# Euler-Maclaurin formula for Casimir Effect

I’ve been reading this http://www.hep.caltech.edu/~phys199/lectures/lect5_6_cas.pdf for a derivation of the Casimir Effect. On page 4 we see that

$$\Delta{E} = \frac{\pi}{L}\left(\sum_{\nu = 1}^{\infty}\nu - \int_{0}^{\infty}\nu d\nu\right).$$

Here natural units are used. The paper then used the Euler-Maclaurin formula to get $$\Delta{E} = -\frac{\pi}{12L}$$

I read the Wikipedia on the formula and it looks like it is a method of evaluate the difference between closely related integrals and sums, but I don’t understand how we can use the formula for this scenario since we start our sum at $$n=1$$ and our integral at $$n=0$$. So my question is, how do we explicitly use the Euler-Maclaurin formula to get our desired result?

• I would suggest you find a better source than the one you link, which appears to be a expository writing project from a student in an undergraduate quantum mechanics class. There is a lot in there that is, while not exactly wrong, not well expressed or explained.
– Buzz
Jul 12, 2022 at 2:25
• Section 2.3.2 of David Tong's QFT notes gives a nice derivation which doesn't use the Euler-Maclaurin formula. Jul 12, 2022 at 2:37

Trick 1: Rewrite the integral so it has the bounds you want.

$$$$\int_0^\infty \nu d \nu = \int_0^1\nu d \nu + \int_1^\infty \nu d \nu = \frac{1}{2} + \int_1^\infty \nu d \nu$$$$

Trick 2: Multiply the integrand and summand by $$1=\lim_{\epsilon\rightarrow 0} e^{- \epsilon \nu}$$

$$$$\Delta E = \frac{\pi}{L} \left(\sum_{\nu=1}^\infty\left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right]\nu -\int_1^\infty \left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right] \nu d \nu - \frac{1}{2}\right)$$$$

Trick 3: Take the limit outside the sum/integral (this is not an obvious step and I suspect might not really be fully rigorous, but ok at a physics level of rigor)

$$$$\Delta E = \frac{\pi}{L} \left[\lim_{\epsilon\rightarrow 0} \left(\sum_{\nu=1}^\infty e^{-\epsilon \nu}\nu -\int_1^\infty e^{-\epsilon \nu} \nu d \nu \right) \right]- \frac{\pi}{2L}$$$$

Trick 4: We apply the Euler-Maclaurin formula to evaluate the difference in the parentheses.

Let's write out the first few terms of the general formula $$$$\sum_{i=a}^b f(i) - \int_a^b f(x) dx = \frac{f(a)+f(b)}{2} + \frac{f'(b) - f'(a)}{12} - \frac{f'''(a) - f'''(b)}{720} + \cdots$$$$ where in our case, $$f(x)=e^{-\epsilon x} x$$, $$a=1$$, and $$b=\infty$$.

The first term yields $$$$\frac{f(a)+f(b)}{2} = \frac{1}{2}$$$$ This exactly cancels the $$-\pi/2L$$ term in $$\Delta E$$ above.

For the second term, we need the derivative $$$$f'(x) = e^{-\epsilon x} - \epsilon e^{-\epsilon x} x$$$$ We will ignore the second term, since in the end we want to take the limit $$\epsilon\rightarrow 0$$, and that term will not contribute. Therefore, $$$$\frac{f'(b) - f'(a)}{12} = -\frac{1}{12} e^{-\epsilon} + {\rm term\ that\ vanishes\ as\ \epsilon\rightarrow 0}$$$$ In the limit $$\epsilon\rightarrow 0$$, this simply becomes $$-1/12$$.

Finally, let's consider the third term. We have $$$$f'''(x) = -\epsilon^2 e^{-\epsilon x}\left(\epsilon x - 3\right)$$$$ Since this will go to zero as $$\epsilon \rightarrow 0$$, we can neglect this term. Similarly, all higher derivatives of $$f$$ (which appear in higher order terms in the Euler-Maclaurin formula) will go to zero as $$\epsilon\rightarrow 0$$.

