Simplified derivation of the Hawking–Unruh temperature for an accelerated observer in vacuum

S. W.

Hawking

, “,” , – ();

S. W.

Hawking

, “,” , – ().

W. G.

Unruh

, “,” , – ().

P. C. W.

Davies

, “,” , – ().

The literature on this subject is vast. For some articles directly relevant for the work presented here see, for instance,

P.

Candelas

and

D. W.

Sciama

, “,” , – ();

T. H.

Boyer

, “,” , – () and

T. H.

Boyer

, “,” , – ();

D. W.

Sciama

,

P.

Candelas

, and

D.

Deutsch

, “,” , – ().

An extensive review is given by

S.

Takagi

, “,” , – (). See Chap. 2 for a review of the Davies–Unruh effect.

N. D. Birrell and P. C. W. Davies, Quantum Fields in Curved Space (Cambridge U.P., New York, 1982).

Note that in order for a detector to remain at a fixed location outside the horizon of a black hole, it must undergo constant acceleration just to remain in place.

W.

Rindler

, “,” , – ().

P. W. Milonni, The Quantum Vacuum (Academic, New York, 1994), pp. 60–64.

I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products (Academic, New York, 1980). We have used integrals 3.761.4 and 3.761.9, $∫0∞xμ−1 sin(ax)dx=[Γ(μ)/aμ]sin(μπ/2)$ and $∫0∞xμ−1 cos(ax)dx=[Γ(μ)/aμ]cos(μπ/2)$ respectively, in the combination of the second plus $i$ times the first. Taken together, both integrals have a domain of definition $a>0,$ $0<Re(μ)<1.$ In Eq. (8) we have $μ=iΩc/a$ with $Re(μ)=0.$ The integrals can be regularized and thus remain valid in the limit $Re(μ)→0$ as can be seen by adding a small imaginary part $−iεa/c,$ $ε>0$ to the frequency Ω so that $μ→μ′≡μ+ε$ and $0<Re(μ′)=ε<1$ is strictly in the domain of definition of the integrals. In the limit of $ε→0$ we have $[Γ(μ+ε)/aμ+ε]ei(μ+ε)π/2→[Γ(μ)/aμ]eiμπ/2$ and thus obtain the same values for the integrals as if we had just set $Re(μ′)=0$ initially. The equivalence of these two approaches to evaluate the integrals occurs because the standard integral form of the gamma function $Γ(z)=∫0∞dt e−ztz−1,$ $Re(z)>0$ can be analytically continued in the complex plane and in fact remains well defined, in particular, for $Re(z)→0,$ $Im(z)≠0,$ which is the case in Eq. (8). See for for example, J. T. Cushing, Applied Analytical Mathematics for Physical Scientists (Wiley, New York, 1975), p. 343.

Reference 10, Sec. 8.332.

Related, although much more involved derivations of the Davies–Unruh effect based on a similar substitution can be found in

L.

Pringle

, “,” , – ();

U. H.

Gerlach

, “,” , – (),

U. H.

Gerlach

, gr-qc/9910097 and

U. H.

Gerlach

, “,” , – ().

We can justify this substitution from the form $φ=∫kμ(x)dxμ$ of the phase of a quantum mechanical particle in curved space–time. See

L.

Stodolsky

“,” , – () and

P. M.

Alsing

,

J. C.

Evans

, and

K. K.

Nandi

, “,” , – (),

P. M.

Alsing

,

J. C.

Evans

, and

K. K.

Nandi

, gr-qc/0010065.

A similar derivation in terms of Doppler shifts appears in the appendix of

H.

Kolbenstvedt

, “,” , – ().

T. Padmanabhan and coauthors also have derived Eq. (9) by similarly considering the power spectrum of Doppler shifted plane waves as detected by the accelerated observer. See

K.

Srinivasan

,

L.

Sriramkumar

, and

T.

Padmanabhan

, “,” , – ();

T.

Padmanabhan

, “,” gr-qc/0311036.

When we later convert a sum over $K$ to an integral, we obtain the Lorentz-invariant measure $dK/ωK$ as a consequence of the $1/ωK$ in Eq. (10). The use of this invariant measure eliminates the need to explicitly transform the frequency term $1/ωK$ in Eq. (14), for instance.

