The Unruh Temperature
University of tennessee, knoxvilleThe Unruh TemperatureFor Uniformly Accelerated ObserversCory R. Thornsberry12/7/2012This paper will show that an observer, accelerating in a vacuum, will detect a thermal bath of particles instead of the vacuum state. It will be shown that the observer will be able to measure a “temperature” associated with those as well as solving for said temperature.The Unruh EffectImagine, for a moment, that an observer is at rest in a vacuum in plain Minkowski Space. Now imagine that we add a second, inertial, observer to our vacuum. Since the two observers are in inertial frames, we would logically expect them to agree that they were both in a vacuum state since the trajectory of the two are related via a Lorentz transformation. But, it was shown by W. G. Unruh in 1976 that “…an accelerated detector, even in flat spacetime, will detect particles in the vacuum”[6]. That is, even though the two observers are moving through the vacuum, the observer that is accelerating will “feel” as though they were moving through a field of thermal particles at temperature Tu. This temperature is known as the Unruh temperature[1].In this paper I will show that the accelerating (Rindler) observer is moving through a thermal bath of particles as well as solving for the Unruh temperature (Tu) of the system. For simplicity, I will examine only the case of an observer moving with a uniform acceleration, a, in the z-direction. Before we are able to approach the Unruh effect, we must first examine the metric we are working in. In this case, it is Minkowski Space. In the following work, I have set c = 1.Since we are only interested in an observer moving in the z-direction we may define our spacetime interval asds2=dt2-dz2[1]We know that Minkowski space is invariant under the boostt→tcoshβ+zsinhβz→tsinhβ+zcoshβ[2]So we may parameterize our coordinates ast(τ)=ρsinhτzτ=ρcoshτ[3]Where ρ and τ are real. When plotted, this parameterization forms a trajectory outside the light-cone, i.e. z2 > t2 (spacelike). Since our interval is spacelike, we know that it is also the proper distance and is a physical value. This trajectory asymptotes to z = ±t, meaning that the observer approaches the speed of light as t → ±∞. From this coordinate transformation, we see thatdt=dρsinhτ+ρdτcoshτdz=dρcoshτ+ρdτsinhτ[4]So our spacetime interval is re-written asds2=dρ2sinh2τ+ρ2dτ2cosh2τ-dρ2cosh2τ-ρ2dτ2sinh2τ[5]Now we use hyperbolic trigonometric identities to see thatds2=ρ2dτ2-dρ2[6]The new spacetime metric is independent of our boost parameter τ. This is the so-called Rindler Space. Moreover, the portion of space we are working in is known as the “Right Rindler Wedge”. Out of convenience, we may perform a second coordinate transformation, ρ=eaξ, τ=aη, such thatt=1aρsinhτ=1aeaξsinhaηz=1aρcoshτ=1aeaξcoshaη[7]Where a is some positive constant[5]. Keeping ξ fixed will create a hyperbolic trajectory. If, instead, we fix τ, a straight line through the origin is produced (see fig 1). As before, we see thatdt=dξeaξsinhaτ+dηeaξcoshaτdz=dξeaξcoshaτ+dηeaξsinhaτ[8]Our newly transformed metric becomesds2=eaξ(dξ2sinh2aτ+dτ2cosh2aτ-dξ2cosh2aτ+dτ2sinh2aτ)[9]And, again, we use standard hyperbolic trigonometric identities to find thatds2=eaξ(dτ2-dξ2)[10]According to the inertial observer, our field φ must satisfy the massless Klein-Gordon equation. Since we are only concerned with motion in the z-direction, we may write it as?μ?μφ=?t2-?z2φ=0[11]Since m=0. This equation has plane wave solutions of the formukt,z=12ke-ik(t-z)[12]Where I have included a normalization constant. Since our field is massless we have thatE=ωk=m2+k2=k[13]The energy of our system can be both positive and negative. We also gather that the positive energy solutions correspond to a momentum (trajectory) in the positive z-direction while the negative solutions correspond to momentum in the negative z-direction. We will define z±=z±t[14]Such that, for “rightward” moving waves (k > 0)ukt,z=12keikz-[15]And similarly for “leftward” moving waves (k < 0)ukt,z=12ke-ikz+[16]So now, for the inertial observer anyway, the massless scalar field, φ, may be expanded asφt,z=0∞dk2π(b(k)uk+b?kuk*)[17]The annihilation and creation operators satisfybk,bk'?