# What is the essence of the Unruh-effect?

The essence of the Unruh effect is basically that coordinate-transformations lead to different excitations/occupation numbers of the quantum fields. Is that statement correct?

So in QFT, while an observer in one frame "sees" a vacuum state, another observer in a different frame sees a highly excited state - a thermal bath of particles. The same happens in the case of diffomorphisms in GR (QFT on curved spacetime background).

Does that mean that the notion of particle or matter/substance is frame-dependent? To put it as naively as possible: in one frame there are certain objects (structures made of particles), while in another frame all of this "disappears" and there is nothing but the vacuum (fluctuations)?

The short answer is yes, the "moral" of the Unruh effect is that the particle content of a given state in a quantum field theory depends on the frame of reference, i.e. on the observer. So yes, what one observer sees as vacuum another describes as being riddled with particles, in a very operational sense (see Unruh-DeWitt detectors). If it seems strange at first I always like to think that this is a great way to understand how the vacuum fluctuations are really important, for the same state seems to have lots of real particles for different observers.

In more detail, one can see what is going on by taking a look at the free massless scalar field case. The equation of motion for the classical field theory is just the wave equation

$\Box \phi=0$.

In the usual cartesian coordinates this looks like

$(\partial_t^2-\partial_x^2-\partial_y^2-\partial_z^2)\phi=0$,

whose solutions are just the plane waves. In quantum field theory you decompose $\phi$ and "promotes" the coefficients to creation and annihilation operators $a_k,a_k^\dagger$ subject to the usual commutation relations.

Great, what about different observers? Well, different observers use different coordinate systems to describe the spacetime. The particular case of an observer with constant acceleration is usually discussed in the Rindler metric $ds^2=e^{2a\xi}(d\tau^2-d\xi^2)-dx^2-dy^2$. In this case the observer with coordinate $\xi=0$ has constant acceleration $a$ in the $z$ direction. Let's repeat the quantization procedure above. The wave equation in this coordinate system is

$[\partial_\tau^2-\partial_\xi^2-e^{2a\xi}(\partial_x^2+\partial_y^2)]\phi=0$.

The transverse directions still have plane-wave solutions, but the when you perform separation of variables the solution for $\xi$ is a modified Bessel function of the third kind with imaginary index (no need to bother with the details here). In any case one still has a complete set of solutions, so one might expand $\phi$ in terms of this functions and "promote" the coefficients to operators $b_k,b_k^\dagger$ satisfying usual commutation relations as before. The difference is that since the mode decomposition is not the same then $a_k\neq b_k$, so that the same state in the Hilbert space is described as containing different particles for the distinct observers.

In even more details, since both ladder operators generate the same Hilbert space then the operators must be linear combinations of one another, i.e.

$a_k=\alpha_k b_k+\beta_k b_k^\dagger$,

and the dagger relation of this expression (Here I'm considering the specific case where both decompositions end up in the same quantum number $k$, this needs not be the case and in the particular case of Unruh effect is not 100% correct). From this is easy to read the situation. The usual vacuum in QFT is defined as the state $|0\rangle$ such that $a_k|0\rangle=0$ for all $k$. The observer in the different frame will define his vacuum the same way, as the state $|0'\rangle$ such that $b_k|0'\rangle=0$ for all $k$.

Now there are really only two options. Either $\beta_k$ is zero, in which case both observers agree on what is the vacuum, or $\beta_k$ is not zero, in which case what one observer think is vacuum (the state annihilated by all lowering operators $a_k$) the other thinks has particles (since it is not annihilated by all $b_k$).

If you start with the inertial coordinate system and perform any Lorentz transformation you end up with different coordinates. It is actually very simple to show that coordinates related by Lorentz transformation always have $\beta_k=0$, so that all inertial observers agree what state is the vacuum. Now for the case of any other observer you will get $\beta_k\neq 0$, so they will describe the usual vacuum as a state with particles. The nature of the particles will depend on the exact functional relationship of the $\alpha_k$ and $\beta_k$. For the particular case of uniform acceleration (Unruh effect) the coefficients imply that the Minkowski vacuum is perceived as having a planckian distribution of particles in the accelerated frame.

