# I Variance of the EM wave equation under Galilean transformation

1. Jun 8, 2017

### Pushoam

• Two very similar threads were merged. Some posts may seem redundant
For using Galilean transformation, I have to assume that speed of light w.r.t. ether frame is c.
W.r.t. ether frame,
E = E0 eik(x-ct)

W.r.t. S' frame which is moving with speed v along the direction of propagation of light,
E' = E0 eik(x'-c't')
Under Galilean transformation,
x' = x-vt,
t' = t,
c' = c -v
So, (x'-c't') = x -ct
thus, E' = E
i.e. under the Galilean transformation, em wave equation is invariant.

Is this correct?
I am asking it here instead of creating a new thread, because I think the topic is similar.
If it is not allowed, sorry for it.

2. Jun 8, 2017

### Staff: Mentor

No, it isn't, because the speed of the wave changes. You go from a wave moving at speed $c$ to a wave moving at speed $c - v$.

(You are also assuming that the wavenumber $k$ does not change. I'm not sure that assumption is justified.)

3. Jun 8, 2017

### Pushoam

Then, x also changes from x to x - vt. And the two changes get canceled in the wave equation.
Assuming that the frequency of an em wave depends on the source generating em wave.
So, the frequency will remain same w.r.t. both reference frames.
Then, ω=ω',
k ≠k'.
k = ω/c,
k' = ω/(c -v)
E = E0ei(kx-ωt) = E0eiω(x/c - t)
E' = E'0eiω((x-vt)/(c -v) - t)
E' ≠ E
Is this correct?
,

4. Jun 9, 2017

### Pushoam

Any em wave could be written as a combination of plane waves.
So, if I show the plane wave solution of em wave equation to be variant under Galilean transformation and then using superposition principle, I can say that any solution of em wave equation is variant under Galilean transformation.
Will this be a proof of variance of em wave equation under Galilean transformation?

For using Galilean transformation, I have to assume that speed of light w.r.t. ether frame is c.
W.r.t. ether frame,
E = E0 eik(x-ct)

W.r.t. S' frame which is moving with speed v along the direction of propagation of light,
E' = E0 eik'(x'-c't')

Under Galilean transformation,
x' = x-vt,
t' = t,
c' = c -v
So, (x'-c't') = x -ct

Assuming that the frequency of an em wave depends on the source generating em wave.
So, the frequency will remain same w.r.t. both reference frames.
Then, ω=ω',
k ≠k'.
k = ω/c,
k' = ω/(c -v)
E = E0 eiω(x/c - t)
E' = E'0 eiω((x-vt)/(c -v) - t)
E' ≠ E
Is this correct?

5. Jun 9, 2017

### Staff: Mentor

The question isn't if E' is different from E. The question is if it satisfies Maxwell's equations in both frames.

6. Jun 9, 2017

### Pushoam

Do you mean that I have to check whether E satisfies Maxwell's equations(or Maxwell's wave equations ?) in ether frame and E' satisfies Maxwell's equations in S' frame?
They have to (as I have taken them as solutions of Maxwell's wave equations in their respective frames).

7. Jun 9, 2017

Yes (and B)

8. Jun 9, 2017

### Pushoam

When you say Maxwell's equations, you mean Maxwell's equations of electrodynamics or Maxwell's wave equation?

9. Jun 9, 2017

### Staff: Mentor

This one. When I say "Maxwell's equations" that is always what I mean.

Last edited: Jun 9, 2017
10. Jun 9, 2017

### Staff: Mentor

Both. The wave equation is derived from the electrodynamic ones. That's why we say that a plane wave moving with speed c is a solution of Maxwell's equations and a plane wave moving at any other speed is not.

The wave equation, from Maxwell's equations, is $(c^2\frac{\partial^2}{\partial{x}^2}-\frac{\partial^2}{\partial{t}^2})E=0$.

$E=E_0e^{ik(x-ct)}$ is a solution to that equation. $E=E_0e^{ik(x-c't)}$ where $c'=c-v$ is not.

11. Jun 10, 2017

### vanhees71

The problem is to define (!) Galileo transformations on the quantities in Maxwell's equations such that these equations stay form-invariant. That turns out to be impossible, and that's how Einstein found the modern interpretation of Lorentz transformations, i.e., that observations clearly show that the Galilei-Newton spacetime is inaccurate and Einstein-Minkoski spacetime is a much better approximation, which is only invalidated by gravity, which needs General Relativity with its completely changed view of spacetime as a dynamical entity.