=δ(k-k')[18]And, as always, we define the vacuum state asbk|0=0[19]So that’s it for the inertial observer. Now we turn our attention to the non-inertial observer. Looking at the Rindler Metric we derived in equation (10), we can see that our new field equation is given by?τ2-?ξ2φ=0[20]Again, we get plane wave solutions of the formukτ,ξ=12ke-ik(τ-ξ)[21]As above, we get a positive energy solution and a negative energy solution. This time, we defineξ±=ξ±τ=1aLnax±[22]So we get as the rightward moving solutionukτ,ξ=12keikξ-[23]And the leftward moving solutionukτ,ξ=12ke-ikξ+[24]We may write the plane waves above in terms of z+ and z-. Ignoring the constant coefficients givese±ikξ?=eLn(az?±iωa)=(az?)±iωa[25]Where, due to our definition of ξ±, we require that z±>0.Fig 1: Minkowsky Space with Rindler CoordinatesRegionz+ = z+tz- = z-tI> 0> 0II> 0< 0III< 0< 0IV< 0 > 0Table 1: Values for Z± vs. Rindler RegionThe way in which we have defined our Rinder coordinates has restricted our hyperbole to Region I of Minkowsky space. Despite our two solutions (eq. 15,16 and eq. 23,24) looking very similar, they differ in the fact that the coordinates (t,z) may vary in all four regions while, on the other hand, our Rindler coordinates (τ,ξ) are now restricted to region I. By observing the table above, we see that equation (25) may be used to “extend” our plane wave solutions into the other three regions.We may analytically extend eiωξ-into region IV where z- > 0. Additionally, we may extend e-iωξ+into region II where z+ > 0. It follows that, since both z+ and z- are less than zero, neither equation may be used to extend our solutions into region III. Therefore, our solutions do not form a complete set. To rectify this problem we simply flip the signs on the time and spatial coordinates (t,z)→(-t,-z). This operation effectively exchanges region I for region III and region II for region IV. Using equations (23,24), we get two Unruh modesuk(1)=12keikξ-=12ke-ik(τ-ξ)0 , Region IRegion IIIuk(2)=012ke-ikξ+'=12keik(τ'-ξ') ,Region IRegion III[26]One has support in region I, the other has support in region III. Notice that the mode with support in region III has an exponential term with a positive time component in its argument. Whereas, the mode with support in region I has a positive component. When we performed the time reversal operation, (t)→(-t), we got1aρsinhaη→-1aρsinhaη=(1aρsinha(-η))[27]This means that our transformed time coordinate, η, now points toward the past. Since we now have all the required parts for the non-inertial observer’s field, we may expand it asφt,z=0∞dk2π(c1(k)uk(1)+c2(k)uk2+c1?(k)uk1*+c2?(k)uk2*)[28]With, again, the vacuum defined asc(r)k|0=0[29]Where r = 1,2 for the two Unruh modes. Now we must use a Bogoliubov transformation to relate the inertial observer’s creation and annihilation operators (b, b?) to the creation and annihilation operators of the non-inertial observer (c, c?). As mentioned earlier, we may use equation (25) to extend our solutions into the other regions of Rindler space. The solutions in region I will also support solutions in region II. We getuk(1)=12k(az-)iωa0 , Region I, IIRegion III, IV[30a]Likewise, solutions in region III will support solutions in region IV. In order to get a solution that looks like the first mode, we take the complex conjugate of the second mode…uk2*=012k(-az+)iωa ,Region I, IIRegion III, IV[30b]The two modes look similar except for the -1iωa. We may easily writez-=-z-eiπ→-z-=z-e-iπ[31]So, we may re-write the second mode asuk2*=12keπωa(az-)iωa[32]Now we define a new solutionUk(1)=uk1+e-πωauk2*=12k(1+e-πωaeπωa)az-iωa=2kaz-iωa[33]Similarly, for the second modeUk(2)=uk2+e-πωauk1*=2kaz+-iωa[34]These two solutions now span all of Minkowski space. Because of this, they form a set of solutions that are just as good as our original inertial observer plane wave solutions, equations (15) and (16). Now we may re-write the inertial observer’s field equation (equation (28)) asφt,z=0∞dk2π(B1(k)Uk(1)+B2(k)Uk2+B1?(k)Uk1*+B2?