I know this is getting really ridiculous but I have yet another question on Lagrangians...(adsbygoogle = window.adsbygoogle || []).push({});

This is our Lagrangian:

[tex]

L=\frac{1}{2}m\dot{\vec{x}}^{2}+e\vec{A}.\dot{\vec {x}}

[/tex]

Using the fact that:

[tex]\vec P= \frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}} + e\vec A[/tex]

and substituting P for [tex]\dot{\vec{x}}[/tex] *

We get this Hamiltonian:

[tex]H=\frac{1}{2m}(\vec P - e\vec A )^2[/tex]

(From * [tex]\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )[/tex], so presumably, the above Hamiltonian is equal to [tex]\frac{m}{2} \dot{\vec{x}}^2[/tex])

The question is to find the Equations of Motion in the Hamiltonian Formalism, ie. we need to determine [tex]\dot{\vec{x}}[/tex] and [tex]\dot{P}_i[/tex]:

We know [tex]\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )[/tex] from above.

Now:

[tex]\dot{P}_i = \frac{\partial H}{\partial x^i} = \frac{1}{2m}\frac{\partial [(\vec P - e\vec A )^2]}{\partial x^i} = \frac{m}{2}\frac{\partial [(\dot{\vec{x}} ^2]}{\partial x^i}=0[/tex]

HOWEVER the answer seems to be this:

[tex]\dot{P}_i = \frac{e}{m}(P-eA)_j \epsilon_{jki} B_k[/tex]

Can someone please explain, why P_dot is non zero?

Thanks...

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# Yet another question on Lagrangians

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