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In my problem we look at a relativistic particle of charge $q$ and mass $m$ in the presence of a second particle of unit charge and mass $M\gg m$, which is fixed at the origin. I need to find an expression for the total energy E of the first particle in terms of canonical coordinates using the Hamiltonian (and ignoring gravitational energy). I also need to afterwards find and expression of E in terms of the conserved angular momentum and polar coordinates.

I'm familiar with the regular classical Hamiltonian (for a time independant potential $U(\vec x)$) where we have $$ H = \frac{1}{2}m\dot{\vec{x}}^2+U(\vec x) = E, $$ where in our case we would have $U(\vec x) = \frac{1}{4\pi\epsilon_0}\frac{q}{|\vec x|}$ (because the charge at the center has unit charge). Written in polar coordinates this would give us $$ E=\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}+U(r), $$ where $l$ is the conserved angular momentum.

My problem is that I don't really know what I have to change in order to take account of the fact that the particle is relativistic and if it would change anything about the angular momentum etc.

Thank you in adnvance for any help and tips!

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    $\begingroup$ The kinetic term becomes $mc^2(\gamma -1)$ $\endgroup$ – Jeanbaptiste Roux Sep 25 '20 at 17:03
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Angular momentum is fine.

Kinetic energy $K$ (the first term in your expressions) is the one you have to change.

Relativistically (and hence correctly): $$ K = (\gamma - 1)m_0 c^2,$$ where $\gamma$ is the Gamma factor $1/\sqrt{1-v^2/c^2}$, $c$ the speed of light, $v$ the speed of the particle, and $m_0$ its rest mass.

At non-relativistic speeds, $v\ll c$, you can Taylor expand $K$ above to get: $$ K \approx \frac{1}{2}m_0 v^2 + \frac{3}{8}\frac{m_0v^4}{c^2} + \frac{5}{16}\frac{m_0 v^6}{c^4} \dots $$ where you'll notice the first term is the classical kinetic energy you started with.

So either you use the relativistic $K$ to begin with, or you use the classical expression and then add perturbations.

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  • $\begingroup$ thank you! I'm still having problems with expressing the final Energy in terms of the angular momentum. Because if I use the hamiltonian with the relativistic kinetic energy and then form the derivative after $\dot{\phi}$, then I get a time-constant quantity which looks like $mr^2\dot{\phi}\gamma^3$ $\endgroup$ – DeltaChief Sep 25 '20 at 20:37

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