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Consider a particle of mass $m$ subjected to the following potential, $$ \vec{A}=\frac{1}{2}(\vec{B}\times\vec{r}) $$ Where the magnetic field $\vec{B}$ is constant.

Can someone prove that $\dot{p}_{i}=0$, $\forall i$?

i.e Prove that

$$ \dot{\vec{p}}=0 $$

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  • $\begingroup$ You'll need to show, $\dot{p}_i = -\frac{\partial \mathcal H}{\partial q_i} = 0$. $\endgroup$
    – JamalS
    Commented Jan 8, 2017 at 16:50
  • $\begingroup$ @JamalS Thank you. It is 0 because $\vec{v}$ only changes if it exists an $\vec{E}$, and here is 0. Also, $\vec{A}$ is constant due to $B$ is constant. My computations are the following. $$ H=\frac{1}{2m}\left(\vec{p} -\frac{q}{2}(\vec{B}\times \vec{r}) \right)^{2} \hspace{30mm} (3) $$ $$ \dot{p}_{j}=-\frac{\partial H}{\partial x_{j}}=-\frac{1}{m}(\vec{p}-\frac{q}{2} (\vec{B}\times \vec{r}))\frac{\partial (\vec{p}-\frac{q}{2} (\vec{B}\times \vec{r}))}{\partial x_j}=+\frac{1}{m}(p_l-\frac{q}{2}\epsilon_{lmn}B_{m}x_{n})(\frac{q}{2}\epsilon_{lmj}B_{m}) \hspace{10mm}(4) $$ What is wrong? $\endgroup$
    – Sergi
    Commented Jan 8, 2017 at 16:54
  • $\begingroup$ @AccidentalFourierTransform okay, sorry. But, is really $\dot{p}_{j}=0$ for $j=1,2,3$? Thank you. $\endgroup$
    – Sergi
    Commented Jan 8, 2017 at 17:06

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The momenta are NOT constant. In the simplest case where $\vec B$ is along $\hat z$ along, the motion will be helicoidal and the momenta in the plane perpendicular to $\vec B$ cannot be constants as this would imply a straight trajectory.

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  • $\begingroup$ Thank you. Are my computatioms right? $$ H=\frac{1}{2m}\left(\vec{p} -\frac{q}{2}(\vec{B}\times \vec{r}) \right)^{2} $$ $$ \dot{p}_{j}=-\frac{\partial H}{\partial x_{j}}=-\frac{1}{m}(\vec{p}-\frac{q}{2} (\vec{B}\times \vec{r}))\frac{\partial (\vec{p}-\frac{q}{2} (\vec{B}\times \vec{r}))}{\partial x_j}=+\frac{1}{m}(p_l-\frac{q}{2}\epsilon_{lmn}B_{m}x_{n})(\frac{q}{2}\epsilon_{lmj}B_{m}) $$ $\endgroup$
    – Sergi
    Commented Jan 8, 2017 at 18:32
  • $\begingroup$ I presume repeated indice are summed over. There is probably a simpler final expression obtained by expanding the scalar product and using vector identities but I don't remember the details. $\endgroup$
    – ZeroTheHero
    Commented Jan 8, 2017 at 19:42
  • $\begingroup$ yes, repeated indice are summed over. If the expression I have obtained is correct, is enough for me. I have found in other source a different expression. Thank you $\endgroup$
    – Sergi
    Commented Jan 10, 2017 at 9:07