- #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 [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!
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!