This document discusses solutions to the differential equation R(r) for Hawking radiation with vanishing mass M and angular frequency ω. It presents: 1) A transformation from r to z and the resulting Legendre form differential equation. 2) A power series solution and the repeated values s = 0, 0. 3) Derivation of the Legendre polynomials as the regular solution and indication of a logarithmic solution. 4) Notation of the Legendre polynomial solution Pμθ and first two examples. It indicates the solution will be continued.

1. Basic Illustration Exercises in Hawking Radiation
(Notes III )
Roa, Ferdinand J. P.
Exercise A.4.4
(page 142 of [1])
Solutions to R(r) with vanishing 𝜔 and M
We take that both M and 𝜔 vanish and by a simple transformation 𝑟 → 𝑧
(15a)
𝑧 =
𝑟 − 𝐺𝑀𝑞
𝐺𝑀𝑞
= −
𝐺𝑀𝑞 − 𝑟
𝐺𝑀𝑞
We transform the given D.E. (12f) for R(r) into its Legendre form
(15b)
𝑑
𝑑𝑧
((1 − 𝑧2 )
𝑑𝑅
𝑑𝑧
) + 𝜇 𝜃
( 𝜇 𝜃 + 1) 𝑅 = 0
There are two regular singular points (RSP): 𝑧0 = −1, +1 that correspond to two values r = 0, 𝐺𝑀𝑞 ,
respectively.
First we may try the substitution
(15c)
𝑦 = 1 − 𝑧
to obtain
(15d)
𝑑
𝑑𝑦
( 𝑦(2 − 𝑦)
𝑑𝑅
𝑑𝑦
) + 𝜇 𝜃
( 𝜇 𝜃 + 1) 𝑅 = 0
and put the solution in power series form
(15e)
𝑅 = ∑ 𝑎 𝑛(𝑠)𝑦 𝑠 +𝑛
∞
𝑛=0
With this power series solution, we can immediately put (15d) into
(15f)
∑[ 𝜇 𝜃
( 𝜇 𝜃 + 1) − ( 𝑠 + 𝑛)( 𝑠 + 𝑛 + 1)] 𝑎 𝑛 𝑦 𝑆+𝑛
+ ∑ 2( 𝑠 + 𝑛)2
∞
𝑛=0
∞
𝑛=0
𝑎 𝑛 𝑦 𝑆+𝑛−1
= 0
Let us note that in (15d) we have an operator

2. (15g)
𝐿 ≡ 𝑦(2 − 𝑦)
𝑑2
𝑑𝑦2
+ 2(1 − 𝑦)
𝑑
𝑑𝑦
+ 𝜇 𝜃
( 𝜇 𝜃 + 1)
This operator as applied on (15e) yields, in the lowest power in y at 𝑛 = 0,
(15h)
𝐿[ 𝑅( 𝑦, 𝑠)] = 2𝑎0 𝑠2
𝑦 𝑠−1
and as the condition
(15i)
𝐿[ 𝑅( 𝑦, 𝑠)] = 0
is imposed for 𝑅( 𝑦, 𝑠) to be a solution, then we obtain the repeated values of 𝑠
(15j)
𝑠 = 0, 0
Let us also note that we have two commuting operators,
𝑑
𝑑𝑠
and 𝐿 so to have, in the lowest power in y at
𝑛 = 0,
(15k)
𝑑
𝑑𝑠
𝐿[ 𝑅( 𝑦, 𝑠)] = 𝐿 [
𝑑
𝑑𝑠
𝑅( 𝑦, 𝑠)] = 4𝑎0 𝑠 𝑦 𝑠−1
+ 2𝑎0 𝑠2
𝑦 𝑠−1
𝑙𝑛𝑦
where to follow as 𝑠 = 0
(15L)
𝑑
𝑑𝑠
𝐿[ 𝑅( 𝑦, 𝑠)] = 𝐿 [
𝑑
𝑑𝑠
𝑅( 𝑦, 𝑠)] = 0
Now, let us take 𝑠 = 0 and write one solution at this s-value in the form,
(16a)
𝑅1
( 𝑦, 𝑠 = 0) = 𝑅1
( 𝑦) = ∑ 𝑎 𝑛(0)𝑦 𝑛
∞
𝑛=0
and to choose( arbitrarily) 𝑎0
(0) = 1. The polynomial L(y) resulting from the operation 𝐿[ 𝑅1
( 𝑦, 𝑠 = 0)] is
obtained in the form
(16b)
𝐿( 𝑦) = ∑[ 𝜇 𝜃
( 𝜇 𝜃 + 1) − ( 𝑛 )( 𝑛 + 1 )] 𝑎 𝑛 𝑦 𝑆+𝑛
+ ∑ 2( 𝑛 )2
∞
𝑛=0
∞
𝑛=0
𝑎 𝑛(0)𝑦 𝑛−1
= 0

