What is the temperature of vacuum?

Question

I wonder what the temperature of vacuum is. Is it the absolute zero？ The vacuum is defined without taking into account the Unruh effect or the CMB，it is just vacuum with nothing in it but quantum fluctuations.

Answer 1

The requirements "without taking into account the Unruh effect" and "taking into account the quantum fluctuations" are contradictory, so in this answer I'll take quantum fluctuations into account by means of the Unruh effect.

Shortly, the answer is: it depends on the spacetime and on the observer.

To define what one means by temperature in the context of quantum fields, one will usually be thinking of: "Is the quantum field's state a thermal state at some temperature $T$?". In other words, is the state given by a density matrix of the form $\rho =e^{- \beta \hat{H}},$ where $\hat{H}$ is the Hamiltonian and $\beta$ is the inverse temperature? (I'm not being rigorous here, a more appropriate discussion would come in the form of KMS states).

The answer depends on the Hamiltonian. However, different observers in spacetime have different notions of proper time and, as a consequence, they have different Hamiltonians: each observer will define a Hamiltonian in their reference frame by using Noether's theorem to their own notion of time translation symmetry (I'm assuming we're only considering observer's whose four-velocities are parallel to some Killing vector field in spacetime). Hence, different observers will have different notions of Hamiltonians.

For inertial observers in Minkowski spacetime, the answer is simple: the temperature vanishes (absolute zero). This isn't particularly surprising, since we fairly often use zero temperature field theory (as opposed to thermal field theory) in Minkowski spacetime.

For accelerated observers, thing get more interesting. The Minkowski vacuum is defined as the unique quantum state with minimum energy as measured by inertial observers. For accelerated observers, we have a different Hamiltonian, so it is actually populated by particles at a thermal spectrum. If the observer has acceleration $a$, then the temperature is given by the Unruh temperature, $T = \frac{a}{2\pi}$ (units with $\hbar = c = 1$).

In more complicated spacetimes, the answer varies. For example, static observers at infinity in Schwarzschild spacetime will measure a temperature $T = \frac{1}{8\pi M}$ ($\hbar = c = G = 1$), where $M$ is the black hole's mass. This is an instance of the Unruh effect in curved spacetimes, which is often confused with the Hawking effect. The difference between them is the field's quantum state. In the Unruh effect, the state is thermal "from everywhere", so you also have particles coming in from infinity. In the Hawking effect, there's particles coming only from the black hole (for this to be a physical quantum state, one can't consider an eternal black hole, but rather a black hole formed by stellar collapse).

There are also other cases. For example, any inertial observer in de Sitter spacetime measures a temperature $T = \sqrt{\frac{\Lambda}{3}}$, as I discussed in this post.

Hence, the answer depends a lot on context. The temperature varies depending on spacetime structure an on the particular observer measuring the temperature.