# Separable PDE for electric field in a cavity

gboff21

## Homework Statement

Here is a photo of a page in Laser Physics by Hooker:
https://www.evernote.com/shard/s245/sh/2172a4e7-63c7-41a0-a0e7-b1d68ac739fc/7ba12c241f76a317a6dc3f2d6220027a/res/642710b5-9610-4b5b-aef4-c7958297e34d/Snapshot_1.jpg?resizeSmall&width=832

I have 3 questions (I'm a bit rusty after a long summer):
1. Is the "c.c." bascially $\textbf{U}(r) e^{i \omega t}$ or is the U actually U*? (Just checking)

2. Why can't we say $\textbf{U}(r) = X(x)Y(y)Z(z)$ instead of separating it into Ux Uy and Uz?

3. Finally: Having gone through the PDE and got an answer with boundary solutions $\textbf{U}(r)_{x} = 0$ at edges, I don't understand where the $\frac{\pi}{L}$ comes from. The π I know is just put in to make the mode numbers simpler later, but how does the L get there?
And how is the x component a cos function? I get a sin!

Thanks for the help!! I'm just a bit rusty!

Bryson

gboff21
Oh, ok here you go (attached)

#### Attachments

• Snapshot_1.jpg
48.3 KB · Views: 393
Gold Member
2021 Award
the +c.c. simply means "plus the conjugate complex expression", i.e.,
$$a+ \text{c.c.}=a+a^*.$$

• 1 person
Bryson
• 