On the Response of Particle Detectors in Vaidya Spacetimes

0.18 MB·English

by  Alex Venditti

On the Response of Particle Detectors in Vaidya
Spacetimes
A.M.Venditti1, C.C.Dyer2
2
DepartmentofPhysics,UniversityofToronto,50St. GeorgeStreet, Toronto,
1
OntarioM5S3H4,Canada1
0
DepartmentofAstronomyandAstrophysics,UniversityofToronto,50St.
2 GeorgeStreet,Toronto,OntarioM5S3H4,Canada2

n DepartmentofPhysicalandEnvironmentSciences,UniversityofTorontoat
a Scarborough,1265MilitaryTrail,Scarborough,OntarioM1C1A4,Canada2
J E-mail: [email protected],[email protected]

2
1
Abstract. Using the formalism of the interaction picture we calculate an

expressionfortheWightmanfunctionforonlythesphericallysymmetricmodesof
]
c aquantum Klein-GordonscalarfieldinageneralVaidyaspacetime withingoing
q nulldust. Itisdemonstratedthatparticledetectorsfollowingtime-liketrajectories
- that are in the ground state at some time outside of the collapsing shell will
r respond independently from the configuration of the ingoing null dust if the
g
responseistakenatanytimeoutsideofthecollapsingshell. Fordetectorsthatare
[
takentobeinthegroundstateatatimeinteriortotheshellitisshownthattheir

responsewilldepend onthe configuration of the ingoingnulldust. Relevance to
1
theinformationlossparadoxisdiscussed.
v
7
6
6
PACSnumbers: 04.62.+v,04.70.Dy
2
.
1
0
Submitted to: Class. Quantum Grav.
2
1
: 1. Introduction
v
i
X In Hawking’s original paper on black hole radiation [1], a general spherically sym-
r metric collapse of matter is considered in which there is no interaction between the
a
quantum field and the classical matter composing the black hole. The trade-off for
considering such a general collapse scenario was that one could not calculate the de-
tailed spectrum of radiation everywhere in the spacetime.
Hawking later demonstrated that if the radiation coming from a black hole was
exactlythermalthenonewouldloseallinformationaboutthematterthatformedthe
black hole [2]. The lose of information was expressedmathematically, in this case, by
the possible non-unitary evolution of quantum states, this is the famous information
paradox.
Therehavebeenseveralattemptstosolvetheinformationloseparadoxbydemon-
strating that the information is not lost at all. One class of these solutions is to
demonstrate that when the back reaction of the emitted radiationis allowedto affect
On the Response of Particle Detectors in Vaidya Spacetimes 2
the mass of the black hole then information is not lost at all, for example see [3],
[4]. There have also been claims that a full theory of quantum gravity is unitary and
therefore information preserving [5].
Otherattemptsclaimsimplythatiftheradiationiscalculatedindetailforspecific
spacetimes then one would find the radiation is not thermal and hence information
can be carried away in the deviations from thermality [6], [7], [8].
A third class of attempts to resolve the information paradox is to demonstrate
that the radiation is exactly thermal and information is lost and to demonstrate that
the laws of physics would still be consistent [9].
Thispaperwilldemonstratethat,intheabsenceofbackreactioninVaidyaspace-
times,aparticledetectorthatfollowsatime-liketrajectorywhichiscoupledtoaspher-
ically symmetric, massless Klein-Gordon field will not be able to distinguish between
the different null dust configurations that collapse to form the black hole. Hence, the
conclusions of this paper are in line with that of [9], [2] and [1] and opposed to the
conclusions of [6], [7], [8] with regards to the thermality of the outgoing radiation.
In this paper the Vaidya metric will be used as a model for a black hole formed
by null dust (see [10]).
ds2 =(1 2m(v)/r)dv2 2dvdr r2dθ2 r2sin2(θ)dϕ2 (1)
− − − −
where v is the ingoing null coordinate that is constant on radially ingoing null
trajectories. We will work with a mass function m(v) that is of the form
0 v <0
m(v)= f(v) 0 v T (2)
 m0 ≤v ≤>T
wheref(v)issomeincreasingfunctionthatgoesfrom0tom0 andT issometimescale.
Physically, the Vaidya class of spacetimes with the above mass function correspond
to aspacetime withasphericallysymmetricshellwithfinite thickness ofingoinglight
collapsing to a point to form a black hole. Outside of the shell the spacetime is
Schwarschildandinsidethe spacetimeisMinkowski. Fromnowonwewillrefertothe
regions v <0, 0<v <T and v >T as regions α, β and γ respectively.
