Plane wave solutions of Maxwell's equations

[Heirot]

In deriving the plane wave solutions of Maxwell's equations in vacuum, one assumes from the very start that the E and B field oscillate with the same frequency omega (cf. Jackson). This is a starting point for all further properties of plane waves. Can one start from two different frequencies, one for E, and one for B and then show that from consistency reasons, they have to be the same?

 

[Born2bwire]

Well... yes you could. But it is trivial. You can decouple the electric and magnetic fields. One such decoupling gives you vector wave equations for the electric and magnetic fields. Doing so allows you to analyze the electric field independently of the magnetic field. However, you always have an identical vector wave equation that requires the same frequency for the other field. Or you can show that they have to be the same by the relationship of the electric and magnetic fields via the curl operator.

In short, you can not truly decouple the two. You can separate them out and work on one or the other independently but you always derive the other field off of this field. So the dependency is still there, just moved to a later point in the evaluation.

 

[Phrak]

In deriving the plane wave solutions of Maxwell's equations in vacuum

Which vacuum condition? Can we still have a vacuum with current but no charge?

 

[Born2bwire]

Which vacuum condition? Can we still have a vacuum with current but no charge?

I think he just means an isotropic medium with unity relative permittivity and permeability. Whether or not it is source-free is immaterial since the sources are localized in the volume.

 

[Phrak]

I think he just means an isotropic medium with unity relative permittivity and permeability. Whether or not it is source-free is immaterial since the sources are localized in the volume.

No. I'm suggesting that there could be current in a particular vacuum solution but no sources. Or maybe I don't know what you mean.

 

[Heirot]

I was thinking of Maxwell's equation without currents and sources with epsilon = mu = 1. Also, I don't understand how the same vector equations for E and B imply the same frequency. All that can be said is that they propagate with the same speed = c.

 

[Vanadium 50]

Given an electric field plane wave, you can calculate the corresponding magnetic field plane wave. Then simply compare the two frequencies.

 

[Born2bwire]

No. I'm suggesting that there could be current in a particular vacuum solution but no sources. Or maybe I don't know what you mean.

I'm not quite sure what you mean. A current is a source. However, since the sources are localized in space, the evaluation of the field in a volume not containing the source is the same as if there were no sources at all (just treat the fields excited by the source as a new incident wave). Oh well, we're all marvelously confused I think.

I was thinking of Maxwell's equation without currents and sources with epsilon = mu = 1. Also, I don't understand how the same vector equations for E and B imply the same frequency. All that can be said is that they propagate with the same speed = c.

The vector wave equations are dependent upon the material properties of the medium (permittivity and permeability) and the frequency of the wave. So if you end up having the same vector wave equations, then it requires that the frequency must be the same since the constants of merit are the same. To be specific, the general vector wave equations are:

$$\nabla \times \overline{\mathbf{\mu}}^{-1} \cdot \nabla \times \mathbf{E}(\mathbf{r}) - \omega^2 \overline{\mathbf{\epsilon}} \cdot \mathbf{E}(\mathbf{r}) = i\omega \mathbf{J}(\mathbf{r}) - \nabla \times \overline{\mathbf{\mu}}^{-1} \cdot \mathbf{M}(\mathbf{r})$$
$$\nabla \times \overline{\mathbf{\epsilon}}^{-1} \cdot \nabla \times \mathbf{H}(\mathbf{r}) - \omega^2 \overline{\mathbf{\mu}} \cdot \mathbf{H}(\mathbf{r}) = i\omega \mathbf{M}(\mathbf{r}) - \nabla \times \overline{\mathbf{\epsilon}}^{-1} \cdot \mathbf{J}(\mathbf{r})$$

