# Maxwell's equations and longitudinal waves

1. Mar 27, 2015

### Ras9

Hi guys, I am having hard times in understanding whether or not the longitudinal electromagnetic waves are solutions to Maxwell's equations. In Cohen-Tannoudji "Introduction to QED" it's stated that by writing the fields as the sum of a longitudinal and transverse part one can show that waves solutions are purely transverse- Given a generic field $$\vec{V}(\vec{r})=\vec{V}_T(\vec{r})+\vec{V}_L(\vec{r})$$ the properties $$\nabla \cdot \vec{V}_T =0$$ $$\nabla \times \vec{V}_L =0$$ are useful, applied to the electric and magnetic field, to show that the longitudinal part of the electric field is connected to the charge distribution, and the one of the magnetic field is 0 while the two transverse parts are connected to waves propagation. Of course this is true only in free space; in waveguides or optical fibers there is also a longitudinal part of one of the two field (TE or TM polarization). So my first question is: where in this reasoning (that seems to work quite fine to me) has been assumed that the fields are in free space?
Moving to the case of confined geometry and following Jackson/Griffiths reasoning one arrives to show that the transverse fields can be expressed as a function of the longitudinal ones: (I will write only the electric one)
$$E_x = {i \over (\omega/c)^2 - k^2} (k \partial_x E_z + \omega \partial_y B_z)$$
$$E_y = {i \over (\omega/c)^2 - k^2} (k \partial_y E_z - \omega \partial_x B_z)$$
Again this reasoning seems pretty general to me and is derived by Maxwell's equations too. I do not understand neither the connection between the two neither how one can have TEM waves if the transverse part depends on the longitudinal one that is supposed to be 0.
It would be clear for me to associate the longitudinal part of the field to the surface charge and current in case of conductors waveguide but then the problem would be on dielectric waveguides.
I think I am (obviously) missing something, and I will appreciate if you can help me! My goal would to derive existence or not of the longitudinal field just working on Maxwell and boundary conditions! Thanks :)

2. Mar 27, 2015

### guedman

only the tranverse solution is a solution in the VACUUM but in material the two kinds can exist

3. Mar 28, 2015

### Ras9

Yes I understand this, but why is it so? Is this property contained in Maxwell's equations?

4. Mar 28, 2015

### Delta²

yes you can prove starting from maxwell's equations and almost pure mathematical reasoning that in the vacuum (where charge and current density equals zero) the electric and magnetic fields satisfy each the wave equation and that are perpendicular to each other and to the propagation vector k.

Last edited: Mar 28, 2015
5. Mar 28, 2015

### Ras9

Yes I agree, and this is basically what I showed in the fist part of my post. What is not clear to me is why if there is confinement in transverse direction there is also longitudinal field and how this results from maxwells

6. Mar 28, 2015

### Delta²

Well not sure, can you post me a link with the case of confined geometry as treated by Jackson/Griffiths. Probably what happens is that the electric field induces charge and current density in the materials that define the confinement and from charge and current density we have the longitudinal fields.

7. Mar 28, 2015

### Ras9

They assume the case of an hollow cilinder made of a perfect conductor. By expressing the fields as $$E=E(x,y)e^{i(kz-\omega t)}$$ (same for B) they derive from Maxwells this 6:
$$\partial_x E_y - \partial_y E_x = i\omega B_z$$
$$\partial_y E_z -ikE_y=i \omega B_x$$
$$ikE_x - \partial_x E_z = i \omega B_y$$
$$\partial_x B_y - \partial_y B_x = -i(\omega/c^2) E_z$$
$$\partial_y B_z -ikB_y=-i (\omega/c^2) E_x$$
$$ikB_x - \partial_x B_z = -i (\omega/c^2) E_y$$
By solving for the transverse component one find the expression I posted in my first post for both the fields. Now if both E_z and B_z are 0 from gauss $$\partial_x E_x + \partial_y E_y =0$$ while from Faraday $$\partial_x E_x - \partial_y E_y =0$$ and "Indeed, the vector E in has zero divergence and zero curl. It can therefore be written as the gradient of a scalar potential that satisfies Laplace's equation. But the boundary condition on E requires that the surface be an equipotential, and since Laplace's equation admits no local maxima or minima, this means that the potential is constant throughout, and hence the electric field is zero-no wave at all."
What I cannot understand is how from the set of those 6 equations I can separate the pure transverse field of the vacuum case and the longitudinal of the confined one. i hope i have been clear and to haven't done a mess!

8. Mar 28, 2015

### Delta²

well seems to me that "what causes the evil" is that we initially assume that the electric field follows this specific form $E=E(x,y)e^{i{(kz-wt)}}$.Is there any justification in the text on why we assume this form and if this form implies somehow that there are current and charge densities on the boundary surface?

9. Mar 28, 2015

### Ras9

No particular reason, simply that the spatial variation of the field happens in the z direction since it's confined in the other two.

10. Mar 28, 2015

### Delta²

well what can i say, all i know is that in the derivation i ve seen that the electromagnetic wave is transverse in vaccum we assume a diferent form for the electric field, that is $E=E_0f(kr-vt)$ where E_0 is a constant vector independent of x,y,z,t and f is any function of one variable.

11. Mar 28, 2015

### Delta²

Seems to me that there arent always pure transverse solutions even in the unconfined vaccum. For example in the electric dipole radiation case, and in polar coordinates, in the far field it might be only the $E_{\theta}$ and $B_{\phi}$ that survive because they fall off in amplitude as 1/r but still there is the component $E_{r}$ which falls off as 1/r^2 and essentialy is a longitudinal component. Ofcourse one will argue that responsible for the radiating energy are the transverse components , while the longitudinal component carries only reactive energy in the near field .

12. Mar 29, 2015

### Ras9

I agree with you. I think that when one says that the field is purely transverse is automatic assuming that the distance from the source charge is such that the longitudinal component is negligible (and is can been seen also by the fact that the purely transverse field is derived from putting to 0 the divergence of the electric field)