The spherically symmetric modes (s-waves) of a complex, massless scalar field
will be quantized in this background. The Wightman function will be calculated
using only the spherically symmetric modes and the response of a particle detector
will be evaluated for a constant r observer. The particle detector response can be
thought of as due only to the spherical modes. It will be shown that the response
rate of the particle detector at a point outside of the collapsing shell will not depend
on the mass function m(v) if the detector is taken to be in its ground state at a null
time outside of the collapsing null dust. Therefore the information about the matter
usedtoformtheblackholecannotbepresentincorrelationsintheoutgoingradiation.
On the Response of Particle Detectors in Vaidya Spacetimes 3
2. Quantization
The action for a complex, massless scalar field in the Vaidya background is
S = d4x√ g∂ φ∂ φ∗gab (3)
a b
Z −
with g taken to be the above metric. The above action will be only linear in time
ab
derivatives of the field. Therefore, the system is a constrained one and this will have
to be taken into account when performing the canonical quantization.
It will be shown that the Hamiltonian for the system can be written in the form
H =H0+H1(v) (4)
The H0 part will be the Hamiltonian for a massless complex scalarfield in flat space-
time. It will be convenient to work in a picture where the states evolve under the
H1(v) part of the Hamiltonian while the operators evolve under the H0 part. Thus
the scalar field operators will evolve as if they were in flat spacetime and the rest of
the time evolution due to the curved geometry will be shunted onto the state.
Assuming no angular dependence of the scalar field (s-waves)we can expand the
field φ as follows
φ(v,r)= a (v)f (r) (5)
k k
Xk
Substituting the above expansion in the action and substituting in the Vaidya
metric the action reduces to
S = dv (m(v)a∗C a a∗B a ia∗A a˙ ) (6)
Z k kl l− k kl l− k kl l
Xk,l
wherethea˙ denotesdifferentiationofa withrespecttotheingoingnulltimev. The
k k
Hermitian matrices A, B and C are defined as
∞
A = 4πi drr2(f ∂ f∗ ∂ f f∗) (7)
kl − Z0 l r k − r l k
∞
B =4π drr2∂ f ∂ f∗ (8)
kl r l r k
Z0
∞
C =8π drr∂ f ∂ f∗ (9)
kl r l r k
Z0
The action can be re-written in a more convenient form by ignoring a boundary
term. The re-written action is
S = dvL(v) (10)
Z
i i
= dv a†Aa˙ + a˙†Aa a†Ba+m(v)a†Ca
Z (cid:18)−2 2 − (cid:19)
where a is a column vector of the modes defined by a := (a1,a2,...)t. The conjugate
momenta is given by the usual formulae
δS i
p = = a†A (11)
δa˙ −2
δS i
p† = = Aa (12)
δa˙† 2
On the Response of Particle Detectors in Vaidya Spacetimes 4
where p is the row vector p := (p1,p2,...) and p† is the corresponding column
vector. Hencetherearetwoinfinite setsofprimaryconstraintsbetweentheconjugate
momenta (p and p†) and the a , a†. In order to quantize this system we follow
k k k k
the Dirac procedure outlined in [11]. We denote the constraints for some conjugate
momenta p from the first set of constraints (13) and p† from the second set of
k k
constraints (14) respectively by
i
G1,k =pk+ 2a†lAlk =0 (13)
Xl
i
G2,k =p†k− 2Aklal =0 (14)
Xl
The canonical Hamiltonian defined by the Legendre transformation is
H =pa˙ +a˙†p† L=a†Ba m(v)a†Ca (15)
− −
where L is the Lagrangian given by (10). Using the primary constraints in equation
(13), (14) we can re-write the Hamiltonian as
H =2pA−1BA−1p† 2m(v)pA−1CA−1p† (16)
−
1 1
+ a†Ba m(v)a†Ca
2 − 2
We nowshowthatthe time derivative ofeachofthe constraintsG1,k, G2,k is equalto
some multiple of themselves. The time evolution is generated by the Hamiltonian
in equation (16). Using the normal Poisson bracket relations a ,p = δ and
k l lk
{ }
a†,p† =δ with all other Poisson brackets zero, the following can be shown:
{ k l} lk
G˙1 = G1,H =iG1(A−1B m(v)A−1C) (17)
{ } −
G˙2 = G2,H = i(BA−1 m(v)CA−1)G2 (18)
{ } − −
where G1 is a row vector and G2 is a column vector. Hence we have a complete and
consistent system of constraints and Hamiltonian.
