# Elementary Explanation of Unruh Effect

1. Oct 17, 2013

### stevendaryl

Staff Emeritus
The Unruh effect is the fact that an accelerated observer doesn't see vacuum, but a thermal bath of particles with an effective temperature proportional the proper acceleration.

The explanations that I've seen for this effect seem very mysterious to me. It seems to involve the fact that the generator for boosts in the Rindler coordinate systems looks just like the Boltzmann factor for thermal physics, once you identify the acceleration with the temperature. But that just seems like slight of hand. I don't understand intuitively the connection between temperature and acceleration.

Can somebody describe in more detail than Wikipedia, why this makes sense?

2. Oct 17, 2013

### dauto

Yes, I agree, it is not intuitive. Another argument that I've seen used is that the zero point energy modes for one reference frame does not match those for a different reference frame, so when you accelerate - that is when you shift between different frames - some of the zero point energy modes in one frame get converted to thermal energy in the other frame. I must say I've never seen the math behind that argument so I can't really tell how sound it is.

3. Oct 17, 2013

### Bill_K

I can briefly describe the derivation given in the book by Birrell and Davies. They consider a massless scalar field φ, whose quanta are the particles to be thermally emitted. They consider a detector moving along an arbitrary world line x(τ). Then they add to the Lagrangian an interaction term m(τ) φ[x(τ)], where m(τ) is the detector's monopole moment, and ask what influence this will have on the field. Well the interaction term is basically a source term, so the field φ produced will be the Green's function G, actually the integral of the Green's function over the worldline.

Now they specialize to a worldline with constant acceleration a. If the detector initially has energy E0 and makes a transition to energy E, the transition probability turns out to be a Planck distribution with ΔE = E - E0 and temperature T = a/2πkB.

"We conclude that the vacuum Green's function for a uniformly accelerated detector is the same as the thermal Green's function for an inertial detector."

Last edited: Oct 17, 2013
4. Nov 14, 2013

### JK423

Bill_K do you know of any argument why one is able to substitute the trajectory x(τ) in the fields argument?
i.e. φ[x,τ] = φ[x(τ)].
Since the x,τ variables are not dynamical in a QFT, the above substitution is somehow an approximation, but i cannot find a justification somewhere.
Any ideas?

5. Nov 14, 2013

### atyy

A system with a temperature is described by statistical mechanics, in which its state is a probabilistic mixture of states.

In QM, the state of a subsystem is characterized by a reduced density matrix, which can be mixed even though the state of the entire system is pure.

The Rindler observer has a horizon, so he only has access to a subsystem. The reduced density matrix corresponding to the Rindler wedge is mixed, and identical to that of a statistical system with a temperature.

A more precise version of this explanation is given in http://arxiv.org/abs/1109.1283.
"An alternate characterization of the Rindler wedge is simply the region of spacetime where the physics is totally controlled by the state of the half space x > 0 at a fixed time ... In fact, the reduced density matrix of the half space is nothing but a thermal state where the Hamiltonian is the generator of Rindler boosts"

A similar explanation is given in http://arxiv.org/abs/gr-qc/9406019.

Last edited: Nov 15, 2013
6. Nov 16, 2013

### JK423

Hi atyy,

From your reply i don't see what the justification for the substitution φ[x,τ] --> φ[x(τ)] actually is. Your reply is mostly concerned, as far as i can understand, of what are the consequences of that substitution and not how this substitution comes about in the first place.

7. Nov 16, 2013

### atyy

As I understand it, that's not the primary argument being made.

The primary argument is that in quantum mechanics, a subsystem can appear thermal, even though the full system is not thermal. This is due to quantum entanglement.

The formal way to see this is by calculating the state of a subsystem from the full state via the reduced density matrix. So take a look at the concept of the reduced density matrix. The Rindler wedge is a subsystem, and if one calculates its reduced density matrix, the reduced density matrix is thermal.

Here is another other example of a quantum subsystem being thermal, even though the full system is pure and has no temperature. http://arxiv.org/abs/1007.3957

Edit: I see you were asking about Bill K's post. The argument I was talking about is different (I think) from Birrell and Davies's.

Last edited: Nov 16, 2013
8. Nov 16, 2013

### JK423

Yes i agree with what you say. But in order to derive what you're saying, you are asumming the following,
x,τ --> x(τ) (i.e. an accelerating trajectory) (1)
and
φ[x,τ] --> φ[x(τ)] (2).
These are the assumptions from which everything you describe follows. But take a look at (2). In QFT x and τ in the field's argument are non-dynamical variables. In (2) they get dynamical, and this is an assumption that has hidden implications. Since x(τ) is the mean position of the observer at time τ, it implies that the field φ is coupled to the observer's position x by some unknown interaction Hamiltonian.

My question is the following. Instead of assuming (2), can we derive it somehow using some approximations? Do we know what the unknown interaction Hamiltonian between φ and x, implied by (2), actually is?

9. Nov 16, 2013

### atyy

Sorry, I replied rather quickly - I think your comments are related to Bill K's, not mine. My argument was different from Bill K's, and I wasn't answering you in my post #5.