I'm having trouble making a conceptual leap in the Lorentz transformation of the energy-momentum 4-vector:

The 4-vector in $S$ is related to $S'$ by, $$ \left( \begin{matrix} E' \\ p'_{x} \end{matrix} \right) = \left( \begin{matrix} \cosh{\vartheta} & \sinh{\vartheta} \\ \sinh{\vartheta} & \cosh{\vartheta} \end{matrix} \right) \left( \begin{matrix} E \\ p_{x} \end{matrix} \right) $$ where $\vartheta$ is the rapidity.

Also,$$ v'_{x} = \frac{p'_{x}}{E'} = \tanh{\vartheta}$$

However, when I try and do the calculation by direct substitution I get, $$ v'_{x} = \frac{e^{2\vartheta}(E+p_{x})-(E-p_{x})}{e^{2\vartheta}(E+p_{x})+(E-p_{x})} $$ which is close to $\tanh{\vartheta}$ but I can't factor out $E$ and $p_{x}$ without doing something weird like mixing $E=\pm(p_{x}^2+m^2)^{1/2}$[edit:, even though this would give me $1/\tanh{\vartheta}$.]

What step am I missing or what conceptual leap do I need to make to get the answer?

Edit: I made a mistake in the last equation due to calculating $E'/p'_{x}$ by accident. Also, I understand that $v'_{x} = \tanh{\vartheta}$ I just can't work out how or why you can say that $(E + p_{x}) = (E - P_{x})$.



That is true, though. The relativistic energy-momentum relation, $E^2 = m^2 + p^2$. So you can use it.

In fact, if you didn't already know about rapidity, this would be a way to discover it: notice that $$E^2 - p^2 = m^2 = E'^2 - p'^2$$ which is reminiscent of the trig identity $$\cosh^2\vartheta - \sinh^2\vartheta = 1 = \cosh^2\vartheta' - \sinh^2\vartheta'$$ I'm sure you can work out from there how to express $E$ and $p$ (and then $v$, if you want) in terms of $m$ and $\vartheta$.

  • $\begingroup$ In fact, I think it's guaranteed that if you have two quantities related like that you can cast them in terms of $\sinh$ and $\cosh$. $\endgroup$ – zeldredge May 2 '15 at 16:53
  • $\begingroup$ I get that it follows from being on-shell and also that $v'_{x} = \tanh{\varphi}$ from $\Lambda{[x]}$, I just can't see why I can't get it to work by direct substitution. $\endgroup$ – bainzo May 2 '15 at 19:05

What's wrong with $E = \pm (p_x^2 + m^2)^{1/2}$?

As I'm not sure if this qualifies as a homework-related question, I'll just give you a hint: try to write energy and momentum in terms of mass, $\beta$ and $\gamma$. It's easier to solve that way.

  • $\begingroup$ It would seem counter-physical to me to use the $E_+, E_-$ solutions at the same time; also I made an error and it wouldn't give me the right solution. $\endgroup$ – bainzo May 2 '15 at 18:33
  • $\begingroup$ @bainzo I didn't say that you have to use both solutions, but you certainly have to use the dispersion relation in some way. You also should make sure that all your formulas are correct (specially the one of the velocity), and how many velocities and rapidities should you use $\endgroup$ – Bosoneando May 2 '15 at 19:00

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