2-10, 11:22 AM
P: 341
Quote Quote by HallsofIvy View Post
No. That's the whole point of "relativity".
Well...I guess that's why he said `*without* SR'.
lenfromkits
#12
Jul2-10, 11:25 AM
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Quote Quote by my_wan View Post
Technically, SR is because the speed of light is constant, the speed of light is not constant because of SR.
my_wan - nice point. You have very smart comments and thanks for helping me on a thread of my own with such logical points of view.
Fredrik
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Jul2-10, 11:51 AM
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Quote Quote by yossell View Post
Well...I guess that's why he said `*without* SR'.
Oops. I guess we all missed that.
yossell
#14
Jul2-10, 12:59 PM
P: 341
@OP,

I think I second austin0's point. I think the answer to your question is that time dilation alone *isn't* enough to get both values back to c. You'll also need the fact that simultaneity is relative, different in different frames. Perhaps an intuitive way to see that this is so is the following:

Imagine a moving carriage with a light-emitter at its centre. Suppose it is moving left to right from a stationary trainspotter's point of view. At some moment, it emits a flash of light, and two light rays travel out to carriage's end points.

From the point of view of someone watching the train go by, the pulse heading left, towards the back of the train, arrives at the back of the train before the pulse going right hits the front of the train. For, in his frame, both light pulses travel at the same speed and, since the train is moving, the backwards travelling one must hit first.

From the point of view a passenger in the train, the train is not moving. Since the emitter is at the midpoint of the carriage, and since the speed of light is constant in his frame, the two pulses must hit the back and the front of the train simultaneously.

The passenger and the trainspotter disagree over which events are simultaneous with which other events. It is this variability is a necessary part of the explanation that trainspotter and passenger both agree on the speed of light
Tantalos
#15
Jul3-10, 05:13 AM
P: 46
Quote Quote by yossell View Post
@OP,

I think I second austin0's point. I think the answer to your question is that time dilation alone *isn't* enough to get both values back to c. You'll also need the fact that simultaneity is relative, different in different frames. Perhaps an intuitive way to see that this is so is the following:

Imagine a moving carriage with a light-emitter at its centre. Suppose it is moving left to right from a stationary trainspotter's point of view. At some moment, it emits a flash of light, and two light rays travel out to carriage's end points.

From the point of view of someone watching the train go by, the pulse heading left, towards the back of the train, arrives at the back of the train before the pulse going right hits the front of the train. For, in his frame, both light pulses travel at the same speed and, since the train is moving, the backwards travelling one must hit first.

From the point of view a passenger in the train, the train is not moving. Since the emitter is at the midpoint of the carriage, and since the speed of light is constant in his frame, the two pulses must hit the back and the front of the train simultaneously.

The passenger and the trainspotter disagree over which events are simultaneous with which other events. It is this variability is a necessary part of the explanation that trainspotter and passenger both agree on the speed of light
Consider this non-relativistic explanation. The photon emitted by the light source on the train must hit the front/back of the carriage, then it is reflected and travelles back to the observer on the train. So whether moving to the front or moving to the back each photon travelles half of the way with speed c+v and the other half with speed c-v. And each photon will reach the observer at the same time. This case is symmetric for the observer in the carriage and he sees both photons reaching him simultanously. But for the trainspotter the ways travelled by the photons will not be symmetric, so he will see a diffrence in time the two events happened.
Phrak
#16
Jul3-10, 05:32 AM
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Quote Quote by lenfromkits View Post
my_wan - nice point. You have very smart comments and thanks for helping me on a thread of my own with such logical points of view.
logical, no. it's a nonsensical dichotomy. True, a theory does not dictate physical reality. However, the velocity of light does not dictate the metric signature of spacetime.
Austin0
#17
Jul3-10, 06:32 AM
P: 1,162
Quote Quote by yossell View Post
@OP,

I think I second austin0's point. I think the answer to your question is that time dilation alone *isn't* enough to get both values back to c. You'll also need the fact that simultaneity is relative, different in different frames. Perhaps an intuitive way to see that this is so is the following:

Imagine a moving carriage with a light-emitter at its centre. Suppose it is moving left to right from a stationary trainspotter's point of view. At some moment, it emits a flash of light, and two light rays travel out to carriage's end points.

From the point of view of someone watching the train go by, the pulse heading left, towards the back of the train, arrives at the back of the train before the pulse going right hits the front of the train. For, in his frame, both light pulses travel at the same speed and, since the train is moving, the backwards travelling one must hit first.

From the point of view a passenger in the train, the train is not moving. Since the emitter is at the midpoint of the carriage, and since the speed of light is constant in his frame, the two pulses must hit the back and the front of the train simultaneously.

The passenger and the trainspotter disagree over which events are simultaneous with which other events. It is this variability is a necessary part of the explanation that trainspotter and passenger both agree on the speed of light

Hi yossell
I like the train context . Maybe I can clarify my point with it.

Assuming the two lightning bolts simultaneous in the track frame. Occuring at the same time according to the proximate track clocks. Also arriving from the location of the other clock after the same elapsed time.

As observed in the track frame the light from the bolt at the front of the train arrives at the back of the train after traveling a shorter distance in the track coordinates due to the motion of the back of the train during transit.
COnversely the light originating at the back of the train travels a greater distance reaching the front relative to the track because the front of the train is moving away from the light in the track frame.
SO if light speed is constant and absolute how can the trains clocks and observers measure the same elapsed time for these paths with different lengths.

And if the train accelerates to a greater inertial velocity, the difference in path lengths will be observed to increase in the track frame but will still be measured as taking equal time in the train frame.
This is the basic enigma and SR tells us the answer is simultaneity.
AT the time of the flashes at equal proper times in the trackF the trains clocks were observed to be running ahead at the back and behind at the front.
Obviously this means that the flash at the front occured earlier according to the train clock there and so when it arrives at the back the clock there will show the time of transit plus the amount it is running ahead.
Likewise the flash at the back will occur later ..so when it arrives at the front it will be elapsed time minus the interval the front clock is running behind.
In practice it works out the desynchronization is exactly such that the two observed times will be equal.
This would of course apply to the train at greater velocity with equivalently greater desynchronization.
This is basic SR and satisfactorily explains the phenomanon as it applies and is observed between frames.
But it is not really a satisfactory explanation for the OP's question because SR says that it is purely relative. According to the trains clocks the tracks are out of synch etc.

But the real basis of the question is putting yourself on the train at various levels of velocity ,without consideration of how it appears to other frames and explaining how the different path lengths are measured with equal dt's.

You can conclude that inertial motion is unreal and there is no actual difference between the different levels of attained velocity and therefore no real difference in light path lengths depending on direction [which I find far from satisfactory]
or
you can conclude that at each of those velocities there is a differing level of actual desynchronization which is undetectable and unquantifiable within the system .
That is my point.
For simplicity I purposely didn't consider dilation or contraction or compare quantitative distance/times between the train and track which would require those factors.
Thanks
my_wan
#18
Jul3-10, 01:42 PM
P: 863
Quote Quote by Phrak View Post
logical, no. it's a nonsensical dichotomy. True, a theory does not dictate physical reality. However, the velocity of light does not dictate the metric signature of spacetime.


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