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I have 2 coordinate systems which move along ##x,x'## axis. I have derived a Lorentz transformation for an ##x## component of momentum, which is one part of an 4-momentum vector ##p_\mu##. This is my derivation:
[itex]
\scriptsize
\begin{split}
p_x &= mv_x \gamma(v_x)\\
p_x &= \frac{m (v_x'+u)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{1 - \left(v_x' + u \right)^2 / c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2}} \\
p_x &= \frac{m (v_x'+u) \left( 1+ v_x' \frac{u}{c^2} \right)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} \right) - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + 2 v_x'u + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{1 + v_x'^2 \frac{u^2}{c^4} - \frac{v_x'^2}{c^2} - \frac{u^2}{c^2} }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left(1 - \frac{u^2}{c^2}\right) \left(1-\frac{v_x'^2}{c^2} \right)}} \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + mu \gamma(v_x') \right] \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + \frac{mc^2 \gamma(v_x') u}{c^2} \right] \\
p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right]
\end{split}
[/itex]
I tried to derive Lorentz transformation for momentum also in ##y## direction, but i can't seem to get relation ##p_y=p_y'## because in the end i can't get rid of ##2v_x'\frac{u}{c^2}## and ##\frac{v_y'^2}{c^2}##. Here is my attempt.
[itex]
\scriptsize
\begin{split}
p_y &= m v_y \gamma(v_y)\\
p_y &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - v_y'^2/c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2}}\\
p_y &= \frac{m v_y' \left( 1 + v_x' \frac{u}{c^2} \right)^2}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4}\right) - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + v_x'^2 \frac{u^2}{c^2} - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} - \frac{v_y'^2}{c^2}}}\\
\end{split}
[/itex]
This is where it ends for me and I would need someone to point me the way and show me, how i can i get ##p_y = p_y'##. I haven't seen any derivation like this (for ##y## component of momentum) on the internet.
Thank you.
[itex]
\scriptsize
\begin{split}
p_x &= mv_x \gamma(v_x)\\
p_x &= \frac{m (v_x'+u)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{1 - \left(v_x' + u \right)^2 / c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2}} \\
p_x &= \frac{m (v_x'+u) \left( 1+ v_x' \frac{u}{c^2} \right)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} \right) - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + 2 v_x'u + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{1 + v_x'^2 \frac{u^2}{c^4} - \frac{v_x'^2}{c^2} - \frac{u^2}{c^2} }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left(1 - \frac{u^2}{c^2}\right) \left(1-\frac{v_x'^2}{c^2} \right)}} \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + mu \gamma(v_x') \right] \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + \frac{mc^2 \gamma(v_x') u}{c^2} \right] \\
p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right]
\end{split}
[/itex]
I tried to derive Lorentz transformation for momentum also in ##y## direction, but i can't seem to get relation ##p_y=p_y'## because in the end i can't get rid of ##2v_x'\frac{u}{c^2}## and ##\frac{v_y'^2}{c^2}##. Here is my attempt.
[itex]
\scriptsize
\begin{split}
p_y &= m v_y \gamma(v_y)\\
p_y &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - v_y'^2/c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2}}\\
p_y &= \frac{m v_y' \left( 1 + v_x' \frac{u}{c^2} \right)^2}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4}\right) - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + v_x'^2 \frac{u^2}{c^2} - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} - \frac{v_y'^2}{c^2}}}\\
\end{split}
[/itex]
This is where it ends for me and I would need someone to point me the way and show me, how i can i get ##p_y = p_y'##. I haven't seen any derivation like this (for ##y## component of momentum) on the internet.
Thank you.