- #1

Nitacii

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- Homework Statement
- Find the solution of Maxwell equations for a source of an unevenly rotating sphere with the surface curret ##\mathbf{J}(t,\mathbf{r}) = H(t) \delta(r-a) \sin \theta \; \mathbf{e}_\phi##

- Relevant Equations
- ##\mathbf{J}(t,\mathbf{r}) = H(t) \delta(r-a) \sin \theta \; \mathbf{e}_\phi##

Hello, Im trying to find the the solution of Maxwell equations for a source of an unevenly rotating sphere with the surface curret ##\mathbf{J}(t,\mathbf{r}) = H(t) j_0 \delta(r-a) \sin \theta \; \mathbf{e}_\phi##, ##H(t)## being an arbitrary function and ##j_0, a## positive constants.

There is a hint to use one of the two Hertz potentials and set the only the zth component non-zero.

The two Hertz potentials are ##\boldsymbol{\Pi}_e## and ##\boldsymbol{\Pi}_m##, which then relate, in the Lorenz gauge, to the electromagnetic potentials by

$$ \phi = - \nabla \cdot \boldsymbol{\Pi}_e, \quad \mathbf{A} = \frac{1}{c^2} \frac{\partial \boldsymbol{\Pi}_e}{\partial_t} + \nabla \times \boldsymbol{\Pi}_m$$

Since from the source term we get ##\mathbf{A} \propto \mathbf{e}_\phi## it is quite obvious, to use just the magnetic potential

$$ \boldsymbol{\Pi}_m = \Pi_m \mathbf{e}_z.$$

Now the magnetic Hertz potential satisfies the equation

$$ \nabla^2 \boldsymbol{\Pi}_m - \frac{1}{c^2} \frac{\partial \boldsymbol{\Pi}_m}{\partial t} = - \mu_0 \mathbf{M}.$$

Where ##\mathbf{M}## is the magnetization, related to the current by the equation ##\nabla \times \mathbf{M} = \mathbf{j}##, it's easy enough to find, that a solution to this is ##\mathbf{M} = H(t) j_0 r \delta(r-a) \cos \theta \; \mathbf{e}_r##.

This all I could do, I have no idea how to solve the equation for the Hertz vector, since it has two components non-zero, if we assume that ##\Pi_m = \Pi_m(t,r)## I get

$$ \cos (\theta ) \left(-\frac{\Pi _m{}^{(2,0)}(t,r)}{c^2}-\frac{2 \Pi _m{}^{(0,1)}(t,r)}{r}\right)=M_r,$$

$$ \sin (\theta ) \left(\frac{\Pi _m{}^{(2,0)}(t,r)}{c^2}+\frac{\Pi _m{}^{(0,1)}(t,r)}{r}+\Pi _m{}^{(0,2)}(t,r)\right)=0.$$

Could anyone give me some tips?

Also I want to emphasise that we are looking at physically relevant solutions, so that at some point in the calculation we probably assume, that the solution is regular at the origin and that it is an outgoing spherical wave at large ##r>>1##

There is a hint to use one of the two Hertz potentials and set the only the zth component non-zero.

The two Hertz potentials are ##\boldsymbol{\Pi}_e## and ##\boldsymbol{\Pi}_m##, which then relate, in the Lorenz gauge, to the electromagnetic potentials by

$$ \phi = - \nabla \cdot \boldsymbol{\Pi}_e, \quad \mathbf{A} = \frac{1}{c^2} \frac{\partial \boldsymbol{\Pi}_e}{\partial_t} + \nabla \times \boldsymbol{\Pi}_m$$

Since from the source term we get ##\mathbf{A} \propto \mathbf{e}_\phi## it is quite obvious, to use just the magnetic potential

$$ \boldsymbol{\Pi}_m = \Pi_m \mathbf{e}_z.$$

Now the magnetic Hertz potential satisfies the equation

$$ \nabla^2 \boldsymbol{\Pi}_m - \frac{1}{c^2} \frac{\partial \boldsymbol{\Pi}_m}{\partial t} = - \mu_0 \mathbf{M}.$$

Where ##\mathbf{M}## is the magnetization, related to the current by the equation ##\nabla \times \mathbf{M} = \mathbf{j}##, it's easy enough to find, that a solution to this is ##\mathbf{M} = H(t) j_0 r \delta(r-a) \cos \theta \; \mathbf{e}_r##.

This all I could do, I have no idea how to solve the equation for the Hertz vector, since it has two components non-zero, if we assume that ##\Pi_m = \Pi_m(t,r)## I get

$$ \cos (\theta ) \left(-\frac{\Pi _m{}^{(2,0)}(t,r)}{c^2}-\frac{2 \Pi _m{}^{(0,1)}(t,r)}{r}\right)=M_r,$$

$$ \sin (\theta ) \left(\frac{\Pi _m{}^{(2,0)}(t,r)}{c^2}+\frac{\Pi _m{}^{(0,1)}(t,r)}{r}+\Pi _m{}^{(0,2)}(t,r)\right)=0.$$

Could anyone give me some tips?

Also I want to emphasise that we are looking at physically relevant solutions, so that at some point in the calculation we probably assume, that the solution is regular at the origin and that it is an outgoing spherical wave at large ##r>>1##

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