I Is this correct? Casimir Effect

  • Thread starter Thread starter raracon
  • Start date Start date
  • Tags Tags
    Casimir effect
Click For Summary
Maxwell's equations indicate that field modes of any frequency exist both between and outside two perfectly conducting plates, leading to infinite zero-point field energy at any finite separation. However, the difference in zero-point energy between the two configurations is finite, resulting in a force described by F = −πhc/480d^4 per unit area, which depends on the separation distance d. The frequency of modes is indeed influenced by the plate separation, contradicting some sources that suggest otherwise. The force formula arises from quantization of vacuum energy due to boundary conditions imposed by the plates. This discussion clarifies the relationship between frequency modes and the Casimir Effect's force calculation.
raracon
Messages
35
Reaction score
14
TL;DR
Quick fact check
"In the example of Fig. 1, Maxwell’s equations allow field modes of arbitrarily large frequency both between the plates and outside them, and therefore the zero-point field energy is infinite when the plates are separated by a finite distance d as well as when they are infinitely far apart. However, the difference in zero-point energy for the two cases is finite, and its dependence on the plate separation d implies a force F = −πhc/480d^4 per unit area."

Fig 1:
CASE.png


Read this in a book about the Casimir Effect. Now my question is if it is correct to say that between the two perfectly conducting plates there can be modes with any frequency? I've read in other books that the frequency is limited by the separation between the plates.

And my other question is how does one explain F = −πhc/480d^4 ?
 
Physics news on Phys.org
A Wiki article https://en.wikipedia.org/wiki/Casimir_effect shows formula, the excitation is quantized by plane boundary condition so lower vacuum energy than outside take place between the plates. The formula of F is also derived there.
 

Thread 'Angular Momentum Vector and Its Magnitude'

For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...
View full post »

Similar threads

I Does quantum tunnelling affect the magnitude of the Casimir Effect?
  • · Replies 21 ·
Replies
21
Views
2K
A In light of new evidence, what might be behind the Casimir Effect?
  • · Replies 46 ·
2
Replies
46
Views
5K
I Is the Archimedes Experiment a groundbreaking test of the Casimir Effect?
  • · Replies 10 ·
Replies
10
Views
3K
I What sense does it make to "weigh" a Casimir cavity?
  • · Replies 9 ·
Replies
9
Views
2K
I Poles in Casimir force as function of frequency & mode
  • · Replies 2 ·
Replies
2
Views
1K
I "Breaking the loop" in Casimir force?
  • · Replies 1 ·
Replies
1
Views
1K
I Casimir effect and vacuum energy and a bit of relativity....
  • · Replies 46 ·
2
Replies
46
Views
7K
I Black Hole Fire Walls, Brick Walls, and the Casimir Effect
  • · Replies 6 ·
Replies
6
Views
2K
I Vacuum energy length scale detectable by the Casimir effect?
  • · Replies 2 ·
Replies
2
Views
1K
I What is the correct explanation of the Casimir force?
  • · Replies 49 ·
2
Replies
49
Views
5K