I’m reading this on the Unruh effect and it is derived by calculating the Bogoliubov coefficients between the Minkowski and Rindler observer. Rindler observers use a different set of modes to describe a vacuum which leads to a different number of particles. However, nowhere in the derivation is the role of the horizon explicitly invoked. It is known that in curved spacetime even without a horizon there will be different numbers of particles for different observers. So what exactly makes the Unruh effect “special”? What’s the role of the horizon?

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    $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Dec 23, 2022 at 9:31

1 Answer 1


I haven't read the notes, but this is a very common calculation, so I assume the author is doing it in the usual way.

In flat spacetime, the role of the horizon is often not that clear because you have a lot of extra symmetry around that makes the calculation a bit more intuitive. It is hidden behind the four-velocities taken for the Rindler observers: the Killing field associated with these four-velocities is what actually generates the horizon, after all. A way to notice how the role of the horizon is relevant is by noticing that you only need to relate two distinct "notions of time" at the horizon to derive the effect: the Killing time (which is associated with the accelerated observers) and the geodesic time (which is associated with inertial observers). This is necessary to understand the generalization of the effect for general spacetimes and is discussed, for example, in Wald's book (I believe in section 5.3).

Shortly, one can study the Bogoliubov transformation only on the horizon and still find that the state is thermal at the Unruh temperature for observers whose four-velocities are aligned with the Killing field. However, in flat spacetime, one can interpret these observers as accelerated observers and get a more concrete interpretation, which makes focusing on the horizon sort of unnecessary.

Notice also that the text you mentioned uses Rindler coordinates, which are only defined on the right Rindler wedge. Hence, the horizon sort of plays a role in restricting the region in which these coordinates are defined.

Tracing Out the Left Wedge

The calculation done in the reference you posted is similar to Hawking's original calculation of the Hawking effect, in which one computes the expected value of the number operator. However, the book by Wald and other of his texts (namely, a 1975 paper and I think Chapter 14 of his GR book) compute explicitly the final state of Hawking radiation. Wald's QFTCS book computes explicitly the Minkowski vacuum as seen by Rindler observers (but you can read the other references for the Hawking effect and make the necessary changes).

When one takes Wald's approach, it is important to notice that the Fock space for Rindler observers can be written as $\mathcal{F}_L \otimes \mathcal{F}_R$, where $\mathcal{F}_L$ is the Fock space for the left Rindler wedge (which is a globally hyperbolic spacetime on its own right) and $\mathcal{F}_R$ is the Fock space for the right Rindler wedge. Accelerated observers only have access to a single wedge, due to the Rindler horizon. Hence, the final state seen by an accelerated observer is not a density matrix on $\mathcal{F}_L \otimes \mathcal{F}_R$, but rather a partial trace of it, which leaves a density matrix on $\mathcal{F}_R$ (assuming the observer is accelerating toward the right).

Hence, at the end of the calculation, one has to trace out the left Rindler wedge to obtain the state seen by the accelerated observer. This is one of the reasons you can get a thermal equilibrium state. Without the partial trace, the state would be pure, and hence it wouldn't be thermal. Once you trace over the left wedge, you are left with a mixed state.

More details on these calculations can be found on the works by Wald that I mentioned. The QFTCS book is a tough read, but I think the 1975 paper can be a bit more accessible, although it is mainly discussing the Hawkign effect instead of the Unruh effect.

  • $\begingroup$ How exactly does one “trace the modes”? I don’t see any kind of tracing in the source I’m reading. $\endgroup$
    – Obama2020
    Dec 22, 2022 at 12:37
  • $\begingroup$ @Obama2020 That's the price of me not reading the source material! hahahaha. I thought he would compute the actual final state (as Wald does), but he computed only the expectation value of the number operator, which is enough to find the Unruh effect. I deleted that last paragraph, as it is not relevant to this calculation. $\endgroup$ Dec 23, 2022 at 7:55
  • $\begingroup$ @Obama2020 I also added a new remark at the end. The horizon also restricts where the Rindler coordinates make sense $\endgroup$ Dec 23, 2022 at 7:56
  • $\begingroup$ I’m still interested in this “tracing” method, as it seems to better cover the role of the horizon. I looked at Wald’s book and I don’t understand it. Do you know of a different good reference that covers it, or could you explain it in detail yourself? $\endgroup$
    – Obama2020
    Dec 25, 2022 at 2:04
  • $\begingroup$ @Obama2020 The complete calculations can be a bit long, so I'm not willing to type them out, but I added some more discussion and references at the end of my answer. I'm not really sure if I know of a reference that is not written by Wald himself (and I don't think so) $\endgroup$ Dec 25, 2022 at 6:39

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