When can I use Helmholtz equation for electromagnetics

[Jeffrey Yang]

The complete Maxwell wave equation for electromagnetic field using the double curl operator "∇×∇×". Only when the transverse condition is hold, this operator can equal to the Laplace operator and form the helmholtz.

My question is what's the condition can we use the helmoltz equation instead of the double curl operator "∇×∇×" one.

If there is no charge (both the free and polarized one), then the electric field is transverse. So even in the piece-wise homogeneous system, we can still use the helmoltz equation in that medium. Is this correct? If this is true, that means once the dielectric function is abrupt changed, we can always use the helmholtz equation to solve the source free mode in one of the homogeneous region and then use the boundary condition to get the overall distribution of the system. Is this right?

So is there any other situation that we have to use the double curl operator "∇×∇×" one apart from the gradual change of the dielectric function?

Thanks a lot

 

[Jano L.]

The complete Maxwell wave equation for electromagnetic field using the double curl operator "∇×∇×". Only when the transverse condition is hold, this operator can equal to the Laplace operator and form the helmholtz.

My question is what's the condition can we use the helmoltz equation instead of the double curl operator "∇×∇×" one.

If there is no charge (both the free and polarized one), then the electric field is transverse. So even in the piece-wise homogeneous system, we can still use the helmoltz equation in that medium. Is this correct? If this is true, that means once the dielectric function is abrupt changed, we can always use the helmholtz equation to solve the source free mode in one of the homogeneous region and then use the boundary condition to get the overall distribution of the system. Is this right?

So is there any other situation that we have to use the double curl operator "∇×∇×" one apart from the gradual change of the dielectric function?

Thanks a lot

Since
$$
\nabla\times\nabla\times \mathbf E = \nabla\nabla\cdot \mathbf E - \Delta \mathbf E
$$
for any field $\mathbf E$, you can replace it by $-\Delta$ if and only if $\nabla \cdot \mathbf E = const(\mathbf x)$, i.e. charge density is independent of position.

This is the case for homogeneous isotropic linear dielectric, as can be seen in the following explanation. Because we have dielectric,
$$
\nabla\cdot \mathbf D = 0
$$

and because of isotropicity and linearity,
$$
\mathbf D = \epsilon \mathbf E
$$
where $\epsilon$ is a constant independent of position. Taking divergence of both sides, we obtain $\nabla \cdot \mathbf E = 0$.

This conclusion is not valid for homogeneous linear dielectric that is anisotropic. For example, quartz has different dielectric constants in different directions. One can use dielectric tensor to describe linear relation between electric displacement field and electric field:

$$
\mathbf D = \mathbf \epsilon \cdot \mathbf E.
$$
Now, if we take divergence of both sides, we obtain the equation

$$
0 = \sum_i \partial_i\left(\sum_k \epsilon_{ik} E_k\right).
$$

Taking into account homogeneity of the medium, we can drop derivatives tensor components:
$$
0 = \sum_i \left(\sum_k \epsilon_{ik} \partial_i E_k\right).
$$
Since the tensor must be symmetric (I think due to laws of thermodynamics), there is a coordinate system where the tensor is diagonal and referring to this system we can write this equation as
$$
0 = \sum_i \left(\epsilon_{i} \partial_i E_i\right)
$$
where $\epsilon_i$ are the principal dielectric constants (in three perpendicular directions, main axes of the crystal). This means that, if all principal dielectric constant are the same, divergence of $\mathbf E$ vanishes, but if they differ, in general the electric field can have non-vanishing divergence.

Thus in anisotropic medium, space charge of bound nature can occur and if it is not of constant density (which may happen if EM waves propagate in the medium), $\nabla\times\nabla\times$ cannot be reduced to -laplacian.

 

[Jeffrey Yang]

Since
$$
\nabla\times\nabla\times \mathbf E = \nabla\nabla\cdot \mathbf E - \Delta \mathbf E
$$
for any field $\mathbf E$, you can replace it by $-\Delta$ if and only if $\nabla \cdot \mathbf E = const(\mathbf x)$, i.e. charge density is independent of position.

This is the case for homogeneous isotropic linear dielectric, as can be seen in the following explanation. Because we have dielectric,
$$
\nabla\cdot \mathbf D = 0
$$

and because of isotropicity and linearity,
$$
\mathbf D = \epsilon \mathbf E
$$
where $\epsilon$ is a constant independent of position. Taking divergence of both sides, we obtain $\nabla \cdot \mathbf E = 0$.

This conclusion is not valid for homogeneous linear dielectric that is anisotropic. For example, quartz has different dielectric constants in different directions. One can use dielectric tensor to describe linear relation between electric displacement field and electric field:

$$
\mathbf D = \mathbf \epsilon \cdot \mathbf E.
$$
Now, if we take divergence of both sides, we obtain the equation

$$
0 = \sum_i \partial_i\left(\sum_k \epsilon_{ik} E_k\right).
$$

Taking into account homogeneity of the medium, we can drop derivatives tensor components:
$$
0 = \sum_i \left(\sum_k \epsilon_{ik} \partial_i E_k\right).
$$
Since the tensor must be symmetric (I think due to laws of thermodynamics), there is a coordinate system where the tensor is diagonal and referring to this system we can write this equation as
$$
0 = \sum_i \left(\epsilon_{i} \partial_i E_i\right)
$$
where $\epsilon_i$ are the principal dielectric constants (in three perpendicular directions, main axes of the crystal). This means that, if all principal dielectric constant are the same, divergence of $\mathbf E$ vanishes, but if they differ, in general the electric field can have non-vanishing divergence.

Thus in anisotropic medium, space charge of bound nature can occur and if it is not of constant density (which may happen if EM waves propagate in the medium), $\nabla\times\nabla\times$ cannot be reduced to -laplacian.

Thanks a lot

 

[Jano L.]

Thanks a lot

You're welcome.