Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with deriving the Casimir Effect?

  1. Nov 29, 2014 #1
    I am at the very last part of a relatively long derivation of the Casimir effect, and I just don't understand the final step D:

    So far, I have derived the ground state energy to be [itex]\langle 0| \hat{H} |0\rangle = \delta (0) \int _{-\infty}^{\infty} dp \frac{1}{2} E[/itex]

    And for a massless field using Plank units and using [itex]E=\sqrt{p^2+m^2}[/itex], then [itex]E=p[/itex]. Between the two parallel plates, only virtual particles of discrete [itex]p[/itex] can exist which are [itex]p=\frac{h}{\lambda}[/itex] and using [itex]\hbar=1[/itex] and [itex]\lambda=\frac{2d}{n}[/itex] where [itex]d[/itex] is the distance between the parallel plates. This means that the summation becomes [itex]\frac{\pi}{2d} \sum\limits_{n=1}^{\infty}n[/itex] which I find by assuming [itex]\sum\limits_{n=1}^{\infty}n=\sum\limits_{n=1}^{\infty}ne^{-an}=-\frac{d}{da}\sum\limits_{n=0}^{\infty}e^{-an}[/itex] Which gave the sum to infinity of [itex]\frac{e^a}{e^a-1)^2[/itex]

    Using the Taylor expansions, the first two terms of this result are [itex]\frac{1}{a^2}-\frac{1}{12}[/itex] and subsequent terms are irrelevant because I take [itex]a\rightarrow 0[/itex] for the sum to become the sum of all natural numbers. Placing [itex]\frac{a\pi}{d}[/itex] for [itex]a[/itex], this gives the sum [itex]\frac{\pi}{d}\sum\limits_{n=1}^{\infty}n[/itex]. So when I put this result back in, I get that the energy density of the vacuum (assuming that [itex]\delta(0)[/itex] corresponds to the volume of space, is [itex]\frac{d}{2\pi a^2}-\frac{\pi}{24d}[/itex].

    Okay, so this is the part that I don't get. To eliminate the infinity, I considered adding a third plate and moving it away to infinity, so the lengths are as shown:

    ___L (from first to third plate)
    |___|____|
    d___L-d

    Now here I want to find the relative energy density between the plates, take the derivative with respect to d to find the force, and move L away to infinity to remove the infinities. My problem is that, looking back over my notes, I have added the energy densities between the first and second plate and second and third plate to give

    [itex]E(d)=\frac{d}{2\pi a^2} -\frac{\pi}{24d} +\frac{L-d}{2\pi a^2}-\frac{\pi}{24(L-d0}[/itex] which works out because then using [itex]F=-E'(d)[/itex] gives the force being [itex]-\left(\frac{\pi}{24d^2} +\frac{\pi}{24(L-d)}\right)[/itex] which works out perfectly to give the force being [itex]\frac{\pi}{24d^2}[/itex] when you move the third plate to infinity away.

    But I just can't figure out why I added the energy densities in my notes? Thank you in advance!
     
  2. jcsd
  3. Dec 5, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Dec 6, 2014 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Maybe this isn't the best analogy but ...

    Consider a window in your house. What happens if you take into account air pressure in the room, and you neglect air pressure outside?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with deriving the Casimir Effect?
  1. Casimir Effect. (Replies: 32)

  2. Casimir effect (Replies: 1)

  3. Casimir Effect (Replies: 1)

  4. Casimir Effect (Replies: 5)

  5. The Casimir effect (Replies: 1)

Loading...