Vector potential

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Vector potential in magnetostatics

Since [math]\nabla\cdot \mathbf{B}=0[/math] is always true, one can think of [math]\mathbf{B}[/math] as a curl of another vector field. (recall [math]\nabla\cdot(\nabla\times A)=0[/math])

[math]{\color{Blue}{B(r)=\nabla\times A(r) }} [/math]

Using vector identities

[math]\nabla\left({1\over R}\right)=-{\nabla}'\left({1\over R}\right)=-{{\hat{\mathbf{R}}}\over{R^2}}=-{\mathbf{R}\over{R^3}}[/math]
[math]\nabla\times(uA)=(\nabla u)\times A+u(\nabla\times A)[/math]

Biot-Savart law can be rearranged as follows:

[math]{{ds'\times\hat{\mathbf{R}}}\over{R^2}}=-ds'\times\nabla\left({1\over R}\right)=\nabla\times\left({ds' \over R}\right)-{{\nabla\times{ds'}}\over R}=\nabla\times\left({ds' \over R}\right) [/math]
[math]\mathbf{B(r)}={\mu_0 \over{4\pi}}{\oint_{C'} {{I' ds' \times \hat{\mathbf{R}}}\over R^2}}={\mu_0 \over{4\pi}}\oint_{C} {{I ds \times\mathbf{R(r)}}\over{R^3}}[/math]
[math]\mathbf{B}={{\mu_0 I'}\over{4\pi}}\oint_{C'} \nabla\times\left({ds'\over R}\right)=\nabla\times\left({{\mu_0 I'}\over{4\pi}}{\oint_{C'} {ds'\over R}}\right) [/math]

Therefore, the vector potential due to current I can be written as:

[math]\mathbf{A(r)}={\mu_0 \over{4\pi}}\oint_{C'} {{I' ds'}\over R}[/math]
[math]\mathbf{A(r)}=\Sigma{{\mu_0 \over{4\pi}}\oint_{C'} {{I' ds'}\over R}}[/math]

Potentials in electrodynamics

When [math]\mathbf{B}=\nabla\times\mathbf{A}[/math] is plugged into Faraday's law, we have [math]\nabla\times\mathbf{E}+\frac{\partial}{\partial t}(\nabla\times\mathbf{A})=0[/math] which leads to [math]\nabla\times\left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right)=0[/math], so

[math]\mathbf{E}=-\mathbf{\nabla} V-{{\partial \mathbf{A}}\over{\partial t}}[/math]

Gauss's law becomes:

[math]{\nabla}^2 V+{\partial\over{\partial t}}(\nabla\cdot \mathbf{A})=-{\rho\over{{\epsilon}_0}} [/math]

Ampere-Maxwell's law becomes:

[math]{\nabla}^2 \mathbf{A}-\mu_0 {\epsilon}_0 {{{\partial}^2 \mathbf{A}}\over{\partial t^2}}-\nabla\left(\nabla\cdot \mathbf{A}+\mu_0 \epsilon_0 {{\partial V}\over{\partial t}}\right)=-\mu_0 \mathbf{J} [/math]

By defining D'Alembertian operator as

[math]\square^2 \equiv \nabla^2 - \mu_0 \epsilon_0 {{\partial}^2\over{\partial t^2}} [/math]

we obtain

[math]\square^2 \mathbf{A} - \nabla L = -\mu_0 \mathbf{J}[/math]
[math]\square^2 V + {\partial L \over \partial t}= - {\rho\over \epsilon_0}[/math]

where we have defined a quantity [math]L\equiv\nabla\cdot \mathbf{A}+\mu_0 \epsilon_0 {{\partial V}\over{\partial t}}[/math].

Ambiguity of vector potential

In magnetostatics,

[math]\nabla\times\mathbf{B} = \nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}=\mu_0\mathbf{J}[/math]

So, if we choose [math]\nabla\times\mathbf{A}=0[/math], we have

[math]\nabla^2\mathbf{A} = -\mu_0\mathbf{J}[/math]

which is a Poisson's equation, whose solution can be read off of it.

[math]\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{R}\, d\tau'[/math]

Gauge transformation

[math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] do not change under the following transformation:

[math]\mathbf{A}' = \mathbf{A} + \nabla\lambda[/math]
[math]V' = V - {\partial\lambda\over\partial t}[/math]

These set of transformation is called gauge transformation.

Coulomb gauge

Most often used in magnetostatics.

[math]\nabla\cdot\mathbf{A} = 0[/math]

This choice of gauge leads us to:

[math]{\nabla}^2 V = -{\rho\over{{\epsilon}_0}} [/math]
[math]{\nabla}^2 \mathbf{A}-\mu_0 {\epsilon}_0 {{{\partial}^2 \mathbf{A}}\over{\partial t^2}}=-\mu_0 \mathbf{J} + \mu_0 \epsilon_0 \nabla\left({{\partial V}\over{\partial t}}\right) [/math]

Lorentz gauge

It treats [math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] field on equal footing, so the whole equation becomes very neat using d'Alembertian.

[math]\nabla\cdot\mathbf{A} = -\mu_0\epsilon_0 {\partial V \over \partial t}[/math]

This choice of gauge leads us to:

[math]\square^2 V = - {\rho\over \epsilon_0}[/math]
[math]\square^2 \mathbf{A} = -\mu_0 \mathbf{J}[/math]