(k)Uk2*)[35]And we may re-write the definition of the inertial observer’s vacuum state (equation (29)) asB(r)k|0=0[36]For the two modes, r=1,2. Now, using the two solutions, equations (33) and (34), we may relate these modes to the modes of the Rindler field equation, equation (28). We see thatc1k=B1k+e-πωaB2?(k)c2k=B2k+e-πωaB1?(k)[37]Now, we use the commutation relationcr(k),cs?(k')=δrs(2π)δ(k-k')[38]To find thatcr(k),cs?(k')=cr(k)cs?(k')-cs?(k')cr(k)[39]=Brk+e-πωaBs?kBs?k'+eπωaBrk'--Bs?k'+eπωaBrk'Brk+e-πωaBs?k[40]=BrkBs?k'+eπωaBrkBrk'+e-πωaBs?kBs?k'+Bs?kBrk'-Bs?k'Brk-e-πωaBs?k'Bs?k--eπωaBrk'Brk-Brk'Bs?k[41]=Brk,Bs?k'-Brk',Bs?k++eπωaBrk,Brk'+e-πωa[Bs?k,Bs?k'][42]The commutators with only creators or annihilators (third and fourth terms) are equal to zero. We may relate the first and second term by switching k’ and k in the second term viaBrk',Bs?k=e-πωaBrk,Bs?k'[43]So, equation (39) becomescr(k),cs?(k')=(1-e-πωa)Brk',Bs?k=δrs(2π)δ(k-k')[44]Brk',Bs?k=11-e-πωaδrs(2π)δ(k-k')[45]Thus,Brk',Bs?k=e-πωa2sinhπωaδrs(2π)δ(k-k')[46]Now definedrk=e-πω2a2sinhπωaBrk[47]As our new annihilation and creation operators. These operators now satisfydrk,ds?k'=δrs(2π)δ(k-k')[48]So now, we may re-write the Rindler modes asc1k=12sinhπωaeπω2ad1k+e-πω2ad2?kc2k=12sinhπωaeπω2ad2k+e-πω2ad1?k[49]These equations are known as the Bogoliubov Transformation. They relate the Rindler and inertial observer modes. Now we assume our system is in the Minkowski Vacuum |0>. ThenNk=c1?kc1k[50]Is the number operator for our system. Let’s look at the expectation value of the number operator.0Nk0=12sinhπωa0|eπω2ad1?k+e-πω2ad2keπω2ad1k+e-πω2ad2?k|0[51]=12sinhπωa0|eπωad1?kd1k+d1?kd2?k++d2kd1k+e-πωad2kd2?k|0[52]Note that term the first and third terms will both annihilate on the vacuum, since d(1)~B(1) and B(1) acting on the vacuum is equal to zero (eq. 36). The expectation value becomes=12sinhπωa0|d1?kd2?k+e-πωad2kd2?k|0[53]=e-πωa2sinhπωa0|d2kd2?k|0[54]=e-πωa2sinhπωa0|d2?kd2k+(2π)δ(0)|0[55]=e-πωa2sinhπωa0|(2π)δ(0)|0[56]=1e2πωa-12πδ(0)[57]This equation looks surprisingly similar to Planck’s LawBT=2hω31ehωkBT-1[58]Comparing the argument of the exponential in the denominator of equation (57) and the argument from the exponential of Planck’s law yieldsTu~ha2πkB[59]That is, for an observer with non-zero acceleration, said observer will experience a temperature while accelerating through the vacuum! Even though the non-inertial observer is moving through a vacuum, they will still observe particles in a thermal bath at temperature Tu.In the previous work, I showed that an observer accelerating (uniformly) in Rindler space will not only detect particles from the vacuum, but will be able to measure the temperature of those particles, Tu~a2π. The temperature the observer will measure is proportional to the acceleration of the observer. At first glance, you may expect that this would violate the law of conservation of energy. But the non-inertial observer is being accelerated by some external force. Some of the energy from the external force is used to create the “thermal bath” of particles which the observer is moving through[3]. If you think of the observer as a “particle detector”, it is possible to show (though I will not derive it in this paper) that the detector will detect the thermal particles in the accelerated frame. Despite being in the Minkowski vacuum, the inertial observer cannot disagree that the detector in the non-inertial frame has detected a particle.ReferencesBièvre, S., Merkli, M. “The Unruh effect revisited”. Class. Quant. Grav. 23, 2006 pp. 6525 – 6542Crispino, L., Higuchi, A., Matsas, G. “The Unruh effect and its applications”, Rev. Mod. Phys. 80, 1 July 2008 pp. 787 – 838Pringle, L. N. “Rindler observers, correlated states, boundary conditions, and the meaning of the thermal spectrum”. Phys. Rev. D. Volume 39, Number 8, 15 April 1989 pp. 2178 – 2186Siopsis, G. “Quantum Field Theory I: Unit 5.3, The Unruh effect”. University of Tennessee Knoxville. 2012 pp. 134 – 140Rindler, W. “Kruskal Space and the Uniformly Accelerating Frame”. American Journal of Physics. Volume 34, Issue 12, December 1966, pp. 1174Unruh, W. G. “Notes on black-hole Evaporation”. Phys. Rev. D. Volume 14, Number 4, 15 August 1976 pp. 870 – 892 ................
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