twice the speed of light

[leveni]

photon A<----[light source]---->photon B

 

Two photons are ejected from a light source at the same time, in opposite linear directions, in a vacuum. They are named photon A and photon B. The light source is stationary and Photon A travels at c in a westerly direction and photon B travels at c in an easterly direction.

 

Question A:

Immediately after being emitted from the light source, is photon A travelling at two times c relative to photon B?

 

In regards to question A: If the light source is stationary, then relative to the light source then the answer should be yes(Photon A is travelling at 2 times c relative to photon B from the light sources point of view).

 

Question B:

But from the point of view of photon B, would photon A be travelling at two times c?

 

In regards to question B: Is this question possible? That is, can photons ever be at rest relative to something else. Do they have rest mass?

[Ophiolite]

QA - No.

 

QB - No.

 

Photons can never be at rest.

[timo]

Immediately after being emitted from the light source, is photon A travelling at two times c relative to photon B? If the light source is stationary, then relative to the light source then the answer should be yes.

Yes, if by "relative to" you mean "in the coordinate system of" and if you define "speed relative to each other" as the difference of velocities.

 

But from the point of view of photon B, would photon A be travelling at two times c? In regards to question B: Is this question possible? That is, can photons ever be at rest relative to something else.

In relativity, there is no point of view of a photon, i.e. no coordinate system in which a photon is at rest. So that question is indeed "not possible" in a sense.

[altergnostic]

Of course in SR you can't ask Q.B because you can't calculate it, and that is because light is never an observer, it is only the tool with which the observer observes other things: light is the c in v/c while what is being observed is the v in v/c.

 

But you can always ask interesting questions and use logic to answer them. Then, logically, one photon would be moving at 2c relative to the other. The only thing you must remember is that this is not observed by either photon: you need a photon being directly detected to observe something.

 

In this case, a third photon being emitted at the surface of one photon and reaching the other directly, else one won't even know the other exists at all. Only the source knows it emited 2 photons and only the source is physically able to determine their relative velocities. An observer at the point of emission can always place two mirrors at x and -x and wait for both photons to return, measure each photon's velocity for the round-trip and calculate the relative speed between them.

 

 

[elfmotat]

Then, logically, one photon would be moving at 2c relative to the other.

 

Um, no. "Logically" the correct value to extrapolate to is c, not 2c. Say a light source emits a photon. Boosting from the rest frame of the light source to a frame traveling at velocity v in the opposite direction of the photon yields a photon velocity in the new frame of:

 

 

This makes sense, because one of the core assumptions of relativity is that light travels at c in all inertial frames.

 

Now if you want to extrapolate this to a nonexistent frame traveling at c relative to the light source (and I don't see why you would to be honest; it doesn't have any physical meaning) then can just take the limit of u as v->c, which is still c.

Edited December 11, 2012 by elfmotat

[leveni]

Thanks elfmotat and altergnostic. My intuition tells me altergnostic is correct but elfmotat has proven, mathematically, the difference in speed between the two photons is c. I guess the next step would be to calculate 'time dilation' and 'length contraction' between the two photons. Would this be right?

[Klaynos]

Thanks elfmotat and altergnostic. My intuition tells me altergnostic is correct but elfmotat has proven, mathematically, the difference in speed between the two photons is c. I guess the next step would be to calculate 'time dilation' and 'length contraction' between the two photons. Would this be right?

The problem you will have is to calculate length contraction you need two valid frames, the lightsource frame is one, but neither of the photon frames is valid in SR.

[J.C.MacSwell]

 

photon A<----[light source]---->photon B

 

Two photons are ejected from a light source at the same time, in opposite linear directions, in a vacuum. They are named photon A and photon B. The light source is stationary and Photon A travels at c in a westerly direction and photon B travels at c in an easterly direction.

 

Question A:

Immediately after being emitted from the light source, is photon A travelling at two times c relative to photon B?

 

In regards to question A: If the light source is stationary, then relative to the light source then the answer should be yes(Photon A is travelling at 2 times c relative to photon B from the light sources point of view).

 

Question B:

But from the point of view of photon B, would photon A be travelling at two times c?

