# Help with Canonical Poisson Brackets & EM Field

• I
• Matthew_
In summary, the Lagrangian for a particle moving in an eletromagnetic field is $$\mathcal{L}=\dfrac{m}{2}(\dot{x}^2_1+\dot{x}^2_2+\dot{x}^2_3)-q\varphi+\dfrac{q}{c}\vec{A}\cdot\dot{\vec{x}}.$$ The conjugated momenta of the generalized coordinates are: $$p_1=m\dot{x}_1-\dfrac{qB}{2c}x_2,$$ $$p_2=m\dot{x}_2+\df Matthew_ We were introduced the lagrangian for a particle moving in an eletromagnetic field (for context, this was a brief introduction before dealing with Zeeman effect) as$$\mathcal{L}=\dfrac{m}{2}(\dot{x}^2_1+\dot{x}^2_2+\dot{x}^2_3)-q\varphi+\dfrac{q}{c}\vec{A}\cdot\dot{\vec{x}}.$$A "velocity-dependent generalized potential" appears. For a constant magnetic field oriented along the ##x_3## axis:$$\vec{A}=\frac{B}{2}(x_1\hat{u}_2-x_2\hat{u}_1).$$Now, the conjugated momenta of the generalized coordinates are:$$p_1=m\dot{x}_1-\dfrac{qB}{2c}x_2,p_2=m\dot{x}_2+\dfrac{qB}{2c}x_1,p_3=m\dot{x}_3.$$It was claimed that canonical poisson brackets hold, namely ##\left\{p_i,p_j\right\}=0##. I have no idea why this is the case tho, since the derivation of the lagrangian with respect of the generalized velocities gives a clear dependence between the coordinates and the adjoint momenta. Evaluating ##\left\{p_1,p_2\right\}## I think I should get something like (summation over j is omitted):$$\left\{p_1,p_2\right\}=\frac{\partial p_1}{\partial x_j}\frac{\partial p_2}{\partial p_j}-\frac{\partial p_1}{\partial p_j}\frac{\partial p_2}{\partial x_j}=\frac{\partial p_1}{\partial x_2}-\frac{\partial p_2}{\partial x_1}=-\frac{qB}{c}\neq 0.$$Is there some reason why this does not work? The canonical Poisson brackets hold by definition, because for phase-space functions ##A## and ##B## by definition$$\{A,B \}=\frac{\partial A}{\partial q_j} \frac{\partial B}{\partial p_j} - \frac{\partial B}{\partial q_j} \frac{\partial A}{\partial p_j}.$$To calculate Poisson brackets you have to express all quantities with the ##q_j## and ##p_j## as independent variables, i.e., you have to eliminate the ##\dot{q}_j## using the ##p_j## and ##q_j## first. So it's the other way, i.e., now the Poisson brackets between the velocities are non-zero now. You have$$\dot{\vec{x}}=\frac{1}{m}(\vec{p}-q \vec{A}/c),$$i.e.,$$[\dot{x}_j,\dot{x}_k]=\frac{1}{m^2} \left [-\frac{q}{c} \frac{\partial A_j}{\partial x_i} \delta_{ki} +\delta_{ij} \frac{q}{c} \frac{\partial A_k}{\partial x_i} \right] = \frac{q}{mc} \left (\frac{\partial A_k}{\partial x_j}-\frac{\partial A_j}{\partial x_i} \right) = \frac{q}{mc} \epsilon_{jki} B_i,$$where I've used that ##\vec{B}=\vec{\nabla} \times \vec{A}##. The Hamiltonian formulation is derived by first calculating the Hamiltonian as a function of the coordinates and canonical momenta,$$H=\vec{x} \cdot \vec{p}-L=\frac{m}{2} \dot{\vec{x}}^2+q \varphi=\frac{1}{2m} (\vec{p}-q/c \vec{A})^2+q \varphi.$$Then the equations of motion are the Hamilton canonical equations,$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}} = \{\vec{x},H \}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}}=\{\vec{p},H \}.$$Note that the canonical momenta ##\vec{p}## are not the mechanical momenta, ##\vec{\pi}=m \dot{\vec{x}}=\vec{p}-q \vec{A}/c##. malawi_glenn and Matthew_ vanhees71 said: The canonical Poisson brackets hold by definition, because for phase-space functions ##A## and ##B## by definition$$\{A,B \}=\frac{\partial A}{\partial q_j} \frac{\partial B}{\partial p_j} - \frac{\partial B}{\partial q_j} \frac{\partial A}{\partial p_j}.$$To calculate Poisson brackets you have to express all quantities with the ##q_j## and ##p_j## as independent variables, i.e., you have to eliminate the ##\dot{q}_j## using the ##p_j## and ##q_j## first. So it's the other way, i.e., now the Poisson brackets between the velocities are non-zero now. You have$$\dot{\vec{x}}=\frac{1}{m}(\vec{p}-q \vec{A}/c),$$i.e.,$$[\dot{x}_j,\dot{x}_k]=\frac{1}{m^2} \left [-\frac{q}{c} \frac{\partial A_j}{\partial x_i} \delta_{ki} +\delta_{ij} \frac{q}{c} \frac{\partial A_k}{\partial x_i} \right] = \frac{q}{mc} \left (\frac{\partial A_k}{\partial x_j}-\frac{\partial A_j}{\partial x_i} \right) = \frac{q}{mc} \epsilon_{jki} B_i,$$where I've used that ##\vec{B}=\vec{\nabla} \times \vec{A}##. The Hamiltonian formulation is derived by first calculating the Hamiltonian as a function of the coordinates and canonical momenta,$$H=\vec{x} \cdot \vec{p}-L=\frac{m}{2} \dot{\vec{x}}^2+q \varphi=\frac{1}{2m} (\vec{p}-q/c \vec{A})^2+q \varphi.$$Then the equations of motion are the Hamilton canonical equations,$$\dot{\vec{x}}=\frac{\partial H}{\partial \vec{p}} = \{\vec{x},H \}, \quad \dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}}=\{\vec{p},H \}.
Note that the canonical momenta ##\vec{p}## are not the mechanical momenta, ##\vec{\pi}=m \dot{\vec{x}}=\vec{p}-q \vec{A}/c##.
Thank you, this was illuminating

vanhees71

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