Distance between two light pulses

[member 743765]

If a source emits a light pulse then waited one second and sent another pulse does the distance between the two pulses remain constant ? If yes is that mean their relative speed is zero? But why when we use lorentz transformation their relative speeds gives us zero over zero but if they travel in opposite directions their relative speed is c which agrees with special relativity?

 

[Ibix]

If a source emits a light pulse then waited one second and sent another pulse does the distance between the two pulses remain constant ?

Yes, assuming they're going in the same direction.

If yes is that mean their relative speed is zero?

No. It means that the difference in their coordinate velocities in your chosen frame is zero, which means that it will be zero in all frames. I usually call this quantity "closing rate" or "separation rate", although I don't think there's a universally accepted term.

The relative velocity of two objects is the velocity measured by one of the other in the rest frame of the first one. Light does not have a rest frame, so "velocity relative to light" isn't defined.

Note that relative velocity and separation rate are always equal in Newtonian physics, but not relativity.

But why when we use lorentz transformation their relative speeds gives us zero over zero but if they travel in opposite directions their relative speed is c which agrees with special relativity?

Lorentz transforms to a frame with speed $v=\pm c$ are not defined. Thus derived formulae such as the velocity transform (which is what I think you are using here) are not valid. That's why you get contradictory answers when you plug in $+c$ and $-c$.

 

[Dale]

But why when we use lorentz transformation their relative speeds gives us zero over zero but if they travel in opposite directions their relative speed is c which agrees with special relativity?

It should be undefined regardless of the direction. Light doesn’t have an inertial rest frame. You may want to check your math with the Lorentz transform for the opposite direction case

 

[PeterDonis]

Yes, assuming they're going in the same direction.

And that we are in flat spacetime (i.e., no gravitating masses are present) and are using an inertial frame.

 

[Dale]

But why when we use lorentz transformation their relative speeds gives us zero over zero but if they travel in opposite directions their relative speed is c which agrees with special relativity?

To follow up on this. The Lorentz transform to an inertial frame moving at velocity $v$ with respect to the unprimed inertial frame is $$ct'=\frac{c t- v x/c}{\sqrt{1-v^2/c^2}}$$$$x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}$$ For an object traveling at velocity $u$ in the unprimed frame we get $x=ut$ which gives $$ct'=\frac{c t- v u t/c}{\sqrt{1-v^2/c^2}}$$$$x'=\frac{u t-vt}{\sqrt{1-v^2/c^2}}$$

So, for the "perspective of light" we would have $v=c$ which gives $$ct'=\frac{c t- u t}{\sqrt{1-c^2/c^2}}=\frac{c t- u t}{0}=undefined$$$$x'=\frac{u t-ct}{\sqrt{1-c^2/c^2}}=\frac{u t-ct}{0}=undefined$$ Regardless of $u$