 

In regards to question B: Is this question possible? That is, can photons ever be at rest relative to something else. Do they have rest mass?

Photons do not have reference frames.

Edited December 12, 2012 by J.C.MacSwell

[md65536]

Yes, if by "relative to" you mean "in the coordinate system of" and if you define "speed relative to each other" as the difference of velocities.

Is this true in all frames (that the photons separate at 2c), or does it imply defining the coordinate system relative to the light source's frame?

 

Or does a separation speed of 2c only occur in frames from which the directions of the photons are collinear?

[timo]

It's true in all reference frames in which the two packets of light move in opposite directions, since only then and only then the difference between the two (3-)velocity vectors with a magnitude of c has a magnitude of 2c. That does include the frame of the light source in this case, because leveni defined the light source to emit in opposite directions.

[md65536]

Yes, if by "relative to" you mean "in the coordinate system of" and if you define "speed relative to each other" as the difference of velocities.

 

Ah, sorry, I got mixed up because "relative to" was used twice maybe to mean different things, each of which you defined, but I took it to mean that the difference of velocities could be defined as "speed in the coordinate system of each other" (but relative to some observer), and now I'm still not sure: Can you define a coordinate system of a photon, relative to another frame of reference? Or is a coordinate system the same as a frame of reference?

[timo]

It's not so complicated, actually. But to be clear:

1) "Coordinate system" is much more general than the special-relativistic "frame of reference". Too general to be of use here, even though I usually prefer it. In the context of this thread (and pretty much any Relativity thread on sfn) you can probably assume the two terms to mean the same.

 

2) In the frame of reference of a source emitting two beams of light to positive x-direction and negative x-direction, respectively, the difference between the light beams velocity is [math]\Delta \vec v = (c,0,0) - (-c,0,0) = (2c,0,0)[/math] (or its negative). The magnitude of the velocity difference is [math] |\Delta \vec v| = 2c[/math].

 

3) In the frame of reference of a source emitting two beams of light into opposite but otherwise arbitrary directions, the magnitude of the velocity difference is again 2c.

 

4) In any frame of reference, in which the two beams move in opposite directions, their relative velocity is 2c. Irrespective of whether this is the frame of the source or not.

 

5) If the two beams do not move in exactly opposite directions, the magnitude of the relative velocity is not equal to 2c.

 

6) Generally, the relative velocity between the beams of light is [math] c (1-\cos x) [/math], where x is the angle between the two velocities. This follows from basic geometry.

 

7) If there exists a frame of reference in which the relative velocity of the two beams is 2c, then there exists another frame of reference in which their relative velocity is smaller than 2c. In the case of case (2), such a "<2c"-System would be one moving in y-direction relative to the light source.

 

8) There is no frame of reference in which either beam of light does not have a velocity with [math]|\vec v| = c [/math] - as pretty much everyone in this thread felt the urge to repeat.

 

In all of the statements above, there is pretty much no change with respect to non-relativistic physics. The big change that Relativity introduces is, in this context, the definition of how velocities transform between different frames of reference (namely not by simply adding the velocity of the new frame to the velocities of the light beams). But within a frame of reference, no one stops you from defining "relative velocity" as the difference of the velocity vectors (whatever that definition may be good for). There is, of course, much more to special relativity than just the transformation rules of velocities upon a change of frame of reference. But I don't believe that any of it matters here.

Edited December 16, 2012 by timo

[elfmotat]

It's not so complicated, actually. But to be clear:

1) "Coordinate system" is much more general than the special-relativistic "frame of reference". Too general to be of use here, even though I usually prefer it. In the context of this thread (and pretty much any Relativity thread on sfn) you can probably assume the two terms to mean the same.

 

2) In the frame of reference of a source emitting two beams of light to positive x-direction and negative x-direction, respectively, the difference between the light beams velocity is [math]\Delta \vec v = (c,0,0) - (-c,0,0) = (2c,0,0)[/math] (or its negative). The magnitude of the velocity difference is [math] |\Delta \vec v| = 2c[/math].

 

3) In the frame of reference of a source emitting two beams of light into opposite but otherwise arbitrary directions, the magnitude of the velocity difference is again 2c.

