- The author considers a Klein-Gordon scalar field in a two-dimensional Rindler space-time background. - They define a two-dimensional action for the scalar field and derive from it the Klein-Gordon equation of motion in this background. - The equation can be solved exactly using imaginary time. The solution is oscillatory with an angular frequency that corresponds to an integral number, suggesting quantization of the scalar field frequencies in this space-time.

1. The Klein-Gordon field in two-dimensional Rindler space-time
Ferdinand Joseph P. Roa
Independent physics researcher
rogueliknayan@yahoo.com
Abstract
The Klein-Gordon scalar in the background of two-dimensional Rindler space-time is considered in this exercise. In an
informal way without resorting to methods of dimensional reduction, a two-dimensional action for the Klein-Gordon
scalar is written with the said background and obtaining from this action the equation of motion for the scalar field. The
equation of motion is solvable exactly in this two-dimensional space-time using imaginary time. In imaginary time, the
solution is oscillatory with a given frequency that corresponds to an integral number.
Keywords: Coordinate Singularity, Series Solution
1 Introduction
The Schwarzschild metric[1]
𝑑𝑆2
= − 𝜂𝑑𝑡2
+ 𝜀 𝑑𝑟2
+ 𝑟2 (𝑑𝜃2
+ 𝑠𝑖𝑛2
𝜃 𝑑𝜙2 )
𝜂 = 𝜀−1
= 1 −
2𝐺𝑀 𝑞
𝑟
(1)
as expressed in the standard coordinates has a (coordinate) singularity[2] at 𝑟 𝐻 = 2𝐺𝑀 𝑞. (Note that in this
entire document we take the square of the speed of light as unity, 𝑐2
= 1 and whenever consistency in
units needed, we insert its appropriate value.) This can be seen crudely from the fact that 𝜀 = ∞ at 𝑟 =
𝑟 𝐻. This coordinate value of 𝑟 at which one piece of the metric is singular defines a horizon[2, 3] that puts
bounds to (1), confining it in a portion of space-time where this metric in that form is sensible. That is, in
rough language say (1) is for all those regions of space-time where 𝑟 > 𝑟 𝐻 and dipping below 𝑟 𝐻 can no
longer be covered by the given metric as expressed in that form.
Figure 1: This is the space-time graph on rt-plane.
𝑑𝑡
𝑑𝑟
= ±
1
1 −
2𝐺𝑀 𝑞
𝑟
(2)
On a space-time graph where one can draw a light-cone bounded by the intersecting lines whose
slopes are given by (2), it can be superficially shown that the region at 𝑟 > 𝑟 𝐻 is not causally connected to
that at 𝑟 < 𝑟 𝐻 . This is so since asymptotically the light-cone closes as 𝑟 𝐻 is approached from the right. As
the light-cone closes there can be no way of connecting a time-like particle’s past to its supposed future
along a time-like path that is enclosed by the light-cone. So any coordinate observer won’t be able to
construct a causal connection between the past and the future for a time-like particle falling into that region
𝑟 < 𝑟 𝐻.
However, such singularity is only a coordinate one specific to the form (1) since expressing the
same metric in suitable coordinates will remove the said coordinate singularity.
For example, from the standard coordinates (𝑡, 𝑟, 𝜃, 𝜙) we can change (1) into
𝑑𝑆2
= − 𝜂𝑑𝑢̃2
+ 2𝑑𝑢̃𝑑𝑟 + 𝑟2 (𝑑𝜃2
+ 𝑠𝑖𝑛2
𝜃 𝑑𝜙2 ) (3)
using the Eddington-Finkelstein coordinate
𝑢̃ = 𝑡 + 𝑟 ∗ (4)
with the Regge-Wheeler coordinate

2. 𝑟 ∗ = 𝑟 + 2𝐺𝑀 𝑞 𝑙𝑛 (
𝑟
2𝐺𝑀 𝑞
− 1)
(5)
Noticeable in (3) is that none of the metric components goes infinite at 𝑟 𝐻 so the singularity
( 𝜀 = ∞ at 𝑟 = 𝑟 𝐻) in (1) is not a case in (3).