First, regarding fermion fields. The argument for inequality of particle descriptions for different observers has its roots at the difference of the classical solutions to the field equations for different coordinate systems and the necessarily different mode decomposition and consequently of the ladder operators. Nothing in this argument depends on the spin of the field in question, so that all the steps can be easily be extended to fermion fields or gauge theories. Therefore the vacuum of a Dirac field in inertial coordinates is a state full of particles for a accelerated observer. Also, since in usual interacting field theory we start with the non-interacting part and treat the couplings perturbatively all this extends to interacting field theories too.

Second, you ask about structures made of particles, specifically if we can make the structures "disappear" in other frames. Now I hope it transpired that the idea of the Unruh effect is that particle is a frame-dependent concept, so you need to define a structure in other way.

Anna v in the comments suggested a bound state. Take a proton for example. It is a bound state of quarks and gluons. Is there an observer that sees this proton as just vacuum? Well the proton is certainly in the confinement phase of QCD, since we do not observe color charge. Arguably the vacuum is just fluctuations of quarks and gluons so it should be a state in the deconfined phase. So in essence the question is if different observer can see different phases of matter.

In quantum field theory we define a phase by the vacuum expectation value of some field operator $\langle \phi\rangle$, the VEV being zero in one phase and not zero in the other. But since the VEV in a scalar it is, by the very definition, invariant under coordinate changes, even non-inertial ones, so that if the VEV is not null in a frame, the it must be not null for all frames. From this we should conclude that if one observer sees the confined phase then all do too. The proton is there for all observers, but different observers can describe it with different amounts of quarks and gluons (this is obviously a very heuristical description, but the order parameter argument should make clear the correctness of it).

• you have not addressed the last paragraph of the question. Is the answer to "there are certain objects (structures made of particles), while in another frame all of this "disappears" negative? I have the impression it is only the vacuum that changes, not dynamically bound structures. Feb 18, 2015 at 19:00
• @annav, yes, I would tend to agree with you on this. In QFT you could define a bound state as a pole in the S-matrix. No matter what frame of reference you choose the pole cannot go away. Therefore all observers agree that there is a bound state, but they can disagree on which particles constitute the bound state. So yes, the definition of particles may be relative, because essentially you are using a different basis for the Hilbert space, but the dynamical features must be the same Feb 18, 2015 at 20:05
• @annav, incidentally I just remmembered a better example. Consider the case of spontaneous symmetry breaking. We define by it by a non null vacuum expectation value $\langle \phi\rangle$. Since the field is a scalar the VEV is an invariant. Therefore if one observer sees a non null VEV all must do too, no matter what they think is the vacuum of the interacting field theory. Repeating the argument with some model of fermion condensate you can argue that Unruh effect cannot deconfine hadrons for instance Feb 18, 2015 at 20:10
• maybe for completeness you could edit and include one of these examples in your answer, as comments may disappear , and it is part of the question. Feb 19, 2015 at 4:17
• @cesaruliana: Thank you for your excellent and very instructive explanation! In your answer you used the example of bosonic fields. Do your arguments also apply to fermionic fields? Could Dirac particles appear/disappear in non-inertial frames? For matter fields I would somehow find the effect even more astonishing. Matter being created/destroyed just by accelerating... Feb 19, 2015 at 11:42

The Unruh effect is better understood in the light of partial trace. Special relativity tells us that an accelerated observer has events horizons, which implies that exist regions in space that this observer can't measure. Quantum mechanics tells us that the proper way to work with this observer is do a partial trace in the total Hilbert space of entire space. The partial trace make the vacuum state of the inertial frame a mixture of state (because the non-zero correlations functions in vacuum state) with well defined only the average energy. The most unbiased way for define the probability distribution in energy state of the mixture is the Boltzman's weight. The temperature could be defined later when we see how energy is related with entropy, and consequently with acceleration (determines the horizon).

What we learn with this effect is that the particle's picture is absolute only in inertial frames.

I expect that future developments in complementarity of black hole and quantum gravity could teach us a "better" look for this effect in terms of quantum information by equivalence principle.

• how can this be applied to the case of rotating observers? supposedly in a rotating frame there is an ergosphere right at $r > c / \omega$ Jun 13, 2018 at 21:50