# Horizon and Unruh radiation for a finite period of acceleration

It's a well known fact that an observer that accelerates at a constant rate from $-c$ at past infinity to $+c$ at future infinity sees a horizon in flat Minkowski spacetime. This is easy to see from a spacetime diagram once you realize that the union of past light cones on such a trajectory has a boundary that divides the spacetime into two regions - one inaccessible by the accelerating observer.

This leads to the classic result of Unruh radiation when one looks at the quantum field theory for such an observer. The horizon plays a crucial role here.

How does one go about determining whether a horizon is seen by a general class of worldines? In particular, is there any reason to believe that a horizon would exist for an observer that is stationary for all time except for a finite period of acceleration and deceleration?

Is there any other class of worldlines other than an indefinitely accelerating observer for which a horizon is known to exist in flat spacetime?

• your comments on Marek's Answer (for which +1) seem to shift the Question much towards the Unruh effect. The Question was and still is about horizons, not about the Unruh effect, but I started to write an Answer about the Unruh effect, duh! The symmetries of a precisely thermal quantum field state are very exacting global constraints; we can't verify that a state we observe is thermal by local measurements. Fell's theorem, which is stated in terms of the folium of a state, is relevant to this (2.2.13 in Haag's "Local Quantum Physics", with a helpful remark). – Peter Morgan Mar 26 '11 at 14:00
• @Peter The question was about horizons but the motivation was to understand Unruh radiation - which I always thought depended crucially on horizons. But now I'm beginning to understand that it's only the thermality of the radiation that depends crucially on the existence of the horizon. Your comment on thermality requiring precise global constraints is interesting - is there an online reference that elaborates on this? I just got back from the library and don't want to make another trip to get Haag again :/ – dbrane Mar 26 '11 at 14:18
• A thermal state is invariant under all translations and under the subgroup of the Lorentz group that leaves a time-like 4-vector invariant (very close to a vacuum). A non-trivial thermal state is different from the vacuum state everywhere. physics.stackexchange.com/questions/4935/… (my Answer there) is just relevant enough to send you to it. A transformation to accelerated coordinates makes the Fourier modes in the new coordinates different in just the right way. Weird transformations are too messy to give the same result. – Peter Morgan Mar 26 '11 at 15:15
• I forgot to @dbrane you. I can't immediately think of any proper web-refs that seem directly useful. Note that I'm somewhat offconventional about QFT, emphasizing a signal processing, non-particle field perspective, which makes Unruh look somewhat different in what I think is an interesting way, but be careful with it. I've consequently, eg, not found the Unruh and Wald interpretation of the Unruh effect in PRD 29,1047(1984) helpful. – Peter Morgan Mar 26 '11 at 15:42

It seems to me that this question is quite trivial, so I hope I am not overlooking anything.

For a horizon to form, the observer needs to be "as light-like as possible" (otherwise the slope of his world-line will be more than 45 degrees and union of his past light-cones will cover all of universe). And because there is no other way of achieving that than just asymptotically by accelerating indefinitely I think your question is resolved.

• I was doodling a bit and thought - what if you have have a worldline with an observer stationary at the origin until $t=0$ after which it starts accelerating. If you draw past light cones at every point, they don't cover the whole space. So there are inaccessible regions for a semi-indefinite accelerator, right? – dbrane Mar 25 '11 at 13:15
• Ok, ignore that comment. The horizon is determined with respect to a congruence of observers, and it's clear that if you include all possible worldines corresponding to different positions at $t=0$, it covers the entire space. Hence, no horizon there. – dbrane Mar 25 '11 at 13:20
• @Marek: your answer has gotten me a bit more confused now. Suppose I'm in a rocket that has been accelerating since past infinity (yes, I'm very old) and I detect Unruh radiation. Does this mean I can't turn off my engine because if I were to do so, I wouldn't be eternally accelerating? Surely this is absurd. I expect therefore that you should see Unruh radiation (or something like it) even for finite time as Lawrence suggests - but I can't see how there can be a horizon. – dbrane Mar 25 '11 at 14:22
• Also, if only the most light-like observers can see the radiation, that would mean that Hawking radiation would be unobservable since there's no one who has been stationary outside a black hole since forever. You get my drift right? Sorry for the comment bombing. – dbrane Mar 25 '11 at 14:24
• @dbrane: yes, that is indeed puzzling and I am not sure how this connects with Unruh radiation at all (mainly because I am not an expert on the topic; I more or less just know it exists)! Is it really the case that you need a horizon to observe the radiation? Isn't acceleration (for e.g. finite time) enough? I know the classical derivation of Unruh effect assumes indefinite acceleration but I am totally clueless as to what happens when you try to compute with arbitrary acceleration. – Marek Mar 25 '11 at 15:39

The question seems to point to the question of whether or not one can “turn on” the Unruh effect. This of course has some experimental relevance, for if we test it with highly accelerated particles then we do have to prepare a system where the acceleration is initiated. So this question is related to other funny problems, such as the physics which generates a black hole. The exact solution is an eternal solution, but an astrophysical black hole is generated by the implosion of a stellar core.

