- #1
jason12345
- 109
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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.
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.