- #1
21joanna12
- 126
- 2
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 \langle 0| \hat{H} |0\rangle = \delta (0) \int _{-\infty}^{\infty} dp \frac{1}{2} E
And for a massless field using Plank units and using E=\sqrt{p^2+m^2}, then E=p. Between the two parallel plates, only virtual particles of discrete p can exist which are p=\frac{h}{\lambda} and using \hbar=1 and \lambda=\frac{2d}{n} where d is the distance between the parallel plates. This means that the summation becomes \frac{\pi}{2d} \sum\limits_{n=1}^{\infty}n which I find by assuming \sum\limits_{n=1}^{\infty}n=\sum\limits_{n=1}^{\infty}ne^{-an}=-\frac{d}{da}\sum\limits_{n=0}^{\infty}e^{-an} Which gave the sum to infinity of \frac{e^a}{e^a-1)^2
Using the Taylor expansions, the first two terms of this result are \frac{1}{a^2}-\frac{1}{12} and subsequent terms are irrelevant because I take a\rightarrow 0 for the sum to become the sum of all natural numbers. Placing \frac{a\pi}{d} for a, this gives the sum \frac{\pi}{d}\sum\limits_{n=1}^{\infty}n. So when I put this result back in, I get that the energy density of the vacuum (assuming that \delta(0) corresponds to the volume of space, is \frac{d}{2\pi a^2}-\frac{\pi}{24d}.
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
E(d)=\frac{d}{2\pi a^2} -\frac{\pi}{24d} +\frac{L-d}{2\pi a^2}-\frac{\pi}{24(L-d0} which works out because then using F=-E'(d) gives the force being -\left(\frac{\pi}{24d^2} +\frac{\pi}{24(L-d)}\right) which works out perfectly to give the force being \frac{\pi}{24d^2} 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!
So far, I have derived the ground state energy to be \langle 0| \hat{H} |0\rangle = \delta (0) \int _{-\infty}^{\infty} dp \frac{1}{2} E
And for a massless field using Plank units and using E=\sqrt{p^2+m^2}, then E=p. Between the two parallel plates, only virtual particles of discrete p can exist which are p=\frac{h}{\lambda} and using \hbar=1 and \lambda=\frac{2d}{n} where d is the distance between the parallel plates. This means that the summation becomes \frac{\pi}{2d} \sum\limits_{n=1}^{\infty}n which I find by assuming \sum\limits_{n=1}^{\infty}n=\sum\limits_{n=1}^{\infty}ne^{-an}=-\frac{d}{da}\sum\limits_{n=0}^{\infty}e^{-an} Which gave the sum to infinity of \frac{e^a}{e^a-1)^2
Using the Taylor expansions, the first two terms of this result are \frac{1}{a^2}-\frac{1}{12} and subsequent terms are irrelevant because I take a\rightarrow 0 for the sum to become the sum of all natural numbers. Placing \frac{a\pi}{d} for a, this gives the sum \frac{\pi}{d}\sum\limits_{n=1}^{\infty}n. So when I put this result back in, I get that the energy density of the vacuum (assuming that \delta(0) corresponds to the volume of space, is \frac{d}{2\pi a^2}-\frac{\pi}{24d}.
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
E(d)=\frac{d}{2\pi a^2} -\frac{\pi}{24d} +\frac{L-d}{2\pi a^2}-\frac{\pi}{24(L-d0} which works out because then using F=-E'(d) gives the force being -\left(\frac{\pi}{24d^2} +\frac{\pi}{24(L-d)}\right) which works out perfectly to give the force being \frac{\pi}{24d^2} 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!