Transformation of k_y in the wave 4-vector

In summary, the conversation discusses the transformation of k_y in the wave 4-vector and its relation to the phase velocity of a plane wave. The speaker realizes their mistake in assuming that the wave vector and phase velocity are always pointing in the same direction in all frames, and this leads to the concept of "relativistically induced optical anisotropy." The paper "Is the phase of plane waves a frame-independent quantity?" is referenced to explain this concept and the correct Lorentz-invariant expression for the phase of waves. The speaker asks if anyone has encountered this effect before.
  • #1
jason12345
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I'm checking how k_y in the wave 4-vector transforms, but not getting what I expect:

The wave 4-vector is defined as [tex](\omega/c,\ \textbf{k} )[/tex] where [tex]\textbf{k} = 2\pi/ \boldsymbol{\lambda},\ \textbf{u}[/tex] is the velocity of propagation of the plane wave

Let s' travel, as usual, along the x-axis of s with velocity v, and k make an angle theta wrt x axis.

[tex]\omega'\ =\ \gamma\omega(1-v/u\ \cos\theta),\ u'_{y'} = u_{y}/\gamma (1-vu_x/c^2)[/tex] are standard results and substituting into

[tex]k'_{y'} = 2 \pi/\lambda'_{ x'}[/tex]

[tex]= \omega'/ u'_{y'}[/tex]

[tex]= \gamma\omega(1-v/u\ \cos\theta)\gamma (1-vu_x/c^2)/u_{y}[/tex]

[tex] = \omega/u_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)[/tex]

[tex] = k_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)[/tex]

Since [tex] k_{y}=k'_{y'}[/tex] then

[tex] (1-vu_{x}/u^2)(1-vu_x/c^2) = 1 - v^2/c^2[/tex]

which isn't generally true.

Where have I gone wrong in my working?

Thanks in advance.
 
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  • #2
I've found where I've gone wrong. I assumed that the wave vector and phase velocity are pointing in the same direction in all frames, when in fact they don't, and the effect is called "relativistically induced optical anisotropy". It means that the wave 4 vector in its usual form is a 4 vector and the phase is a Lorentz scalar only for a phase velocity equal to c, as outlined in this paper:

Is the phase of plane waves a frame-independent quantity?

The invariance of the phase of plane waves among inertial frames is investigated in some details. The reason that eventually led the author of a recent EPL letter [EPL \textbf{79}, 1006 (2007)] to a spurious conclusion of the non-invariance of the phase of waves has been identified -- it is the ignorance of the effect of relativistically-induced optical anisotropy in the analysis of the problem. It is argued that the Lorentz-invariant expression for the phase of waves should be taken in the form $\Phi=\mathbf{k\cdot r}-\mathbf{k\cdot u}/c$, instead of the widely-used expression $\Phi=\mathbf{k\cdot r}-\omega t$ which has a limited validity.

http://arxiv.org/abs/0801.3149

Had anyone come across this effect before?

Regards
 

FAQ: Transformation of k_y in the wave 4-vector

What is the significance of the transformation of ky in the wave 4-vector?

The transformation of ky in the wave 4-vector is significant because it allows us to understand how the momentum and energy of a particle changes when observed from different frames of reference. This transformation is a crucial concept in special relativity and helps us to reconcile the differences between classical mechanics and the laws of physics at high speeds.

How is the transformation of ky calculated in the wave 4-vector?

The transformation of ky is calculated using the Lorentz transformation equations, which take into account the relative velocity between the two frames of reference. The transformation of ky is determined by the ratio of the velocity of the particle in the original frame to the velocity of the particle in the new frame.

What is the difference between the transformation of ky and the transformation of kx in the wave 4-vector?

The transformation of ky and the transformation of kx differ in that they correspond to different components of the wave vector. Ky represents the y-component of the wave vector, while kx represents the x-component. The transformation equations for these components are similar, but take into account the different orientations of the frames of reference.

How does the transformation of ky affect the frequency of a wave?

The transformation of ky does not directly affect the frequency of a wave. However, it does affect the wavelength of the wave, which in turn affects the frequency through the wave equation (frequency = speed/wavelength). As the frame of reference changes, the wavelength changes, and thus the frequency may appear different in different frames.

Can the transformation of ky be applied to all types of waves?

Yes, the transformation of ky can be applied to all types of waves, including electromagnetic waves, sound waves, and matter waves. The concept of a wave vector and its transformation is fundamental to the understanding of wave phenomena in physics and can be applied to any type of wave.

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