3. Note that we can make the shift 𝑛 → 𝑛 − 1. As 𝑅1
( 𝑦, 𝑠 = 0) must satisfy the condition 𝐿[ 𝑅1
( 𝑦, 𝑠 = 0)] =
0 and with the shift 𝑛 → 𝑛 − 1 in the first major term in L(y), we get the recurrence relation between
𝑎 𝑛(0) coefficients ( ∀𝑛 ≥ 1 ).
(16c)
𝑎 𝑛
(0) =
𝑛( 𝑛 − 1) − 𝜇 𝜃 (𝜇 𝜃 + 1)
2𝑛2
𝑎 𝑛−1
(0)
By repeated use of this recurrence relation we can write any coefficient 𝑎 𝑚(0) in terms of 𝑎0
(0) = 1
(16d)
𝑎 𝑚
(0) = −𝜇 𝜃
( 𝜇 𝜃 + 1)[(1)(2) − 𝜇 𝜃
( 𝜇 𝜃 + 1)] [(2)(3) − 𝜇 𝜃
( 𝜇 𝜃 + 1)]
×
[(3)(4) − 𝜇 𝜃
( 𝜇 𝜃 + 1)] ⋯ [( 𝑚 − 1) 𝑚 − 𝜇 𝜃
( 𝜇 𝜃 + 1)]
(2) 𝑚 (𝑚!)2
The series 𝑅1((1 − 𝑧), 𝑠 = 0) can terminate at the qth term as when 𝑎 𝑞 = 0, given ( 𝑞 − 1) 𝑞 =
𝜇 𝜃
( 𝜇 𝜃 + 1), where 𝑞 = 𝜇 𝜃 + 1. The recurrence formula (16c) shows that higher terms following the qth
term vanish also.
In crude form therefore, we write the Legendre polynomial for solution (16a) as
(16e)
𝑃𝜇 𝜃
= 𝑅1
𝜇 𝜃
((1 − 𝑧), 𝑠 = 0) = ∑ 𝑎 𝑛
𝜇 𝜃
(0)(1 − 𝑧) 𝑛
𝜇 𝜃
𝑛=0
𝑎0
𝜇 𝜃 (0) = 1, 𝑓𝑜𝑟 ∀𝜇 𝜃
For example, we have the first two of these polynomials:
(16f1)
𝑃1 = 𝑎0
1
+ 𝑎1
1 (1 − 𝑧) = 𝑧
𝑎1
1
(0) = −1
𝑎0
1
(0) = 1
𝑛𝑜𝑡𝑒𝑑: 𝑎2
1
(0) = 0
𝑃2 =
1
2
(3𝑧2
− 1)
𝑎0
2
(0) = 1
𝑎1
2 (0) = −3
𝑎2
2
(0) = 3
2⁄

4. Let us delve into the linearly independent (logarithmic) solution.
From the power series solution (15e) we can derive
(17a)
𝜕
𝜕𝑠
𝑅( 𝑦, 𝑠) = 𝑅( 𝑦, 𝑠) 𝑙𝑛𝑦 + ∑
𝜕𝑎 𝑛
( 𝑠)
𝜕𝑠
∞
𝑛=0
𝑦 𝑠 + 𝑛
and reflect this on the previous result ((15L)) that
(17b)
𝑑
𝑑𝑠
𝐿[ 𝑅( 𝑦, 𝑠 = 0)] = 𝐿 [
𝑑
𝑑𝑠
𝑅( 𝑦, 𝑠 = 0)] = 0
to identify the linearly independent solution
(17c)
𝑅2
( 𝑦, 𝑠 = 0) =
𝜕
𝜕𝑠
𝑅( 𝑦, 𝑠 = 0) = 𝑅( 𝑦, 𝑠 = 0) 𝑙𝑛𝑦 + ∑
𝜕𝑎 𝑛
(0)
𝜕𝑠
∞
𝑛=0
𝑦 𝑛
where in all powers in y, 𝑅2
( 𝑦, 𝑠 = 0) must also satisfy
(17d)
𝐿[ 𝑅2
( 𝑦, 𝑠 = 0)] = 0
Our convenient choice for 𝑅( 𝑦, 𝑠 = 0) are the Legendre polynomials, 𝑃𝜇 𝜃
, each satisfies
(17e)
𝐿[ 𝑃𝜇 𝜃
] = 0
in the substitution (15c).
Note that it is convenient to arbitrarily choose
(17f)
𝜕𝑎0
(0)
𝜕𝑠
= 0
for 𝑛 ≥ 1. Applying 𝐿[ 𝑅2
( 𝑦, 𝑠 = 0)] = 0, we get
(17g)
2𝑏1
(0) − 𝑎0
(0) + ∑[( 𝜇 𝜃
( 𝜇 𝜃 + 1) − ( 𝑛 )( 𝑛 + 1 ) ) 𝑏 𝑛
(0) − 𝑎 𝑛
(0) ] 𝑦 𝑛 +
∞
𝑛=1
∑ 2𝑛2
𝑏 𝑛
(0) 𝑦 𝑛−1 = 0
∞
𝑛=2
and as to be noted we can make the shift 𝑛 → 𝑛 − 1 in the second major term.

5. We stick unto the condition that the coefficients of 𝑦 𝑛
(𝑛 = 0,1, 2, 3, …) must vanish so as
consequences we have
(17h)
𝑏1
(0) =
1
2
𝑎0
(0) =
1
2
, 𝑎0
(0) = 1
and for 𝑛 ≥ 2,
𝑏 𝑛
(0) =
1
2𝑛2
[ 𝑎 𝑛−1
(0) − ( 𝜇 𝜃
( 𝜇 𝜃 + 1) − ( 𝑛 )( 𝑛 − 1 ) ) 𝑏 𝑛−1
(0)]
in which recurrence relations between coefficients 𝑏 𝑛
(0) in closed form are impossible.
We may write (17c) in the form
(17i)
𝑄 𝜇 𝜃
( 𝑧) = 𝑃𝜇 𝜃
( 𝑧) 𝑙𝑛(1 − 𝑧) + ∑ 𝑏 𝑛
(0)(1 − 𝑧) 𝑛
∞
𝑛=1
in the substitution (15c) and to be noted in this substitution is that the solution is sensible only within the
interval 0 ≤ 𝑟 ≤ 2𝐺𝑀𝑞
Note: To be continued.
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