I Help with Canonical Poisson Brackets & EM Field

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The discussion focuses on the Lagrangian formulation of a particle in an electromagnetic field and the implications for canonical Poisson brackets. The Lagrangian includes a velocity-dependent generalized potential, and the conjugate momenta are derived, revealing a relationship between momenta and coordinates. There is confusion regarding the claim that canonical Poisson brackets hold, as calculations suggest that the brackets between momenta are non-zero due to dependencies on the coordinates. The Hamiltonian is then derived, leading to the conclusion that canonical momenta differ from mechanical momenta, emphasizing the need to express quantities in terms of independent variables for accurate Poisson bracket calculations. The conversation concludes with a clarification of these concepts in the context of electromagnetic fields.
Matthew_
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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?
 
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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##.
 
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Likes 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
 
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