In this paper, we deal with the Klein- Gordon scalar as dipped very near the horizon but not
having completely fallen into those regions at 𝑟 < 𝑟 𝐻. We consider that near the horizon we can make the
substitution[2]
𝑥2
8𝐺𝑀 𝑞
= 𝑟 − 2𝐺𝑀 𝑞
(6)
Hence, in approximate form we write (1) as
𝑑𝑆2
≈ −(𝜅𝑥)2
𝑑𝑡2
+ 𝑑𝑥2
+ 𝑟𝜅
2
𝑑Ω2
(7)
𝜅 =
1
4𝐺𝑀 𝑞
where 𝑟𝜅 = 1/2 𝜅 is the approximate radius of a two-sphere 𝑆2
: 𝑟 𝜅
2 𝑑Ω2
and we think of the (3+1)-
dimensional space-time ascribed to metric (7) as a product of a two-dimensional Rindler space-time
𝑑𝑆(𝑅)
2
≈ −(𝜅𝑥)2
𝑑𝑡2
+ 𝑑𝑥2
(8)
and that of the two-sphere. We give to this two dimensional Rindler space-time the set of coordinates
𝑥 𝜇
= {𝑥0
= 𝑡, 𝑥1
= 𝑥} (9.1)
with t as the real time to be transcribed into an imaginary time by
𝑡 → 𝜏 = −𝑖𝑡 (9.2)
2 Two-dimensional action
In this Rindler space-time we just write a two dimensional action for our scalar field
𝑆 𝐶 = ∫ 𝑑2
𝑥 √−𝑔 (
1
2
𝑔 𝜇𝜔
(𝜕 𝜇 𝜑𝑐)(𝜕 𝜔 𝜑𝑐) +
1
2
𝑀2
𝜑 𝐶
2
) (10)
The metric components in this action are those belonging to the two-dimensional Rindler space-time given
by metric form (8). This is rather an informal way without having to derive it from an original 3 + 1
dimensional version that would result into (10) through the process of a dimensional reduction[4] with the
Rindler space-time as background. Anyway, on the way the solution exists for the resulting equation of
motion to be obtained from the given action and this solution is oscillatory as taken in the imaginary time.
Taking the variation of (10) in terms of the variation of our classical scalar 𝜑 𝑐would yield the
equation of motion
1
√−𝑔
𝜕𝜇(√−𝑔 𝑔 𝜇𝜔
𝜕 𝜔 𝜑𝑐) − 𝑀2
𝜑 𝑐 = 0
(11.1)
or with (8) as the said background we have explicitly
𝜕1
2
𝜑𝑐 +
1
𝑥
𝜕1 𝜑𝑐 −
1
𝜅2 𝑥2
𝜕0
2
𝜑𝑐 = 𝑀2
𝜑 𝑐
(11.2)
We take that the solution is variable separable, 𝜑𝑐 = 𝜒(𝑥)𝑇(𝑡)so that (11.2) could be written into
two independent equations
1
𝜒
(𝜕1
2
𝜒 +
1
𝑥
𝜕1 𝜒) −
1
𝜅2 𝑥2
𝜇 𝐸
2
= 𝑀2
(12.1)
and
1
𝑇
𝜕0
2
𝑇 = −
1
𝑇
𝜕𝜏
2
𝑇 = 𝜇 𝐸
2
𝜇 𝐸
2
> 0 (12.2)
Later, the constant 𝜇 𝐸 is to be identified as the angular frequency 𝜔 in the imaginary time (9.2).

3. 3 The solution
The differential equation (12.1) is satisfied by a series solution of the following form
𝜒(𝑚) =
1
√ 𝑥
∑
1
𝑥 𝑛
(𝑎 𝑛 exp(𝑀𝑥) + 𝑏 𝑛 exp(−𝑀𝑥) )
𝑚
𝑛=0
(13)
This series solution corresponds to an integral number 𝑚 and the series stops at the 𝑚𝑡ℎ term. (As
a cautionary let us not confuse 𝑚 with 𝑀. The latter is the mass of our scalar field in units of per length.)
The (𝑚 + 1)𝑡ℎ term and all other higher terms vanish as the 𝑎 𝑚+1 and 𝑏 𝑚+1 coefficients are terminated.
That is, 𝑎 𝑚+1 = 𝑏 𝑚+1 = 0. Each coefficient 𝑎 𝑛 is given by this recursion formula
𝑎 𝑛 =
(2𝑛 − 1)2
− (2𝜅−1
𝜇 𝐸)2
8𝑀𝑛
𝑎 𝑛−1
(14.1)
and each 𝑏 𝑛 by
𝑏 𝑛 = −
(2𝑛 − 1)2
− (2𝜅−1
𝜇 𝐸)2
8𝑀𝑛
𝑏 𝑛−1
(14.2)
These formulas are defined for all 𝑛 ≥ 1 and with these the vanishing of those 𝑚 + 1 coefficients would
imply that
𝜇 𝐸(𝑙) = (𝑙 +
1
2
)𝜅 (14.3)
𝑙 = 0, 1, 2, 3, … , 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 (14.4)
𝜅 = (4𝐺𝑀 𝑞)
−1
(14.5)
(There in (14.3) we have relabeled 𝑚 as 𝑙 and this includes zero as one of its parameter values.
