This document provides exercises on Hawking radiation using scalar field theory in the Kruskal spacetime. It asks the student to find the radial equation for the scalar field and show that near the horizon, the field takes the form of ingoing and outgoing waves that are analytic in different coordinate systems. The document then derives the Klein-Gordon equation in Schwarzschild coordinates and uses separation of variables to obtain approximate solutions near the horizon. It shows that ingoing waves are regular on the future horizon but outgoing waves are not, and vice versa for the past horizon.

1. Basic Illustration Exercises in Hawking Radiation
(Notes I )
Roa, Ferdinand J. P.
Exercise A.4.4
(page 142 of [1])
A scalar field Φ in the Kruskal spacetime satisfies Klein-Gordon equation (KGE)
022
=Φ−Φ MD . (1)
Given that, in static Schwarzshild coordinates, Φ takes the form
( ) ( )ϕθω ,exp)( mll YtirR −=Φ , (2)
( )ϕθ,mlY - Spherical harmonics
find the radial equation satisfied by )(rRl . Show that near the horizon at qGMr 2= ,
( )*exp~ riω±Φ , r* - Regge-Wheeler radial coordinate. Verify that ingoing waves are analytic in
Kruskal coordinates on future horizon
+
H , but not in general on past horizon
−
H , conversely for
outgoing waves. Given that both M and ω vanish, show that
)()()( zQBzPArR lllll += . (3)
)(zPl - Legendre polynomials
)(zQl - linearly independent solution
q
q
M
Mr
z
−
=
lA , lB - constants
Hence, show that there are no non-constant solutions that are both regular on horizon
−+
∪= HHH ,
bounded at infinity.
Answers
Review Entries
The action for the scalar field is given by[7]
LxdSS ∫= 4
, (4)
where the Lagrangian L takes the form[7, 8, 9]
[ ])(2))((
2
1
Φ+Φ∂Φ∂−= VggL ωµ
ωµ
. (5)
We are using the metric signature (- + + + ) and assume that the potential )(ΦV is of the form
22
2
1
)( Φ=Φ MV . (6)
Varying (4) with respect to the scalar field yields

2. Φ
















Φ∂
∂−
Φ
+Φ
Φ∂
= ∫∫ δ
δ
δ
δ
δ
δ
δ
δ
σδ
µ
µ
µσ
µ
µ
µµ )()(
4 LL
xd
L
dS
B
A
x
x
S . (7)
Note that in this variation the metric fields µωg are independent of the variation of Φ . Thus, obtaining
(7). Upon applying the boundary condition, 0)()( =Φ=Φ BA δδ , the first integral term in (7) vanishes
and together with the condition that action be stationary, wherein 0=SSδ , we get the Euler-Lagrange
equation
0
)(
=








Φ∂
∂−
Φ µ
µ
δ
δ
δ
δ LL
. (8)
We substitute (5) in (8) to arrive at the equation of motion for the scalar field
[ ] 0)(
1 2
=Φ−Φ∂−∂
−
Mgg
g
ω
µω
µ (9a)
or in its covariant form
[ ] 0)( 2
=Φ−Φ∂∇ Mg ω
µω
µ . (9b)
Note that it is easy to prove that with metric compatible connections we have
[ ] [ ])()(
1
Φ∂∇=Φ∂−∂
−
ω
µω
µω
µω
µ ggg
g
(10)
and by metric compatible connections we mean those connections that are derived from the metric, where
covariant differentiations satisfy 0=∇ ωλ
µ g and that the connections are symmetric with respect to the
interchanges of their lower indices,
σ
νµ
σ
µν Γ=Γ [6].
Klein-Gordon Field in Schwarzschild Space-time
The equation of motion (9a) for the Klein-Gordon field is the field equation for the said field in
curved space-time endowed with a metric µνg other than the flat Minkowski’s metric µνη . Using the
Schwarzschild metric [metric signature (- + + + )]
22222222
sin φθθεη drdrdrdtdS +++−= , (11a)
r
GM q2
1
1
−==
ε
η (11b)
we transcribe (9a) as
[ ] Φ=Φ∂−∂
−
2
)(
1
Mgg
g
ω
µω
µ :
( )[ ] ( )[ ] Φ=Φ∂+Φ∂∂+Φ∂∂+Φ∂− 22
222
2
2
2
0
sin
1
sin
sin
111
M
rr
r
r
rr φθθ
θ
θ
θ
η
η
. (12a)
By separation of variables[10],

3. ( ) )()()()(,,,0
φψθφθ Θ=Φ rRtTrx (12b)
we put:
221
φφ µψ
ψ
−=∂ , (12c)
( )[ ] )1(
sin
sin
sin
11
2
2
+−=−Θ∂∂
Θ
θθ
φ
θθ µµ
θ
µ
θ
θ
, (12d)
22
0
1
ω−=∂ T
T
(12e)
and
η
ωµµ
η θθ
2
2
2
2
2
)1(11
−=
+
−