In order to quantize this system we observe that
G1,G2 =iA=0 (19)
{ } 6
These constraints are called “second class”. To quantize the system the Dirac
procedure will be used where the commutation relations of the observables are given
by
[fˆ,gˆ]=i f,g (20)
DB
{ }
where fˆ and gˆ are the operators corresponding to the classical variables f and g
respectively. f,g is the Dirac bracket given by the following formula
DB
{ }
f,g = f,g f,G Fab G ,g (21)
DB a b
{ } { }− { } { }
Xa,b
wheretheG areallofthesecondclassconstraintsandFabistheinverseofthematrix
a
G ,G .
a b
{ }
Following the same procedure as seen in [8] we use the principal axis transfor-
mation ([12]) to choose the basis a so that A is the identity matrix and C is some
k
diagonal matrix with real entries denoted by λ for each mode a . The variables
k k
On the Response of Particle Detectors in Vaidya Spacetimes 5
a are complex so a new basis can be chosen so that the negative eigenvalues of the
k
matrix A become equal to 1 by re-scaling the modes a by i. The zero eigenvalues of
k
the matrix A can be ignoredas any mode with a corresponding zero eigenvalue for A
would have no time derivative term appearing in the action for that mode (10) and
would thus be non-dynamical.
NowthattheAmatrixistheidentitywecandealwithasinglesetofmodesa ,a†,
k k
p ,p†. TheDiracbracketisreadilycalculatedbetweeneachpairofthesevariablesand
k k
the commutation relations are given by the rule in (19). The commutation relations
are
i
[aˆ ,pˆ] = δ (22)
k l kl
2
aˆ ,pˆ† =0 (23)
k l
h i
aˆ ,aˆ† = δ (24)
h k li − kl
aˆ†,pˆ =0 (25)
k l
h i
i
aˆ†,pˆ† = δ (26)
k l 2 kl
h i
1
pˆ ,pˆ† = δ (27)
k l 4 kl
h i
The Hamiltonian in terms of the new basis which diagonalizes A and C is given
as
1
Hˆ = 2pˆB pˆ† + aˆ†B aˆ (28)
l lk k 2 k kl l
Xk,l
1
2m(v)λ pˆ pˆ† m(v)aˆ†aˆ λ
− k k k− 2 k k k
Now we divide this Hamiltonian up into the flat spacetime part and the curved
spacetime part as in equation (4) with
1
Hˆ0 = 2pˆlBlkpˆ†k+ 2aˆ†kBklaˆl (29)
Xk,l
1
Hˆ1 = −2m(v)λkpˆkpˆ†k− 2m(v)aˆ†kaˆkλk (30)
Xk,l
To work in the interaction picture we demand that the operators become time
dependent and evolve under Hˆ0. Using the commutation relations obtained above
we solve the Heisenberg equations for the fundamental operators.
1 i
ipˆ˙k =[pˆk,Hˆ0]= (cid:18)2pˆl− 4aˆ†l(cid:19)Blk (31)
Xl
1
iaˆ˙k =[aˆk,Hˆ0]= Bkl(cid:18)ipˆ†l − 2aˆl(cid:19) (32)
Xl
ipˆ˙†k =[pˆ†k,Hˆ0]= Bkl(cid:18)−12pˆ†l − 4iaˆl(cid:19) (33)
Xl
iaˆ˙†k =[aˆ†k,Hˆ0]= (cid:18)ipˆl+ 12aˆ†l(cid:19)Blk (34)
Xl
On the Response of Particle Detectors in Vaidya Spacetimes 6
Takinglinearcombinationsoftheaboveequationstheycanbesolvedrelativelyeasily.
The general solution is
aˆ= 2ieiBvJˆ†+2iDˆ† (35)
−
pˆ=Jˆe−iBv+Dˆ (36)
where Jˆ† and Dˆ† are column vectors of operator integration constants. To be clear
Jˆ† = (ˆj†,ˆj†,...)t so that the acting on the vector of operators also acts on each
1 2
†
individual operator.