With the inclusion of zero, the lowest vanishing (𝑚 + 1) coefficients given 𝑚 = 0 would be 𝑎1 = 0 and
𝑏1 = 0 so that all other higher terms with their corresponding coefficients vanish. Then in this particular
case, the series solution only has the 0th terms with their coefficients 𝑎0 and 𝑏0 that are non-zero.)
Given (14.3), we can write the differential equation (12.1) as
𝜕1
2
𝜒(𝑙) +
1
𝑥
𝜕1 𝜒(𝑙) −
1
4𝑥2
(2𝑙 + 1)2
𝜒(𝑙) = 𝑀2
𝜒(𝑙)
(14.6)
whose solution is in the series form (13).
Put simply, the differential equation (12.2) has as solution the following function of the imaginary
time 𝜏
𝑇(𝜏) = 𝐴𝑐𝑜𝑠𝜔(𝑙) 𝜏 + 𝐵𝑠𝑖𝑛𝜔(𝑙) 𝜏
(14.7)
Here we have identified the separation constant 𝜇 𝐸 as the angular frequency 𝜔 in the solution above. That
is, 𝜇 𝐸(𝑙) = 𝜔(𝑙), and given (14.3), we find that this oscillatory solution has an angular frequency that
corresponds to an integral number 𝑚.
We can choose to set 𝑎0 = 𝑏0 so that 𝑏 𝑛 = (−1) 𝑛
𝑎 𝑛 and at 𝑚 = 0 , implying 𝑎1 = 0 and
𝑏1 = 0, we have
𝜒(0) =
2𝑎0 𝑐𝑜𝑠ℎ𝑀𝑥
√ 𝑥
(15.1)
with 𝜔(0) = 𝜅/2 .
Going back to the recursion relations (14.1) and (14.2), we must take note that these fail when the scalar
field is massless since these are singular at 𝑀 = 0. In the massless case we may consider 𝑎0 = 𝑏0, so as a
consequence, 𝑏 𝑛 = (−1) 𝑛
𝑎 𝑛 and
𝜒(𝑚) =
2𝑎0
√ 𝑥
+
2
√ 𝑥
∑
𝑎 𝑛
𝑥 𝑛
𝑚
𝑛 =2
(15.2)
So in the massless case we apply (14.6) with 𝑀 = 0 to (15.2) to get a form of constraint on the coefficients
𝑎 𝑛 and this is given by
∑ [2 (𝑛 +
1
2
)
2
− 2 (𝑚 +
1
2
)
2
]
𝑚
𝑛 =0
𝑎 𝑛 𝑥−𝑛−5/2
= 0
(15.3)

4. From this constraint we can choose only one coefficient 𝑎𝑙 to be non-zero, while all other coefficients
𝑎 𝑛 ≠ 𝑙 to be zero. Example, if we choose 𝑎0 to be the only non-zero, then 𝑚 = 0, and all other coefficients
𝑎 𝑛 ≠ 0 to be zero. For every non-zero coefficient 𝑎𝑙, there corresponds an angular frequency in the form of
(14.3). Eventually then, following this condition in the constraint, series (15.2) would only be made up of a
term with the non-zero coefficient,
𝜒(𝑚) =
2𝑎 𝑚
√ 𝑥
1
𝑥 𝑚
(15.4)
This satisfies the differential equation in the same form (14.6) for a given integral value of 𝑚 with 𝑀 = 0.
4 Conclusions
Taking the Klein-Gordon field as a classical scalar, we have shown that its two-dimensional equation of
motion in Rindler space-time has a series solution that can terminate at a certain term. As a consequence of
this termination the angular frequency (14.3) with the identification 𝜇 𝐸(𝑙) = 𝜔(𝑙), given (9.2) seems to
have values that correspond to integral values of 𝑙. This is seemingly suggestive that the classical scalar
field can already appear quantized in terms of its angular frequency or can have a spatial mode given by
(13) that corresponds to an integral value of 𝑚. We also have a curious result as manifest in (14.3) that in
the imaginary time, the scalar field can oscillate at frequencies that are odd multiples of the surface gravity
over four pi, 𝑓(𝑙) = (2𝑙 + 1)𝜅/4𝜋, where surface gravity is 𝜅 = 𝑐3
/4𝐺𝑀 𝑞.
Some details of spatial solution
Two-dimensional scalar action defined in the background of two-dimensional Rindler space-time
We have arbitrarily defined a two-dimensional scalar action (10), given the fundamental line element (8) of
the two-dimensional Rindler space-time. This action we explicitly write as
𝑆 𝐶 = ∫ 𝑑𝑡 ∫ 𝑑𝑥
1
2
(−
1
𝜅𝑥
(𝜕𝑡 𝜑𝑐)2
+ 𝜅𝑥 (𝜕 𝑥 𝜑𝑐)2
+ 𝑀2
𝜅𝑥 𝜑 𝐶
2
)
(16)
As earlier stated this Rindler space-time is endowed with a set of coordinates (9.1). Thus, the classical
scalar can have two degrees of freedom and its motion is along 𝑥0
= 𝑡 (the time direction) and one spatial
direction 𝑥1
= 𝑥.
