What happens when two light beams approach each other.

[albertrichardf]

Hello.
Suppose there are two objects that are traveling at the speed of light, and that the observer is in an inertial frame of reference.
These two objects are say, a light-second apart, and are approaching each other. From the frame of the observer, each object covers half a light-second before meeting each other. However, problems arise if a transformation is used.
Since any inertial reference frame is as good as any other, it should be possible to transform the measurements from the observer's frame to that of one of the light speed frames.
But then, the v^2/c^2 term in the transformation becomes one, and gamma becomes indeterminate. Because of that it becomes unclear what would happen in the c-frame.

This problem arises any time you have an object traveling at light speed approaching you, even if you are not traveling at light speed yourself. For instance, if there was a train approaching a beam of light, because you don't have to deal with gamma in the train's frame, you could calculate when you would meet the light beam, but from the light beam's point of view you can't.

Could this arise from the nature of traveling at c? The transformations then do not apply because from such an object's point of view, the spacetime distance is always zero.
Thank you for the answers.

 

[robphy]

The Lorentz Transformation preserves the causal character of vectors: timelike to timelike, spacelike to spacelike, lightlike to lightlike.
So timelike-to-lightlike is not achievable by the Lorentz Transformation... for any relative velocity in the transformation.

 

[albertrichardf]

Right, so essentially, I am taking a timelike and spacelike vector and transformating it into a lightlike vector, which cannot be done through the Lorentz transformation?

 

[Nugatory]

Since any inertial reference frame is as good as any other, it should be possible to transform the measurements from the observer's frame to that of one of the light speed frames.

There is no such thing as a lightspeed frame. That would be a frame in which the light was at rest and the observer was approaching at speed $c$; but light is moving at speed $c$ in all frames, so it can't be at rest in any frame.

This comes up so often that it's a FAQ: https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

 

[albertrichardf]

There is no such thing as a lightspeed frame. That would be a frame in which the light was at rest and the observer was approaching at speed $c$; but light is moving at speed $c$ in all frames, so it can't be at rest in any frame.

This comes up so often that it's a FAQ: https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

Oh right. that makes sense. I forgot that light would still travel at c no matter what the inertial frame was.
As an aside I did check the FAQ, but I either dismissed that one or did not see it

 

[Battlemage!]

Just for laughs and giggles:

\frac{c + c}{1+\frac{c^2}{c^2}} = \frac{2c}{2} = c

Of course I know this is not valid since these equations apply to inertial observers, and light has no inertial reference frame where it is at rest. In fact it becomes obvious it's not valid when you switch the signs:

\frac{c - c}{1-\frac{c^2}{c^2}} = \frac{0}{0} = ~~:(

In fact next time I see someone try that (I did it myself earlier in the year) I'll just post the second one, with the minus sign.

 

[albertrichardf]

Just for laughs and giggles:

\frac{c + c}{1+\frac{c^2}{c^2}} = \frac{2c}{2} = c

Of course I know this is not valid since these equations apply to inertial observers, and light has no inertial reference frame where it is at rest. In fact it becomes obvious it's not valid when you switch the signs:

\frac{c - c}{1-\frac{c^2}{c^2}} = \frac{0}{0} = ~~:(

In fact next time I see someone try that (I did it myself earlier in the year) I'll just post the second one, with the minus sign.

Yeah that is actually what I tried and that is from what came my original question. I just found out now, from the two previous replies that the equation does not work for light