2
$\begingroup$

The derivation of the Casimir effect in $1 + 1$ approximates, say, the left mirror and the asymptotic "in" region of spacetime as two fixed mirrors separated by a spacial distance of $L \to \infty$, with the resulting modes of the free scalar quantum field being of the form of standing waves. The right mirror is then a free boundary which disturbs the field depending on its position. I read somewhere that this is an "infrared regularization", but cannot find this source again and was hoping to clarify whether this is true, why it is true if so, and whether it is an accurate approximation of two plates in free space.

To calculate the Casimir energy, the standard method is to introduce a damping term into the summation of wave numbers such that there is a "cut-off" at some high frequency, and then the renormalized energy density is simply the difference of the energy density of the region between the left and right mirrors (as their separation $d \to 0$), and the free space of this model, which is where the free mirror is halfway between the boundaries such that as $L \to \infty$, the boundary effects become negligible. This is described extensively in literature as the damping term is an ultraviolet regularization, but again, approximating the Casimir effect as two fixed and one free mirrors in the first place is rarely discussed and I would like some insight as to why this a "good" model. The only discussion I have come across describe it as an "infrared regularization" and hence my confusion.

$\endgroup$

1 Answer 1

0
$\begingroup$

The answer to the question is yes.

If the "true" free space (no mirror at $L$ approximating the asymptotic "in" region) energy density is calculated, it will be a density of modes intergration over all $k > 0$. This introduces a divergent term of $k < 1$, which is an infrared divergence.

As the Casimir effect is the desired result, and physically meaningful expectation values must be finite, the approximation of a mirror at $L \to \infty$ is ok because only the divergent term is being neglected. The continuous sum of $k$ in the "mirror approximated" free space is now regularized as a discrete sum, because there is a minimum $k$ due to the Dirichlet boundary conditions now imposed.

$\endgroup$
1
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    May 4, 2023 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.