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The electric field (for example) in a source free volume obeys the following equations for monochromatic waves according to Maxwell:

$$ ( \nabla^2 + k^2) \vec{E}=0 \tag{1}$$

and

$$ \mathrm{div}\vec{E} = 0 \tag{2}$$

Now there is the claim in books about optics, that this can be (for example if we dont care about polarization) approximated by the Helmholtz equation

$$ (\nabla ^2 + k^2) u = 0 \tag{3}$$

usually it is alluded to that each component of the orig. eq. fulfills this condition (true), so if one component is predomenant, than one just takes this. But I think, this could violate badly the divergence free condition.

Some books claim, that from a solution of the scalar Helmholtz equation one can construct a solution of the maxwell equation by setting

$$ \vec {E} = \nabla \times ((x,y,z) u) \tag {4} $$

which is true, but somehow misses the point, since this u is in no way an approximation to $E$.

So in what sense is $u$ related to $E$ in the sense that it gives an approximation? To clarify: the question is not about eq. 4, which I believe is a false approach in some literature.

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The art of going from scalar optics to a full solution of the Maxwell equations is known as vector optics; it is a very wide field with multiple different methods, and ultimately which one you choose depends on the situation. Ultimately, the underlying theme is your

Question: Given a solution $u(\mathbf r)$ of the scalar wave equation $\frac{\partial^2u}{\partial t^2}=c^2\nabla^2 u$, how can one obtain a full set of vector solutions to the Maxwell equations?

As you note, taking $\mathbf E(\mathbf r) = \mathbf r \, u(\mathbf r)$ is not going to cut it, and neither is the constant-polarization solution $\mathbf E(\mathbf r) = \mathbf E_0 u(\mathbf r)$, which will fail to have a zero divergence if $u(\mathbf r)$ has any dependence along $\mathbf E_0$, so you need to look elsewhere.

For this, there's a number of things you can try. (Most of these assume that you start with a solution of the Helmholtz equation $(\nabla^2 +k^2)u=0$ rather than with full time dependence, but you can roll back from there through a Fourier transform if you really need to.) So, some techniques to manufacture vector solutions of the Helmholtz equation out of known solutions:

  • Multiply a scalar solution $u(\mathbf r)$ by a constant polarization $\mathbf E_0$ to get $\mathbf E(\mathbf r) = \mathbf E_0 u(\mathbf r)$.
  • Take the gradient $\mathbf E(\mathbf r) = \nabla u(\mathbf r)$ of a scalar solution $u(\mathbf r)$.
  • Take the cross product of an existing vector solution $\mathbf E_0(\mathbf r)$ with some constant vector $\mathbf c$ to get $\mathbf E(\mathbf r)= \mathbf c\times\mathbf E_0(\mathbf r)$.
  • Take the curl of an existing vector solution $\mathbf E_0(\mathbf r)$ to get $\mathbf E(\mathbf r)= \nabla \times\mathbf E_0(\mathbf r)$.

Depending on the situation, most of these won't work, but some of these might work (and indeed, you might need to daisy-chain several of these methods to get the solutions you want; thus, one set of lecture notes in my back archive uses things like $\nabla\times\nabla\times\left(\mathbf c \, u(\mathbf r)\right)$ as one of its solutions). Choosing a method that will give you a solution that is simple to work with and appropriate to your problem can be a bit of an art, and again it depends on the situation. Similarly, depending on the regime (like, say, a loosely focused gaussian beam), several of the techniques above which look like they don't work (like a constant polarization) can work rather well as approximations - but only on a case-by-case basis.

What this means, ultimately, is that the relationship between the solutions of vector optics and the simpler solutions of scalar optics will depend on the problem, and there is no one-size-fits-all description of that relationship. This then percolates to the fact that generally speaking $u(\mathbf r)$ is not an approximation for $\mathbf E(\mathbf r)$: instead, it is an auxiliary quantity on the way up the ladder towards a full vector solution of the Maxwell equations.

