Lecture Thirteen: Theory of Relativity

13.1 Static, isotropic, metric

In this section, I want to find a static, isotropic, metric. What does this mean? Starting from the metric interval,

ds^{2} = g_{μν}dx^{μ}dx^{ν},

we want to find a set of coordinates x^{μ} in which the g_{μν} do not depend on time t (i.e. the metric is static) and in which ds^{2} depends only on rotational invariants of the spatial coordinates (i.e. the metric is isotropic). We can form the following rotational invariants from the coordinates:

\begin{matrix}dx^{2} \\ x · dx \\ x^{2} ≡ r^{2} \end{matrix} \} (x are the spatial coordinates)

and so the general form of a static, isotropic, metric must be

ds^{2} = A(r)dt^{2} − B(r)dt x · dx − C(r)(x · dx)^{2} − D(r)dx^{2},  (1)

where A, B, C and D are arbitrary function of the radius r. The metric must have the form (1) because we seek a metric that is quadratic in the coordinate differentials that must be rotationally invariant. Now transform to spherical polar coordinates

x^{1} = r sin θ cos ϕ,
x^{2} = r sin θ sin ϕ,
x^{3} = r cos θ.

Since

x^{2} = r^{2}, x · dx = rdr,
dx^{2} = dr^{2} + r^{2}dθ^{2} + r^{2} sin^{2} θdϕ^{2},

the general metric (1) looks like

ds^{2} = A(r)dt^{2} − B(r)dt rdr − C(r)r^{2}dr^{2} − D(r) [dr^{2} + r^{2}dθ^{2} + r^{2} sin^{2} θdϕ^{2}]  (2)

in spherical polar coordinates. Now define a new time coordinate

t′ = t + Φ(r),
dt′ = dt + dΦ(r),

and lets look at the first two terms in (2)

A(r) [dt′ − dΦ]^{2} − B(r) [dt′ − dΦ] rdr

= A(r)dt^{′2} + A(r) (\frac{dΦ}{dr})^{2} dr^{2} − \underbrace{2A(r)dt′ \frac{dΦ}{dr}dr} − \underbrace{B(r)rdt′dr} +B(r)\frac{dΦ}{dr}r^{2}dr.

So if we set

\frac{dΦ}{dr} = \frac{rB(r)}{2A(r)} our metric will become diagonal ,

ds^{2} = A(r)dt^{′2} − [C(r) − \frac{1}{4} \frac{B^{2}(r)}{A(r)}] r^{2}dr^{2} − D(r) [dr^{2} + r^{2}dθ^{2} + r^{2} sin^{2} θdϕ^{2}] .

If we redefine the radius r

r^{′2} = D(r)r^{2}

then we can write the metric as

ds^{2} = α(r′)dt^{′2} − β(r′)dr^{′2} − [r^{′2}dθ^{2} + r^{′2} sin^{2} θdϕ^{2}],

and there is no need to retain the primes on the variables so we can rewrite the metric as

ds^{2} = α(r)dt^{2} − β(r)dr^{2} − [r^{2}dθ^{2} + r^{2} sin^{2} θdϕ^{2}].  (3)

The general static, isotropic, metric is thus specified by two functions functions, α(r) and β(r) which we need to determine by solving the Einstein field equations.

The components of the metric are

\begin{aligned}g_{00} & = & g_{tt} & = & α(r) \\ g_{11} & = & g_{rr} & = & −β(r) \\ g_{22} & = & g_{θθ} & = & −r^{2} \\ g_{33} & = & g_{ϕϕ} & = & −r^{2} sin^{2} θ \end{aligned}  \begin{aligned}g^{00} & = & g^{tt} & = & 1/α(r) \\ g^{11} & = & g^{rr} & = & −1/β(r) \\ g^{22} & = & g^{θθ} & = & −1/r^{2} \\ g^{33} & = & g^{ϕϕ} & = & −\frac{1}{r^{2} sin^{2} θ} \end{aligned}

