Comments on anomaly versus WKB methods for calculating Unruh radiation* Douglas Singleton, CSU...
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Transcript of Comments on anomaly versus WKB methods for calculating Unruh radiation* Douglas Singleton, CSU...
Comments on anomaly versus WKB methods for calculating Unruh radiation*
Douglas Singleton, CSU Fresno and PFUR
“String Field Theory and Related Aspects” Moscow, Russia April 16th, 2009
*Work in collaboration with V. Akhmedova, T. Pilling, and A. de GillPhysics Letters B 673 (2009) 227-231
(arXiv:08083413)
Reduction 3+1 1+1 and gravitational anomaly A scalar field in some gravitational background has an action, S:
Expanding and integrating reduces this to 1+1
Chiral theories in 1+1 have a gravitational anomaly*
Consistent Anomaly Covariant Anomaly
*Alvarez-Gaume and Witten Nucl. Phys. B 234, 269 (1984) Bertlmann and Kohlprath, Ann. Phys. 288, 137 (2001)
nm
inzimymn eert
,
),(
)(
2
1][ 4 ggxdS
mnmnnm
ggxdS
)(2
1][
,
2
N
ggT H
1
96
1)( R
gT H
96
1)(
Cancellation of the anomaly via flux
Vary the 1+1 action with parameters λµ =(λt , λr)
The action is not invariant under this general variation because of the anomaly.
Break up energy-momentum tensor as
Combine this with the variation λµ =(λt , λr) and require δS=0 and you get
TgxdS 2
)()(1 )()(
HOHH rrTrrTT
g
NTT
rtr
tHrtO
)()(g
NTT
rrr
rHrrO
)()(
Cancellation of the anomaly via flux
If Ntr≠0 then one needs Tr
(O)t≠0. Assume a 2D Planckian distribution. Then
Ntr = Φ yields the temperature for a given spacetime, but not the spectrum which is
assumed.
2
0/)0( 1212
1T
e
EdET
TErt
Hawking temperature via anomalies For a Schwarzschild black hole one
finds a flux of
This gives the (correct) Hawking
temperature of T=1/8πM
28
1
96
1
96 MN r
ttr
trrt
Rindler Spacetime
)21()21(
222
ar
drdtards
222 '')'1( drdtards
The Rindler metric has two well known forms
The two forms are related by the transformation
arar 21'1
Rindler metrics: different temperatures
For the first set of coordinates (r,t) the flux Ntr is
Apparently give correct Unruh temperature T=a/2π
For the second set of coordinates (r’,t’) the flux Ntr is
Which gives an incorrect Unruh temperature of T=a/2π√2
4896
2aN r
ttr
trrt
9696
2aN r
ttr
trrt
Rindler metrics: zero temperatures However (i) Nt
r =constant (ii) anomaly is zero (iii) zero Unruh temperature
For the covariant anomaly this is even easier to see since the 2D Ricci scalar vanishes R=0
The anomaly method fails (in its simplest form) for Rindler
N
ggT H
1
96
1)(
Rg
T H
96
1)(
De Sitter spacetime: split result De Sitter spacetime emits Gibbons-Hawking radiation
With temperature T=1/2πα
The consistent anomaly does give this temperature
The covariant anomaly is zero since R=const.
The two anomaly methods give different answers for de Sitter.
Rg
T H
96
1)(
2
2
22
2
22
1
1
r
drdt
rds
WKB/tunneling calculation of Unruh temperature Use φ(x)~exp[i S(x)/h] one finds the Hamilton-
Jacobi form of Klein-Gordon
Split action as S(x)=Et+S0(x). Solution
S0=∫prdr.
Imaginary S0(x) the quasi-classical decay and temperature given via
0))(( 2 mSSg
TEpdree /Im
)Im(
drp
ET
r
Im(S0) for 1st form of Rindler metric For the first form of the Rindler metric S0 is [with (+) outgoing and (-) ingoing]
Imaginary contribution comes from contour integration around r=-1/2a. The contour is parametrized as r=-(1/2a)+εeiθ
A round trip gives iπE/a which gives twice the Unruh temperature (a/π instead of a/2π)
a
Eidr
ar
armES
2)21(
)21(22
0
Im(S0) for 2nd form of Rindler metric The second form of the Rindler metric appears to give the correct answer
This can’t be correct since the two metrics are related by a coordinate transformation
The contour is also transformed to a quarter circle r’=-(1/a)+√εeiθ/2
a
Eidr
ar
armES
')'1(
)'1( 222
0
arar 21'1
Resolution: temporal contribution The Both forms of Rindler metric give twice the Unruh temperature
The Rindler spacetime is obtained from ds2 =- dT2 + dR2 via
r>-1/2a
r<-1/2a
Crossing the horizon involves an imaginary time change tt-iπ/2a so Im(EΔt)=-πE/2a. For a round trip Im(EΔt)=-πE/a
ata
arR cosh
21 at
a
arT sinh
21
ata
arT cosh
|21| at
a
arR sinh
|21|
Spatial + temporal contribution
Spatial + temporal contribution gives correct Unruh temperature via
2//)Im()Im(
a
aE
E
aE
E
tE
E
drp
ET
r
Emission/Absorption Probability The probability for emissions/absorption is Pa,e~|φin,out|2~|exp[2iSin,out(x)]|
Need Pa=1
Without temporal piece (will give Probability>1 for large enough E)
With temporal piece
aEa eP /
1|| 0)2/2/(2 eeP aEiaEiia
Canonical Invariance Physical quantities should be canonically invariant
Note: 2Im(S0)=2Im∫ p dr is not canonically invariant so that Γ~exp[2Im(S0)] is not a proper observable [B.D. Chowdhury, hep-th/0605197]
But is canonically invariant. o
i
i
o
r
r
r
r
inout drpdrppdr
Summary/Conclusions Neither anomaly method works for Rindler spacetime/Unruh radiation.
The gravitational WKB method works for Rindler spacetime/Unruh radiation, but has both spatial and time contributions.
The gravitational WKB/tunneling problem has some distinct features: time contribution and ingoing and outgoing probabilities for tunneling are not equal.
Acknowledgments Work partially supported through a 2008-2009 Fulbright Scholars Grant
Canonical Invariance The proper, observable decay rate is then
For the Rindler metric in=out so numerically both give the same answer
There are cases when there is a difference such as the Painleve-Gulstrand formof the Schwarzschild metric
pdre
Im
o
i
o
i
r
r
in
r
r
out drpdrppdr 22
drdrdrdt
r
Mdt
r
Mds 2222 2
22
1
Painleve-Gulstrand case
r
MmE
rM
drE
r
M
rM
drS
CC
21
21
22
1
220
The spatial part of the action is now
The two integrals have the same magnitude imaginary contributions.
Thus the ingoing and outgoing probabilities are not equal in this case (or for any case if the temporal piece is taken into account).