An accelerated observer will measure a thermal background of radiation on the Rindler chart, which defines wedges in flat spacetime. The Rindler wedge is a region an accelerated observer may completely causally interact with. Consider the Cartesian coordinate chart $$ds^2~=~-dT^2~+~dX^2~+~dY^2~+~dZ^2,$$ for $0~<~X~<~\infty$ and $-X~<~T~<~X$. This slice of spacetime is a Rindler wedge, and the accelerated observer is on the transformed coordinates, $$T~=~x~\mathrm{sinh}(t),~X~=~x~\mathrm{cosh}(t),~y~=~y,~Z~=~z,$$ so the Minkowski line element is $$ds^2~=~-x^2dt^2~+~dx^2~+~dy^2~+~dz^2,~~0~<~x~<~\infty,~-\infty~<~t,~y,~z~<~\infty.$$

On the Rindler chart projections of null geodesics onto a spatial hypersurface are semicircular arcs. A highly accelerated observer will observe a laser beam bent into an arc, just as a projectile on this frame will move on a curve. A Killing vector defines a foliation of spatial surfaces orthogonal to static frames. This is a metric on these spatial surfaces according to a conformal transformation of the original metric, defined as the Fermat metric. For a fixed time $t~=~0$ the line element $$ds^2~=~g_{tt}dt^2~+~g_{ij}dx^idx^j,$$ gives the Fermat line element $$d\sigma^2~=~{{g_{ij}}\over{g_{tt}}}dx^idx^j.$$ The Killing vector $\xi_t~=~\partial_t$ defines a foliation of these surfaces so the static spacetime is a trivial vacuum solution to the Einstein field equation. The Fermat metric for Rindler observers is $$d\sigma^2~=~x^{-2}(dx^2~+~dy^2~+~dz^2),~0~<~x~<~\infty.~-\infty~<~\{y,~z\}~<~\infty,$$ which is the hyperbolic three space ${\cal H}^3$ in the upper half of the space. The projection on $z~=~0$ is the Poincare half place ${\cal H}^2$ with semicircular geodesics. This space is a conformal map from the plane ${\mathbb R}^2$.

The hyperbolic geometry of the spacetime reduced to two dimensions gives the propagator $$G(X,~X^\prime)~=~-{1\over{4\pi^2}}{1\over{(X^\prime_0~-~X_0~-~i\epsilon)^2~-~(X^\prime_i~-~X_i)^2}}$$ for coordinate index $i$ chosen. This propagator has the value $$G(X,~X^\prime)~=~-{1\over {16\pi^2(\alpha~+~1)}}\mathrm{cosech}^2(\Delta t/\alpha~-~i\epsilon/\alpha),$$ for $\alpha$ determined by the hyperbola $X^2~-~T^2~=~\alpha^2$. This propagator satisfies a harmonic condition $(\Delta~-~\lambda)G(X,~X^\prime)$ $=~\delta(X,~X^\prime)$, for $\lambda$ an eigenvalue. This propagator has a thermal or heat kernel interpretation. The thermal radiation is what the accelerated observer detects in the Rindler wedge.

This spacetime is very interesting, for it has an analogue with de Sitter and anti de Sitter spacetimes. Another characteristic is that this is eternal. The exact solution assumes a Rinder wedge for an accelerated mass or observer in Minkowski spacetime that is perpetual. This is analogous in a way to the eternal black hole solution, which is somewhat fictional. However, the eternal solution approximates very closely a final state of a black hole formed from a collapse. In a similar manner if one accelerates a particle to $a$ at a time much earlier than one performs and experiment then the exact theory above will serve as an approximation.

I feel somewhat compelled to write an addendum to this.

Unruh radiation is something which is observed only by the accelerated observer. An inertial observer watching an accelerated mass would detect none of this radiation at all. The accelerated mass starts out at $v~=~-c~+~-\epsilon$ approaches the origin where the split horizon of the Rindler wedge exists at $v~-~0$ and then accelerates off to $v~=~c~-~\epsilon$ in the other direction. The horizon exists at a distance $d~=~a/c^2$, for $a$ the acceleration, from the accelerated mass. If you were on this frame this horizon would “appear” behind your direction of motion. It would be a region from which no information can be received.

