Lagrangian: electron in magnetic field (what are the conserved quantities?)

Question

Given an electron subjected to a magnetic field $\vec{B} = (0,0,B_0)$

I have chosen the Lagrangian to be $L = T-V = \frac{1}{2}\,m\,\Vert\mathbf{\dot r}\Vert^2+ e\,\mathbf{\dot r}\boldsymbol \cdot\mathbf{A}$.

In between question: $V = -e\,\mathbf{\dot r}\boldsymbol \cdot\mathbf{A}$ is a generalised potential usually given by $V = e\,(\phi-\mathbf{\dot r}\boldsymbol \cdot\mathbf{A})$. Whatsoever I only get the right Hamiltonian same as on the sheet if I use the shortened potential. Why is that? $\phi= 0?$

With the vector potential set on $\mathbf{A} = (-B_0/2\,y, B_0/2\,x,0)$ the total Lagrangian becomes:

$L = \frac{1}{2}\,m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2 \right) -e\,\dot{x}\,B_0/2\,y+e\,\dot{y}\,B_0/2\,x$.

In order to get more information on conserved quantities I use Euler-Lagrange-Equations: $\dfrac{\mathrm{d}}{\mathrm{dt}}\partial_{\textstyle \dot{r}}L-\partial_{\textstyle {r}} L= 0$:

$m\,\ddot{x}- e\,B_0/2\,\dot{y}-e\,\dot{y}\,B_0/2 = 0$

$m\,\ddot{y}+ e\,B_0/2\,\dot{x}+e\,\dot{x}\,B_0/2 = 0$

$m\,\ddot{z} \qquad \qquad \qquad \qquad =0$

Final question: what are the conserved quantities now? Based on the last line I get $m\,\dfrac{\mathrm{d}}{\mathrm{dt}}\dot{z} = 0$. Hence momentum in $z$-direction is conserved. But what else? There must be another quantity.

Answer 1

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Hint :

First note that \begin{equation} \mathbf E\e\m\bl\nabla\phi\m\dfrac{\partial \mathbf A}{\partial t} \tl{a} \end{equation} Since in our case \begin{equation} \mathbf E\e\bl 0 \quad \texttt{and} \quad \dfrac{\partial \mathbf A}{\partial t}\e\bl 0 \tl{b} \end{equation} we have \begin{equation} \bl\nabla\phi\e\bl 0 \quad \bl\implies \quad \phi\e \texttt{constant} \tl{c} \end{equation} Adding any constant in the Lagrangian $\,L\,$ doesn't affect the equations of motion so without loss of generality we set $\,\phi\e 0$.

I think that using a 3-dimensional version of the Beltrami identity a constant of the motion would be \begin{equation} \mathbf{\dot r}\bl\cdot\dfrac{\partial L}{\partial \mathbf{\dot r}}\m L\e \texttt{constant} \tl{1} \end{equation}

Also, looking in the first two Euler-Lagrange equations what are our thoughts about the complex function \begin{equation} \mathrm w \e x\p i\,y \tl{2} \end{equation} Could we translate these equations to a simple 2nd order linear differential equation with respect to $\,\mathrm w\,$? (try to relate this with the axial symmetry of @LSS's answer).

$\hebl$

Responds to OP comments

First, for the $\,\texttt{constant}\,$ of equation \eqref{1} we have \begin{equation} \begin{split} &\mathbf{\dot r}\bl\cdot\dfrac{\partial L}{\partial \mathbf{\dot r}}\m L \e \texttt{constant}\quad \bl\implies\\ &\mathbf{\dot r}\bl\cdot\underbrace{\plr{m\,\mathbf{\dot r}\p e\mathbf A }}_{\partial L/\partial \mathbf{\dot r}}\m\underbrace{\plr{1/2\,m\,\Vert\mathbf{\dot r}\Vert^2\p e\,\mathbf{\dot r}\bl\cdot\mathbf{A}}}_{L}\e \texttt{constant}\quad \bl\implies \end{split} \nonumber \end{equation} \begin{equation} \Vert\mathbf{\dot r}\Vert\e \texttt{constant} \tl{R-1} \end{equation} the well-known result that the magnetic force as perpetually normal to the velocity of the particle doesn't change the magnitude of this velocity and doesn't produce work keeping unchanged its non-relativistic kinetic energy.

Second, for the complex function $\,\mathrm w\e x\p i\,y \,$ of equation \eqref{2} we have from the first two Euler-Lagrange equations \begin{align} \ddot{x} & \e \p\omega\,\dot{y} \tl{R-2a}\\ \ddot{y} & \e \m\omega\,\dot{x} \tl{R-2b}\\ \texttt{where}\quad \omega & \e\dfrac{e\,B_0}{2m} \tl{R-2c} \end{align} So, \begin{equation} \ddot{x}\p i\,\ddot{y}\e\m i\,\omega\plr{\dot{x}\p i\,\dot{y}} \tl{R-3} \end{equation} or \begin{equation} \ddot{\mathrm w}\e\m i\,\omega\,\dot{\mathrm w} \tl{R-4} \end{equation} etc...

