# 4-momentum and an $y$ component of momentum

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:

$$\scriptsize \begin{split} p_x &= mv_x \gamma(v_x)\\ p_x &= \frac{m (v_x'+u)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{1 - \left(v_x' + u \right)^2 / c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2}} \\ p_x &= \frac{m (v_x'+u) \left( 1+ v_x' \frac{u}{c^2} \right)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\ p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\ p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} \right) - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\ p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + 2 v_x'u + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\ p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - u^2 \right] / c^2 }} \\ p_x &= \frac{mv_x'+mu}{\sqrt{1 + v_x'^2 \frac{u^2}{c^4} - \frac{v_x'^2}{c^2} - \frac{u^2}{c^2} }} \\ p_x &= \frac{mv_x'+mu}{\sqrt{\left(1 - \frac{u^2}{c^2}\right) \left(1-\frac{v_x'^2}{c^2} \right)}} \\ p_x &= \gamma \left[mv_x' \gamma(v_x') + mu \gamma(v_x') \right] \\ p_x &= \gamma \left[mv_x' \gamma(v_x') + \frac{mc^2 \gamma(v_x') u}{c^2} \right] \\ p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right] \end{split}$$

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.

$$\scriptsize \begin{split} p_y &= m v_y \gamma(v_y)\\ p_y &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - v_y'^2/c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2}}\\ p_y &= \frac{m v_y' \left( 1 + v_x' \frac{u}{c^2} \right)}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\ p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\ p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4}\right) - v_y'^2\right]/c^2}}\\ p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + v_x'^2 \frac{u^2}{c^2} - v_y'^2\right]/c^2}}\\ p_y &= \frac{m v_y'}{\gamma \sqrt{1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} - \frac{v_y'^2}{c^2}}}\\ \end{split}$$

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 know I am very close.

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$p_{y}=p'_{y}$ only if $u=u_{x}$ and $v_{y}=0$ – TMS Dec 4 '12 at 13:57
I stated this at the beginning. – 71GA Dec 6 '12 at 22:54
This is honestly a pretty laborious way of doing things. Is there a reason you're avoiding just using the Lorentz transformations (which make the result obvious)? Also, you write $p_y = m v_y \gamma(v_y)$, but $\gamma$ depends on the total velocity $v$, not just the y-component. – Muphrid Dec 6 '12 at 23:28
@Muphrid, you could suggest an alternative in an answer. – Magpie Dec 7 '12 at 7:04
@Muphid It is $p=mv \gamma(v)$ in general. This is what i am sure of. But isn't it like this $p=mv_x \gamma(v_x)$ (gamma is a function of a speed which stands next to $m$) for a component? Was i just lucky above to sucessfully derive first equation i.e. for $p_x$? I mean lucky because i should have used $\gamma(v)$ but i used $\gamma(v_x)$. But because $v_x$ is the whole speed it worked out for $x$ direction. It is not working for $y$ direction because $v_y$ isn't the whole speed. It is actually zero. So i should use $p_y = m v_y \gamma(v_x)$? – 71GA Dec 7 '12 at 8:43

Well this is how $p_y$ part of a four-momentum is put together.

$$\scriptsize \begin{split} p &= m v \gamma(v)\\ &\Downarrow\\ p_y &= m v_y \gamma(v) = m v_y \gamma \left( \sqrt{v_x^2 + v_y^2 + v_z^2}\right) = m v_y \gamma \left( \sqrt{v_x^2 + 0 + 0}\right) = m v_y \gamma(v_x) =\\ &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - \frac{\left(v_y' + u\right)^2}{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2}}} = \frac{mv_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\frac{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2}{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2}}}=\\ &= \frac{mv_y' \left(1 + v_x' \frac{u}{c^2}\right)}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2\right] / c^2}} = \frac{mv_y'}{\gamma \sqrt{\left[c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2\right] / c^2}}=\\ &= \frac{mv_y'}{\gamma \sqrt{\left[c^2 \left(1 + 2 v_x' \frac{u}{c^2} + {v_x'}^2 \frac{u^2}{c^4}\right) - {v_x'}^2 - 2 {v_x'}u - u^2\right] / c^2}}=\\ & = \frac{mv_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + {v_x'}^2 \frac{u^2}{c^2} - {v_x'}^2 - 2 {v_x'}u - u^2\right] / c^2}}= \frac{mv_y'}{\gamma \sqrt{\left[c^2 + {v_x'}^2 \frac{u^2}{c^2} - {v_x'}^2 - u^2\right] / c^2}}=\\ & = \frac{mv_y'}{\gamma \sqrt{1 + {v_x'}^2 \frac{u^2}{c^4} - \frac{{v_x'}^2}{c^2} - \frac{u^2}{c^2}}}= \frac{mv_y'}{\gamma \sqrt{\left(1 - \frac{u^2}{c^4}\right) \left(1-\frac{{v_x'}^2}{c^2}\right)}}= mv_y' \gamma(v_x')\\ \end{split}$$

In our case $v_x' = v'$ and we can modify last part of this big equation so that we get:

$$\scriptsize \begin{split} p_y= mv_y' \gamma(v')\\ \end{split}$$

Now we can see Lorentz tr. and reverse Lorentz tr. which are:

$$\scriptsize \begin{split} &\boxed{p_y=p_y'} ~~~\boxed{p_y'=p_y}\\ \end{split}$$

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nice work! I saw some people suggest above that this was the long way of doing things. I would like them to show what they are if that is the case, too. – Magpie Dec 10 '12 at 18:15
This is very clear and i used only simple math. Now it is easy to put together 4-momentum. What i have to do is only put equations for $W$, $p_x$, $p_y$ and $p_z$ one above the other and make a matrix form from which i can see Lorentz matrix $\Lambda$. – 71GA Dec 10 '12 at 20:11
My momentum 4 vector is still sketchy to judge but I don't understand why someone would say there's a better way and not show it. – Magpie Dec 10 '12 at 22:37