Hello,

I am having a problem taking the constancy of the speed of light as a fact. I know that this is widely accepted but cannot get my head around it. If two photons speed away from eachother at c then they must be going faster than c relative to one another. It seems to me that the warping of spacetime to account for what we observe is an easy way out. Can anyone give me an anology to help me understand.

Thanks in advance.

ash

If 2 bodies move away from each othe at relativistic velocities u and v, their relative velocity, is (u+v) / (1 + uv/c^2) by velocity addition for relativistic speeds derived from Lorentz transformations.

if u and v are conventional speeds then it is simply (u+v)/(1+0) = u+v.

suppose the two bodies being photons then u = v = c.

we get their relative velocity as 2c / (1+c^2/c^2) = c.

The relative velocity of photons moving away from each other is always c.

Thank you for your response. I do understand the formula but isn't the formula just a construct of our observations. I guess I am asking for a more qualitative explanation. If two photons are going in the opposite direction at c. How can their relative velocity be c?

Thanks again.

Because when you're travelling at C, time stands still, thus they must be moving away from each other at the speed of light.

yayacatfight:

<i>I do understand the formula but isn't the formula just a construct of our observations.</i>

Yes. All established physics is just a construct based on observations. What more do you want?

<i>If two photons are going in the opposite direction at c. How can their relative velocity be c?</i>

Relative velocity is what one photon sees when it looks at the other one. The formula given before lets you calculate that.

The relativistic addition of velocities does not apply to photons.

ryans,

The relativistic addition of velocities does not apply to photons.


Are Yyou sure you said what you wanted to say or how it is going to be interpreted?

If you pick the statement apart it can be interpreted correctly but on a quick read and by someone asking the question asked, I would expect it to be shear confusion.

I think what you wanted to express is that "Actual velocity must be calculated using the Velocity Addition Formula".

Photons don't emit photons.

ryans:

Please explain.

photons don t emit photons, this is true. but what do you mean, the relativistic velocity addition formula doesn t apply to photons?

for one thing, the relativistic addition formula is for finding the velocity in a particular rest frame. it is difficult to make any meaning out of the idea of a rest frame of a photon. usually it is said that photons simply do not have rest frames. is this what you mean?

Yes Lethe, this is exactly what I mean. The velocity addition formulas don't apply to photons in the sense that they deifine the velocity addition formula's via the axiom

The speed of light is constant for all observers.

You cannot construct a frame of reference for which the photon is motionless, and thus you cannot say for example, how fast is say photon A receeding from photon B, in B's frame of reference.

But here's a good one for you. An observer who is say standing above a decay event which produces 2 photons, will measure the 2 photons to be moving away from each other at 2c.

That is sure to stuff some people up, and also definately bring questions from Mac and Tom.

ryans,

That was actually the sort of situation I was referring to. Two photons moving in opposite directions (according to Relativity) is not 2c but c per the VAF.

The rest frame is the observer not either of the photons.

ryans,

Originally posted by ryans
But here's a good one for you. An observer who is say standing above a decay event which produces 2 photons, will measure the 2 photons to be moving away from each other at 2c.


The original question is same. photons moving away from each other in opposite direction by, could be as simple as, partial reflection of a light pulse or pulse emitted by a light source filtered thro exactly opposite directions.

As for the observer in rest frame their relative velocity is between 0 to 2c depending on the directions of the photons. it can't be more or less than that range. how the constancy of c is violated.?

Besides the orginal post asked for an anology and i took 2 bodies moving away from each other at u and v. i resorted to VAF for velocities < c and then for what if u=v=c, ofcourse using VAF at c is in effect circularly referring to the basic axiom of constant c, as the conclusion. still i feel, though that is logically incorrect, factually consistent.

As lethe and you said in another post , Maxwell's equations shows that EM waves need to probagate at constant c, may be a more appropriate answer for the original question of constancy of c.

All I am saying is the relativistic addition of velocities is not valid for light(photons). Relativity is not violated by the fact that the 2 photons move apart from each other at 2c, as all observers must see the light moving with velocity c. You cannot construct a reference frame which is at rest w.r.t. either photon. Any frame of reference will see the velocity at which the photons receed from one another as 2c. However, the important thing to note is that information is still propogated at velocity c, as the photons carry away information about the location that they were created.

P.S. If anybody has any questions regarding entanglement from the process outlined, please do raise them here as they will detract from the arguement. Post them in a new thread.

Fair enough. I agree with what you've said, ryans.

