# Could the cosmological constant be due to vacuum fluctuations in a box, i.e., in a finite universe?

Assumption: If the universe were a finite box whose boundary is the cosmological horizon, then there would be a zero-point energy inside that box.

Consequence 1: This zero-point energy would be given by the size of the box. The calculated energy value is very similar to the measured cosmological constant.

Consequence 2: The zero-point energy would have been larger when the universe was smaller. The cosmological constant would not be a constant, but decay in time.

Question: Could that be the case?

• "The calculated energy value is very similar to the measured cosmological constant." Do you have a source? – Javier Mar 6 at 20:26
• Experimentally, Lambda is very near 1/L^2, where L is the distance to the cosmological horizon. – frauke Mar 26 at 6:40

If quantum fields are restricted to a finite box of dimension $$L$$, then this changes the computation of the zero point energy because the fluctuations exist at discrete frequencies, $$\omega_n = \frac{2\pi n}{L}$$ Following Ford (in the case of a 1-dimensional 'box') in https://journals.aps.org/prd/pdf/10.1103/PhysRevD.11.3370, the zero-point energy with a frequency cutoff is $$E_0 = \sum_n \omega_n \exp(-\alpha \omega_n)= \frac{L}{2\pi \alpha^2} -\frac{\pi}{6L} + {\rm positive\, powers\, of\,} \alpha$$ Following the Casimir prescription, the idea is to subtract the vacuum density appropriate to the unconstrained topology from this energy (which includes it), $$\tilde{E}_0 = E_0 - E_0(L \rightarrow \infty) = -\frac{\pi}{6L}$$ Note that the energy is inversely related to the dimension of the box, and is negative.
The cosmological constant $$\lambda$$ is the observed vacuum energy density.
The zero point energy predicted by Quantum Field Theory is much large than $$\lambda$$ - it can be $$120$$ orders of magnitude larger depending on the assumptions.