THE RETURN of two opposite travelling photons

[dubiousraves]

Your question is rather confusing because of the words "THE RETURN" in your title. I can only assume that you mean a return to the subject rather than that the two photons are returning to each other. So if that is what you meant, then in a frame in which two photons are emitted from a common source, it is true that after one year, the distance between them is 2 light years, and you are correct that we cannot consider a frame of reference in which a photon is at rest, but we can still use the formula that Einstein gave in section 5 of his 1905 paper for "The Composition of Velocities" and see what velocity we would get if we plugged in c for both v and w in his equation:



And we would get c as the resultant speed.

However, if we want to follow the rules, we could change your question slightly to what would be the relative speed between a photon emitted in one direction and an observer traveling at under c in the other direction?

In fact, Einstein even makes this point when he says:

It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain


If you don't see why this is true, just multiply the top and bottom of the fraction by c:

c(c+w)/c(1+w/c) = c(c+w)/(c+cw/c) = c(c+w)/(c+w) = c

He is making the point that no matter what the speed of the observer is, the photon is still traveling at c relative to him.

Does this all make perfect sense to you?

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

 

[ghwellsjr]

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

The equations would have come through in the quote if you had just clicked on the QUOTE button in the lower right corner of the post you wanted to quote. It would also have provided a link back to the original post so that you wouldn't have to bother saying it was post #3. Here, like this:

Your question is rather confusing because of the words "THE RETURN" in your title. I can only assume that you mean a return to the subject rather than that the two photons are returning to each other. So if that is what you meant, then in a frame in which two photons are emitted from a common source, it is true that after one year, the distance between them is 2 light years, and you are correct that we cannot consider a frame of reference in which a photon is at rest, but we can still use the formula that Einstein gave in section 5 of his 1905 paper for "The Composition of Velocities" and see what velocity we would get if we plugged in c for both v and w in his equation:

And we would get c as the resultant speed.

However, if we want to follow the rules, we could change your question slightly to what would be the relative speed between a photon emitted in one direction and an observer traveling at under c in the other direction?

In fact, Einstein even makes this point when he says:

It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain

If you don't see why this is true, just multiply the top and bottom of the fraction by c:

c(c+w)/c(1+w/c) = c(c+w)/(c+cw/c) = c(c+w)/(c+w) = c

He is making the point that no matter what the speed of the observer is, the photon is still traveling at c relative to him.

Does this all make perfect sense to you?

Now to answer your question, no, this post is not about time dilation or length contraction. The purpose of the post was to show that the closest possible calculation of the relative speed between two photons going in the opposite direction is c and not 2c.

 

[harrylin]

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

It's unclear to me what you mean with "specifically addressed". Time dilation and length contraction as well as clock synchronization are inherent in the Lorentz transformation, from which that equation is derived. So, without time dilation and length contraction that equation would be different, if that is what you mean; they are needed to find c for the one-way speed of light after clock synchronization.

 

[dubiousraves]

Sorry, I don't see any quote button. Perhaps it doesn't exist in Google Chrome?

Anyway, I posted my question on this thread because harrylin, the esteemed commentator here, suggested the answer to my question could be found in the post I referenced. It does seem my query would fit more comfortably in the time dilation thread, but I'll state my question here anyway and kick it over to wherever you experts think it belongs.

To wit: What are the time-dilation and space-contraction reasons for c being constant for an observer not only traveling in the same direction of a light beam (say at .75c, where I understand the time-contraction reasons for the traveler seeing the light beam at c), but for an observer heading directly toward a light beam coming at him?

Thanks.

 

[harrylin]

Sorry, I don't see any quote button. Perhaps it doesn't exist in Google Chrome?

Anyway, I posted my question on this thread because harrylin, the esteemed commentator here, suggested the answer to my question could be found in the post I referenced. It does seem my query would fit more comfortably in the time dilation thread, but I'll state my question here anyway and kick it over to wherever you experts think it belongs.

To wit: What are the time-dilation and space-contraction reasons for c being constant for an observer not only traveling in the same direction of a light beam (say at .75c, where I understand the time-contraction reasons for the traveler seeing the light beam at c), but for an observer heading directly toward a light beam coming at him?

Thanks.

Yes, you asked: "what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this? "

That is closer to the topic of this thread. See also post #7 as well as my earlier reply which was just before yours here.

 

[dubiousraves]

It's unclear to me what you mean with "specifically addressed". Time dilation and length contraction as well as clock synchronization are inherent in the Lorentz transformation, from which that equation is derived. So, without time dilation and length contraction that equation would be different, if that is what you mean; they are needed to find c for the one-way speed of light after clock synchronization.

Harrylin, I just saw your answer, after I posted my question somewhat differently. And now I see the quote button! Anyway, you originally sent me to that equation, and while I think I understand how it works at solving for different speeds by an observer, I guess I'm confused about why c is constant for an observer heading into a light beam, as opposed to catching up to one. I'd like to know the reasons for this based on different inertial frames in space-time. Thanks.