M
rdr
dR
r
dr
d
rR
. (12f)
Approximate Solutions for R(r) near the Horizon, qH GMr 2=
We transform (12f) into a form adapted with Regge-Wheeler radial coordinate








−+= 1
2
ln2*
q
q
GM
r
GMrr (12g)
and obtain
2
2
2
*2
2
*
)1()2(2111
M
r
R
r
GMr
R
R
R
r
q
r +
+
=





+∂
−
+∂ θθ µµ
ω
η
. (12h)
Note that near the horizon, qH GMr 2= , the quantity η/1 takes on very large values as when
( ) 02 ≈− qGMr so that we can ignore the RHS of (12h) and the other terms that are effectively
ignorable. Thus, from (12h) we have as our effective differential equation for R(r) near the horizon as
022
* =+∂ RRr ω , (13a)
which has the solution
*)exp(*)exp(*)( 0201 riRriRrR ωω +−= (13b)
where 01R and 02R are constants. Note also that (12e) has the solution
)exp()exp()( 0201 tiTtiTtT ωω +−= (13c)
with 01T and 02T constant.
Near the horizon, we can combine solutions (13b) and (13c) such that we can define
*~ rtu += (13d.1)
as our in-falling coordinate, while
*~ rtv −= (13d.2)
as the out-going coordinate. That is, near the horizon, we have as our effective solutions
( )[ ] )~exp(*exp 00 viArtiArt ωω −=−−=Φ +++
, (13e.1)
which is the out-going wave and
( )[ ] )~exp(*exp 00 uiArtiArt ωω −=+−=Φ +−−
(13e.2)
as the in-falling wave.

4. We can approximate that both )~(vrt
+
Φ and )~(urt
−
Φ travel along the out-going null-like path
+
γ
and in-going null-like path
−
γ , respectively.
+
γ :
+
−=− aηχ (13f.1)
−
γ :
−
=+ aηχ (13f.2)
+
H : πηχ −=− , future event horizon (13g.1)
−
H : πηχ −=+ , past event horizon (13g.2)
+
ℑ : πηχ =+ , future null infinity (13h.1)
−
ℑ : πηχ =− , past null infinity (13h.2)
u ′=+ ~2ηχ , uu ~~tan =′ (13i.1)
v ′−=− ~2ηχ , vv ~~tan =′ (13i.2)
In the graph, we see that along tconsrtu tan*~ =+= , the infalling null path can go from the
past null infinity ),*( −∞→+∞→ℑ−
tr and hit the future event horizon
),*( +∞→−∞→+
trH . The infalling wave )~(urt
−
Φ is regular on both
−
ℑ and
+
H , while
±∞→v~ on these surfaces so the out-going wave )~(vrt
+
Φ is not regular on these surfaces.

5. While along tconsrtv tan*~ =−= , the out-going null path can go from past event horizon
),*( −∞→−∞→−
trH and hit the future null infinity ),*( +∞→+∞→ℑ+
tr . We also see in
here that the out-going wave )~(vrt
+
Φ is regular on both
−
H and
+
ℑ , while ±∞→u~ on these surfaces
so that )~(urt
−
Φ is not regular on these surfaces.
We can try putting the out-going null-like path
+
γ very close to the future event horizon
+
H by
parametrizing
+
γ in terms of u~ , where we restrict u~ to span its values only within the interval
0~ <<∞− u . Our trial choice for va ′=+ ~2 as parametrized in terms of u~ would be
)~ln(2~ uGMv −−= , 0~ <u (14a)
taking note of (13i.2). By confining the values of u~ within the interval 0~ <<∞− u , we prohibit the
infalling null path to span near the future null infinity where out-going wave must exist there while
infalling wave must not. Given the chosen form of v~ as parametrized in terms of u~ , the out-going null
path can be near both the surfaces
+
H and
−
ℑ since as 0~ →u , 2/~ π+→′v , which is at the future
event horizon; while as −∞→u~ , 2/~ π−→′v , which is at past null infinity.
So by parametrizing v~ in terms of u~ , the out-going wave )~(vrt
+
Φ is now expressed in the form
)]~ln(2exp[)~( 0 uGMiAurt −=−Φ ++
ω (14b)
which holds only for all 0~ <u , while keeping 0)~( =Φ+
vrt in those regions where 0~ >u . We might
consider the Fourier components of this wave in the positive frequency modes 0>′ω and workout how
these components are related to those in the negative frequency modes 0<′ω . Since the out-going wave
does not exist in those regions where 0~ >u , these Fourier components are simply given by
)~exp()]~ln(2exp[~)(
~
0
0 uiuGMiudAtr ωωω ′−−=′Φ ∫∞−
++
(14c)
For the positive frequency modes, we consider the contour in the upper left quadrant of the complex z-
plane. The branch cut is where 0>x in the complex plane, iyxz += so that this contour taken in the
upper left quadrant has paths that connect the following main points )0,0(O , )0,( RA − and ),0( RB .
The direction of each path is as indicated by the associated directional arrow. The contour integral would
then be given by
)exp()]ln(2exp[)exp()( zizGMidzzizfdz ωωω ′−−=′− ∫∫ ΓΓ
(14d.1)
)]ln(2exp[)( zGMizf −= ω (14d.2)