Let us assume that the mass starts accelerating at a time $t~=~0$. This would be a case where only the upper half of the Rindler wedge exists. We now assume that the particle accelerates for a long period of time, so it approaches $v~=~c~=~\epsilon$. The motion asymptotes to a null line, and the path is half of a hyperbola. The absence of the bottom null asymptote will have less and less influence on what happens later on. This is a physical statement. First off if an inertial observer is moving at a velocity $u$ in the direction of the accelerated mass, this mass will appear to approach this observer and then recede. For $u$ large enough this accelerated mass would appear to come at this observer with a velocity $v’~=~-c~+~\delta$ where $\delta~>~\epsilon$, but not by “that much.” The accelerated mass will then approach this observer, stop and accelerate outwards. So “modulo” this small error, this situation is a good approximation to the eternal accelerated case.

The comparison to a black hole is with the Penrose diagram for the Schwarzschild metric. The physical black hole removes the bottom half of the diagram which corresponds to a “white hole.” The Rindler wedge is a near horizon approximation to a black hole. We do not expect any strange loss of Hawking radiation from a physical black hole to result from this

The particle horizon partitions regions of the Minkowski spacetime so the accelerated observer receives signals from only a portion of the spacetime. This causal partition means the accelerated observer measures the vacuum state as being a mixed state of thermal bosons. The partition removes a unitary description of the vacuum for the accelerated observer. So what the inertial observer witnesses as a pure vacuum states with zero temperature, the accelerated observer witnesses as a vacuum plus thermal distribution of radiation.

We now have the problem of detecting this. The inertial observer does not see the Unruh radiation. The accelerated observer does. The temperature is $T~\simeq~10^{-21}K/cm-sec^{-2}$ So you need a pretty serious acceleration to observe anything on that frame. Clearly a real observer can’t be on the frame. So not only do we need a start for the acceleration, we need a stop as well. The object on the accelerated frame will absorb this Unruh radiation, and if the inertial observer can “catch” the object its temperature may be measured. If this object is a Bose Einstein condensate of Ba ions that are accelerated enormously through a high $\vec E$ field region, say in a capacitor, the Unruh radiation might be enough to change the phase of the BEC.

I wrote a paper a fair number of years ago which got “honorable mention” for the Gravity Essay Prize where I proposed just this. Unfortunately nobody seemed interested in actually doing the experiment --- which is what I wrote this paper for!

Clearly we can’t have an eternal accelerated object, so we must work with approximate situations. However, some physical reasoning should strongly suggest that a removal of the $\infty$ or eternal aspects of the Unruh results can be made into a small perturbation.

• Thanks Lawrence. But what I'm really interested in is whether a horizon exists in the non-eternal Rindler case. It's possible to demonstrate the existence of a horizon in the case of a non-eternal astrophysical black hole, but I don't see how one can do this for non-eternal Rindler. – dbrane Mar 25 '11 at 14:14
• As I indicated the eternal solution should serve as a reasonable approximation to a situation where one measures the physics at a time long after you turned the acceleration on. A plausible criterion for that would be to measure things at a time $T~>>~1/\nu$, for $\nu$ the frequency of radiation at the peak of the black body curve. So I would say that horizons would exist. – Lawrence B. Crowell Mar 25 '11 at 14:28
• I don't know - I'm not convinced at all that a horizon should exist, though I can see why there should be particle detection. – dbrane Mar 25 '11 at 15:06
• neither do i, it seems to me that any observer that is inertial in both the asymptotic past and future will affect any event in the Minkowski plane, and will be affected by all events in it. This seems to apply even if the observer has a constant non-linear acceleration (i.e: rotating around a center) but it is otherwise translationally inertial – lurscher Mar 25 '11 at 15:44
• ...unless the absence of an horizon is inconsequential to the presence or not of some form of thermal radiation? – lurscher Mar 25 '11 at 16:24

Since you're really interested in the horizon as part of an approach to understanding the Unruh effect, I will address that instead of the initial Question. What experimental resources are we to consider an accelerating observer A to have? If A can make only local measurements in the laboratory La in the diagram, then (in the Wightman or Haag-Kastler axiomatic backgrounds) A can make exactly the same deductions about the state from their measurements as can an observer who can use the laboratory L0. Because of the Reeh-Schlieder theorem, what is at space-like separation from the two laboratories is equally accessible to both (in the sense that both can get as close as they like to knowing what the statistics of measurement results would be everywhere in Minkowski space by making precise enough measurements in their local laboratory, on the assumption that the Wightman axioms hold or that the Haag-Kastler axioms hold). [It would have been better not to have drawn the two laboratories the same shape, because the shapes are not relevant to the Reeh-Schlieder argument, but it's uploaded now.]