We will no longer (or perhaps in the much later portion of this draft) present here the details of varying this
action in terms of the variation of the classical scalar to arrive at the equation of motion (11.1) or as
explicitly given by (11.2).
The spatial series solution
In tackling the differential equation (11.2), we assumed a variable separable solution in the form of
𝜑 𝑐 = 𝜒(𝑥)𝑇(𝑡) so decomposing (11.2) into equations (12.1) and (12.2). Most of our effort here is to work
on the details involved in solving (12.1).
Given the variable separable solution, we were able to write the spatial part of (11.2) as (12.1) and re-
writing that here as
𝑥2
𝜕1
2
𝜒 + 𝑥 𝜕1 𝜒 − (𝜇 𝐸 𝜅−1)2
𝜒 = 𝑀2
𝑥2
𝜒
(17.1)
In considering the solution in series form as given by (13), we have actually decomposed 𝜒 into two
separate components
𝜒(𝑚) = 𝜒1 + 𝜒2 = 𝜒1(𝑚) + 𝜒2(𝑚)
(17.2)
(Again, we must note that 𝑚 here is an integral number and must not be confused with mass 𝑀 of the
scalar field.) We plug (17.2) into (17.1) and collect like terms we then write two separate differential
equations from (17.1).
𝑥2
𝜕1
2
𝜒1 + 𝑥 𝜕1 𝜒1 − (𝜇 𝐸 𝜅−1)2
𝜒1 = 𝑀2
𝑥2
𝜒1
𝜒1 = 𝜒1(𝑚)
(17.3)

5. where
𝜒1 = 𝜒1(𝑚) = 𝑄1(𝑚)(𝑥) 𝑒 𝑀𝑥
=
1
√ 𝑥
∑
1
𝑥 𝑛
𝑎 𝑛 exp(𝑀𝑥)
𝑚
𝑛=0
(17.4)
while the other part is given by
𝑥2
𝜕1
2
𝜒2 + 𝑥 𝜕1 𝜒2 − (𝜇 𝐸 𝜅−1)2
𝜒2 = 𝑀2
𝑥2
𝜒2
𝜒2 = 𝜒2(𝑚)
(17.5)
to which belongs the other solution
𝜒2 = 𝜒2(𝑚) = 𝑄2(𝑚)(𝑥) 𝑒− 𝑀𝑥
=
1
√ 𝑥
∑
1
𝑥 𝑛
𝑏 𝑛 exp(−𝑀𝑥)
𝑚
𝑛=0
(17.6)
We proceed from (17.3), given (17.4) to get the following differential equation for 𝑄1(𝑚).
𝑥2
𝑄 ′′1(𝑚) + 2𝑀𝑥2
𝑄 ′1(𝑚) + 𝑥𝑄 ′1(𝑚) + 𝑀𝑥𝑄1(𝑚) − (𝜇 𝐸 𝜅−1)2
𝑄1(𝑚) = 0
𝜇 𝐸𝐹 = 𝜇 𝐸 𝜅−1
(17.7)
The form of 𝑄1(𝑚) is already evident in (17.4) and substituting this solution in (17.3) would give us the
recurrence relations between the coefficients 𝑎 𝑛.
Proceeding, we write
∑ [(𝑛 +
1
2
)
2
− 𝜇 𝐸𝐹
2
] 𝑎 𝑛 𝑥−(𝑛+ 1/2)
𝑚
𝑛=0
− ∑ 2𝑀𝑛
𝑚
𝑛 = 0
𝑎 𝑛 𝑥
−(𝑛−
1
2
)
= 0
(17.8)
In the first major summation, we make the shift 𝑛 → (𝑛 − 1) so that (17.8) can be re-written into the
following form
∑ [((𝑛 −
1
2
)
2
− 𝜇 𝐸𝐹
2
) 𝑎 𝑛−1 − 2𝑀𝑛𝑎 𝑛] 𝑥−(𝑛 − 1/2)
𝑚
𝑛=1
= 0
(17.9)
For this to be satisfied for all coefficients we must have the recurrence relations between adjacent
successive coefficients and such relations are already given by (14.1). The series solution (17.4) can be
terminated so that it will consist only of series of terms up to the 𝑙𝑡ℎ place with 𝑚 relabeled as 𝑙. This can
be done by setting the 𝑎 𝑛 = 𝑙 +1 coefficients to zero.