6. As given in the figure, this contour is split into three integral paths
+′−+′−=′− ∫∫∫Γ
)exp()()exp()()exp()( zizfdzzizfdzzizfdz
ABOA
ωωω
)exp()( zizfdz
BO
ω′−∫ (14d.3)
Path from O to A is at 0=y so that
)exp()]ln(2exp[)exp()(
0
xixGMidxzizfdz
ROA
ωωω ′−−−=′− ∫∫ −
. (14d.4)
Path from A to B can be taken as that along the quarter circle, RzCR =: , then
)exp()()exp()( zizfdzzizfdz
CRAB
ωω ′−=′− ∫∫ . (14d.5)
Path from B to O is at 0=x so that
)exp()]ln(2exp[)exp()exp()(
0
yyGMidyGMizizfdz
R
BO
ωωωπω ′−=′− ∫∫ (14d.6)
Taking note that since (14d.2) does not have the singularities inside the upper left quarter circle,
then the contour vanishes,
0)exp()( =′−∫Γ
zizfdz ω (14d.7)
and we invoke that in the limit ∞→R , the integral path for A to B vanishes
0)exp()(lim =′−∫∞→
zizfdz
CRR
ω (14d.8)

7. Following these limiting conditions as ∞→R , integral equation (14d.3) yields
)exp()ln2exp()exp()exp()]ln(2exp[
0
0
yyGMidyGMixixGMidx ωωωπωω ′−=′−− ∫∫
∞
∞−
(14d.9)
Next, we come to the contour for the negative frequency modes.
Again, take note that the branch cut is on those regions where 0>x in the complex z-plane, iyxz += .
Our contour here, like the one we have above has three main integral paths that connect the points
)0,0(O , )0,( RA − and ),0( RB − , and the direction of each path is as indicated by a given directional
arrow. The contour thus consists of three integral paths
+′+′=′ ∫∫∫Γ
)exp()()exp()()exp()( zizfdzzizfdzzizfdz
RCOA
ωωω
)exp()( zizfdz
BO
ω′∫ (14d.10)
Path from O to A is at 0=y so that
)exp()]ln(2exp[)exp()(
0
xixGMidxzizfdz
ROA
ωωω ′−−=′ ∫∫ −
(14d.11)
Path from A to B can be taken as that along the quarter circle RzCR =: so that
)exp()()exp()( zizfdzzizfdz
RCAB
ωω ′=′ ∫∫ . (14d.12)
Path from B to O is at 0=x so that

8. )exp()]ln(2exp[)exp()exp()(
0
yyGMidyGMizizfdz
RBO
ωωωπω ′−=′ ∫∫ −
(14d.13)
We take note also that for the negative frequency modes )(zf does not have the singularities
inside the lower left quarter circle, then the contour vanishes
0)exp()( =′∫Γ
zizfdz ω (14d.14)
and also with the limiting condition that as ∞→R ,
0)exp()(lim =′∫∞→
zizfdz
RCR
ω . (14d.15)
Following these, contour (14d.10) gives
)exp()ln2exp()exp()exp()]ln(2exp[
00
yyGMidyGMixixGMidx ωωωπωω ′−=′− ∫∫ ∞−∞−
(14d.16)
To proceed, let us take (14d.16) and rotate x
iyyix −=−→ )2/exp( π . (14e.1)
This rotation transforms the left-hand side of (14d.16) into
)exp())ln(2exp()exp()exp()]ln(2exp[
00
xixGMidxGMyyGMidyi ωωωπωω ′−=′− ∫∫ ∞−∞−
(14e.2)
Take (14d.9) and rotate x
iyyix =→ )2/exp( π (14e.3)
so that whole of (14d.9) transforms as
)exp()ln2exp()exp()]ln(2exp[
0
0
yyGMidyyyGMidy ωωωω ′−=′ ∫∫ ∞−
∞
. (14e.4)
The we substitute (14e.4) in (14e.2) to get
)exp())ln(2exp()exp()exp()]ln(2exp[
0
0
xixGMidxGMyyGMidyi ωωωπωω ′−=′ ∫∫ ∞−
∞
(14e.5)
and in turn substitute this in (14d.9), yielding the required result

9. )exp())ln(2exp()2exp()exp()]ln(2exp[
00
xixGMidxGMxixGMidx ωωωπωω ′−−=′−− ∫∫ ∞−∞−
(14e.6)
Note that if we identify
)~exp()]~ln(2exp[~)(
~
0
0 uiuGMiudBtr ωωω ′−=′−Φ ∫∞−
++
(14e.7)
as in comparison with (14c) so that in view of (14e.6), we can relate the Fourier components in the
positive frequency modes to those Fourier components in the negative frequency modes,
)(
~
)2exp()(
~
0
0
ωωπω ′−Φ−=′Φ +
+
+
+
trtr GM
B
A
. (14e.8)
Note: To be continued.
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