jason12345
- 108
- 0
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 (\omega/c,\ \textbf{k} ) where \textbf{k} = 2\pi/ \boldsymbol{\lambda},\ \textbf{u} 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.
\omega'\ =\ \gamma\omega(1-v/u\ \cos\theta),\ u'_{y'} = u_{y}/\gamma (1-vu_x/c^2) are standard results and substituting into
k'_{y'} = 2 \pi/\lambda'_{ x'}
= \omega'/ u'_{y'}
= \gamma\omega(1-v/u\ \cos\theta)\gamma (1-vu_x/c^2)/u_{y}
= \omega/u_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)
= k_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)
Since k_{y}=k'_{y'} then
(1-vu_{x}/u^2)(1-vu_x/c^2) = 1 - v^2/c^2
which isn't generally true.
Where have I gone wrong in my working?
Thanks in advance.
The wave 4-vector is defined as (\omega/c,\ \textbf{k} ) where \textbf{k} = 2\pi/ \boldsymbol{\lambda},\ \textbf{u} 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.
\omega'\ =\ \gamma\omega(1-v/u\ \cos\theta),\ u'_{y'} = u_{y}/\gamma (1-vu_x/c^2) are standard results and substituting into
k'_{y'} = 2 \pi/\lambda'_{ x'}
= \omega'/ u'_{y'}
= \gamma\omega(1-v/u\ \cos\theta)\gamma (1-vu_x/c^2)/u_{y}
= \omega/u_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)
= k_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)
Since k_{y}=k'_{y'} then
(1-vu_{x}/u^2)(1-vu_x/c^2) = 1 - v^2/c^2
which isn't generally true.
Where have I gone wrong in my working?
Thanks in advance.