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I've been given the lagrangian

$L(\vec{x},\dot{\vec{x}}) = -mc^2 \sqrt{1-\frac{\dot{\vec{x}}}{c^2}} + \frac{e}{c} \vec{A} \cdot \dot{\vec{x}} - e \phi$

and I have to determine the hamiltonian and the equations of motion as well as proving that the time derivative of the relativistic kinetic energy is equal to

$\frac{dE_{\text{kin}}}{dt} = e \vec{E} \cdot \dot{\vec{x}}$.

I've done the first part and got for the Hamiltonian

$H = c \sqrt{m^2 c^2 + (\vec{p} - \frac{e}{c} \vec{A}})^2 + e \phi$

which then gives me the equations of motion

$\dot{\vec{p}} =-\frac{\partial H}{\partial t} = \frac{e}{c} (\vec{\nabla} \cdot \vec{A}) \dot{\vec{x}} - e \vec{\nabla} \phi$

$\dot{\vec{x}} = \frac{\vec{p} - (e/c) \vec{A}}{\sqrt{m^2 + (1/c^2)(\vec{p} - (e/c)\vec{A})}}$.

Then the kinetic energy is given as

$E_{\text{kin}} := \frac{mc^2}{\sqrt{1-(\dot{\vec{x}}^2/c^2)}}$

but I have been stuck trying to find its time derivative since hours. I figured that the total relativistic energy was equal to the hamiltonian and this led me to

$E_{\text{rel}} = H = c\sqrt{m^2c^2 + (\vec{p}-(e/c)\vec{A}}+e\phi = c\sqrt{m^2c^2 + \vec{p}_{\text{mec}}^2} + e\phi = mc^2 \gamma + e\phi = E_{\text{kin}}+e\phi$

$\implies E_{\text{kin}} = H - e\phi$

$\implies \frac{d E_{\text{kin}}}{dt} = \frac{dH}{dt} - e \frac{d \phi}{dt}$

At this point I've used the lagrangian to calculate the time derivative of the hamiltonian since

$\frac{dH}{dt} = -\frac{dL}{dt}$

which is

$\frac{dH}{dt} = -m \gamma \dot{\vec{x}} \cdot \ddot{\vec{x}} - (e/c) \dot{\vec{x}} (\dot{\vec{x}} \cdot \vec{\nabla}) \vec{A} - (e/c) \dot{\vec{A}} + e \frac{d \phi}{dt}$

so

$\dot{E}_{\text{kin}} = \vec{p} \cdot \ddot{\vec{x}} - \frac{e}{c} \big( \dot{\vec{x}} (\dot{\vec{x}} \cdot \vec{\nabla}) \vec{A} + \dot{\vec{A}} \big)$.

I think I've been wandering in circles since I still need $\phi$ for the electric field and the first term is again a problem. Does anybody have a hint before I lose five more hours?

Thanks a lot in advance, your help is appreciated!

Julien.

1 Answers 1

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Nevermind i found the answer. I wasn't that far:

$E_{\text{kin}}= H - e\phi \implies \dot{E}_{\text{kin}}= \dot{H} - e \dot{\phi}$

$=c \frac{d}{dt}(\sqrt{m^2c^2 + \vec{p_{\text{mech}}}})+e\dot{\phi} - e\dot{\phi}$

$=c \frac{\vec{p_{\text{mech}}} \cdot \dot{\vec{p_{\text{mech}}}}}{\sqrt{m^2c^2+\vec{p_{\text{mech}}}^2}}$

$= \frac{\dot{\vec{x}} \gamma}{\sqrt{c^2+\dot{\vec{x}}^2 \gamma^2}} (e \vec{E} + (e/c)\dot{\vec{x}}\times \vec{B})$

$= \dot{\vec{x}} e \vec{E} + \frac{e}{c} \dot{\vec{x}} \cdot (\dot{\vec{x}} \times \vec{B})$

$= e \vec{E} \cdot \dot{\vec{x}}$

I've skipped a few calculation steps, but that should be enough to help anyone else searching for the answer. The second term disappears in the last step because $\dot{\vec{x}}$ is perpendicular to $\dot{\vec{x}} \times \vec{B}$.