A local laboratory, however, cannot tell whether the state of the universe is in the folium of the vacuum state, in the folium of a thermal state, or in the folium of some other less symmetrical state; the folium of a highly symmetric state is the set of states that are in the Hilbert space that is constructed by the GNS construction from the vacuum state (resp. from a thermal or some other state) over the net of algebras of observables associated with regions of Minkowski space.

I take your discussion about the Unruh effect that is prompted by the earlier Answers to boil down to the fact that it's only if our laboratory is not confined to a finite region of space-time that we can make any global claims about the state of the universe, despite the Reeh-Schlieder theorem. We could only possibly determine whether the state is in the folium of a thermal state instead of in the folium of the vacuum state if our laboratory is infinite in extent. Having available as our experimental resource a particle that interacts with the vacuum state in an ad-hoc way for the whole of an accelerating trajectory, as is proposed in Unruh and Wald's Phys. Rev. A 29,1047(1984), even though it's infinite in extent, seems unlikely to be enough to make such a fine-grained determination.

I don't plan on putting the terse comments that I made earlier on the Question into this Answer, however they do inform the perspective a little if you choose to think about the Unruh effect in something like the way I have done here. The above is essentially tentative, something I've been mulling for several years, so I'd be glad to have suggestions for how anyone here would fill the gaps or of what I haven't taken into account that will shoot it down.

Finally, thanks @dbrane for the Question. I held off giving it +1 because although it seemed definitely useful it didn't seem quite clear enough, but when I realized how much it was stimulating, I decided that it was better for me than had it been clear.

PS. Hans Halvorson's paper on "Algebraic Quantum Field Theory" (with Michael Mueger), http://arxiv.org/abs/math-ph/0602036, is a good on-line resource, although in many ways it takes you one step higher than Haag. His treatment of the concept of a folium is more sophisticated than Haag introduces in his Local Quantum Physics, in terms of quasi-equivalence of states. Hans is one of the few Philosophers of Physics who can really come up to scratch with Mathematicians on AQFT.

• Infinite extent laboratories? May be somewhat ambiguous concept - space or time infinite, not consistent with the Equivalence Principle and thus not transferable to GR (which is where I assume that all this is leading). – Roy Simpson Mar 26 '11 at 22:46
• @Roy, Yes, there could be seen to be an element of reductio ad absurdam to infinite extent laboratories. Part of why AQFT has gone to local observables, I 'spose, incidentally making the folia of the thermal and vacuum states experimentally indistinguishable. The tradition for QFT on CSTs is to introduce enough constraints that there is a concept of generators of translations, which is enough for us to talk about translation invariant states as an ideal, with vacuum and thermal states as examples. With said constraints, a nontrivial connection doesn't prevent my discussion. Good comment, +1. – Peter Morgan Mar 27 '11 at 12:12
• At the moment, this answer isn't particularly useful for me since I have zero familiarity with axiomatic QFT. The way I see it (probably naively), a detector can make enough local observations (particle clicks at different wavelengths) to conclude over a long period of time that the spectrum registering is a thermal one. Is there a somewhat more physical way of putting your argument? (Perhaps, a detector confined to an infinitesimal region cannot detect particles of arbitrarily large wavelength and hence establish the thermality of the spectrum at the large wavelength tail?) – dbrane Mar 29 '11 at 13:15
• I think your "perhaps" is a good enough way to think about this. Note that thinking in terms of pure frequencies is inherently infinitely nonlocal, and also singular, since a Fourier mode is not square integrable, but exactly how much we should care about mathematical niceties in QFT is moot. – Peter Morgan Mar 29 '11 at 13:54

If Unruh effect is a real process it is obvious that turning off the acceleration will not make the temperature suddenly skip to zero. We can wonder what happens for an eternal inertial observer who is only accelerated during the period [-L , L]

Rovelli gives an answer for an observer with such a finite life. https://arxiv.org/abs/gr-qc/0212074 As the notion of temperature is only well adapted for equilibrium Rovelli generalizes it.

we then read that it evolves during his life even if the acceleration is constant. it gives the usual value for τ=0 eq 54 reads $β(τ) = 2π/La^2 \sqrt{1 + a^2 L^2} − ch aτ$

here β is the inverse temperature.