Imposing the commutation relations between the aˆ, aˆ†, pˆ and pˆ†’s we find that
the only noncommuting pair of variables from the set (Jˆ,Jˆ†,Dˆ ,Dˆ†) are Jˆ and Jˆ†.
a a a a a a
The following commutation relations are found
1
Jˆ,Jˆ† = δ (37)
a b 4 ab
h i
In particular the operators Dˆ and Dˆ† commute with everything. From now on we
a a
chooseDˆ =0. Withthis choicetheflatspacetimeHamiltonianHˆ0 willnotdependon
time (v), as it would if Dˆ =0, which is a physically reasonable requirement. Further,
the choice Dˆ =0 is necess6aryto have the same relations between the operatorspˆand
aˆasintheclassicalrelations(13)and(14)inabasiswherethematrixAistheidentity.
In terms of the above solutions to the Heisenberg equations we can re-write Hˆ0
as
Hˆ0 = 4JˆlBlkJˆk† (38)
Xk,l
Since B is hermitian it can be diagonalizedby performing a unitary transformation
lk
on the operators Jˆ. To this end we re-define the operators as Jˆ† = U†χˆ†/2, where
k
U† is unitary. The flat spacetime Hamiltonian becomes
Hˆ0 = ωk χˆ†kχˆk+1 (39)
Xk (cid:16) (cid:17)
wheretheω aretherealeigenvaluesofthematrixB and[χˆ ,χˆ†]=δ holding. Itwill
k l k lk
be shown in section (3) that ω >0. Therefore,the χˆ operators are the annihilation
k k
operators of positive energy particles with their adjoint being the creation operators.
The “curved” or time-dependent part of the Hamiltonian becomes
Hˆ1 = m(v)χˆ†exp(iΩv)Λ˜exp( iΩv)χˆ m(v)Tr(Λ˜) (40)
− − −
where here Λ˜ is the C matrix in equation (9) in the new basis which diagonalizes the
matrix B. Ω is a diagonal matrix with the real diagonal entries being the ω that are
k
in the Hˆ0 part of the Hamiltonian. Specifically we have
Ω:=UBU† =diag[ω1,ω2,...] (41)
Λ˜ :=(UCU†)t (42)
The state of the Klein-Gordon field will be taken to be given by
χˆ 0>=0 (43)
k
|
Hence 0> is the vacuum for the operators χ and χ†. It will be shown below, when
| k k
we calculate the φˆ field explicitly, that this state is the unique state such that the
particle detectors of constant r observers will have a null response when m(v) = 0.
On the Response of Particle Detectors in Vaidya Spacetimes 7
For the mass function (2) the state has this interpretation in region α. It is the most
naturalchoiceofvacuumstateforspacetimesoftheform(1)with(2)holdingbecause
in region α the observers are in a Minkowski spacetime and completely causally dis-
connected from the non vacuum region of spacetime (see figure 1).
In the picture we are working in the state of the Klein-Gordon field evolves as
v
Ψ(v)>= exp( i Hˆ1(v′)dv′)Ψ(v0)> (44)
| T − Z |
v0
where isthetimeorderingoperator. Ifwepicktheinitialstatetobethevacuumwith
T
respect to the above creation and annihilation operators (i.e. χˆk Ψ(v0) >= 0) then
wecanseefromthe formofHˆ1 thatthestate evolvesintimeonly|bythe phasefactor
exp(i v m(v′)Tr(Λ˜)) which means that the vacuum state vector is mathematically
v0
thesaRmeatalltimes. NoticehoweverthatageneralstatefromtheFockspaceformed
bythecreationoperatorswillnotevolveonlybyaphasefactorandhencetherewillbe
a change in the state due to the Vaidya background. This can be seen by expanding
equation (44) to first order to obtain
v
Ψ(v)> Ψ(v0)> i Hˆ1(v′)dv′ Ψ(v0)> (45)
| ≈| − Z |
v0
So now if we have a one particle state |Ψ(v0)>=χˆ†k|0> at time v =v0 then to first
order we will have a linear combination of one particle states given by
v
Ψ(v)> 1+i dv′m(v′)Tr(Λ˜) Ψ(v0)>+ (46)
| ≈ (cid:18) Z (cid:19)|
v0
v
i dv′m(v′)exp(iΩv′) Λ˜ exp( iΩv′) χˆ† 0>
Z lp pj − jk l|
pX,j,l v0
3. Response of particle detectors
The responserateofaparticledetector ata givenfrequencyν asit followssome path
through spacetime is given by
τ−τ0
F˙ (ν)=2 ds ( exp( iνs)G+(x(τ),x(τ s))) (47)
τ
Z0 ℜ − −
where x(τ) is the path through spacetime (see [13], [14]) and denotes the realpart.
