Calculating geodesic equation from Hamiltonian in presence of EM

[Jokar]

I have a Hamiltonian

$$
H = \frac{1}{2} g^{\alpha \beta}\left(p_\alpha- A_\alpha\right)\left(p_\beta- A_\beta\right)
$$

I want to calculate the equation of motion. How can I calculate the equation of motions

$$
\frac{dx^\mu}{d\tau} = g^{\mu\nu}(p_\nu - A_\nu)
$$

This one is straight forward. However, how can I calculate

$$
\frac{dp^\mu}{d\tau} -\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = g^{\mu\nu} F_{\nu\beta} p^\beta
$$

Can someone please help me with the derivation or give me some reference?

 

[Yuras]

I never saw derivation of geodesic from Hamiltonian, but it seems to work indeed. Where did you find this?

Regarding how to proceed with the derivation. You already used the first equation, namely
$$\frac{dx^\alpha}{d\tau}=\frac{\partial H}{\partial p_\alpha}$$
Now use the second one
$$\frac{dp_\alpha}{d\tau}=-\frac{\partial H}{\partial x^\alpha}$$
Note that both $g$ and $A$ depend on $x$, while $p$ doesn't.

Though the derivation actually splits into two separate parts - one for geodesic equation (no force) and other for the Lorentz force. And it's easier to do these parts individually, so I'd suggest to start without EM field, namely with $H=\frac{1}{2}g^{\alpha\beta} p_\alpha p_\beta$ (you should get the usual Euler-Lagrange equation for geodesic) and only than add the field.

I didn't go through the calculation to the very end myself, but I'm more or less confident I can do it. So feel free to ask if you have issues with particular steps.

 

[Jokar]

Thanks. I am also doing the same thing. But I think there are some steps in between.

If you take the derivative then you will get

$$
\frac{dp_\gamma}{d\tau} = -\frac{1}{2}
\frac{\partial {g^{\alpha\beta}}}{\partial x^\gamma}
(p_\alpha - A_\alpha)(p_\beta - A_\beta) + g^{\alpha\beta}(p_\alpha - A_\alpha)\frac{\partial A_\beta}{\partial x^\gamma}
$$

What next? How can get the Christoffel symbol and the electromagnetic tensor. Am I doing anything wrong?

 

[Jokar]

I never saw derivation of geodesic from Hamiltonian, but it seems to work indeed. Where did you find this?

Regarding how to proceed with the derivation. You already used the first equation, namely
$$\frac{dx^\alpha}{d\tau}=\frac{\partial H}{\partial p_\alpha}$$
Now use the second one
$$\frac{dp_\alpha}{d\tau}=-\frac{\partial H}{\partial x^\alpha}$$
Note that both $g$ and $A$ depend on $x$, while $p$ doesn't.

Though the derivation actually splits into two separate parts - one for geodesic equation (no force) and other for the Lorentz force. And it's easier to do these parts individually, so I'd suggest to start without EM field, namely with $H=\frac{1}{2}g^{\alpha\beta} p_\alpha p_\beta$ (you should get the usual Euler-Lagrange equation for geodesic) and only than add the field.

I didn't go through the calculation to the very end myself, but I'm more or less confident I can do it. So feel free to ask if you have issues with particular steps.

Any help?

 

[Yuras]

You have two equations. Now you use the first one to eliminate $p$ from the second one. Note the convenient $p_\mu-A_\mu$ in the second one waiting to be replaced with the first one. Also when replacing $p$ on the left side, you'll peek up another derivative of the metric (it will go to the Christoffel symbol together with the one of the right) and a derivative of $A$ (it will make the EM tensor together with the other one). Collect few terms, do a bit of algebra and you are almost there.

 

[Jokar]

Thanks. I did not replace it in the left side. So I was not getting the answer. Thank you very much.