To be a bit more precise, $u(\mathbf r)$ isn't just an auxiliary quantity ─ it's usually something that you can derive from the full vector solution, normally by just taking one of its components, since if you have a full vector solution $\mathbf E(\mathbf r)$ then all of its cartesian components will be wave-equation solutions. That means that while $u(\mathbf r)$ isn't sufficient to construct a $\mathbf E(\mathbf r)$, it is in a sense necessary, and often we can get much of the relevant insight about our desired vector solutions just from the behaviour of the scalar solutions in the geometry of interest. Thus, even when you know that it is only a partial description of your system, it is still useful to study the dynamics of scalar waves in whatever geometry you're working.

On the other hand, however well you understand the scalar-wave dynamics, you still need to construct a full vector solution, and if you want a rough-and-ready most-common-case answer for how this gets done, though, it turns out to be the first one above, which you mentioned in the question: the constant-polarization solution $$\mathbf E(\mathbf r) = \mathbf E_0 u(\mathbf r).$$ It is easy to see that this is a solution to the wave equation, but as you say,

this could violate badly the divergence free condition.

However, "badly" is a relative term, and - particularly if the characteristic dimension $a$ of the problem are large compared to the wavelength - it can work just fine as an approximation. If you really want a 'true' solution in those cases, you can find them out as above, and then you recover the scalar-optics constant-polarization solution as the limit $\lambda\ll a$ of the true solutions. (In this connection, see my answer to this question and its references for examples of vector optics in action.) However, most often, it's really not necessary to go through all that hassle, because the scalar-optics solution works perfectly fine to explain experiment, and that is the ultimate goal.

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I saw in one of your comments that you were looking for literature on Fourier optics, this is a paper that I can recommend: Vector Fourier optics of anisotropic materials by Robert R. McLeod and Kelvin H. Wagner.

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  • $\begingroup$ Hello Stephen, and welcome to Physics.SE! Note that we tend to discourage link-only answers. While you don't yet have the reputation to post a comment (where this sort of content belongs), it would be much appreciated if you add an explanation of this content and how the OP could use it to answer his/her question. Thanks in advance! $\endgroup$ – Dave Coffman May 11 '17 at 2:22
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Helmholtz Equation

The general wave equation takes the form: $$ \nabla^{2} \Psi - \frac{1}{c^{2}} \frac{\partial^{2} \Psi}{\partial t^{2}} = -4 \ \pi \ f\left( \mathbf{x}, t \right) \tag{1} $$ for source distribution $f\left( \mathbf{x}, t \right)$ and $c$ is the speed of light in vacuum. If we take the Fourier transform of both $\Psi\left( \mathbf{x}, t \right)$ and $f\left( \mathbf{x}, t \right)$ under the linear approximation (i.e., $\partial_{t}$ $\rightarrow$ -i$\omega$), then we find: $$ \left( \nabla^{2} - \frac{(-i \ \omega)^{2}}{c^{2}} \right) \Psi\left( \mathbf{x}, \omega \right) = -4 \ \pi \ f\left( \mathbf{x}, \omega \right) \tag{2} $$ which is the inhomogeneous Helmholtz wave equation.

In the absence of sources $f\left( \mathbf{x}, \omega \right)$ $\rightarrow$ 0 and for free modes we know that $k$ = $\omega$/$c$, then we see that Equation 2 reduces to $$ \left( \nabla^{2} + k^{2} \right) \Psi\left( \mathbf{x}, \omega \right) = 0 \tag{3} $$ which is the standard form of the Helmholtz wave equation.

Wave Equation

Using Maxwell's equations, we know that we can define $\mathbf{B} = \nabla \times \mathbf{A}$ (i.e., from $\nabla \cdot \mathbf{B} = 0$). We can then use Faraday's law to show that: $$ \nabla \times \left( \mathbf{E} + \frac{\partial \mathbf{A}}{\partial t} \right) = 0 \tag{4} $$ We know from vector calculus than any curl-free vector can be written as the gradient of a scalar, so we define: $$ \begin{align} \mathbf{E} + \frac{\partial \mathbf{A}}{\partial t} & = -\nabla \phi \tag{5a} \\ \mathbf{E} & = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t} \tag{5b} \end{align} $$ which is the standard form for the electric field.

