Conclusion about wave propagation in SR given L dot S = N = L' dot S'

[PhDeezNutz]

Homework Statement

I'm having trouble rendering latex on this website so I attached a typed up picture of my question and attempt at a solution (See below)

Relevant Equations

(See below)

The problem I am having is "What can you conclude about wave prorogation in SR given the results?". The best I can come up with is that the number of wave planes N crossing a section of spacetime in either frame is the same. The section may be bigger or smaller depending on which frame you're in but there's always N waves crossing regardless of which frame you're in.

I'm using intuitive 3-space notions of flux and applying it to minkowski spacetime and that may very well be a mistake.

Any hints on how to interpret L dot delta s = L' dot \delta s' = N?

Thank you in advanced for any help.

 

[PhDeezNutz]

I think we can conclude (assuming S' is obtained from S through a Lorentz boost in the x-direction)

$L_0 c \Delta t = L'_0 \Delta t' , L_1 \Delta x = L'_1 \Delta x', L_2 \Delta y = L'_2 \Delta y', L_3 \Delta z = L'_3 \Delta z'$

Concentrating on the last 3 statements I think this result says that the component of the wave vector along the observation direction is the same in both frames. More details to follow.

 

[PhDeezNutz]

My previous post is definitely wrong. I think the answer is trivial...we were asked to show that L' can be obtained from L via a Lorentz transform and then what we could conclude from that. I think the answer is obvious (and kind of trivial)...only the component of the wave vector along the boost direction changes. That's the definition of a lorentz transform.

unless I was supposed to interpret L \dot S = L' \dot S' = N instead. I'm going to email my instructor and ask about this.

 

[PhDeezNutz]

I asked my instructor said "the response is that the equation keeps the same form in any coordinate system. So it is written in covariant form including invariance to Lorentz transformations".

I'll have to think about all that.

 

[PhDeezNutz]

I guess the interpretation is that L dot \delta S is a true Lorentz scalar.