The sources are the electric and magnetic currents, J and M respectively. Now, if we assume an isotropic homogeneous medium then,

$$\nabla \times \nabla \times \mathbf{E}(\mathbf{r}) - k^2 \mathbf{E}(\mathbf{r}) = i\omega\mu \mathbf{J}(\mathbf{r}) - \nabla \times \mathbf{M}(\mathbf{r})$$
$$\nabla \times \nabla \times \mathbf{H}(\mathbf{r}) - k^2 \mathbf{H}(\mathbf{r}) = i\omega\epsilon \mathbf{M}(\mathbf{r}) - \nabla \times \mathbf{J}(\mathbf{r})$$

Where the wavenumber, k, is equal to \omega\sqrt{\epsilon\mu}. When we assume plane wave solutions we assume solutions of the type:

$$\mathbf{E}(\mathbf{r}) = \mathbf{E}_0 e^{i\mathbf{k}\cdot\mathbf{r}}$$

where the magnitude of the wavevector k is the wavenumber.

So even though we have decoupled the electric and magnetic fields mathematically, the common wavenumber in the two equations ensures that both fields must share the same frequency.

 

[Born2bwire]

Took longer to do my edit than I thought it would so I'll just put it up as a new post.

Quick note, we did assume that the fields were time harmonic with some frequency \omega. This allows us to convert the time derivatives of the fields to be factors of i\omega. In a way, we started enforcing that the fields had to have the same frequency from the beginning. Again, it just goes to show as to what I stated earlier. The fields are invariable coupled and this is always true throughout the physics of the problem. So we can obfuscate this coupling with fancy mathematics but we can't remove it.

I guess if we allow that they can have different frequencies then the results would be:

$$\nabla \times \nabla \times \mathbf{E}(\mathbf{r}) - \omega_E\omega_H \epsilon \mu \mathbf{E}(\mathbf{r}) = i\omega_H\mu \mathbf{J}(\mathbf{r}) - \nabla \times \mathbf{M}(\mathbf{r})$$
$$\nabla \times \nabla \times \mathbf{H}(\mathbf{r}) - \omega_E\omega_H\epsilon\mu \mathbf{H}(\mathbf{r}) = i\omega_E\epsilon \mathbf{M}(\mathbf{r}) - \nabla \times \mathbf{J}(\mathbf{r})$$

In a source-free region then the right hand sides are zero and we see that our new wavenumber is

$$k' = \sqrt{\omega_E\omega_H\epsilon\mu}$$

The group velocity, which we know is to be c and must be the same for the electric and magnetic fields, is taken as

$$v_g^{-1} = \frac{\partial k}{\partial \omega} = \sqrt{\epsilon\mu} = \frac{\partial \omega_E\omega_H}{\partial \omega} \frac{\sqrt{\epsilon\mu}}{2\sqrt{\omega_E\omega_H}}$$

So therefore,

$$\frac{\partial \omega_E\omega_H}{\partial \omega} = \frac{1}{2\sqrt{\omega_E\omega_H}}$$

Thus,

$$\omega = \sqrt{\omega_E\omega_H}$$

Since we stipulated that the electric field already has a frequency of \omega_E, then \omega_E = \omega_H = \omega for it all to be consistent.

 

[Heirot]

Oh, I see it now, thanks!

 

[Born2bwire]

Yeah, and another further more underlying reason is simply due to the fact that the sources induce both the electric and magnetic fields. In the above, the magnetic currents are fictitious (but they have computational value) and thus we see that the electric currents are both the source of the electric and magnetic fields. Thus, the frequency of the fields are both set by the same electric currents.

 

[yungman]

In deriving the plane wave solutions of Maxwell's equations in vacuum, one assumes from the very start that the E and B field oscillate with the same frequency omega (cf. Jackson). This is a starting point for all further properties of plane waves. Can one start from two different frequencies, one for E, and one for B and then show that from consistency reasons, they have to be the same?

I thought if varying E or H field, it automatically generate the counter H or E resp!??

If you have an E1 with freq f1 and B2 with freq f2. They will generate H1 of f1 and E2 of f2 resp. There is no varying E or H alone. These are two complete different plane waves and has nothing to do with each other.

Yes, you can go through all the calculation on the interference between the two and get all the mixing effect like post #9 from Born2Wire. But they are completely different.

