Light Speed Question: How Can 2 Light Particles Approach Each Other at 1xC?

[fpayam82]

there's something I don't understand.

Assume we have two light particles that approach each other. (obviously with light speed)
Logically since each of them are traveling with the speed of light they should approch to each other with 2xC, but according to relativity theory they approach with1xC.

How is this possible ?

 

[Elroch]

Actually, you can't really define the relative speed of one photon to the other, because it is not in an inertial frame. In every inertial frame, both photons appear to travel at speed c.

 

[fpayam82]

ok, but you did understand my question.
I you turn your flash light towards me and I do towards you how fast do the two light beems reach each other ?
1xC or 2xC ? I guess 1xC, but why ?

 

[smithhs]

You see the distance between the two light beams to be decreasing at a speed c: If the distance between the photons if x, then you see dx/dt=c.

In special relativity the velocity addition formula isn't just v1+v2=v3

 

[Matterwave]

You see the distance between the two light beams to be decreasing at a speed c: If the distance between the photons if x, then you see dx/dt=c.

In special relativity the velocity addition formula isn't just v1+v2=v3

You would see the distance between the beams decreasing at speed 2c. If you put a particle in between the beams, you would see one beam of light approach that particle at c, and the other approach from the opposite direction at c, and therefore the distance between the two light beams must be decreasing at 2c (in YOUR frame of reference).

The OP is asking for "in the frame of reference of a photon", which doesn't exist.

A better question would be if 2 elementary particles are approaching each other at .9999c, what speed would one see the other move at. The answer is very close to c, and not close to 2c.

 

[fpayam82]

I know in special relativity the velocity addition formula isn't just v1+v2=v3 .
But what I fail to understand is why ??
How come that V1+V2 is not V3 if V1 and V2 are traveling at light speed.

 

[harrylin]

there's something I don't understand.

Assume we have two light particles that approach each other. (obviously with light speed)
Logically since each of them are traveling with the speed of light they should approch to each other with 2xC, but according to relativity theory they approach with1xC.

How is this possible ?

Your logic is perfectly correct. The essential point of relativity theory is that the speed of light is c as determined with any standard reference system. Consequently, according to relativity theory two light rays in opposite direction will approach each other at 2c, as measured with any standard reference system. And this was well understood from the very start; for example Einstein phrased it as follows for a light ray relative to a moving object:

the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v

- http://www.fourmilab.ch/etexts/einstein/specrel/www/

Note that light is not really "particles" as light has no rest mass and relativity models light as waves; thus "wave packets" comes closer to the standard meaning of words.

 

[harrylin]

I know in special relativity the velocity addition formula isn't just v1+v2=v3 .
But what I fail to understand is why ??
How come that V1+V2 is not V3 if V1 and V2 are traveling at light speed.

When using a single reference system, the normal addition rule applies. However, when you add velocities that are measured with different reference systems then you need the "velocity addition formula" - which is a misnomer, for it is in fact a velocity transformation formula.
It's like with financial transactions: as long as you use the same currency you may use the normal addition rule, but when you add dollars to euros then you should use a currency converter equation.