ℜ
τ isa propertime ofthe detectorthatparameterizesthe pathx(τ). Also,the particle
detector is in the ground state at τ0. First, we must evaluate the Wightman function
G+(x(τ),x(τ s))=<0φˆ(x(τ))φˆ(x(τ s))0>.
− | − |
The scalar field operators are given by
φˆ(v,r)= aˆ (v)f (r) (48)
k k
Xk
We have solved for the operator coefficients aˆ in equation (35). The φˆ(v,r) field
k
operators are then given by
φˆ(v,r)= 2if (r)exp(iBv) Jˆ† (49)
− l la a
Xl,a
On the Response of Particle Detectors in Vaidya Spacetimes 8
where exp(iBv) is a matrix. If we now make the substitution Jˆ† = 1/2U∗ χˆ†, which
a ba b
diagonalizes the Hamiltonian Hˆ0, we end up with the field operator as
φˆ(v,r)= if (r)U∗ exp(iΩv) χˆ† (50)
− l cl cb b
Xb,c,l
Defining g (r) = f U∗ and substituting this into the above expression, we finally
c l cl
obtain.
φˆ(v,r)= ig (r)exp(iω v)χˆ† (51)
− k k k
Xk
In order to solve explicitly for the g (r) modes we use the fact that the complex
k
field φ(v,r) is a solution of the wave equation determined by the H0 operator. It
is not difficult to see that this will be the wave equation in the Vaidya background
with m(v) = 0, i.e. the wave equation in Minkowski spacetime. By substituting the
expansion (51) into the wave equation for the Vaidya background with m(v) = 0 we
obtain that the equation
r2∂2g (r)+2iω r2∂ g (r)+2r∂ g (r)+2iω rg (r)=0 (52)
r k k r k r k k k
must hold for each mode g (r). The general solution to this equation is
k
C(k) exp( 2iω r)
k
g (r)= +D(k) − (53)
k
r r
whereC(k)andD(k)arearbitrarycomplexconstants. Wewishthemodesweexpand
in to be physical and hence regular at r = 0 in Minkowski spacetime. The only way
to achieve this is to choose C(k)= D(k) so that the divergence there cancels.
−
The expression for the matrix B is given by (8) in terms of the original f (r)
k
modes. The diagonalized matrix Ω is given in terms of B as
Ω = U B U∗ (54)
lk la ab kb
Xa,b
Using the definition of the matrix B and the definition of the modes g (r) we obtain
k
that
∞
Ω =4π drr2∂ g∗(r)∂ g (r) (55)
lk r l r k
Z0
As acheck,the modes(53)withC(k)= D(k),whensubstitutedintothe righthand
−
side of (55) should be diagonal in l and k. Substitution of the modes
1 exp( 2iω r)
k
g (r)=D(k)(− + − ) (56)
k
r r
into this integral can be shown to reduce to the single term
∞
Ω =16πD∗(l)D(k) dr ω ω exp(2i(ω ω )r) (57)
lk l k l k
Z0 −
where ω and ω are real numbers. To show this matrix is diagonal we will put our
l k
systeminaboxoffinitesizesothatwecanindexourmomentummodeswithintegers.
The radius of our system will be R and this will be the upper bound of the integral
in (57). The discrete limit is obtained by making the replacements
πl πk
ω ,ω , (58)
l k
→ R R
On the Response of Particle Detectors in Vaidya Spacetimes 9
where on the right hand side l and k are integers. The integral (57) becomes
R π2 2π 2π
16πD∗(l)D(k) dr lk cos( (l k)r)+isin( (l k)r) (59)
Z0 R2 (cid:20) R − R − (cid:21)
This is easily shown to be equal to
16π3
l2δ D(l)2 (60)
lk
R | |
RecallthattheelementsofΩ areω onthediagonal. Usingthisinthediscretelimit
lk k
we have that the above expression must be equal to πl/R, in the discrete limit. This
fixes the the magnitude of the complex constant to be
1
D(l) = (61)
| | 4π√l
The expression(60) is manifestly positive. Therefore, only the modes ω >0 need to
k
be considered when we calculate the Wightman function.
The quantizedcomplexKlein-Gordonfieldin(51)isnotHermitianandtherefore
not a valid observable. We will use the Hermitian observable that is formed from
equation (51) which is given by the following expansion
i
φˆ(v,r) = − (g (r)exp(iω v)χˆ† g∗(r)exp( iω v)χˆ ) (62)
ℜ(cid:16) (cid:17) Xk 2 k k k− k − k k
with the g given by equation (56) and (61). The above Hermitian field obeys the
k
correct equations of motion and the correct commutation relations and is therefore
the real, quantum, spherically-symmetric Klein-Gordon field operator.
