I have a little trouble in demonstrating that the group velocity of an e.m. wave transform (in special relativity, by passing from an inertial frame to another) like the velocity of a particle with mass.(adsbygoogle = window.adsbygoogle || []).push({});

So, [tex] \left( \frac{w}{c}, k_x, k_y, k_z\right) = \left( \frac{w}{c}, \vec k \ \right)[/tex] is a 4-vector, where [tex]w[/tex] is [tex]2\pi\nu[/tex] and [tex]\vec k[/tex] is the wave vector. So, if another inertial frame moves with velocity [tex]v[/tex] in the direction of the [tex]x[/tex] axis, we have

[tex]\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}[/tex]

[tex]dw' = \gamma\left(dw-v\cdot dk_x\right)[/tex]

[tex]dk_x' = \gamma\left(dk_x-\frac{v}{c^2}dw\right)[/tex]

[tex]dk_y' = dk_y[/tex]

Let be [tex]u_x = \frac{dw}{dk_x}[/tex], [tex]u_y = \frac{dw}{dk_y}[/tex].

So, I find [tex]u_x' = \frac{dw'}{dk_x'} = \frac{u_x-v}{1-\frac{v\cdot u_x}{c^2}}[/tex] (correct, because it's the same transformation for the velocity in [tex]x[/tex] of a particle with mass), and

[tex]u_y' = \frac{dw'}{dk_y'} = \gamma\frac{dw-v\cdot dk_x}{dk_y} = \gamma\left(u_y-\frac{dk_x}{dk_y}\cdot v\right) = \gamma\left( u_y-v\cdot\frac{dw}{dk_y}\cdot\frac{1}{\frac{dw}{dk_x}}\right) = \gamma\left(u_y-v\cdot u_y\cdot\frac{1}{u_x}\right) = \gamma\cdot u_y\left(1-\frac{v}{u_x}\right)[/tex]

But this is different from the relationship for particle with mass velocity, that is

[tex]u_y' = \frac{u_y}{\gamma\cdot \left(1-v\cdot\frac{u_x}{c^2}\right)}[/tex]

Where's the problem with this derivation? How can I find the right relationship?

Maybe the problem is with the fact that I should do the partial derivation

[tex]u = \frac{\partial w}{\partial \vec k}[/tex]

(I know the transform must be the same for group velocity of an e.m. wave AND for velocity of particles with mass). Please, if you can, show the passages of the derivation.

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# Little problem with group velocity

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