Lorentz transf. of a spherical wave in Euclidean space

[Rumo]

This thread is not about the lorentz invariance of the wave equation: $$\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}-\Delta \Phi = 0$$

It is about an interesting feature of a standing spherical wave:
$$A\frac{\sin(kr)}{r}\cos(wt)$$

It still solves the wave equation above, when it is boosted in the following way:
$$z' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(z-vt)$$
and
$$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(t-\frac{vz}{c^2})$$
This means that a transformation, which looks like the lorentz transformation, is needed for a moving standing spherical wave to still solve the wave equation. It is important to notice, that this takes place in an Euclidean space!

Source: http://arxiv.org/abs/1408.6195

I would like to know, what you think about this. Including the paper above. The paper was written for educational purposes, because it shows, that the lorentz transformation can arise in an Euclidean space.

 

[Rumo]

I calculated this with Mathematica. I would have uploaded the notebook, but I can't.
Hence I made a screenshot. You can calculate this easily with any program of your choice.
Just replace c with w/k.

 

[vanhees71]

I only don't understand the first sentence in #1. It's of course all about the Lorentz transformation, which is a symmetry of the wave equation.

[OT for the admins] Why can't one upload an nb (Mathematica notebook) file? Wouldn't this be very nice for those among us, who have Mathematica at hand? I never thought to upload a Mathematica notebook so far, but it's a nice idea, isn't it?

 

[Rumo]

I only don't understand the first sentence in #1. It's of course all about the Lorentz transformation, which is a symmetry of the wave equation.

Thanks for your reply!
Isn't the lorentz transformation transforming the spatial and time coordinates? Hence, I would use the chain rule to show the invariance. The Φ would not be important at all, just a solution to the wave equation. It would describe how Φ would look like from the point of view of a boosted observer!?

But in this case the spatial and time coordinates are not transformed.

$$\Phi = A\frac{\sin(k\sqrt{x^2+y^2+(\gamma(z-vt))^2})}{\sqrt{x^2+y^2+(\gamma(z-vt))^2}}\cos(w\gamma(t-\frac{vz}{c^2})$$
is a solution of the wave equation. v is the velocity of the 'standing' spherical wave, relative to the e.g. ideal gas.

 

[vanhees71]

Both space and time are transformed. The spherical standing wave carries 0 momentum in the old frame and a momentum in z-direction in the new one.

 

[Rumo]

Both space and time are transformed. The spherical standing wave carries 0 momentum in the old frame and a momentum in z-direction in the new one.

But in this case, it is not a coordinate transformation!
It is a moving spherical standing wave.
I still use the same coordinates to differentiate.
x, y, z, t. not x, y, z', t'!