The affine connection is, by definition,

Γ_{μν}^{λ} = \frac{1}{2}g^{λρ} (\frac{∂g_{ρμ}}{∂x^{ν}} + \frac{∂g_{ρν}}{∂x^{μ}} − \frac{∂g_{μν}}{∂x^{ρ}} )

and so the components are:

Γ_{00}^{0} = 0,
Γ_{00}^{i} = \frac{1}{2}g^{iρ} \{\frac{−∂g_{00}}{∂x^{ρ}} \} ⟹ Γ_{00}^{r} = −\frac{1}{2}g^{rr} \frac{∂g_{00}}{∂x^{r}} = \frac{1}{2} \frac{1}{β(r)} \frac{dα(r)}{dr},
Γ_{0i}^{0} = \frac{1}{2}g^{0ρ} \{\frac{∂g_{ρ0}}{∂x^{i}} + \frac{∂g_{ρi}}{∂x^{0}} − \frac{∂g_{0i}}{∂x^{ρ}} \}
= \frac{1}{2}g^{00} \{\frac{∂g_{00}}{∂x^{i}} \} ⟹ Γ_{0r}^{0} = \frac{1}{2} \frac{1}{α(r)} \frac{dα(r)}{dr},
Γ_{ij}^{0} = 0,
Γ_{ii}^{i} = \frac{1}{2}g^{iρ} \{\frac{∂g_{ρi}}{∂x^{i}} + \frac{∂g_{ρi}}{∂x^{i}} − \frac{∂g_{ii}}{∂x^{ρ}} \}
= \frac{1}{2}g^{ii} \{\frac{∂g_{ii}}{∂x^{i}} \} ⇒ Γ_{rr}^{r} = \frac{1}{2} \frac{1}{β(r)} \frac{dβ(r)}{dr},
Γ_{θθ}^{r} = \frac{1}{2}g^{rr} \{\frac{∂g_{rθ}}{∂x^{θ}} + \frac{∂g_{rθ}}{∂x^{θ}} − \frac{∂g_{θθ}}{∂x^{r}} \}
= −\frac{1}{2} \frac{1}{β(r)}2r ⟹ Γ_{θθ}^{r} = −\frac{r}{β(r)},

Γ_{ϕϕ}^{r} = −\frac{1}{2}g^{rr} \frac{∂g_{ϕϕ}}{∂r} = −\frac{r sin^{2} θ}{β(r)},
Γ_{θr}^{θ} = \frac{1}{2}g^{θθ} \{\frac{∂g_{θθ}}{∂x^{r}} \} = \frac{1}{r},
Γ_{ϕϕ}^{θ} = \frac{1}{2}g^{θθ} \{−\frac{∂g_{ϕϕ}}{∂x^{θ}} \} = − sin θ cos θ,
Γ_{ϕr}^{ϕ} = \frac{1}{2}g^{ϕϕ} \{\frac{∂g_{ϕϕ}}{∂x^{r}} \} = \frac{1}{2}(−)\frac{1}{r^{2} sin^{2} θ}(−)2r sin^{2} θ = \frac{1}{r},
Γ_{ϕθ}^{ϕ} = \frac{1}{2}g^{ϕϕ} \{\frac{∂g_{ϕϕ}}{∂x^{θ}} \} = \frac{1}{2}(−)\frac{1}{r^{2} sin^{2} θ}(−)r^{2} sin θ cos θ = \frac{cos θ}{sin θ}.

Summarising, we end up with nine non-vanishing components of Γ:

(1)  Γ_{rr}^{r} = \frac{β′}{2β} ,
(2)  Γ_{ϕϕ}^{r} = \frac{−r sin^{2} θ}{β},
(3)  Γ_{rθ}^{θ} = \frac{1}{r}
(4)  Γ_{ϕr}^{ϕ} = \frac{1}{r}
(5)  Γ_{0r}^{0} = \frac{α′}{2α}
(6)  Γ_{θθ}^{r} = −\frac{r}{β}
(7)  Γ_{00}^{r} = \frac{α′}{2β}
(8)  Γ_{ϕϕ}^{θ} = − sin θ cos θ
(9)  Γ_{ϕθ}^{ϕ} = \frac{cos θ}{sin θ},

where the primes on α and β denote derivatives with respect to r,

β′ = \frac{dβ}{dr} , α′ = \frac{dα}{dr}.