Please tell me whether I am correct.

 

[Meir Achuz]

Curl E~-dB/dt. If B~exp(-iwt), then E~exp(-iwt).

 

[Born2bwire]

I thought if varying E or H field, it automatically generate the counter H or E resp!??

If you have an E1 with freq f1 and B2 with freq f2. They will generate H1 of f1 and E2 of f2 resp. There is no varying E or H alone. These are two complete different plane waves and has nothing to do with each other.

Yes, you can go through all the calculation on the interference between the two and get all the mixing effect like post #9 from Born2Wire. But they are completely different.

Please tell me whether I am correct.

You can't label a causality on it. It comes down simply that if you have a time-varying magentic field then there is a time-varying electric field (and vice-versa). One does not cause the other as they are both produced simultaneously by the sources. Jefimenko's Equations help make this explicit.

 

[yungman]

You can't label a causality on it. It comes down simply that if you have a time-varying magentic field then there is a time-varying electric field (and vice-versa). One does not cause the other as they are both produced simultaneously by the sources. Jefimenko's Equations help make this explicit.

I thought that is what the two Maxwell's curl equations meant.

$$\nabla X \vec E = -\frac {\partial \vec B}{\partial t}$$

means in the presence of varying B field, there exist a circulating E field.

And vise versa. So you make one the other one follow. I don't see the difference, they co-exist.

For example, a moving magnetic, the varying B absolute causing the E to exist.

 

[Born2bwire]

I thought that is what the two Maxwell's curl equations meant.

$$\nabla X \vec E = -\frac {\partial \vec B}{\partial t}$$

means in the presence of varying B field, there exist a circulating E field.

And vise versa. So you make one the other one follow. I don't see the difference, they co-exist.

For example, a moving magnetic, the varying B absolute causing the E to exist.

You are introducing a causal relationship. That is, the varying magnetic field causes the varying electric field. This is not true. They both exist simultaneously (and considering the underlying physics we should understand that sources produce fields, the fields do not produce fields themselves). A source does not produce a magnetic field nor an electric field. It would be more accurate to say that a source produces an electromagnetic field. Only in the static case do the two become decoupled. Maxwell's Equations provide the relationship between the fields but one cannot say that the right hand side of one equation produces the left hand side any more than the reverse. One can reformulate Maxwell's Equations to a more explicit formulation. Jefimenko's Equations is one such way that explicitly shows how the sources produce the fields. Under these equations, we see that any time varying sources produce simultaneously an electric and magnetic field. So one cannot state that one causes the other, they exist together.

 

[yungman]

OK, thanks.

 

[Phrak]

I'm not quite sure what you mean. A current is a source. However, since the sources are localized in space, the evaluation of the field in a volume not containing the source is the same as if there were no sources at all (just treat the fields excited by the source as a new incident wave). Oh well, we're all marvelously confused I think.

I may see what's going on. I'm using the so called microscopic Maxwell equations. I'm not familiar with the macroscopic variety you are using. In the micro variant there may be currents in a region of space but no charge.

Next. This,

$$\nabla \times \nabla \times \mathbf{E}(\mathbf{r}) - k^2 \mathbf{E}(\mathbf{r}) = i\omega\mu \mathbf{J}(\mathbf{r}) - \nabla \times \mathbf{M}(\mathbf{r})$$

is a bit confusing. Where one term has a k², I expect to see an omega² on another term. A typo, maybe? --or is it because the temporal domain is suppressed.

 

[Born2bwire]

I may see what's going on. I'm using the so called microscopic Maxwell equations. I'm not familiar with the macroscopic variety you are using. In the micro variant there may be currents in a region of space but no charge.

Next. This,

$$\nabla \times \nabla \times \mathbf{E}(\mathbf{r}) - k^2 \mathbf{E}(\mathbf{r}) = i\omega\mu \mathbf{J}(\mathbf{r}) - \nabla \times \mathbf{M}(\mathbf{r})$$

is a bit confusing. Where one term has a k², I expect to see an omega² on another term. A typo, maybe? --or is it because the temporal domain is suppressed.

