# Relative speed of two photons

## Homework Statement

What is the speed of a photon with respect to another photon if:
1. the two photons are going in the same direction.
2. they are going in opposite direction?
2. The attempt at a solution
I think the answer to the first question should be zero and to the second one be 2xC; C⇒speed of light.
But at the same time i think that the speed of anything can't exceed C, thus my second answer should be wrong. I also feel the we cannot have an inertial frame which is travelling at the speed C as then the Newtonian laws of motion will fail.
I don't know much about Relativity, and am pretty confident about some flaws in my approach.
Thank You:)

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PeroK
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## Homework Statement

What is the speed of a photon with respect to another photon if:
1. the two photons are going in the same direction.
2. they are going in opposite direction?
2. The attempt at a solution
I think the answer to the first question should be zero and to the second one be 2xC; C⇒speed of light.
But at the same time i think that the speed of anything can't exceed C, thus my second answer should be wrong. I also feel the we cannot have an inertial frame which is travelling at the speed C as then the Newtonian laws of motion will fail.
I don't know much about Relativity, and am pretty confident about some flaws in my approach.
Thank You:)
The question is ambiguous. You are right that there is no frame of reference for a photon. But, by "speed" the question may mean "separation speed", being the speed at which two objects (or photons) converge or diverge in a given reference frame.

tnich
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## Homework Statement

What is the speed of a photon with respect to another photon if:
1. the two photons are going in the same direction.
2. they are going in opposite direction?
2. The attempt at a solution
I think the answer to the first question should be zero and to the second one be 2xC; C⇒speed of light.
But at the same time i think that the speed of anything can't exceed C, thus my second answer should be wrong. I also feel the we cannot have an inertial frame which is travelling at the speed C as then the Newtonian laws of motion will fail.
I don't know much about Relativity, and am pretty confident about some flaws in my approach.
Thank You:)
This is a tricky question. A photon moves at the speed of light, and the speed of light is invariant, meaning it is the same in all reference frames. So in the reference frame of photon A, what is the speed of photon B?

Ray Vickson
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The question is ambiguous. You are right that there is no frame of reference for a photon. But, by "speed" the question may mean "separation speed", being the speed at which two objects (or photons) converge or diverge in a given reference frame.
"Separation speed" can exceed ##c##, even for non-photons.

In some inertial frame O, say particle A is moving with velocity ##v_A =(.75 c, 0,0)## while B is moving with velocity ##v_B = (-.75 c, 0, 0).## In frame O the separation speed is ##1.5 c##, but the separation speed of B as seen from the rest frame of A is less than ##c##; ditto for A as seen by B.

For photons there are no "rest frames". An observer cannot ride along on photon A and then make measurements for photon B.

Orodruin
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So in the reference frame of photon A, what is the speed of photon B?
There is no "reference frame of a photon".

David Lewis and MeAndMyLucidLife
What if you replaced photon with "2 protons with speeds of say .999999 c"?
It seems that is what the question implies ( reference frames moving at speed c).

Nugatory
Mentor
What if you replaced photon with "2 protons with speeds of say .999999 c"?
It seems that is what the question implies ( reference frames moving at speed c).
That is a sensible question with a very interesting and important answer, but the answer and the thought process used to find that answer tell us nothing about the situation when the speed is 1.0c instead of .99999999.....c

PeroK
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That is a sensible question with a very interesting and important answer, but the answer and the thought process used to find that answer tell us nothing about the situation when the speed is 1.0c instead of .99999999.....c
On the other hand: ##1.0c = 0.999 \dots c##.

vela and SammyS
Nugatory
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On the other hand: ##1.0c = 0.999 \dots c##.
That is true, and I'm guilty of (knowingly) perpetuating a sloppiness from the post I was replying to. Getting to a rigorous statement is beyond the scope of this homework thread though.

MeAndMyLucidLife and PeroK
Orodruin
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What if you replaced photon with "2 protons with speeds of say .999999 c"?
It seems that is what the question implies ( reference frames moving at speed c).
Reference frames do not "move at speed c". First of all, all motion is relative. Second, no matter how close to c a reference frame moves relative to another, it is still not anywhere close to be "moving at c".

kimbyd
Gold Member
Lots of people in this thread have pointed out that reference frames can't move at the speed of light, but I'm not sure anybody has pointed out why that is the case.

The fundamental problem is time.