 

[dubiousraves]

Yes, you asked: "what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this? "

That is closer to the topic of this thread. See also post #7 as well as my earlier reply which was just before yours here.

Sorry, but I don't see how post #7 answers my question. I mean, it might, but I don't understand how it does. I understand that c is constant for all reference frames. I guess my question is why? If given the example of an observer traveling .75c in the same direction as a light beam, my understanding of why the observer sees the light beam at c is based on the idea that space-time is slowed/contracted for the observer, relative to the light beam, so that he sees the beam as c. But if I try to apply the same principle to the case of an observer traveling .75c TOWARD a light beam coming at him, I don't understand how it works out that he seems the light beam at c.

Can anyone explain this without merely telling me that it's already been explained? Thanks.

 

[harrylin]

Sorry, but I don't see how post #7 answers my question. I mean, it might, but I don't understand how it does. I understand that c is constant for all reference frames. I guess my question is why? If given the example of an observer traveling .75c in the same direction as a light beam, my understanding of why the observer sees the light beam at c is based on the idea that space-time is slowed/contracted for the observer, relative to the light beam, so that he sees the beam as c. But if I try to apply the same principle to the case of an observer traveling .75c TOWARD a light beam coming at him, I don't understand how it works out that he seems the light beam at c.

Can anyone explain this without merely telling me that it's already been explained? Thanks.

OK, I somehow misunderstood that you wanted to know how to calculate it for such a case.
If instead you want to know how the different effects keep the speed of light invariant, then the answer to that belongs in the "speed of light constant" thread, because that is there mistaken as meaning "speed of light invariant". I'll answer there!

 

[ghwellsjr]

...
What are the time-dilation and space-contraction reasons for c being constant for an observer not only traveling in the same direction of a light beam (say at .75c, where I understand the time-contraction reasons for the traveler seeing the light beam at c), but for an observer heading directly toward a light beam coming at him?
Thanks.

You keep talking about an observer "seeing the light beam at c" and similar statements. You have to come to grips with the fact that no one can see how fast a light beam goes. It doesn't matter if the light beam is traveling away from you or towards you, it's always the same problem. It's not like watching the progress of a massive object like a bullet or a rocket ship for which we can use light to watch them, either coming or going.

But we can't do that with light. If a flash of light is heading towards us, we can have no knowledge or awareness of it's presence until it finally reaches us and then we have no opportunity to measure its speed of approach because it's all over. If you don't recognize this fact, then I'd like you to tell me what you mean when you ask about seeing the light beam coming towards you and how you would measure its speed.

The same problem exists for a flash of light that is traveling away from you. You cannot observe its progress. It's gone. You can't shine another light on it like you could with a bullet and observe the reflections to see its progress.

If you read the first part of Einstein's paper that I referenced in the post you quoted, you will see that Einstein addresses this issue and points out that since we can only measure the round trip speed of light by using a ruler, a timer and a mirror. Thus, if we want to measure the speed of a flash of light coming towards us, we would start the timer when it got to us, let the flash hit the mirror which is beyond us and stop the timer when the reflection got back to us. Then we use the ruler to measure how far away the mirror was, double that value (because the light has to traverse that distance twice) and divide by the time interval we got on the timer and that value will always come out to be the universal constant c, as Einstein pointed out in his paper.

But this doesn't help us determine the speed of the light that was coming toward us, only the combined round trip speed. And we can never know if the time it took for the light to reach the mirror is the same as the time it took for the light to get back. Einstein then reasons that since this knowledge is hidden from us by nature itself, we are free to assign any combination of times to the two parts of the trip so why not pick the one where they are equal? That's what Einstein's second postulate does for us.

So I hope you can see that the same problem exists for a flash of light coming towards us as it does for a flash of light going away from us. You said you understood how the flash going away could be traveling at c but you didn't yet understand how a flash coming towards us could also be traveling at c. Yet, I don't see how you could really understand the one without also understanding the other. So if you still aren't clear, please help me understand how you understand that a flash going away from us is traveling at c.

By the way, I have recast your problem from a beam of light to a flash of light because if the light is present all the time, you will have no "markings" on it with which to pinpoint its progress. You can always turn a beam of light into a flash with a shutter so if you are bothered by how I recast your problem, then just put in a shutter that opens when you start the timer and stop the timer when the reflected image of the opening of the shutter gets back to you.

 

[harrylin]

[..] If a flash of light is heading towards us, we can have no knowledge or awareness of it's presence until it finally reaches us and then we have no opportunity to measure its speed of approach because it's all over. [..]
The same problem exists for a flash of light that is traveling away from you. You cannot observe its progress. It's gone. You can't shine another light on it like you could with a bullet and observe the reflections to see its progress. [..]

While that is somewhat true, it can lead to the kind of misunderstanding as in https://www.physicsforums.com/showthread.php?t=643225. SR doesn't map observers but reference systems, and in that thread I emphasize that even real reference systems can record the progression of a light flash over a certain distance.