 

4) In any frame of reference, in which the two beams move in opposite directions, their relative velocity is 2c. Irrespective of whether this is the frame of the source or not.

 

5) If the two beams do not move in exactly opposite directions, the magnitude of the relative velocity is not equal to 2c.

 

6) Generally, the relative velocity between the beams of light is [math] c (1-\cos x) [/math], where x is the angle between the two velocities. This follows from basic geometry.

 

7) If there exists a frame of reference in which the relative velocity of the two beams is 2c, then there exists another frame of reference in which their relative velocity is smaller than 2c. In the case of case (2), such a "<2c"-System would be one moving in y-direction relative to the light source.

 

8) There is no frame of reference in which either beam of light does not have a velocity with [math]|\vec v| = c [/math] - as pretty much everyone in this thread felt the urge to repeat.

 

In all of the statements above, there is pretty much no change with respect to non-relativistic physics. The big change that Relativity introduces is, in this context, the definition of how velocities transform between different frames of reference (namely not by simply adding the velocity of the new frame to the velocities of the light beams). But within a frame of reference, no one stops you from defining "relative velocity" as the difference of the velocity vectors (whatever that definition may be good for). There is, of course, much more to special relativity than just the transformation rules of velocities upon a change of frame of reference. But I don't believe that any of it matters here.

 

You're using a confusing definition of "relative velocity." What you're describing is separation velocity. "Relative velocity" is traditionally defined as the the speed of one object from the rest frame of another.

[timo]

 

You're using a confusing definition of "relative velocity." What you're describing is separation velocity. "Relative velocity" is traditionally defined as the the speed of one object from the rest frame of another.

See post #2.

Edited December 17, 2012 by timo

[elfmotat]

See post #2.

 

Yes, I'm well aware that photons don't have a rest frame. That's why everyone has been saying that that relative velocity of one photon with respect to another is undefined.

[timo]

I was referring the other one of the two sentences in post #2. Possibly should have made that more clear . EOD for me.

[altergnostic]

Yes, I'm well aware that photons don't have a rest frame. That's why everyone has been saying that that relative velocity of one photon with respect to another is undefined.

...but why is it undefined? Mathematically, we can't define it, and we can't define because we can't divide by zero... Circular. We can't define it because we use c as a tool to measure other things, it is clear that we can't measure the tool with the tool itself, hence c/c doesn't work in our equations.

But that doesn't mean that one photon does not travel at some speed with respect to another, or that it has no inertial frame itself, or that a body travelling at c wrt some background can't observe another body travelling at any v<c relative to him or anything like that. Of course there IS a relative v between both photons, even if we can't define the photon's frame or if this realisation has any practical meaning.

The point is that one photon is unable to measure the speed of the other. Think about it, we determine the speed of light with a roundtrip measurement. If one photon A traveling in the +x direction emits a second photon B in the -x direction towards a mirror, B bounces back but never reaches A, since now they are both travelling at c in the +x direction and are some distance apart. It's as simple as that. Information would have to travel faster than light between the two photons otherwise one can't determine the speed of the other, and if such thing was possible, we would have v/n>c and whatever is carrying the information between photons has become the tool with which we would make measurements, and now it is this thing that moves at v>c that has the undefined frame. It is a matter of operation, not a physical property. You can play with any number in the denominator and see what happens if you define a frame for the photon, or any body moving at c, and you find that it is theoretically possible to build a frame for light, although that would violate c as a speed limit and bring further complications. Anyway, it really is a valid exercise and helps to understand precisely why we can't ask what would be the measurements of a photon in any given situation, and why is it that the photon frame is undefined.

 

I hope I have properly expressed myself. Cheers.

[elfmotat]

...but why is it undefined?

 

Because "relative velocity" is defined as the velocity of one object in the rest frame of another, and there's no such thing as a rest frame for a photon. That's just the way it is, despite your objections.

 

As timo was explaining, the separation velocity between two parallel photons is 2c. That's the only kind of velocity you can define between photons.

 

 

 

I was referring the other one of the two sentences in post #2. Possibly should have made that more clear . EOD for me.

 

"QA - No.

 

QB - No."

 

???