$\hebl$

ADDENDUM A

Motivated by the Euler-Lagrange-Equations of the post and @hft's answer I realize the conservation of the linear momentum \begin{equation} \mathbf p_{\mathbf q}\e m\,\mathbf{\dot r}\m e\,\mathbf r\x\mathbf B \e m\,\mathbf{\dot r}\m e\,\mathbf r\x\plr{\bl\nabla\x\mathbf A} \tl{A-1} \end{equation} Now I try to find if this is the generalized momentum conjugate to a cyclic coordinate $\,\mathbf q\,$ the latter satisfying : \begin{equation} \dfrac{\partial L}{\partial \mathbf{\dot q}}\e\mathbf p_{\mathbf q}\e m\,\mathbf{\dot r}\m e\,\mathbf r\x\mathbf B\qquad \texttt{and} \qquad \dfrac{\partial L}{\partial \mathbf q}\e 0 \tl{A-2} \end{equation}

$\hebl$

ADDENDUM B

We will express the Lagrangian in cylindrical coordinates as suggested by @LSS's answer. \begin{equation} \mathbf r\e \begin{bmatrix} \:x\:\vp\\ \:y\:\vp\\ \:z\:\vp \end{bmatrix} \e \begin{bmatrix} \:\rho\cos\phi\:\vp\\ \:\rho\sin\phi\:\vp\\ z\vp \end{bmatrix}\quad\bl\implies\quad \mathbf{\dot r}\e \begin{bmatrix} \:\:\:\dot{\!\!\rho}\cos\phi\m\rho\:\:\dot{\!\!\phi}\sin\phi\:\vp\\ \:\:\:\dot{\!\!\rho}\sin\phi\p\rho\:\:\dot{\!\!\phi}\cos\phi\:\vp\\ \dot z\vp \end{bmatrix} \tl{B-01} \end{equation} So \begin{equation} \Vert\mathbf{\dot r}\Vert^2\e\:\:\dot{\!\!\rho}^{\:2}\p\rho^2\:\:\dot{\!\!\phi}{}^{\;2}\p\dot{z}^2 \tl{B-02} \end{equation} Also \begin{equation} \mathbf{\dot r}\bl\cdot\mathbf A\e \dfrac{B}{2} \begin{bmatrix} \:\:\:\dot{\!\!\rho}\cos\phi\m\rho\:\:\dot{\!\!\phi}\sin\phi\:\vp\\ \:\:\:\dot{\!\!\rho}\sin\phi\p\rho\:\:\dot{\!\!\phi}\cos\phi\:\vp\\ z\vp \end{bmatrix}^{\bl\top} \begin{bmatrix} \:\m\rho\sin\phi\:\vp\\ \:\hphantom{\m}\rho\cos\phi\:\vp\\ 0\vp \end{bmatrix}\e \dfrac{B}{2}\rho^2\:\:\dot{\!\!\phi} \tl{B-03} \end{equation} From equations \eqref{B-02},\eqref{B-03} we have the following expression of the Lagrangian in cylindrical coordinates \begin{equation} L\plr{\rho,\phi,z,\:\:\dot{\!\!\rho},\:\:\dot{\!\!\phi},\dot z}\e\frac12 m\plr{\:\:\dot{\!\!\rho}^{\:2}\p\rho^2\:\:\dot{\!\!\phi}{}^{\;2}\p\dot{z}^2}\p\dfrac{e\,B}{2}\rho^2\:\:\dot{\!\!\phi} \tl{B-04} \end{equation} In above Lagrangian the coordinates $\:\phi,z\:$ are cyclic, that is the Lagrangian is independent of them, so the conjugate momenta are conserved as shown also in the Euler-Lagrange equations of motion below \begin{equation} \left. \begin{cases} \dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}}\m\dfrac{\partial L}{\partial \rho}\e0\\ \dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\phi}}}\m\dfrac{\partial L}{\partial \phi}\e0\\ \dfrac{\mathrm d}{\mathrm dt}\plr{\dfrac{\partial L}{\partial \dot{z}}}\m\dfrac{\partial L}{\partial z}\e0 \end{cases} \right\} \quad\bl\implies\quad \left. \begin{cases} \dfrac{\mathrm d}{\mathrm dt}\plr{m\:\:\dot{\!\!\rho}} \e m\rho\:\:\dot{\!\!\phi}{}^{\;2}\p e\,B\rho\:\:\dot{\!\!\phi}\vphantom{\plr{\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}}}\\ \dfrac{\mathrm d}{\mathrm dt}\plr{m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2 }\e0\\ \dfrac{\mathrm d}{\mathrm dt}\plr{m\dot{z}}\e0 \end{cases} \right\} \tl{B-05} \end{equation} The conjugate momenta are \begin{align} p_\rho&\e\dfrac{\partial L}{\partial \:\:\dot{\!\!\rho}}\e m\:\:\dot{\!\!\rho} \tl{B-06.1}\\ p_\phi&\e\dfrac{\partial L}{\partial \:\:\dot{\!\!\phi}}\e m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2 \tl{B-06.2}\\ p_z&\e\dfrac{\partial L}{\partial \dot z}\e m\dot z \tl{B-06.3} \end{align} The conserved momenta conjugate to the cyclic coordinates, angle $\,\phi\,$ and cartesian $\,z$, are the angular momentum $\,p_\phi\,$ and the linear momentum $\,p_z\,$ respectively \begin{equation} \dot{p}_\phi\e0\,,\qquad \dot{p}_z\e0 \tl{B-07} \end{equation} Note that the angular momentum $\,p_\phi\,$ if expressed by the initial cartesian coordinates yields \begin{equation} p_\phi\e m\rho^2\:\:\dot{\!\!\phi}\p\dfrac{e\,B}{2}\rho^2\e m\plr{x\dot y\m y\dot x}\p\dfrac{e\,B}{2}\plr{x^2\p y^2} \tl{B-08} \end{equation} in agreement with @Thomas Fritsch's answer, the item \begin{equation} m\rho^2\:\:\dot{\!\!\phi}\e m\plr{x\dot y\m y\dot x} \tl{B-09} \end{equation} being the angular momentum of the particle with respect to the $\,z\m$axis.