𝑎 𝑛 = 𝑙 +1 =
(2𝑙 + 1)2
− (2𝜅−1
𝜇 𝐸)2
8𝑀(𝑙 + 1)
𝑎𝑙 = 0
(17.10)
As earlier stated in the conclusion the consequence of having a terminated series solution is that the
frequencies are odd multiples of the surface gravity over four pi
𝑓(𝑙) = (2𝑙 + 1)
𝜅
4𝜋
(17.11)
It should already be clear from (12.2) that 𝜇 𝐸 is the angular frequency 𝜔(𝑙) in the imaginary time.
Given the vanishing coefficient (17.10), we can slightly re-write (17.7) as
𝑥2
𝑄 ′′1(𝑙) + 2𝑀𝑥2
𝑄 ′1(𝑙) + 𝑥𝑄 ′1(𝑙) + 𝑀𝑥𝑄1(𝑙) =
(2𝑙 + 1)2
4
𝑄1(𝑙)
(17.12)
with

6. 𝑄1(𝑙)(𝑥) =
1
√ 𝑥
∑
𝑎 𝑛
𝑥 𝑛
𝑙
𝑛=0
(17.13)
Test for 𝑙 = 0:
𝑄1(0)(𝑥) =
𝑎 0
√ 𝑥
(18.1)
𝑥𝑄 ′1(0) = −
1
2
𝑎 0
√ 𝑥
𝑥2
𝑄 ′′1(0) =
3
4
𝑎 0
√ 𝑥
(18.2.1)
𝑥2
𝑄 ′1(0) = −
1
2
𝑎 0√ 𝑥 𝑥𝑄1(0) = 𝑎 0√ 𝑥 (18.2.2)
It is relatively easy to see that (18.1) and the results in (18.2.1) and (18.2.2) satisfy their given differential
equation
𝑥2
𝑄 ′′1(0) + 2𝑀𝑥2
𝑄 ′1(0) + 𝑥𝑄 ′1(0) + 𝑀𝑥𝑄1(0) =
1
4
𝑄1(0)
(18.3)
Test for 𝑙 = 1:
𝑄1(1)(𝑥) =
𝑎 0
√ 𝑥
−
1
𝑀
𝑎 0
√𝑥3
(19.1)
𝑥𝑄1(1) = 𝑎 0√ 𝑥 −
1
𝑀
𝑎 0
√ 𝑥
(19.2.1)
𝑥𝑄1(1)
′
= −
1
2
𝑎 0
√ 𝑥
+
3
2𝑀
𝑎 0
√𝑥3
(19.2.2)
𝑥2
𝑄1(1)
′
= −
1
2
𝑎 0√ 𝑥 +
3
2𝑀
𝑎 0
√ 𝑥
(19.2.3)
𝑥2
𝑄1(1)
′′
=
3
4
𝑎 0
√ 𝑥
−
15
4𝑀
𝑎 0
√𝑥3
(19.2.4)
Solution (19.1) given for 𝑙 = 1 along with (19.2.1), (19.2.2), (19.2.3) and (19.2.4) satisfies the following
corresponding differential equation
𝑥2
𝑄 ′′1(1) + 2𝑀𝑥2
𝑄 ′1(1) + 𝑥𝑄 ′1(1) + 𝑀𝑥𝑄1(1) =
9
4
𝑄1(1)
(19.3)
Test for 𝑙 = 2:
𝑄1(2) =
𝑎 0
√ 𝑥
−
3
𝑀
𝑎 0
√𝑥3
+
3
𝑀2
𝑎 0
√𝑥5
(20.1)
𝑥𝑄1(2) = 𝑎 0√ 𝑥 −
3
𝑀
𝑎 0
√ 𝑥
+
3
𝑀2
𝑎 0
√𝑥3
(20.2)
𝑥𝑄1(2)
′
= −
1
2
𝑎 0
√ 𝑥
+
9
2𝑀
𝑎 0
√𝑥3
−
15
2𝑀2
𝑎 0
√𝑥5
(20.3)

7. 𝑥2
𝑄1(2)
′
= −
1
2
𝑎 0√ 𝑥 +
9
2𝑀
𝑎 0
√ 𝑥
−
15
2𝑀2
𝑎 0
√𝑥3
(20.4)
𝑥2
𝑄1(2)
′′
=
3
4
𝑎 0
√ 𝑥
−
45
4𝑀
𝑎 0
√𝑥3
+
105
4𝑀2
𝑎 0
√𝑥5
(20.5)
For this case the differential equation to be satisfied is given by
𝑥2
𝑄 ′′1(2) + 2𝑀𝑥2
𝑄 ′1(2) + 𝑥𝑄 ′1(2) + 𝑀𝑥𝑄1(2) =
25
4
𝑄1(2)
(20.6)
So far our tests for 𝑄1(𝑙) in the integral values 𝑙 = 0, 1, 2 prove that (17.12) holds true for all these cited
positive integral values of 𝑙. For the moment let us assert that these tests are sufficient enough to claim that
(17.12) is satisfied in all 𝑙 ∈ ℤ ∶ 𝑙 > 0 .