[md65536]

It is a matter of operation, not a physical property. You can play with any number in the denominator and see what happens if you define a frame for the photon, or any body moving at c, and you find that it is theoretically possible to build a frame for light, although that would violate c as a speed limit and bring further complications.

 

No, it is physically meaningless to speak of a frame for light, if you properly deal with all the details of SR (and if you don't, you'll probably derive contradictions). Any way you look at it, such a frame doesn't make sense. Mathematically; or imagining riding on a wave of light; or considering length contraction and time dilation in the limit of v as it approaches c (there's no proper time for light); or considering measurable properties of a photon---no proper treatment will give you a meaningful frame for light.

 

Yes, you could imagine something false by ignoring some details, just like you could imagine anything else that's physically meaningless.

 

Your example---that you couldn't measure c relative to a photon with round-trip timing---is consistent with the principle that the photon doesn't have an observational frame of reference, but it doesn't define it.

[altergnostic]

 

No, it is physically meaningless to speak of a frame for light, if you properly deal with all the details of SR (and if you don't, you'll probably derive contradictions). Any way you look at it, such a frame doesn't make sense. Mathematically; or imagining riding on a wave of light; or considering length contraction and time dilation in the limit of v as it approaches c (there's no proper time for light); or considering measurable properties of a photon---no proper treatment will give you a meaningful frame for light.

 

Yes, you could imagine something false by ignoring some details, just like you could imagine anything else that's physically meaningless.

 

Your example---that you couldn't measure c relative to a photon with round-trip timing---is consistent with the principle that the photon doesn't have an observational frame of reference, but it doesn't define it.

I agree that there no physical meaning for light to have a frame. But the more you play with theory and equations the more you understand it. You named a few examples of different ways you can try to determine a frame for light, and none of them work, and I think it is important to really try them out, so one understands precisely why they don't work, instead of simply taking other's word for it, don't you agree?

 

And as to your last sentence, it doesn't define it, but I personally think the most fundamental reason (not the only reason) why we can't possibly determine a frame for light is because we use light as the tool to determine frames to start with. It is like trying to weight a scale with the scale itself, or screwing a screw with a screw. I think you can reach the same conclusion by imagining light with mass and volume and inventing a smaller and faster particle to measure it, then you will be able to find a frame for light, but not for the imaginary particle. The "what if" exercise is always a good exercise, I believe. What if there is more then one universe? What if the constants were not always constants? What if there's an aeher? What if the photon was not the smallest and fastest particle? These are physical questions, maybe untrue and imaginary, but physical nonetheless.

 

Anyway, my intention was simply to advocate curiosity and exploration, so one understands for himself why is it that giving a frame to light is meaningless, since it is counter-intuitive, otherwise we wouldn't see so many people asking questions about it and having a hard time understanding it.

[Delta1212]

And as to your last sentence, it doesn't define it, but I personally think the most fundamental reason (not the only reason) why we can't possibly determine a frame for light is because we use light as the tool to determine frames to start with.

This is not, strictly speaking, true. Light isn't used. The speed 'c' is. We happen to call that 'the speed of light' but we could equally call it 'the speed of gravity' or any number of other names that have nothing to do with light. The reason that we cannot determine a frame for light is that it travels at c, and anything traveling at c does not experience time or distance within its "frame". We're not talking about trying to measure the precise length of a yard using a yardstick. We're talking about a fundamental property of the universe that was derived from measurements obtained by experiment and observation.

 

Nothing traveling at c will experience time or distance in the way that we think of those things, and if there's no time and no distance, you can't measure speeds from such a frame. Let's say we're somehow riding on one of these two separating photons. We look back and attempt to measure the relative speed of the other. Well, we can't actually look back because in this impossible hypothetical, we'd be frozen in time, but let's say we can somehow move anyway. What would we be measuring? You'd need to measure how far the other photon is. Then wait a set amount of time and measure how far away it is again. Then divide the distance traveled by the elapsed time. But traveling at c we don't have any measurement of how far the other photon has traveled, and we don't have any time elapsing in order to mark of the distance even if we did.

 

There is no frame from the perspective of light not because light is defined as the measuring stick, but because the properties being measured don't exist for anything traveling at that speed.

