1.1) One may use any reasonable equation to obtain the dimension of the questioned quantities. Especially, it is.. independent of the Planck constant h which is characteristic of qua[r]
(1)1.1) One may use any reasonable equation to obtain the dimension of the questioned quantities
I) The Planck relation is hν =E ⇒ [h][ν]=[E] ⇒ [h]=[E][ν]−1 =ML2T−1 (0.2)
II) [c]= LT−1 (0.2)
III) = 2 ⇒ [G]=[F][r2][m]−2 =M−1L3T−2 r
m m G
F (0.2)
IV) = ⇒ [ ]=[ ]−1[ ]= −2 −1 K T ML E K
K
E Bθ B θ (0.2)
1.2) Using the Stefan-Boltzmann's law,
θ σ = Area Power
, or any equivalent relation, one obtains: (0.3)
]
[ ]
[ ]
[ = −2 −1⇒ = −3 −4
K MT T
L E
K σ
σ (0.2)
1.3) The Stefan-Boltzmann's constant, up to a numerical coefficient, equals ,
δ γ β α
σ =h c G kB where α,β,γ,δ can be determined by dimensional analysis Indeed, ,
] [ ] [ ] [ ] [ ]
[σ = hα c β G γ kB δ where e.g [ ] − − =MT K
σ
(
1) (
1) (
2) (
2 1)
2 ,4
3 − − α − β − − γ − − δ α−γ+δ α+β+γ+ δ −α−β− γ−δ −δ
− K = MLT LT M LT MLT K =M L T K
MT
(0.2) The above equality is satisfied if,
− = −
− = − − − −
= + + +
= + − ⇒
,
, 2
,
,
δ
δ γ β α
δ γ β α
δ γ α
(Each one (0.1)) ⇒
= =
− =
− =
,
,
,
δ γ β α
(Each one (0.1))
⇒
3
4 h c
kB =
σ
2.1) Since A , the area of the event horizon, is to be calculated in terms of mfrom a classical theory of relativistic gravity, e.g the General Relativity, it is a combination of
(2)independent of the Planck constant h which is characteristic of quantum mechanical phenomena
γ β α
m c G A=
Exploiting dimensional analysis,
(
) (
α)
β γ α γ α β α βγ β
α − − − − + + − −
= =
⇒ =
⇒ 3
] [ ] [ ] [ ]
[A G c m L M LT LT M M L T
(0.2) The above equality is satisfied if,
= − −
= +
= + − ⇒
,
,
,
β α
β α
γ α
(Each one (0.1)) ⇒
= − =
= ,
,
,
γ β α
(Each one (0.1))⇒
2
c G m A= 2.2)
From the definition of entropy
θ
dQ
dS = , one obtains [S]=[E][θ]−1 =ML2T−2K−1(0.2)
2.3) Noting η=S A, one verifies that,
= =
= =
− − − − − + + + + + − − − −
, ]
[ ] [ ] [ ] [ ] [
, ]
][ [ ] [
2 2
2
δ δ γ β α δ γ β α δ β α δ
γ β α
η η
K T
L M
k c h G
K MT A
S
B
(0.2)
Using the same scheme as above,
⇒
=
− = − − − −
= + + +
= + + −
,
, 2
, 2
3
,
δ
δ γ β α
δ γ β α
δ β α
(Each one (0.1))
= =
− =
− = ⇒
,
,
,
,
δ γ β α
(Each one (0.1))
thus,
3
h G
k c B =
η (0.1)
(3)The first law of thermodynamics is dE=dQ+dW By assumption, dW =0 Using the definition of entropy,
θ
dQ
dS= , one obtains, ,
0 +
= dS
dE θH (0.2) + (0.1), for setting dW =0
Using,
= =
, ,
2 mc E
m ch
k G
S B
[(0.1) for S]
one obtains,
1
1 −
−
= = =
dm dS c dE
dS dS
dE H
θ (0.2)
Therefore,
m k G
h c
B H
1
1
=
θ (0.1)+(0.1) (for the coefficient)
3.2) The Stefan-Boltzmann's law gives the rate of energy radiation per unit area Noting that E=mc2 we have:
= = =
− =
,
2
3
4
4 ,
, /
mc E
c G m A
h c
k
A dt
dE
B
H
σ
σθ
⇒ ,
2
2
3
4
c G m m k G
h c h c
k t
d m d c
B B
−
= (0.2)
⇒
16
2
m G
h c t
d m d
−
= (0.1) (for simplification) + (0.2) (for the minus sign)
3.3)
By integration: 16
1
2
m G
h c t
d m d
−
= dt
G h c dm
m
∫
∫
=−⇒
2
16 (0.3)
, 16
3 ) ( )
( 2
4
3
t G
h c m
t
m − =−
⇒ (0.2) + (0.2) (Integration and correct boundary values)
At t=t* the black hole evaporates completely:
) (t* =
m (0.1) 4
2 *
3 16
m h c
G t =
⇒ (0.2)+(0.1) (for the coefficient)
3.4) CV measures the change in E with respect to variation of θ
(4)
= = =
m k G
h c
c m E
d E d C
B V
1
, ,
3
θ
θ
⇒ m2 h c
k G
C B
V =− 0.1)+(0.1) (for the coefficient)
4.1) Again the Stefan-Boltzmann's law gives the rate of energy loss per unit area of the black hole A similar relation can be used to obtain the energy gained by the black hole due to the background radiation To justify it, note that in the thermal
equilibrium, the total change in the energy is vanishing The blackbody radiation is given by the Stefan-Boltzmann's law Therefore the rate of energy gain is given by the same formula
=
+ −
= ,
4
c m E
A A
t d
E d
B
σθ σθ
⇒ 8 3
(
)
42 2 1
16 c h k m
G m G hc dt
dm
B Bθ +
−
= (0.3)
4.2)
Setting =0 dt dm
, we have:
(
)
1 16
2 *
8 2 *
= +
− k m
h c
G m G hc
B
Bθ (0.2)
and consequently,
B B
k G
h c m
θ
1
3 * =
(0.2)
4.3)
− −
= ⇒
= *4
4
2 *
3
1 16
2 m
m m
G hc dt
dm m
k G
h c
B B
θ (0.2)
4.4) Use the solution to 4.2,
B B
k G
h c m
θ
1
3 * =
(0.2) and 3.1 to obtain, B
B m
k G
h c
θ
θ = * =
3
*
2 (0.2)
One may also argue that m*corresponds to thermal equilibrium Thus for m=m*the black hole temperature equalsθB
Or one may set =−
(
*4 − 4)
A=0 td E d
B
θ θ
σ to get θ* =θB
(0.2)
(0.2)
(5)4.5) Considering the solution to 4.3, one verifies that it will go away from the
equilibrium (0.6)
⇒
− −
= *4
4
2
1
m m m
G hc dt
dm
< ⇒
<
> ⇒
>
0 *
*
dt dm m
m
dt dm m
m