# The role of the separation constant when solving the wave equation for electromagnetic waves & Cut off wave number

I have the following wave equation that I need help to solve via separation of variables: $$\nabla^2 \mathbf{E} + k^2 \mathbf{E} = 0$$ Where E is the electric field and k is the wave number

Using separation of variables for each component:

$$E_i(x,y,z) = f(x)g(y)h(z) \text{ , where i = x , y , z}$$

For a single component:

$$(f_{xx})(g)(h) + (f)(g_{yy})(h) + (h_{zz}) + k^2(fgh) = 0$$

or $$f_{xx}/f + g_{yy}/g + h_{zz}/h + k^2 = 0$$

My physics reader skips a bunch of steps and writes it as 3 separate equations:

$$f_{xx}/f = -k_x$$ $$g_{yy}/g = -k_y$$ $$h_{zz}/h = -k_z$$

and defines the wave number vector k. I think we assume a complex solution since we get a final result of: $$E = e^{ - \mathbf{k} \circ \mathbf {r} }$$

My engineering reader writes the same equations as:

$$f_{xx}/f = -k_x$$ $$g_{yy}/g = -k_y$$ $$h_{zz}/h = \gamma$$

where gamma is the propagation constant and gives the following equation

$$-k_x - k^y + \gamma^2 = - k^2$$

Question 1: why can the propagation constant (a complex number) have a different sign? For I thought: $$-k^2 = - || \mathbf{k} ||^2 = -(k_x^2 + k_y^2 + \gamma^2)$$

What in the mathematics is going on?

Question 2: Why is the wave equation not written/solved using the following form

$$\nabla^2 \mathbf{E} + k_c^2 \mathbf{E} = 0$$

if other source state $$k_{cut}^2 = \gamma^2 + k^2$$ where gamma is the propagation constant, k is the wavenumber and kc is the cut off wave number?

For the record, they simplified it to and then skipped to the final result, leaving me out of the derivation

$$\nabla^2 \mathbf{E} + (\gamma^2 + k^2)^2 \mathbf{E} = 0$$

I do not get the 'correct' answer, which is:

$$H_x = \frac{-1}{K_c^2}( \gamma \frac{d}{dx} H_z + j \omega \epsilon \frac{d}{dy} E_z)$$

Question 3: when do we use:

$$\nabla^2 \mathbf{E} + k^2 \mathbf{E} = 0$$

versus

$$\nabla^2 \mathbf{E} + \gamma^2 \mathbf{E} = 0$$

Notation:

$$\gamma = \alpha + j \beta$$

$$k = - j \alpha + \beta$$

1. alpha is the attenuation constant
2. beta is the phase constant / "propagation wave number"

Edits As requested. The book is elements of electromagnetism

• Are you sure $-k_x - k^y + \gamma^2 = - k^2$ is correct?
– Gert
Commented May 31, 2021 at 21:54
• yes! check the reader that i've posted Commented May 31, 2021 at 23:09

Separation of variables allows one to introduce three constants (which may be imaginary) which obey $$\alpha^2 + \beta^2 + \gamma^2 = -k^2.$$ One could of course define these three constants differently so that the signs change. They choose to write $$-k_x^2-k_y^2 + \gamma^2 = -k^2$$ because they anticipate that the boundary conditions they will impose later will be such that with this choice of sign, the constants $$k_x,k_y$$ and $$\gamma$$ will be real.

As your book says in equation (12.7): $$-k_{x}^2-k_{y}^2+\gamma^2 = -k^2$$ by multiplying with $$-1$$ and taking the square root this is equivalent to: $$k = \sqrt{k_{x}^2+k_{y}^2-\gamma^2}$$ notice you made a sign error for $$\gamma$$ in your equation. Now if: $$k_{x}^2+k_{y}^2 < \gamma^2$$ then $$k$$ is imaginary otherwise it is real. The existence of $$j$$ (square root of -1) or its absence affects the sign of $$k$$. Now keep in mind $$k$$ is not the prorogation constant itself, $$\gamma$$ is, as it is mentioned in your book below equation (12.8c). Same logic applies for why $$\gamma$$ can have it's sign changed.

We always use the first equation (Helmholtz equation) you mentioned the one with $$k$$ notice equation (12.4). Keep in mind equation (12.7) for how $$k$$ relates to the other constants.

Here: $$k_{cut}^2 = k_{x}^2+k_{y}^2$$ the interpretation is the "cutoff wave number" for a certain mode in a wave guide. But this is specifically true for rectangular wave guides. This results when $$\gamma = 0$$ for such waves guides from equation (12.7) and you can use it to find the cut off frequencies for specific modes. Now if you keep my answer for your 3rd question in mind you know you must use $$k$$ in the Helmholtz equation NOT $$k_{cut}$$. If you assume:

$$\textbf{E} = \hat{\textbf{E}}(x,y)e^{-\gamma z}$$

which you can confirm from looking at equations (12.9) but here I am only considering the propagation in the +z direction. Lets solve for the $$E_{z}$$ component for instance then:

$$\frac{\partial^2 E_{z}}{\partial x^2} + \frac{\partial^2 E_{z}}{\partial y^2} + \frac{\partial^2 E_{z}}{\partial z^2} + k^2E_{z} = 0$$

$$\frac{\partial^2 \hat{E_{z}}}{\partial x^2} + \frac{\partial^2 \hat{E_{z}}}{\partial y^2} + \gamma^2 \hat{E_{z}} + k^2\hat{E_{z}} = 0$$

$$\frac{\partial^2 \hat{E_{z}}}{\partial x^2} + \frac{\partial^2 \hat{E_{z}}}{\partial y^2} +\hat{E_{z}} * (k_{x}^2+k_{y}^2) = 0$$

then you solve for $$\hat{E_{z}}$$ and then you can find every other component by using the decoupling equations from the Maxwell laws.