Going much further we will now deal with the other part of the solution that is given by (17.6) that must
satisfy (17.5) to yield the differential equation for 𝑄2(𝑚). Upon substitution of (17.6) into (17.5) we would
obtain the said differential equation for 𝑄2(𝑚)
𝑥2
𝑄 ′′2(𝑚) − 2𝑀𝑥2
𝑄 ′2(𝑚) + 𝑥𝑄 ′2(𝑚) − 𝑀𝑥𝑄2(𝑚) = (𝜇 𝐸 𝜅−1)2
𝑄2(𝑚)
𝜇 𝐸𝐹 = 𝜇 𝐸 𝜅−1
(21.1)
We read off the series form of 𝑄2(𝑚) from (17.6) and plug this in (21.1) to arrive at an expression that will
put a constraint on the coefficients involved.
∑ [(𝑛 +
1
2
)
2
− 𝜇 𝐸𝐹
2
] 𝑏 𝑛 𝑥−(𝑛+ 1/2)
𝑚
𝑛=0
+ ∑ 2𝑀𝑛
𝑚
𝑛 = 0
𝑏 𝑛 𝑥
−(𝑛−
1
2
)
= 0
(21.2)
Then making the shift 𝑛 → (𝑛 − 1) in the first summation from the left to re-write (21.2) into the
following form
∑ [((𝑛 −
1
2
)
2
− 𝜇 𝐸𝐹
2
) 𝑏 𝑛−1 + 2𝑀𝑛𝑏 𝑛] 𝑥−(𝑛 − 1/2)
𝑚
𝑛=1
= 0
(21.3)
From this we get the recurrence relation between successive 𝑏 𝑛 coefficients and this is already given by
(14.2). Likewise, we can also terminate (17.6) up to the 𝑙𝑡ℎ place with 𝑚 relabeled as 𝑙. This is done by
setting all higher coefficients 𝑏 𝑛 = 𝑙 +1 equal to zero, 𝑏 𝑛 = 𝑙 +1 = 0.
𝑏 𝑛 = 𝑙 +1 = −
(2𝑙 + 1)2
− (2𝜅−1
𝜇 𝐸)2
8𝑀(𝑙 + 1)
𝑏𝑙 = 0
(21.4)
The consequence of this termination is already expressed in (17.11) and we can re-write (21.1) as
𝑥2
𝑄 ′′2(𝑙) − 2𝑀𝑥2
𝑄 ′2(𝑙) + 𝑥𝑄 ′2(𝑙) − 𝑀𝑥𝑄2(𝑙) =
(2𝑙 + 1)2
4
𝑄2(𝑙)
(21.5)
given with a terminated series solution
𝑄2(𝑙)(𝑥) =
1
√ 𝑥
∑
𝑏 𝑛
𝑥 𝑛
𝑙
𝑛=0
(21.6)

8. Test for 𝑙 = 0:
(22.1)
𝑄2(0)(𝑥) =
𝑏0
√ 𝑥
(22.1)
𝑥𝑄2(0) = 𝑏0√ 𝑥
(22.2)
,
𝑥𝑄 ′2(0) = −
1
2
𝑏0
√ 𝑥
(22.3)
𝑥2
𝑄 ′2(0) = −
1
2
𝑏0√ 𝑥
(22.4)
𝑥2
𝑄 ′′2(0) =
3
4
𝑏0
√ 𝑥
(22.5)
Corresponding differential equation for this 𝑙 = 0 case is given by
𝑥2
𝑄 ′′2(0) − 2𝑀𝑥2
𝑄 ′2(0) + 𝑥𝑄 ′2(0) − 𝑀𝑥𝑄2(0) =
1
4