It is certainly possible to create something that looks like a hypothetical rest frame that moves at the speed of light. This is easy to do since we don't need to think about mass. But if you look at the properties of that hypothetical rest frame you have a problem: there is no time coordinate. You can see this by examining time dilation in the limit as ##v -> c##, which shows that no matter how much time passes in an inertial frame, no time ever passes in this hypothetical speed-of-light frame. That lack of time means that most of the questions we normally associate with a reference frame become nonsensical. For instance, velocity can't be measured because time doesn't pass, so questions about "relative velocity" are pointless.

So this problem cannot be answered because the "reference frame" of a photon isn't a proper reference frame at all (because it has no time coordinate).

Orodruin, Delta2, MeAndMyLucidLife and 1 other person
strangerep
[...] S this problem cannot be answered because the "reference frame" of a photon isn't a proper reference frame at all (because it has no time coordinate).
Note, however, that the original problem statement did not mention the phrase "reference frame".

@MeAndMyLucidLife:
As stated, the original problem can be answered sensibly in terms of limits. I.e., take the usual addition rule for Lorentz boosts in a given direction (with ##|v| < c##), and take a limit as ##v## approaches ##c##.

E.g., consider bodies "A,B" moving at colinear velocities ##-v## and ##v## respectively relative to "O". The velocity ## v_{OA}## of O relative to A is ##+v##, and the velocity ##v_{BO}## of B relative to O is also ##+v##. Hence the velocity ##v_{BA}(v)## of B relative to A in this scenario is $$v_{BA}(v) ~=~ \frac{v + v}{1 + v^2/c^2} ~=~ \frac{2v}{1 + v^2/c^2}~.$$ Now take the limit: $$\lim_{v\to c^-} v_{BA}(v) ~=~ \lim_{v\to c^-} \frac{2v}{1 + v^2/c^2} ~=~ c ~.$$ More rigorously, one should talk about ##\epsilon##'s and ##\delta##'s, but the above is sufficient "at the usual level of rigor in theoretical physics".

Briefly, this approach is valid because these transformations depend on the underlying Lie group theory (wherein the group is a well-behaved differentiable manifold). ##|v|=c## happens to be a fixed point in the group, but we can still talk about sensibly in terms of limits because of the underlying theory.

Last edited:
MeAndMyLucidLife
kimbyd
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Note, however, that the original problem statement did not mention the phrase "reference frame".

@MeAndMyLucidLife:
As stated, the original problem can be answered sensibly in terms of limits. I.e., take the usual addition rule for Lorentz boosts in a given direction (with ##|v| < c##), and take a limit as ##v## approaches ##c##.

E.g., consider bodies "A,B" moving at colinear velocities ##-v## and ##v## respectively relative to "O". The velocity ## v_{OA}## of O relative to A is ##+v##, and the velocity ##v_{BO}## of B relative to O is also ##+v##. Hence the velocity ##v_{BA}(v)## of B relative to A in this scenario is $$v_{BA}(v) ~=~ \frac{v + v}{1 + v^2/c^2} ~=~ \frac{2v}{1 + v^2/c^2}~.$$ Now take the limit: $$\lim_{v\to c^-} v_{BA}(v) ~=~ \lim_{v\to c} \frac{2v}{1 + v^2/c^2} ~=~ c ~.$$ More rigorously, one should talk about ##\epsilon##'s and ##\delta##'s, but the above is sufficient "at the usual level of rigor in theoretical physics".

Briefly, this approach is valid because these transformations depend on the underlying Lie group theory (wherein the group is a well-behaved differentiable manifold). ##|v|=c## happens to be a fixed point in the group, but we can still talk about sensibly in terms of limits because of the underlying theory.
Nope, this doesn't work. You've only avoided talking about a reference frame, rather than actually avoiding it altogether. It still has the same exact problem of time I mentioned.

The only way to make the problem correct is to consider the "velocity of photon A relative to photon B" in a reference frame that is neither A's nor B's. In that case the answer is pretty trivial: just add or subtract their velocities as you would any two objects given their relative direction of motion. This estimated velocity wouldn't be anything real per se, but it would be the time rate of change of distance between the two photons as measured by an observer.

strangerep
Nope, this doesn't work. You've only avoided talking about a reference frame, rather than actually avoiding it altogether.
Of course I've avoided talking about a "reference frame of a photon". That's my point: by re-expressing the statement in terms of limits, things become better defined. I avoided the ill-defined concept of "reference frame of a photon" by using something else that is well-defined, just as in ordinary differential calculus, one replaces the ill-defined division by ##h=0## by the limit of a well-defined ratio.

All the answers are pretty convincing!

Things which I learnt are:
We can't assume an inertial frame with respect to a photon!
So we cannot take any inertial frame into consideration.
I think this is because time and space are not defined for a photon.