Putting this together: $$\begin{eqnarray} \Delta E &=& \frac{\pi}{L} \left[\lim_{\epsilon\rightarrow 0} \left(\sum_{\nu=1}^\infty e^{-\epsilon \nu}\nu -\int_1^\infty e^{-\epsilon \nu} \nu d \nu \right) \right]- \frac{\pi}{2L} \\ &=& \left(\frac{\pi}{2L} - \frac{\pi}{12 L} + 0 \right) - \frac{\pi}{2L} \\ &=& -\frac{\pi}{12 L} \end{eqnarray}$$ as desired.

• 1. Your sums over $n$ should be sums over $\nu$. 2. Staring from the formula in Trick 3, we can also directly evaluate the sum and integral, e.g., recognizing them as the derivatives with respect to $\epsilon$ of a geometric series and an elementary integral. This gives for the argument of the limit: $e^{-\epsilon}\bigl((1-e^{-\epsilon})^{-2} - \epsilon^{-2} - \epsilon^{-1}\bigr)$. Then the limit as $\epsilon \to 0$ is $5/12$ (consistent with the approach shown). Jul 12, 2022 at 12:37
• @nanoman 1. Thanks, fixed. 2. Yes, but the question was about how to use the Euler-Maclaurin formula. Jul 12, 2022 at 13:04

I want to add something to @Andrews answer. They gave a good answer which is technically correct, but I think there should be at least some things added to Trick 3.

In the present answer there are some "mathematical non-rigorous" steps and the author claims that this is okay for physicists. The true reason why this is okay is that the expression, $$$$\Delta E = \frac{\pi}{L} \left(\sum_{n=1}^\infty\nu -\int_1^\infty \nu d \nu - \frac{1}{2}\right)\qquad(1)$$$$ was wrong to begin with. And the reason that $$1 = \left[\lim_{\epsilon\rightarrow 0}e^{-\epsilon \nu}\right]$$ solves this problem is not only a mathematical trick but a physical argument which makes sense out of eq. (1). Let us remember what kind of system we are trying to describe. There are two conducting metal plates which result in boundary conditions for the electromagnetic field, because we know that the charges within the plate will move along the plate until the field satisfies the boundary conditions. But there is more physics than that! What about small wavelengths? Will the plates be able to shield electromagnetic waves of a shorter wavelength than let's say the inter-molecular distance within the plate? No, the photons of such a wavelength will just penetrate the metal plate, hence the boundary condition is violated for high-energy photons. This physical fact yields a UV-cutoff natural to the system! There is no mathematical non rigorous interchange of limits needed, but the initial formula was wrong physics-wise! This goes even so far that the calculation is "independent" of the regulator scheme you choose to cut-off the short wavelength waves. For a conceptual discussion of the matter at hand see e.g. "Quantum Field Theory and the Standard Model" by M.D. Schwartz.

• Good clarification! Jul 12, 2022 at 12:00
• So essentially for high values of $\nu$ (corresponding to high energy photons) shouldn’t contribute to the total energy since they leave the plates, and this is described by the fact that $e^{-\epsilon\nu}$ decreases as $\nu \rightarrow \infty$? Also, why can we talk about photons in between the plates if we are dealing with a vacuum (no particles present)?
– user310742
Jul 14, 2022 at 16:43
• @Obama2020 Yes, if you wish you can imagine it the following (maybe oversimplifying) way. The electrons cannot adjust "fast-enough" to guarantee the boundary condition. About your second question, it is not at all "clear" if the Casimir force really is due to "real" vacuum fluctuations or if it is just another way to account for the Van-der-Waals like interaction between the electrons in the conducting plates. Hence, the "dispersion-force"-type interaction (based on Coulomb interaction) could be seen as mediated by photons. See e.g. physics.stackexchange.com/questions/11544 Aug 30, 2022 at 12:18