Answer 2

The system has axial symmetry, that is, if you rotate around the z axis the system remains the same.

I would recommend you to go to the cylindrical frame. In fact, if you do that, you will realize that the expression is independent of $\phi$.

Answer 3

$m\,\ddot{x}- e\,B_0/2\,\dot{y}-e\,\dot{y}\,B_0/2 = 0$

$m\,\ddot{y}+ e\,B_0/2\,\dot{x}+e\,\dot{x}\,B_0/2 = 0$

$m\,\ddot{z} \qquad \qquad \qquad \qquad =0$

Final question: what are the conserved quantities now? Based on the last line I get $m\,\dfrac{\mathrm{d}}{\mathrm{dt}}\dot{z} = 0$. Hence momentum in $z$-direction is conserved. But what else? There must be another quantity.

There are lots of conserved quantities. For example, any quantity that is a function of only the z momentum is conserved.

In addition, based on the first two equation that you wrote, we can also see that:

$\frac{d}{dt}\left( m\,\dot{x}- e\,B_0/2\,y-e\,y\,B_0/2 \right)= 0$

and

$\frac{d}{dt}\left(m\,\dot{y}+ e\,B_0/2\,x+e\,x\,B_0/2 \right)= 0\;.$

Note: I'm assuming, for example, that your notation $B_0/2\,y$ means $y$ times $B_0$ divided by 2, not $B_0$ divided by two and divided by $y$. Your notation is a little confusing here.

So... I'll just combine a couple terms for you and define a couple more constant quantities: $$ \alpha = \left( m\,\dot{x}- e\,B_0\,y \right) $$ and $$ \beta = \left(m\,\dot{y}+ e\,B_0\,x \right) $$

Answer 4

So you have got these Euler-Lagrange equations: $$\begin{align} m\ddot{x}&=eB_0\dot{y} \\ m\ddot{y}&=-eB_0\dot{x} \\ m\ddot{z}&=0 \end{align}$$

1. The magnetic force is known to do no work on the particle. So we can guess, the kinetic energy $$E_\text{kin}=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)$$ should be conserved. This can indeed be verified by using the Euler-Lagrange equations: $$\begin{align} \frac{d}{dt}E_\text{kin} &=\frac{d}{dt}\left(\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)\right) \\ &=\ ... \text{ (I leave the calculation to you)} \\ &=0 \end{align}$$ and hence $$E_\text{kin}=\text{const}$$

2. The system has a rotational symmetry around the $z$-axis. And hence there should be a conserved quantity related to this symmetry. Without having the magnetic field (i.e. with $B_0=0$) this conserved quantity would be the $z$-component of angular momentum: $$L_z=m(x\dot{y}-y\dot{x}).$$ But with the magnetic field this is not true anymore, as can be checked with the Euler-Langrange equations: $$\begin{align} \frac{d}{dt}L_z &=\frac{d}{dt}(m(x\dot{y}-y\dot{x})) \\ &=\ ... \text{ (I leave the calculation to you)} \\ &=-eB_0(x\dot{x}+y\dot{y}) \end{align}$$ This is obviously not zero, and hence $L_z$ is not conserved. But luckily we can rewrite the right side as a total time derivative $$\frac{d}{dt}L_z=-\frac{d}{dt}\left(\frac{1}{2}eB_0(x^2+y^2)\right)$$ and then bring everything to the left side $$\frac{d}{dt}\left(L_z+\frac{1}{2}eB_0(x^2+y^2)\right)=0$$ So we have found the conserved quantity $$L_z+\frac{1}{2}eB_0(x^2+y^2)=\text{const}.$$

3. You can repeat the procedure above with the $x$- and $y$-components of momentum ($p_x=m\dot{x}$ and $p_y=m\dot{y}$) to find two more conserved quantities (the same two as given in @hft's answer): $$p_x-eB_0y=\text{const}$$ $$p_y+eB_0x=\text{const}$$