So, What should be the proper answer statement for the question in which two photons are travelling in opposite directions?

And if both the photons are travelling in the same direction, then can't we see the motion with respect to either of the photon?
Assume the light wave: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > ; where ~ denotes a single photon.

Now considering any two photons from this wave train, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ >

Now these two photons will always be moving along the same line of path and with equal velocities ie. C.
So now, can't we say that the relative speed between them is ZERO?

Thank You:-)

PeroK
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All the answers are pretty convincing!

Things which I learnt are:
We can't assume an inertial frame with respect to a photon!
So we cannot take any inertial frame into consideration.
I think this is because time and space are not defined for a photon.

So, What should be the proper answer statement for the question in which two photons are travelling in opposite directions?

And if both the photons are travelling in the same direction, then can't we see the motion with respect to either of the photon?
Assume the light wave: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > ; where ~ denotes a single photon.

Now considering any two photons from this wave train, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ >

Now these two photons will always be moving along the same line of path and with equal velocities ie. C.
So now, can't we say that the relative speed between them is ZERO?

Thank You:-)
"Relative" speed has a precise meaning, which is the speed of one object in the reference frame of the other. As speed is distance/time and neither is defined for a photon, then relative speed of any object is not defined for a photon.

"Separation speed" is defined as above and hence is defined within any inertial frame for two photons, and is either 0 or 2c in these cases.

Cutter Ketch and MeAndMyLucidLife
PeroK
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PS in physics you must be careful not to confuse everyday meanings with precise physics meanings. E.g. energy and power.

This also applies to words like "relative", which has a precise meaning in this case.

MeAndMyLucidLife
strangerep
We can't assume an inertial frame with respect to a photon!
So we cannot take any inertial frame into consideration.
I think this is because time and space are not defined for a photon.
That last statement is incorrect. Rather, it's because a photon has no rest frame, as others have explained.

So, What should be the proper answer statement for the question in which two photons are travelling in opposite directions?
[...]
these two photons will always be moving along the same line of path and with equal velocities ie. C.
So now, can't we say that the relative speed between them is ZERO?
Keep in mind that this is a homework thread. Therefore we're not supposed to simply give you the answers, but rather to offer hints to help you reach the answer for yourself.

Indeed, I think I might have given too much of a hint in my post #12, which already contains both the answers ,if you study and apply it. I.e., don't guess, and don't try to reason "intuitively". Use the math of special relativity, in this case the velocity addition formula.

MeAndMyLucidLife
"Relative" speed has a precise meaning, which is the speed of one object in the reference frame of the other. As speed is distance/time and neither is defined for a photon, then relative speed of any object is not defined for a photon.

"Separation speed" is defined as above and hence is defined within any inertial frame for two photons, and is either 0 or 2c in these cases.
Now i got it!
Thanks for making my concepts and terminology clearer...
Thanks a lot!

jbriggs444
Homework Helper
Of course I've avoided talking about a "reference frame of a photon". That's my point: by re-expressing the statement in terms of limits, things become better defined. I avoided the ill-defined concept of "reference frame of a photon" by using something else that is well-defined, just as in ordinary differential calculus, one replaces the ill-defined division by ##h=0## by the limit of a well-defined ratio.
The limit of the thing is not always the thing at the limit. Especially when the thing at the limit is undefined.

kimbyd
Gold Member
Of course I've avoided talking about a "reference frame of a photon". That's my point: by re-expressing the statement in terms of limits, things become better defined. I avoided the ill-defined concept of "reference frame of a photon" by using something else that is well-defined, just as in ordinary differential calculus, one replaces the ill-defined division by ##h=0## by the limit of a well-defined ratio.
It still doesn't work because it leads to an inconsistency in the case where the two photons are traveling in the same direction.

strangerep
It still doesn't work because it leads to an inconsistency in the case where the two photons are traveling in the same direction.
What inconsistency?

strangerep
The limit of the thing is not always the thing at the limit. Especially when the thing at the limit is undefined.
I know that. That's why I was re-interpreting the original question in terms of approaching a limit rather than "at the limit point".

kimbyd
Gold Member
What inconsistency?
The observed speed of the other photon would be zero. It's not possible for photons to have a speed anything other than c, for any observer.

Really, there are only two ways to interpret this problem:
2. It doesn't mean that we should take the difference in speed in a photon's reference frame at all, but by an outside observer.

Edit:
Also, you are using a reference frame of a photon. You can't do that, because there's no time coordinate, and thus no way to measure velocity at all. You avoid the word reference frame, but that doesn't change the math you're using.

strangerep