Inorderto calculatethe Wightmanfunction usingthe interactionpicturewe will
first calculate the Feynman correlation function using the formula given in [15].
< φˆF(vn,rn)...φˆF(v1,r1) > (63)
⊘|T |⊘
= <0|Tφˆ(vn,rn)...φˆ(v1,r1)exp(−i −∞∞dvHˆ1(v))|0>
<0|T exp(−i −∞∞dvHˆ1(Rv))|0>
R
The state > signifies the vacuum state of the full Hamiltonian Hˆ = Hˆ0 +Hˆ1(v)
|⊘
and 0 > is the vacuum state of the “free” Hamiltonian Hˆ0 and φˆF(v,r) denote the
|
scalar field with the full Hamiltonian. Note that the contribution from the integrals
at cancel since Hˆ1(v) = 0 for v < 0 (in region α). Also, the Tr(Λ˜) term in Hˆ1(v)
∞
cancels. For the case of two φˆ factors the expression becomes for v2 >v1
< φˆF(v2,r2)φˆF(v1,r1) > (64)
⊘| |⊘
v2
=<0φˆ(v2,r2) exp( i dvη(v))φˆ(v1,r1)0>
| T − Z |
v1
η(v) m(v)χˆ†exp(iΩv)Λ˜exp( iΩv)χˆ
≡− −
where wehaveusedthe definitionofthe vacuumstate(43)to evaluatethe denomina-
tor in (63).
Atthispointwecanspecifythemeaningofthevacuumstategivenin(43). Using
the general formula for the Wightman function (63) we can obtain the Wightman
On the Response of Particle Detectors in Vaidya Spacetimes 10
functioninMinkowskispacetimeforthesphericallysymmetricmodesoftherealscalar
field operator given in (62). In the continuum limit it is expressed by the integral
1 ∞ dω
G+(t,r;t′,r′)= exp( iω(t t′))sin(ωr)sin(ωr′)(65)
16πrr′ Z0 ω − −
where we have used that v = t+r in Minkowski spacetime and we have used the
relations (58) and the relation
1
dω (66)
R →Z
Xk
to go to the continuum limit. By plugging (65) into equation (47) it can be verified
that constant r observers in Minkowski spacetime in the state (43) will have a null
response for their particle detectors (see appendix for details).
We note that the v = const surfaces are not Cauchy and thus there is a con-
cern that the field evolution will not be unique if we specify the evolution in terms of
these surfaces. For Vaidya metrics of the form (1), (2) we have that for constant v as
r the metric becomes Minkowskispacetime. The limitr caneffectively be
→∞ →∞
reproduced by considering m(v) = 0. Therefore an alternate characterization of the
vacuum (43) is that it is the vacuum state such that constant r particle detectors do
not click on past null infinity (see figure 2). Pastnull infinity is a Cauchy surface and
thus specifying the state of the field there is sufficient to determine the evolution of
the field uniquely throughout the spacetime.
Another way to see that the evolution of the Klein-Gordon field is unique is to
notethatimposingtheconditionthatφˆbefiniteatr =0aswellaspickingthemodes
to be positive frequency with respect to the inertial time t in Minkowski spacetime
fixes all arbitrary constants present in the mode solutions to the Klein-Gordon equa-
tion.
The proper time of the constant r observers will be used to parametrize the
paths in formula (47). In order to evaluate the response rate of a particle detector
for a constant r observer it is convenient to re-write the equation (47) in terms of an
integral over v instead of the proper time of the time-like particle detector. To this
end, make the substitution p = τ s so that we now integrate over the variable p
−
insteadofs. τ is aconstantrepresentingthe propertime atwhichthe responseofthe
detector is evaluated at.
τ
F˙ (ν)=2 dp (exp( iν(τ p)))G+(x(τ),x(p)) (67)
τ
Z ℜ − −
τ0
Since τ and p are both proper times we can use the line element (1) to solve for them
in terms of the null time v.
2m(v)
dτ = 1 dv (68)
r − r
The proper times as functions of the null time are denoted as τ =τ(v) andp=τ(v′).
In the integral over the proper time it is p that is integrated over, so in terms of null
time we will be integrating over v′. The final formula is then given by
v
F˙ (ν)=2 dv′ (ξ(v′,r)G+(v,r;v′,r)) (69)
v
Z ℜ
v0