Of course you agree, I am right. But Mac doesn't seem to get the point.

ryans,

Relativity is not violated by the fact that the 2 photons move apart from each other at 2c

Doesn't Relativity state that nothing can move faster than c.?

ash

Originally posted by ryans
All I am saying is the relativistic addition of velocities is not valid for light(photons). Relativity is not violated by the fact that the 2 photons move apart from each other at 2c, as all observers must see the light moving with velocity c. You cannot construct a reference frame which is at rest w.r.t. either photon. Any frame of reference will see the velocity at which the photons receed from one another as 2c. However, the important thing to note is that information is still propogated at velocity c, as the photons carry away information about the location that they were created.

P.S. If anybody has any questions regarding entanglement from the process outlined, please do raise them here as they will detract from the arguement. Post them in a new thread.

wait a second... are you saying that the fact that the photons seperate at 2c is because they do not follow the relativistic velocity addition formula?

the formula says nothing about rates of seperation, and the same statement applies to electrons. if two electrons are going 0.9c, then they are seperating at 1.8c, although neither electron is going at that speed with respect to the other.

in fact, while i see your point, that since the photon doesn t have a rest frame, it is meaningless to ask what is the one photons velocity with respect to the other, there are ways to try and make sense of this question, like suppose instead it is a particle with a vanishingly small mass and take the limit as mass -->0. if you do so, you see that the velocity addition formula works swimmingly, exactly as everneo calculated.

my point, the 2c seperation rate has nothing to do with the velocity addition formula, and certainly does not imply that the formula does not apply to photons.

Originally posted by lethe
wait a second... are you saying that the fact that the photons seperate at 2c is because they do not follow the relativistic velocity addition formula?


Yeah, things are getting confused here. It is true that if one observes two photons moving apart that the distance between them will be measured to grow at a rate of 2c, but as you point out later, this has nothing to do with the velocity addition formula.

Massive objects--which are definitely subject to the formula--exhibit the same kind of phenomenon. If two rocket ships take off from their home planet in opposite directions at a speed of 0.9c, the distance between them will be measured (by someone on the planet) to grow at a rate of 1.8c.

Tom

Have you tryed to use the VAF for 2 photons travelling in opposite directions. Put in c for both velocities and see what you get.

u'=(u+v)/(1+uv/c^2)

Let u=v=c

u'=(c+c)/(1+c^2/c^2)
u'=2c/2
u'=c

Tom

Yes that is correct Tom.

However this stipulates that a photon has a rest frame, which it doesn't.

So yes the VAF gives the correct value for photons, but it is on the presumption that photons have a rest frame. By definition photons do not have a rest frame.

Originally posted by ryans
Yes that is correct Tom.

However this stipulates that a photon has a rest frame, which it doesn't.

So yes the VAF gives the correct value for photons, but it is on the presumption that photons have a rest frame. By definition photons do not have a rest frame.

ryans, i think this is a bit unreasonable.

O.K. I am going off into pedantics. I will accept the VAF holds for photons theoretically, but no rest frame can be ascribed to them. The truth is that velocity calculations for photons come from Maxwell's equations, not relativity, as when time dilation and length contraction formula's are applied to photons, there is trouble to be had. Strictly, relativity is a theory describing the kinematics of massive bodies.

Originally posted by ryans
Have you tryed to use the VAF for 2 photons travelling in opposite directions. Put in c for both velocities and see what you get.
VAF too is relative or what..? ;)

Originally posted by ryans
Yes that is correct Tom.

However this stipulates that a photon has a rest frame, which it doesn't.

So yes the VAF gives the correct value for photons, but it is on the presumption that photons have a rest frame. By definition photons do not have a rest frame.

Err....I am not the one claiming that photons move apart at 2c, and that this has something to do with the velocity addition formula. You are.

You said:


All I am saying is the relativistic addition of velocities is not valid for light(photons). Relativity is not violated by the fact that the 2 photons move apart from each other at 2c, as all observers must see the light moving with velocity c. You cannot construct a reference frame which is at rest w.r.t. either photon. Any frame of reference will see the velocity at which the photons receed from one another as 2c. However, the important thing to note is that information is still propogated at velocity c, as the photons carry away information about the location that they were created.


To that I responded that there is no need to mention the velocity addition formula. When two photons move apart, the rate at which the distance grows is 2c, but that is not the velocity of any object, so there is no need to invoke the velocity addition formula. Indeed, I even pointed out that you can get this pseudo-superluminal effect with massive bodies.

When two photons move apart, the rate at which the distance grows is 2c, but that is not the velocity of any object

That is where I was hung up. Thanks.

