Derivation of the Casimir energy in flat space

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 2K views
highflyyer
Messages
27
Reaction score
1
I am trying to understand the derivation of the Casimir energy from https://en.wikipedia.org/wiki/Casim...f_Casimir_effect_assuming_zeta-regularization.

At one point, the derivation writes the following:

The vacuum energy is then the sum over all possible excitation modes ##\omega_{n}##. Since the area of the plates is large, we may sum by integrating over two of the dimensions in ##k##-space. The assumption of periodic boundary conditions yields
$$\langle E \rangle=\frac{\hbar}{2} \cdot 2 \int \frac{A dk_x dk_y}{(2\pi)^2} \sum_{n=1}^\infty \omega_n$$


What does the size of the plates have to do with being able to integrate over two of the dimensions in ##k##-space? I don't quite see the connection.

Thank you so much in advance for any comments.
 
Physics news on Phys.org
For example they don't need to care for diffraction. And they don't care about what is happening around corners since that part is very small when you have large plate.
 
If you have periodic boundary conditions in a cuboid volume, the momenta are quantized: ##p_j=\frac{2 \pi}{L_j} n## with ##n \in \mathbb{Z}##. For large ##L_j## you can often approximate a sum over ##p_j## by an integral
$$\sum_{p_j} \rightarrow \frac{L}{2 \pi \hbar} \int_{\mathbb{R}} \mathrm{d}p_j.$$
Here you assume that the plates are very large compared to their distance. Thus you can use the approximation of the sum by an integral for the momentum components along the plates but not for the one perpendicular to it.
 
  • Like
Likes   Reactions: highflyyer and Demystifier