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- Thread starter Salmone
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In summary, after setting ##\vec{E}=0## and solving for ##\vec{\nabla} \phi##, one can find that ##\phi=\text{const}## and that the only possible vector potential is ##\vec{A}=\frac{1}{2} \vec{A} \times \vec{x}##.f

- #1

- 101

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- #2

- 23,737

- 14,404

$$m \ddot{\vec{x}}=e (\vec{E}+\vec{v} \times \vec{B}),$$

where

$$\vec{E}=-\partial_t \vec{A} -\vec{\nabla} \phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$

For the homogeneous magnetic field you can use

$$\vec{A}=\frac{1}{2} \vec{B} \times \vec{x}, \quad \phi=0.$$

- #3

- 101

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I know how Lagrangian and Hamiltonian are derived in the case of a charged particle inside an electromagnetic field, I don't know how to derive the Hamiltonian in the case of a charged particle in a magnetic field of type ##\vec{B}=(0,0,B)##. Why do I have to use that Gauge transformation? If ##\vec{E}=0##, then what I obtain is ##\vec{\nabla} \phi=0##.

$$m \ddot{\vec{x}}=e (\vec{E}+\vec{v} \times \vec{B}),$$

where

$$\vec{E}=-\partial_t \vec{A} -\vec{\nabla} \phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$

For the homogeneous magnetic field you can use

$$\vec{A}=\frac{1}{2} \vec{B} \times \vec{x}, \quad \phi=0.$$

- #4

- 23,737

- 14,404

$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$

describes the same homogeneous ##\vec{B}##-field, for any arbitrary scalar field, ##\chi##.

- #5

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- #6

- 23,737

- 14,404

$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \phi=0, \quad \vec{\nabla} \times \vec{A}=\vec{B}=\text{const}.$$

It's obvious that you can make the potentials time independent. Then from the first equation you have ##\phi=\text{const}##. Then it's easy to check that a possible choice for the vector potential is

$$\vec{A}=\frac{1}{2} \vec{A} \times \vec{x}.$$

- #7

- 101

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Okay, thank you so much.

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