The Ricci tensor is, by definition,

R_{μκ} = \frac{∂Γ_{μλ}^{λ}}{∂x^{κ}} − \frac{∂Γ_{μκ}^{λ}}{∂x^{λ}} + Γ_{μλ}^{η}Γ_{κη}^{λ} − Γ_{μκ}^{η}Γ_{λη}^{λ}.

It requires quite a lot of tedious (but simple) algebra to work out the components of the Ricci tensor. I will work out one of these components in full so that you can see what is involved and then state the results for the others:

R_{rr} = \frac{∂Γ_{rλ}^{λ}}{∂x^{r}} − \frac{∂Γ_{rr}^{λ}}{∂x^{λ}} + Γ_{rλ}^{η}Γ_{rη}^{λ} − Γ_{rr}^{η}Γ_{λη}^{λ}.

Term [1] \frac{∂Γ_{rλ}^{λ}}{∂x^{r}} = \frac{∂Γ_{rr}^{r}}{∂r} + \frac{∂Γ_{rθ}^{θ}}{∂r} + \frac{∂Γ_{rϕ}^{ϕ}}{∂r} + \frac{∂Γ_{r0}^{0}}{∂r} ,
Term [2] \frac{∂Γ_{rr}^{λ}}{∂x^{λ}} = \frac{∂Γ_{rr}^{r}}{∂r} ,
Term [3] Γ_{rλ}^{η}Γ_{rη}^{λ} = Γ_{rλ}^{0}Γ_{r0}^{λ} + Γ_{rλ}^{i}Γ_{ri}^{λ}
= (Γ_{r0}^{0})^{2} + Γ_{rλ}^{r}Γ_{rr}^{λ} + Γ_{rλ}^{θ}Γ_{rθ}^{λ} + Γ_{rλ}^{ϕ}Γ_{rϕ}^{λ}
= (Γ_{r0}^{0})^{2} + (Γ_{rr}^{r})^{2} + (Γ_{rθ}^{θ})^{2} + (Γ_{rϕ}^{ϕ})^{2},
Term [4] Γ_{rr}^{η}Γ_{λη}^{λ} = Γ_{rr}^{r}Γ_{λr}^{λ} = Γ_{rr}^{r} [Γ_{rr}^{r} + Γ_{θr}^{θ} + Γ_{ϕr}^{ϕ} + Γ_{0r}^{0}] .

Hence, summing these terms

R_{rr} = \frac{∂Γ_{0r}^{0}}{∂r} + \frac{∂Γ_{θr}^{θ}}{∂r} + \frac{∂Γ_{ϕr}^{ϕ}}{∂r} + (Γ_{0r}^{0})^{2} + (Γ_{rr}^{r})^{2} + (Γ_{rθ}^{θ})^{2} + (Γ_{rϕ}^{ϕ})^{2} − Γ_{rr}^{r} [Γ_{rr}^{r} + Γ_{θr}^{θ} + Γ_{ϕr}^{ϕ} + Γ_{0r}^{0}] ,

and so evaluating this expression using the components of the affine connection that we worked out before

R_{rr} = −\frac{1}{2} \frac{α^{′2}}{α^{2}} + \frac{1}{2} \frac{α′′}{α} − \frac{2}{r^{2}} + \frac{1}{4} \frac{α^{′2}}{α^{2}} + \frac{1}{4} \frac{β^{′2}}{β^{2}} + \frac{1}{r^{2}} + \frac{1}{r^{2}} − \frac{β′}{2β} [\frac{β′}{2β} + \frac{1}{r} + \frac{α′}{2α} + \frac{1}{r}] ,