Follow up question:

When we talk of a photon moving at c, what are we using as a reference for the velocity. Is it the observer? If so, let's make the other photon travelling in the opposite direction the observer. What does it observe?

yayacatfight,

When two photons move apart, the rate at which the distance grows is 2c, but that is not the velocity of any object

That is where I was hung up. Thanks.

Follow up question:

When we talk of a photon moving at c, what are we using as a reference for the velocity. Is it the observer? If so, let's make the other photon travelling in the opposite direction the observer. What does it observe?


ANS: v = c via the VAF.


ryans,


Why is it I have to keep telling you you are wrong?.:D

I don't disagree with any of the arguements made here. They are all correct. That is why it is called Relativity.

When we talk of a photon moving at c, what are we using as a reference for the velocity. Is it the observer?


You can use yourself as a reference, since you can always regard yourself as stationary.


If so, let's make the other photon travelling in the opposite direction the observer. What does it observe?


This is the exact same situation as I just described. The observer sees one photon moving in one direction at c, and the other photon moving in the opposite direction at c. The distance between the photons is seen to grow at a rate 2c.

Mac I am simplt following your arguement about using relativity to prove relativity (although really this is the onlt way it can be done, it is a self contained theory). I am not disagreeing with anybody, all predictions are correct, but you have to be careful when using the formula's.

One of the axioms of SR IS

c is constant for all observers. Now I agree with you on the point that you cannot prove this axiom using the formula's of relativity which is based on this axiom.

Thus since VAF is derived from this axiom, you cannot prove this axiom by going, look at this axiom, I 've proved it by applying it to light, because the speed of light cannot be determined from SR.

P.S. I agree with all of you, but be careful when applying formula's to situations.

A common example is the misuse of the uncertainty principle.

Originally posted by yayacatfight
Follow up question:

When we talk of a photon moving at c, what are we using as a reference for the velocity. Is it the observer? If so, let's make the other photon travelling in the opposite direction the observer. What does it observe?
back to square 1. people don't agree with that a photon can observe the other photon. in essential a photon is a blind bugger. it seems it could not observe anything else (why anything else? - it is not sure/aware of its own existence) for it has no time, no frame of reference of its own, no mass etc.. just to preserve c to be constant, yeah.:D any further questions, whether its genuine or not, would label you as a crack-pot. best of luck.

Edit :
If you forget for a moment that photon does not have a rest frame then VAF gives the vlocity of one photon observed by the other photon that is moving in opposite direction, as c. hey, yayacatfight, you are preparing the ground for another round of catfight...

We have 2 airplanes flying in opposite direction, each plane is travelling 800 kms per hour, as seen by a stationary observer. The gap, as seen by the observer expands at 1600 kms per hour (hey isn't this is faster than the speed of sound?!).

Will we hear a sonic boom?


OFFCOARSE NOT!!! From the observer point of view (or hear) The soundbarrier is not broken, by any of the planes.

UNLESS THE OBSERVER MOVES AS WELL FAST TOWARDS THE DIRECTION OF AN INCOMING PLANE, He will hear a boom as he passes the plane. But suppose we put the observer in the other plane, the planes move in opposite directions, so even if one plane breaks the soundbarrier, the soundfront will not reach the observer, so even if you think you break causality in nature, nature will find a way to hide it from you:p

Vortexx,

I am not disagreeing with your above post nor am I suggesting an aether but:

Your comment about the production of a sonic boom by two planes moving apart at a collective speed of greater than the speed of sound, and then compared to light:

That conjures up the idea of an aether. Sound is produced between the airplane and a background medium (air). Using this logic for light is suggetive of a background medium (aether).

Don't think that is what you wanted to do.:D

Yes your analogy with the aether may be correct, but may be also severly misinterpreted. A sonic boom is a reult of absolute velocity, namely with a medium which it is travelling through. A sonic boom is not a function of relative velocity.

I forget what I was going to say, I'll get back to you.:)

Originally posted by Vortexx

UNLESS THE OBSERVER MOVES AS WELL FAST TOWARDS THE DIRECTION OF AN INCOMING PLANE, He will hear a boom as he passes the plane.

I don't think this is accurate. A change in the position of the observer will not create a sonic boom around the plane...A sonic boom would be caused by the plane moving through the air faster than the speed of sound. moving the observers frame of reference will not change the relative speed difference between the plane and the air surrounding it.

When talking about a sonic boom and an observer, you actually have two step of observation. The plane is being obsevered by the air molecules, which are in turn being observed by our observer's ear. The difference in speed between the 1 level observer (the air) and the plane much be greater than the speed of sound to create a sonic boom. our observer's ear doesn't come into the equation anywhere.