[md65536]

I agree that there no physical meaning for light to have a frame. But the more you play with theory and equations the more you understand it. You named a few examples of different ways you can try to determine a frame for light, and none of them work, and I think it is important to really try them out, so one understands precisely why they don't work, instead of simply taking other's word for it, don't you agree?

Not really... I agree it's helpful and interesting.

 

It's like trying out different examples in SR and asking "will this example work too?" It's interesting to see that it does work out, but it's better to realize that SR is consistent and not question whether the examples will work, because until you see why they all will, you can forever question it with more examples.

 

There is no way in which a frame of light has meaning. If you came across one, that would be a problem that would need to be looked at and resolved, but I don't think it's "important" to look for such problems. It's like saying "It's important to think about ways that reality might not be consistent, so that we can make sure that reality is actually consistent!" You can assume that it is, until you have a real problem.

 

 

Agreed that they're good exercises to think about these things. I agree with your second paragraph. Our measurements of time and distance are defined by light c, and our experience of reality is defined by how we measure it, so it's definitely related to not being able to define measurements in terms of "what light sees".

 

 

I agree with you pretty much, but just nit-pick on some of the details. I don't think that understanding relativity through examples is the right way to go, unless you can generalize from the examples and make further predictions from it. And that's usually not the case... often people will understand eg. the results twin paradox, but they'll not know why it's so, and so with the next example that comes along they'll have to relearn or be told how it works out. Best I think is to understand it in terms of the theory and the maths, and understand that it is consistent. Then, any new example can be worked out and predicted, and each example only reassures you that it all works out, rather than every new example being a new mystery to puzzle over.

 

In this case, lack of a frame of light comes from math. Any examples of what that means, physically, are consistent with the math. So, unless you can deduce all of the other possible examples from an inability to measure return trips, then the return-trip example is not a sufficient explanation for why light can't have a frame of reference.

Edited December 18, 2012 by md65536

[altergnostic]

Well, first of all, the statement that it is c and not light that makes a frame undefinable is backwards. The value for c comes from light and exists only for light, nothing else moves at c. Light and c are the same thing, but the number is secondary, it could be any other number and the theory would still be valid.

 

Our measurements of time and space are defined by c only because it is the speed of light, not the other way around. The fact that we measure with light is more fundamental than the fact that light happens to move at c.

 

In the end I think this is a matter of practice vs. theory, purely academical. In practice, if you want to perform operations and move on, surely it seems a waste of time to wonder about imaginary examples, but if your goal is theoretical development, than I think it really is the way to go. It is how Einstein worked himself. Didn't he say that imagination was more important than intelligence?

 

Anyway, I believe this has become digression from the main subject already and we probably don't even disagree on anything important here.

[Delta1212]

Well, first of all, the statement that it is c and not light that makes a frame undefinable is backwards. The value for c comes from light and exists only for light, nothing else moves at c. Light and c are the same thing, but the number is secondary, it could be any other number and the theory would still be valid.

This is not technically true. All massless particles travel at c and, as I alluded to earlier, gravity does as well. We first came across c through studying light, which is why it's called the speed of light, but that's not the last place we've encountered it. They are not the same thing.

[md65536]

In the end I think this is a matter of practice vs. theory, purely academical. In practice, if you want to perform operations and move on, surely it seems a waste of time to wonder about imaginary examples, but if your goal is theoretical development, than I think it really is the way to go. It is how Einstein worked himself. Didn't he say that imagination was more important than intelligence?

I think it proves my point, that we know the special theory of relativity in terms of the Lorentz transformation and e=mc^2, rather than say the theory of what it would be like to ride on a wave of light. As Delta1212 points out, the theory implies more than just the examples or experiments that lead to it.

 

I definitely agree it's important to imagine these things, especially as Einstein did with things that have never been thoroughly thought out before, and even when it's nothing new it's still not a waste of time to do so. Sorry to carry the digression further. But to try to bring it back to the topic, it is in theory that light cannot be measured from any meaningful "point of view of light". That all experiments or examples agree with it is what makes a theory accepted.

Edited December 19, 2012 by md65536