Hmmm... well we are assuming time-harmonic solutions so there is an implicit factor of exp(-i\omega t}. The k^2 term has an \omega^2 factor in it. Otherwise as I look at the equation it looks correct to me.

 

[Phrak]

Hmmm... well we are assuming time-harmonic solutions so there is an implicit factor of exp(-i\omega t}. The k^2 term has an \omega^2 factor in it. Otherwise as I look at the equation it looks correct to me.

OK, I guess that makes sense, then. But the temporal dependence is missing, so I assume E(x) = E(x, t=t₀).

The disconcerting part is that we talk in almost two distinct languages. And it's only electromagnetism! You have B's, E's, H's, M's, etc. whereas I jump immediately from the microscopic Maxwell equations to four dimensional (and complex) antisymmetric tensors with lower index notation. The harmonic equations in vacuum look like (*+i)d*d*d\zeta=0 to me.

 

[Born2bwire]

OK, I guess that makes sense, then. But the temporal dependence is missing, so I assume E(x) = E(x, t=t₀).

The disconcerting part is that we talk in almost two distinct languages. And it's only electromagnetism! You have B's, E's, H's, M's, etc. whereas I jump immediately from the microscopic Maxwell equations to four dimensional (and complex) antisymmetric tensors with lower index notation. The harmonic equations in vacuum look like (*+i)d*d*d\zeta=0 to me.

Ja ja. It's implicitly assumed that in the above equations that,
$$\mathbf{E}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})e^{-i\omega t}$$

Hmmm...

(*+i)d*d*d\zeta=0

Crazy gibberish.

I did forget to mention that the charge is found via the continuity equation. It is inherently included as part of the divergence part of the currents. Still, when we do computation it is sometimes better in terms of numerical stability to enforce the continuity equation explicitly.

 

[Phrak]

Crazy gibberish.

It's just different notation. Although yours is surely more useful in application, I'm sure.

Charge continuity is especially concise: $$dJ = \partial_\mu J_\nu - \partial_\nu J_\mu = 0$$

I did forget to mention that the charge is found via the continuity equation. It is inherently included as part of the divergence part of the currents. Still, when we do computation it is sometimes better in terms of numerical stability to enforce the continuity equation explicitly.

I vaguely recall something about using or attempting to use the charge continuity equation to simplify derivation of the wave equations in E and B. What did you mean?

 

[Born2bwire]

It's just different notation. Although yours is surely more useful in application, I'm sure.

Charge continuity is especially concise: $$dJ = \partial_\mu J_\nu - \partial_\nu J_\mu = 0$$

I vaguely recall something about using or attempting to use the charge continuity equation to simplify derivation of the wave equations in E and B. What did you mean?

Yes, I remember the dark days of graduate EM in the physics department where we wrestled with Jackson and his cryptic four vectors. I always found the difference between the electrical engineering department's and physics department's versions of graduate EM courses rather interesting.

We mainly do electromagnetics and so we are concerned with mainly time-varying signals. In this case the time-varying charge produces the irrotational currents (since the divergence of the current is equal to the charge scaled by frequency). So for us, the only sources are electric (and fictitious magnetic when it suits us) currents. For the most part we can solve for the fields by working with computional methods based upon the aforementioned electric and magnetic vector wave equations without including any explicit charge terms.

Theoretically this is fine but we run into numerical troubles at low frequencies. There are a couple reasons for this but one reason is that the contribution from the irrotational currents (the charge) starts to overpower the contributions from the currents. That is, the contribution to the field from the scalar potential \phi overpowers the contribution from the vector potential A. So there are a couple of low frequency methods that try to deal with this and some of them need to explicitly enforce the continuity equation for added numerical stability (This is in a formulation that explicitly separates the currents and charges and we solve for them as independent variables. As such we need to make sure that certain conditions, like charge continuity and even charge conservation, are held.).