,Student Solutions Manual: Modern General Relativity
Mike Guidry
This document gives the solutions for select problems (those marked by ***)
at the ends of chapters for the first edition of Modern General Relativity:
Black Holes, Gravitational Waves, and Cosmology by Mike Guidry (Cam-
bridge University Press, 2019). Unless otherwise indicated, literature refer-
ences, equation numbers, figure references, table references, and section num-
bers refer to the print version of that book.
,1 Introduction
1.3 This question is ambiguous, since it does not specify whether the curvature is that of
the surface itself (which is called intrinsic curvature) or whether it is the apparent curvature
of the surface seen embedded in a higher-dimensional euclidean space (which is called
the extrinsic curvature). In general relativity the curvature of interest is usually intrinsic
curvature. Then the sheet of paper can be laid out flat and is not curved, the cylinder is
also flat, with no intrinsic curvature, because one can imagine cutting it longitudinally
and rolling it out into a flat surface, but the sphere has finite intrinsic curvature because it
cannot be cut and rolled out flat without distortion. The reason that the cylinder seems to
be curved is because the 2D surface is being viewed embedded in 3D space, which gives
a non-zero extrinsic curvature, but if attention is confined only to the 2D surface it has no
intrinsic curvature. This is a rather qualitative discussion but in later chapters methods will
be developed to quantify the amount of intrinsic curvature for a surface.
1
, Coordinate Systems and
2
Transformations
2.1 Utilizing Eq. (2.31) to integrate around the circumference of the circle,
s
Z +R 2
dy
I I
2 2 1/2
C = ds = (dx + dy ) = 2 dx 1 + ,
−R dx
subject to the constraint R2 = x2 + y2 , where the factor of two and the limits are because x
ranges from −R to +R over half a circle. The constraints yield dy/dx = −(R2 − x2 )−1/2 x,
which permits the integral to be written as
s
R2
Z R
C=2 dx .
−R R 2 − x2
Introducing a new integration variable a through a ≡ x/R then gives
Z +1
da
C = 2R √ = 2π R,
−1 1 − a2
since the integral is sin−1 a. In plane polar coordinates the line element is given by Eq.
(2.32) and proceeding as above the circumference is
I I
C= (dr2 + r2 d ϕ 2 )1/2
ds =
s 2
Z 2π Z 2π
2
dr
= dϕ r + =R d ϕ = 2π R,
0 dϕ 0
where r = R has been used, implying that dr/d ϕ = 0.
2.4 Using the spherical coordinates
u1 = r u2 = θ u3 = ϕ
defined through Eq. (2.2) and the results of Example 2.2,
e 1 ·ee1 = 1 e 2 ·ee2 = r2 e 3 ·ee3 = r2 sin2 θ ,
while all non-diagonal components vanish. Thus the metric tensor is
1 0 0
gi j = 0 r 2 0 .
2 2
0 0 r sin θ
The corresponding line element is
ds2 = dr2 + r2 d θ 2 + r2 sin2 θ d ϕ 2 ,
2
Mike Guidry
This document gives the solutions for select problems (those marked by ***)
at the ends of chapters for the first edition of Modern General Relativity:
Black Holes, Gravitational Waves, and Cosmology by Mike Guidry (Cam-
bridge University Press, 2019). Unless otherwise indicated, literature refer-
ences, equation numbers, figure references, table references, and section num-
bers refer to the print version of that book.
,1 Introduction
1.3 This question is ambiguous, since it does not specify whether the curvature is that of
the surface itself (which is called intrinsic curvature) or whether it is the apparent curvature
of the surface seen embedded in a higher-dimensional euclidean space (which is called
the extrinsic curvature). In general relativity the curvature of interest is usually intrinsic
curvature. Then the sheet of paper can be laid out flat and is not curved, the cylinder is
also flat, with no intrinsic curvature, because one can imagine cutting it longitudinally
and rolling it out into a flat surface, but the sphere has finite intrinsic curvature because it
cannot be cut and rolled out flat without distortion. The reason that the cylinder seems to
be curved is because the 2D surface is being viewed embedded in 3D space, which gives
a non-zero extrinsic curvature, but if attention is confined only to the 2D surface it has no
intrinsic curvature. This is a rather qualitative discussion but in later chapters methods will
be developed to quantify the amount of intrinsic curvature for a surface.
1
, Coordinate Systems and
2
Transformations
2.1 Utilizing Eq. (2.31) to integrate around the circumference of the circle,
s
Z +R 2
dy
I I
2 2 1/2
C = ds = (dx + dy ) = 2 dx 1 + ,
−R dx
subject to the constraint R2 = x2 + y2 , where the factor of two and the limits are because x
ranges from −R to +R over half a circle. The constraints yield dy/dx = −(R2 − x2 )−1/2 x,
which permits the integral to be written as
s
R2
Z R
C=2 dx .
−R R 2 − x2
Introducing a new integration variable a through a ≡ x/R then gives
Z +1
da
C = 2R √ = 2π R,
−1 1 − a2
since the integral is sin−1 a. In plane polar coordinates the line element is given by Eq.
(2.32) and proceeding as above the circumference is
I I
C= (dr2 + r2 d ϕ 2 )1/2
ds =
s 2
Z 2π Z 2π
2
dr
= dϕ r + =R d ϕ = 2π R,
0 dϕ 0
where r = R has been used, implying that dr/d ϕ = 0.
2.4 Using the spherical coordinates
u1 = r u2 = θ u3 = ϕ
defined through Eq. (2.2) and the results of Example 2.2,
e 1 ·ee1 = 1 e 2 ·ee2 = r2 e 3 ·ee3 = r2 sin2 θ ,
while all non-diagonal components vanish. Thus the metric tensor is
1 0 0
gi j = 0 r 2 0 .
2 2
0 0 r sin θ
The corresponding line element is
ds2 = dr2 + r2 d θ 2 + r2 sin2 θ d ϕ 2 ,
2
