# Circular Waveguide equation help

1. May 11, 2014

### shayaan_musta

Hello experts!

I am looking for the proof of the following equation:
$\frac{∂^{2}E}{∂r^{2}}$+$\frac{1}{r}$$\frac{∂Ez}{∂r}$+$\frac{1}{r^{2}}$$\frac{∂^{2}Ez}{∂ø^{2}}$+q²Ez=0

I think this equation is somehow related to the cylindrical waveguides. Right?

I am looking for it and I am unable to find any material on the internet. I googled but unsuccessful. Please if any one has any material then provide me or share with me here.
Thank you all

2. May 11, 2014

### jasonRF

3. May 16, 2014

### Claude Bile

Looks like the equation that emerges from expressing the Laplacian of the Helmholtz equation in cylindrical coordinates.

Claude.

4. May 27, 2014

### shayaan_musta

Is there any notes or book which have this proof in post#1?

5. May 27, 2014

### jasonRF

Yes, the waveguide chapter (chapter 9) in the link posted in my first reply. Did you bother to look at it?

6. May 28, 2014

### shayaan_musta

Yes, because I and the whole students of my class are studying this subject on their own because our teacher is so much old even we can't understand we he is trying to say us. We can't understand their words.

I am unable to find my equation from the chapter 9. That's why I did the post again and again you replied same. Can you point out the equation number in the chapter 9?

Thanks a lot.

7. May 29, 2014

### vanhees71

It's not too difficult to prove from Maxwell's Equations. First of all the equation refers to the special case of waves with harmonic time dependence. Since you can build any form of time dependence from it, using Fourier series or Fourier integrals that's not much of a restriction, and you can use the exponential form of the series or integral which simplifies the equations a lot. Thus we assume that the electromagnetic field is of the following form
$$\vec{E}(t,\vec{x})=\vec{E}_0(\vec{x}) \exp(-\mathrm{i} \omega t), \quad \vec{B}=\vec{B}_0(\vec{x}) \exp(-\mathrm{i} \omega t).$$
Here, I use the usual physicisist's convention with a minus sign in the exponent. Usually engineers use the opposite sign convention, but everything is of course equivalent at the end.

Now we plug this ansatz into Maxwell's equations for free fields, i.e., for vanishing charge and current densities. They read
$$\vec{\nabla} \cdot \vec{E}=0 \; \Rightarrow \; \vec{\nabla} \cdot \vec{E}_0=0, \qquad(1)$$
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0 \; \Rightarrow \; \vec{\nabla} \times \vec{E}_0 -\mathrm{i} k \vec{B}_0=0, \qquad (2)$$
$$\vec{\nabla} \cdot \vec{B}=0 \; \Rightarrow \; \vec{\nabla} \cdot \vec{B}_0=0, \qquad (3)$$
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=0 \; \Rightarrow \; \vec{\nabla} \times \vec{B}_0 + \mathrm{i} k \vec{E}_0=0. \qquad (4)$$
I used the abbreviation $k=\omega/c$. To get an equation for the electric field alone we take the curl of (2) and use (3):
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E}_0)-\mathrm{i} k \vec{\nabla} \times \vec{B}_0=\vec{\nabla} \times (\vec{\nabla} \times \vec{E}_0)-k^2 \vec{E}_0=0. \qquad(5)$$
Now for cartesian (and only cartesian!) coordinates you can write
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{E}_0)=\vec{\nabla}(\vec{\nabla} \cdot \vec{E}_0)-\Delta \vec{E}_0=-\Delta \vec{E}_0, \qquad(6)$$
where in the last step we made use of (1).

Plugging this into (5) we find the Helmholtz equation
$$\Delta \vec{E}_0+k^2 \vec{E}_0=0.$$
As emphasized above, to use this form of the equation you must use Cartesian coordinates. To write it in orthonormalized curvilinear coordinates, you either must go back to the equation (5) and work out the "double curl operator" or you rewrite everything in cartesian coordinates first. The latter way is often faster, because it already made use of the divergenceless of the electric field.

Let's do this calculation for cylinder coordinates as this is needed for waveguides with cross sections of circular symmetry (e.g., for the coax cable, which is the most important practical example).

We simply need to express the "curvilinear basis vectors" in terms of cartesian coordinates, i.e., use the definitions
$$\vec{e}_r=\vec{e}_x \cos \varphi+\vec{e}_y \sin \varphi, \quad \vec{e}_{\varphi}=-\vec{e}_x \sin \varphi+\vec{e}_y \cos \varphi,$$
In cylinder coordinates the third basis vector is simply the cartesian $\vec{e}_z$.

Now we write
$$\vec{E}_0=\vec{e}_r E_r + \vec{e}_{\varphi} E_{\varphi} + \vec{e}_z E_z = \vec{e}_x (E_r \cos \varphi - E_{\varphi} \sin \varphi)+\vec{e}_y (E_r \sin \varphi + E_{\varphi} \cos \varphi) + \vec{e}_z E_z.$$
Now you can apply the Laplace operator in terms of cylinder coordinates, which is a standard formula you can find in any proper textbook about vector calculus. Without further difficulties. For the $z$ component it's of course trivial, because this is already a Cartesian coordinate. This gives your equation, which is written in somewhat simplified form (an also note that you must write $E_z$ everytwhere):
$$\frac{1}{r} \frac{\partial}{\partial r} \left (r \frac{\partial E_z}{\partial r} \right ) + \frac{1}{r^2} \frac{\partial^2 E_z}{\partial \varphi^2} + \frac{\partial^2 E_z}{\partial z^2} + k^2 E_z=0. \qquad (7)$$

Now, since the problem is translation invariant in $z$ direction you can further write
$$\vec{E}_0(\vec{x})=\vec{E}_1(x,y) \exp(\mathrm{i} k_z z)+\vec{E}_2(x,y) \exp(-\mathrm{i} k_z z).$$
Plugging this ansatz into (7) you see that both $E_{1z}$ and $E_{2z}$ fulfill your equation with
$$q^2=k^2-k_z^2.$$

8. May 30, 2014

### jasonRF

vanhees71 gave a nice derivation. Also, in the link section 9.1 has the derivation - including showing how the z-components of E and H are all that are needed since the transverse components can be computed in terms of the z components (of course TM modes have Hz=0, and TE modes have Ez=0). The first part of Section 9.1 does the general derivation, then at the end of section 9.1 there are parts with headings like "Cartesian coordinates" and "cylindrical coordinates." The exact equation you posted is 9.1.23. It sounds like you could benefit from working through that entire section with pencil and paper in hand.

9. Oct 14, 2016

### JezuzStardust

Sorry for brining this thread up again, but I have made an interesting observation. The assumption that is made in the end

which basically says that

$$\frac{\omega^2}{c^2} - k_z^2 > 0$$

is made in all sources I have found about circular waveguide without ever motivating it. I believe that this assumption is made without too much hassle because in most source the case of a rectangular waveguide is solved before and for that case it is necessary that this condition holds.

However, if one solves the circular waveguide from scratch, there is no reason a priori to assume that the condition above holds. If one goes through the motion without this assumption then it can be seen that if $$\omega^2/c^2 - k_z^2 < 0$$ then instead of getting the Bessel equation, we get the modified Bessel equation. One can then quickly see that the solutions to the latter equation cannot fulfil the boundary conditions (one of the solutions blows up at the axis and the other is nonzero away from the axis so it cannot fulfil the boundary condition).