Because we use a one-dimensional model, the factor $V/2π$ appears instead of the more familiar $V/(2π)3.$ In other works, our volume $V$ here is really just a length.

If we use the same gamma function integrals as in Ref. 10, we will have for the Dirac case $μ=iΩc/a+1/2,$ with $Re(μ)=1/2$ clearly in their domain of definition $0<Re(μ)<1.$

S. Weinberg, Gravitation and Cosmology (Wiley, New York, 1972), pp. 365–370.

C. W. Misner, K. S. Thorne, and J. A. Wheeler, Gravitation (Freeman, San Francisco, 1973), Chap. 6, pp. 163–176.

Reference 5, p. 101, and

P.

Candelas

and

D.

Deutsch

, “,” , – ().

This result can be understood as follows. If the observer was traveling at constant velocity $v$ in the positive $z$ direction, we would Lorentz transform the spinor in the usual special relativistic way via the operator $Ŝ(v)=exp(γ0γ3ξ/2)$ where $tanh ξ=v/c.$ See J. D. Bjorkin and S. D. Drell, Relativistic Quantum Mechanics (McGraw–Hill, New York, 1964), pp. 28–30. From Eq. (3) we have $v/c=tanh(aτ/c)$ so that $ξ=aτ/c,$ yielding the spinor Lorentz transformation to the instantaneous rest frame of the accelerated observer.

For simplicity, we have chosen the spin up wave function as an eigenstate of $Ŝ(τ)$ with eigenvalue $exp(aτ/2c).$ The exact spatial dependence of the accelerated (Rindler) spin up wave function is more complicated than this simple form, although both have the same zero bispinor components. See W. Greiner, B. Müller, and J. Rafelski, Quantum Electrodynamics in Strong Fields (Springer, New York, 1985), Chap. 21.3, pp. 563–567;

M.

Soffel

,

B.

Müller

, and

W.

Greiner

, “,” , – ().

The spinor Lorentz transformation $Ŝ(τ)$ does not mix spin components. Thus, for example, a spin up Minkowski state remains a spin up accelerated (Rindler) state. We can therefore drop the constant spinor |↑〉 from our calculations and retain the essential, new time-dependent modification $exp(aτ/2c)$ to the plane wave for our Dirac “wave function.”

Reference 5, Sec. 2, in particular Eqs. (2.7.4) and (2.8.8).

S. A.

Fulling

, “,” , – () and

Aspects of Quantum Field Theory in Curved Space-Time (Cambridge U.P., New York, 1989).

This viewpoint is also taken in a different derivation of the Unruh–Davies effect in Ref. 9.

The unaccelerated Minkowski vacuum $|0M〉$ is unitarily related to the Rindler vacuum $|0R〉⊗|0L〉$ via $|0M〉=Ŝ(r)|0R〉⊗|0L〉,$ where $Ŝ(r)$ is the squeezing operator $Ŝ(r)=exp[r(âRâL−âR†âL†)].$ The subscripts $R$ and $L$ denote the right $(z>0,z>|t|)$ and left $(z<0,|z|>|t|)$ Rindler wedges, respectively, which are regions of Minkowski space–time bounded by the asymptotes $t=±z.$ $|0R〉$ is the Fock state of zero particles in the right Rindler wedge and $|0L〉$ is the Fock state of zero particles in the left Rindler wedge. Note that the orbit of the accelerated Rindler observer given by Eq. (4) is confined to the right Rindler wedge. Because the right and left Rindler wedges of Minkowski space–time are causally disconnected from each other, the creation and annihilation operators $âR†,$ $âR$ and $âL†,$ $âL$ live in the right and left wedges, respectively, and mutually commute with each other, that is, $[âR,âL†]=0,$ etc. Because, physical states that live in the right wedge have zero support in the left wedge (and vice versa), they are described by functions solely of the operators $âR†,$ $âR$ appropriate for the right wedge, that is, $|ΨR〉=f(âR,âR†)|0R〉⊗|0L〉=|ψR〉⊗|0L〉.$ It is in this sense that we can speak of $|0R〉$ as the vacuum for the right Rindler wedge, and similarly $|0L〉$ as the vacuum for the left Rindler wedge.