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).