𝑄2(0)
(22.6)
Test for 𝑙 = 1:
𝑄2(1)(𝑥) =
𝑏0
√ 𝑥
+
1
𝑀
𝑏0
√𝑥3
(23.1)
𝑥𝑄2(1) = 𝑏0√ 𝑥 +
1
𝑀
𝑏0
√ 𝑥
(23.2)
𝑥𝑄2(1)
′
= −
1
2
𝑏0
√ 𝑥
−
3
2𝑀
𝑏0
√𝑥3
(23.3)
𝑥2
𝑄2(1)
′
= −
1
2
𝑏0√ 𝑥 −
3
2𝑀
𝑏0
√ 𝑥
(23.4)
𝑥2
𝑄2(1)
′′
=
3
4
𝑏0
√ 𝑥
+
15
4𝑀
𝑏0
√𝑥3
These satisfy the following corresponding differential equation
𝑥2
𝑄 ′′2(1) − 2𝑀𝑥2
𝑄 ′2(1) + 𝑥𝑄 ′2(1) − 𝑀𝑥𝑄2(1) =
9
4
𝑄2(1)
(23.5)
Test for 𝑙 = 2:
𝑄2(2) =
𝑏0
√ 𝑥
+
3
𝑀
𝑏0
√𝑥3
+
3
𝑀2
𝑏0
√𝑥5
(23.6)
𝑥𝑄2(2) = 𝑏0√ 𝑥 +
3
𝑀
𝑏0
√ 𝑥
+
3
𝑀2
𝑏0
√𝑥3
(23.7)

9. 𝑥𝑄2(2)
′
= −
1
2
𝑏0
√ 𝑥
−
9
2𝑀
𝑏0
√𝑥3
−
15
2𝑀2
𝑏0
√𝑥5
(23.8)
𝑥2
𝑄2(2)
′
= −
1
2
𝑏0√ 𝑥 −
9
2𝑀
𝑏0
√ 𝑥
−
15
2𝑀2
𝑏0
√𝑥3
(23.9)
𝑥2
𝑄2(2)
′′
=
3
4
𝑏0
√ 𝑥
+
45
4𝑀
𝑏0
√𝑥3
+
105
4𝑀2
𝑏0
√𝑥5
(23.10)
For this case (𝑙 = 2) the differential equation to be satisfied is given by
𝑥2
𝑄 ′′2(2) − 2𝑀𝑥2
𝑄 ′2(2) + 𝑥𝑄 ′2(2) − 𝑀𝑥𝑄2(2) =
25
4
𝑄2(2)
(23.11)
Massless Case
Continuing to the massless case, we take (17.12) for the differential equation of 𝑄1(𝑙) that is given for
𝑀 = 0.
𝑥2
𝑄 ′′1(𝑙) + 𝑥𝑄 ′1(𝑙) =
(2𝑙 + 1)2
4
𝑄1(𝑙)
(24.1)
In (24.1), we apply (17.13) and see what constraint for the coefficients 𝑎 𝑛 we would have in the massless
case. Using (17.13) we obtain the following results
𝑥𝑄 ′1(𝑙) = ∑ − (𝑛 +
1
2
)
𝑙
𝑛=0
𝑎 𝑛 𝑥
−(𝑛+
1
2
)
(24.1.1)
𝑥2
𝑄 ′′1(𝑙) = ∑ (𝑛 +
1
2
)
𝑙
𝑛=0
(𝑛 +
3
2
) 𝑎 𝑛 𝑥−(𝑛+
1
2
)
(24.1.2)
and write (24.1) as
∑((2𝑛 + 1)2
− (2𝑙 + 1)2)
𝑙
𝑛=0
𝑎 𝑛 𝑥−(𝑛+
1
2
)
= 0
(24.2)
This resulting equation (see (15.3) above) is satisfied with a constraint that there is only one 𝑎 𝑛=𝑙
coefficient that is non-zero (𝑎 𝑛=𝑙 ≠ 0), while all coefficients 𝑎 𝑛≠𝑙 vanish. That is, ∀𝑎 𝑛≠𝑙 = 0. Thus, in
the massless case the series solution consists only of one term that corresponds to the non-vanishing
coefficient.