We can use something called the Lorentz gauge to show that: $$ \nabla \cdot \mathbf{A} + \frac{1}{c^{2}} \frac{\partial \phi}{\partial t} = 0 \tag{6} $$ which derives from the arbitrariness of the vector and scalar potentials $\mathbf{A}$ and $\phi$, respectively.

We can then use Equation 6 and Maxwell's equations (in the source-free limit, i.e., $\rho = 0$ and $\mathbf{j}$ = 0) to show that: $$ \begin{align} \nabla^{2} \phi - \frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}} & = 0 \tag{7a} \\ \nabla^{2} \mathbf{A} - \frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} & = 0 \tag{7b} \end{align} $$

Upon inspection, one can see that a Fourier transform and similar approach as used to derive Equation 3 above will result in a similar form for Equations 7a and 7b to the standard Helmholtz equation.

In the source-free Coulomb gauge (i.e., $\phi$ = 0 and $\nabla \cdot \mathbf{A} = 0$), we know that $\mathbf{E} = -\tfrac{\partial \mathbf{A}}{\partial t}$. Then we can use the fact that partial derivatives commute to end with: $$ \frac{\partial}{\partial t} \left( \nabla^{2} \mathbf{E} - \frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \right) = 0 \tag{8} $$ and after a Fourier transform and few steps of algebra we can show that: $$ \left( \nabla^{2} + k^{2} \right) \mathbf{E} = 0 \tag{9} $$

Your Question

If we can define $\mathbf{E} = \nabla \times \mathbf{u}$, then it follows from vector calculus that $\nabla \cdot \mathbf{E} = 0$. If we take the divergence of Equation 5b above, we find: $$ -\nabla^{2} \phi - \frac{\partial}{\partial t} \left( \nabla \cdot \mathbf{A} \right) = 0 \tag{10} $$ and we use Equation 6 above to replace the $\nabla \cdot \mathbf{A}$ term to find: $$ \nabla^{2} \phi - \frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}} = 0 \tag{11} $$ which is the wave equation for the electric potential. Thus, by defining the electric field as the curl of some other field, we can re-derive the wave equation (i.e., Equation 7a above). We can then follow similar steps to those shown above to get the Helmholtz equation.

Answer

One of the simplest physical examples for the meaning of $\mathbf{u}$ above would come from a form of Ohm's law given by: $$ \mathbf{E} = \eta \ \mathbf{j} \tag{12} $$ where $\eta$ is some scalar resistivity and $\mathbf{j}$ is the current density. One could then use Ampere's law to show that $\mathbf{E} \propto \nabla \times \mathbf{B}$, which represents a similar form to $\mathbf{E} = \nabla \times \mathbf{u}$. Thus, unit-wise the function $\mathbf{u}$ would be magnetic field times resistivity divided by the permeability (i.e., $\mathbf{u} = \tfrac{\eta \ \mathbf{B}}{\mu_{o}}$), or energy per unit charge.

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  • $\begingroup$ Why the down vote? $\endgroup$ – honeste_vivere May 10 '17 at 16:20
  • $\begingroup$ the downvote wasnt me. Everything up to eq. 9 seems unrelated to the question (nice derivation though). So what is u? $\endgroup$ – lalala May 10 '17 at 16:27
  • $\begingroup$ So you are looking for some sort of physical quantity or parameter in nature that one can measure? $\endgroup$ – honeste_vivere May 10 '17 at 16:58
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    $\begingroup$ @lalala - I updated my answer to provide a specific, physical, example that would provide the mathematical form similar to what you ask. However, this would not be relevant to an electromagnetic wave propagating in free space. Though I imagine Emilio's answer is more general. $\endgroup$ – honeste_vivere May 10 '17 at 18:10

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