Hence,

R_{rr} = \frac{1}{2} \frac{α′′}{α} − \frac{α′}{4α} (\frac{β′}{β} + \frac{α′}{α} ) − \frac{1}{r} \frac{β′}{β} ,  (4a)

and, in a similar way, one can show that the rest of the components of the Ricci tensor are,

R_{θθ} = −1 − \frac{r}{2β} (\frac{β′}{β} − \frac{α′}{α} ) + \frac{1}{β} ,  (4b)
R_{ϕϕ} = sin^{2} θ R_{θθ},  (4c)
R_{00} = −\frac{α′′}{2β} + \frac{1}{4} (\frac{α′}{β} ) (\frac{β′}{β} + \frac{α′}{α} ) − \frac{1}{r} (\frac{α′}{β} ) ,  (4d)
R_{μν} = 0, μ ≠ ν.  (4e)

We showed in the last lecture that Einstein’s field equations in vacuum are

R_{μν} = 0.

Hence all of the components of the Ricci tensor (4a) to (4d) must vanish. First, let’s eliminate α′′ from (4a) and (4d),

\frac{R_{rr}}{β} + \frac{R_{00}}{α} = −\frac{α′}{4αβ} (\frac{β′}{β} + \frac{α′}{α} ) + \frac{α′}{4αβ} (\frac{β′}{β} + \frac{α′}{α} ) − \frac{1}{rβ} (\frac{β′}{β} + \frac{α′}{α} ) = 0,

i.e.,

\frac{β′}{β} + \frac{α′}{α} = 0,

which requires that

α(r)β(r) = constant.

Recall that the functions α and β appear in the metric as

ds^{2} = α(r)dt^{2} − β(r)dr^{2} − r^{2}dθ^{2} − r^{2} sin^{2} θdϕ^{2}

and so if we want to match on to the Minkowski metric as r → ∞, the functions α and β must satisfy

\begin{matrix}α(r) → 1 \\ β(r) → 1 \end{matrix} \} as r → ∞, therefore α(r) = 1/β(r).

Now look at the other components:

R_{θθ} = −1 − \frac{r}{2}α (−2\frac{α′}{α} ) + α
= −1 + rα′ + α = 0,  so \frac{d (rα(r))}{dr} = 1,  i.e. rα(r) = r + constant.  (5)

R_{rr} = \frac{1}{2} \frac{α′′}{α} + \frac{1}{r} \frac{α′}{α} = \frac{(rα′′ + 2α′)}{2rα} = \frac{1}{2rα} \frac{dR_{θθ}}{dr} ,

and so setting R_{θθ} = 0 over all space guarantees that R_{rr} = 0. Now, in the weak field limit we know that

α(r) → 1 + \underbrace{2ϕ} .

↗

Newtonian potential

So, a long way from a mass M

α(r) → 1 − \frac{2GM}{r}

hence the constant in (5) must be \underline{−2GM} and so

α(r) = 1 − 2\frac{GM}{r}

Our metric is therefore

ds^{2} = [1 − \frac{2GM}{r} ] dt^{2} − [1 − \frac{2GM}{r} ]^{−1} dr^{2} − r^{2}dθ^{2} − r^{2} sin^{2} θdϕ^{2}.  (6)

This is called the Schwarzchild metric after its discoverer. This metric is important in General Relativity. It is an exact solution of the field equations and represents the metric around a spherical compact object of mass M. By construction, the metric reproduces the Newtonian potential ϕ = −GM/r. However, since the metric (6) is an exact solution of the field equations, it is valid in the strong gravitational field limit GM/r ≫ 1. We can therefore use the metric (6) to describe the orbits of planets around the Sun (a weak field application of GR) or to describe the orbits of particles around a compact object such as a black hole (a strong field application of GR). We will investigate both of these problems in the next few lectures.