𝑄1(𝑙) =
𝑎 𝑙
√𝑥2𝑙 + 1
(24.3)
Test for 𝑙 = 0:
𝑄1(0)(𝑥) =
𝑎 0
√ 𝑥
(24.4.1)
𝑥𝑄 ′1(0) = −
1
2
𝑎 0
√ 𝑥
(24.4.2)
𝑥2
𝑄 ′′1(0) =
3
4
𝑎 0
√ 𝑥
` (24.4.3)
These satisfy their given differential equation

10. 𝑥2
𝑄 ′′1(0) + 𝑥𝑄 ′1(0) =
1
4
𝑄1(0)
(24.4.4)
Test for 𝑙 = 1:
𝑄1(1)(𝑥) =
𝑎 1
√𝑥3
(24.5.1)
𝑥𝑄1(1)
′
= −
3
2
𝑎 1
√𝑥3
(24.5.2)
𝑥2
𝑄1(1)
′′
=
15
4
𝑎 1
√𝑥3
(24.5.3)
Their corresponding differential equation is given by
𝑥2
𝑄 ′′1(1) + 𝑥𝑄 ′1(1) =
9
4
𝑄1(1)
(24.5.4)
Test for 𝑙 = 2:
𝑄1(2) =
𝑎 2
√𝑥5
(24.6.1)
𝑥𝑄1(2)
′
= −
5
2
𝑎 2
√𝑥5
(24.6.2)
𝑥2
𝑄1(2)
′′
=
35
4
𝑎 2
√𝑥5
(24.6.3)
It is straightforward to check that these satisfy the following differential equation
𝑥2
𝑄 ′′1(2) + 𝑥𝑄 ′1(2) =
25
4
𝑄1(2)
(24.6.4)
To be noted that this constraint in the massless case will also lead to (17.11) if we are to reflect on (17.7)
upon the setting of 𝑀 = 0 to give us
𝑥2
𝑄 ′′1(𝑙) + 𝑥𝑄 ′1(𝑙) = (𝜇 𝐸 𝜅−1)2
𝑄1(𝑙)
(24.7)
For the second part of the solution in the massless case we continue from (21.5) with 𝑀 = 0. So here we
write that as
𝑥2
𝑄 ′′2(𝑙) + 𝑥𝑄 ′2(𝑙) =
(2𝑙 + 1)2
4
𝑄2(𝑙)
(24.8)
Not surprisingly, this is identical to (24.1).
With (21.6) at hand we get the following results
𝑥𝑄 ′2(𝑙) = ∑ − (𝑛 +
1
2
)
𝑙
𝑛=0
𝑏 𝑛 𝑥−(𝑛+
1
2
)
(24.9.1)
and

11. 𝑥2
𝑄 ′′2(𝑙) = ∑ (𝑛 +
1
2
)
𝑙
𝑛=0
(𝑛 +
3
2
) 𝑏 𝑛 𝑥−(𝑛+
1
2
)
(24.9.2)
Given these results along with (21.5), we write (24.8) as
1
4
∑((2𝑛 + 1)2
− (2𝑙 + 1)2)
𝑙
𝑛=0
𝑏 𝑛 𝑥−(𝑛+
1
2
)
= 0
(24.9.3)
from which we infer constraint regarding our coefficients 𝑏 𝑛 and it follows that the only non-zero
coefficients are those at 𝑛 = 𝑙 that is, 𝑏 𝑛 = 𝑙 ≠ 0, while those at 𝑛 ≠ 𝑙 must vanish, 𝑏 𝑛 ≠ 𝑙 = 0. As a
consequence our series solution in the massless case would only consist of a single term containing this
non-vanishing coefficient.
𝑄2(𝑙) =
𝑏 𝑙
√𝑥2𝑙 + 1
(24.10)
Test for 𝑙 = 0:
𝑄2(0)(𝑥) =
𝑏0
√ 𝑥
(24.11.1)
𝑥𝑄 ′2(0) = −
1
2
𝑏0
√ 𝑥
(24.11.2)
𝑥2
𝑄 ′′2(0) =
3
4
𝑏0
√ 𝑥
` (24.11.3)
The differential equation of these is given by
𝑥2
𝑄 ′′2(0) + 𝑥𝑄 ′2(0) =
1
4
𝑄2(0)
(24.11.4)
Test for 𝑙 = 1:
𝑄2(1)(𝑥) =
𝑏1
√𝑥3
(24.12.1)
𝑥𝑄2(1)
′
= −
3
2
𝑏1
√𝑥3
(24.12.2)
𝑥2
𝑄2(1)
′′
=
15
4
𝑏1
√𝑥3
(24.12.3)
Their corresponding differential equation is given by
𝑥2
𝑄 ′′2(1) + 𝑥𝑄 ′2(1) =
9
4
𝑄2(1)
(24.12.4)
Test for 𝑙 = 2:
𝑄2(2) =
𝑏2
√𝑥5
(24.13.1)

12. 𝑥𝑄2(2)
′
= −
5
2
𝑏2
√𝑥5
(24.13.2)
𝑥2
𝑄2(2)
′′
=
35
4
𝑏2
√𝑥5
(24.13.3)
It is straightforward to check that these satisfy the following differential equation
𝑥2
𝑄 ′′2(2) + 𝑥𝑄 ′2(2) =
25
4
𝑄2(2)
(24.13.4)
[To be continued…]
References
[1] J. Foster, J. D. Nightingale, A SHORT COURSE IN GENERAL RELATIVITY, 2nd
edition copyright
1995, Springer-Verlag, New York, Inc.,
[2] P. K.Townsend, Blackholes , Lecture Notes, http://xxx.lanl.gov/abs/gr-qc/9707012
[3] S. M. Carroll, Lecture Notes on General Relativity, arXiv:gr-qc/9712019
[4] Kaluza-Klein Theory, http://faculty.physics.tamu.edu/pope/ihplec.pdf