## Lecture 14: The Einstein Field Equations and Derivation of Newton's Law

14. The Einstein Field Equations and Derivation of Newton's Law

Einstein's field equations show how the sources of gravitational fields alter the metric. They can actually be motivated by Newton's law for gravitational potential , with which we begin this discussion.

First, Newton's law postulates the existence of a certain scalar field , called gravitational potential which exerts a force on a unit mass given by

F =       (classical gravitational field)

Further, satisfies

2 = .() = 4G     ...     (I)
Div(gravitational field) = constant mass density

where is the mass density and G is a constant. (The divergence theorem then gives the more familiar F = = GM/r2 for a spherical source of mass M -- see the exercise set.) In relativity, we need an invariant analogue of (I). First, we generalize the mass density to energy density (recall that energy and mass are interchangeable according to relativity), which in turn is only one of the components of the stress-energy tensor T. Thus we had better use the whole of T.

Question What about the mysterious gravitational potential ?

Answer That is a more subtle issue. Since the second principle of general relativity tells us that particles move along geodesics, we should interpret the gravitational potential as somehow effecting the geodesics. But the most fundamental determinant of geodesics is the underlying metric g. Thus we will generalize to g. In other words, Einstein replaced a mysterious "force" by a purely geometric quantity. Put another way, gravity is nothing but a distortion of the local geometry in space-time. But we are getting ahead of ourselves...

Finally, we generalize the (second order differential) operator to some yet-to-be-determined second order differential operator . This allows us to generalize (I) to

(g**) = kT**,

where k is some constant. In an MCRF, (g) is some linear combination of gab,ij, gab,i and gab, and must also be symmetric (since T is). Examples of such a tensors are the Ricci tensors Rab, gabR, as well as gab. Let us take a linear combination as our candidate:

Rab + gabR + gab = kTab ... (II)

We now apply the conservation laws Tab|b = 0, giving

(Rab + gabR)|b = 0     ...     (a)

since gab|b = 0 already (Exercise Set 8 #4). But in Lecture 9 we also saw that

 (Rab - 12 gabR)|b = 0, ...     (b)

where the term in parentheses is the Einstein tensor Gab. Calculating (a) - (b), using the product rule for differentiation and the fact that gab|b = 0, we find

 ( + 12 )gabR|b = 0,

giving (upon multiplication by g**)

 ( + 12 )R|j = 0,
which surely implies, in general, that must equal - 1/2. Thus, (II) becomes

Gab + gab = kTab.

Finally, the requirement that these equations reduce to Newton's for v/c << 1 tells us that k = 8 (discussed below) so that we have

 Einstein's Field Equations Gab + gab = 8Tab

The constant is called the cosmological constant. Einstein at first put = 0, but later changed his mind when looking at the large scale behavior of the universe. Later still, he changed his mind again, and expressed regret that he had ever come up with it in the first place. The cosmological constant remains a problem child to this day . We shall set it equal to zero in what follows.

Solution of Einstein's Equations for Static Spherically Symmetric Stars

In the case of spherical symmetry, we use polar coordinates (r, , , t) with origin thought of as at the center of the star as our coordinate system (note it is singular there, so in fact this coordinate system does not include the origin) and restrict attention to g of the form

g** =  grr 0 0 grt 0 r2 0 0 0 0 r2sin2 0 grt 0 0 -gtt
,

or

ds2 = 2grt dr dt + grr dr2 + r2 d2 + r2 sin d2 - gt dt2,

where each of the coordinates is a function of r and t only. In other words, at any fixed time t, the surfaces = const, = const and r = const are all orthogonal. (This causes the zeros to be in the positions shown.)

Question Explain why the non-zeros terms have the above form.

Answer For motivation, let us first look at the standard metric on a 2-sphere of radius r: (see Example 5.2(d))

g** =  r2 0 0 r2 sin2
.

If we throw r in as the third coordinate, we could calculate

g** =  1 0 0 0 r2 0 0 r2 sin2
.

Moving into Minkowski space, we have

 ds2 = dx2 + dy2 + dz2 - dt2 = dr2 + r2(d2 + sin2 d2) - dt2,
giving us the metric

Minkowski Space Metrtic in Polar Coordinates

g** =
 1 0 0 0 0 r2 0 0 0 0 r2sin2 0 0 0 0 -1

For the general spherically symmetric stellar medium, we can still define the radial coordinate to make g = r2 (through adjustment by scaling if necessary). Further, we take as the definition of spherical symmetry, that the geometry of the surfaces r = t = const. are spherical, thus foring us to have the central 22 block.

For static spherical symmetry, we also require, among other things, (a) that the geometry be unchanged under time-reversal, and (b) that g be independent of time t. For (a), if we change coordinates using

(r, , , t) ' (r, , , -t),

then the metric remains unchanged; that is, = g. But changing coordinates in this way amounts to multiplying on the left and right (we have an order 2 tensor here) by the change-of-coordinates matrix diag (1, 1, 1, -1), giving

** =  grr 0 0 -grt 0 r2 0 0 0 0 r2sin2 0 -grt 0 0 -gtt
.

Setting = g gives grt = 0. Combining this with (b) results in g of the form

g** =  e2 0 0 0 0 r2 0 0 0 0 r2sin2 0 0 0 0 -e
,

where we have introduced the exponentials to fix the signs, and where = (r), and = (r). Using this version of g, we can calculate the Einstein tensor to be (see the exercise set!)

Einstein Tensor for Static Spherically Symmetric Stars

G** =
 2r 'e-4 - 1 r2 e2 (1-e-2)
0  0 0
0
 e-2[''+(')2 + ' r - '' - 'r ]
0 0
0 0
G

sin2
0
0 0 0
 1 r2 e-2 ddr [r(1-e-2)]

We also need to calculate the stress energy tensor,

Tab = (+p)uaub + pgab.

In the static case, there is assumed to be no flow of star material in our frame, so that u1 = u2 = u3 = 0. Further, the normal condition for four velocity, "u, u' = -1, gives

[0, 0,  0,  u4]  e2 0 0 0 0 r2 0 0 0 0 r2sin2 0 grt 0 0 -e2
 0 0 0 u4
= -1,

whence

u4 = e-,

so that T44 = (+p)e-2 + p(-e-2) (note that we are using g** here). Hence,

T**= ( + p)u*u* + pg**
=
 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (+p)e-2)
+ p  e2 0 0 0 0 r2 0 0 0 0 r2sin2 0 0 0 0 -e2
=
 pe2 0 0 0 0 pr2 0 0 0 0 pr2sin2 0 0 0 0 e-2
.

(a) Equations of Motion Tab|b = 0

To solve these, we first notice that we are not in an inertial frame (the metric g is not nice at the origin; in fact, nothing is even defined there!) so we need the Christoffel symbols, and use

 Tab|b = Tabxb + kabTkb + bbkTak,

where

 hpk = 12 glp gklxh + glhxk - ghkxl .

Now, lots of the terms in Tab|b vanish by symmetry, and the restricted nature of the functions. We shall focus on a = 1, the r-coordinate. We have:

T1b|b = T11|1 + T12|2 + T13|3 + T14|4 ,

and we calculate these terms one-at-a-time.

 a = 1, b = 1: T11|1 = T11x1 + 111T11 + 111T11.

To evaluate this, first look at the term 111:

 111 = 12 gl1(g1l,1 + gl1,1 - g11,l) = 12 g11(g11,1 + g11,1 - g11,1) (because g is diagonal, whence l = 1) = 12 g11(g11,1) = 12 e-2e2.2'(r) = '(r)

Hence,

 T11|1 = dpdr e-2 + (- 2p'(r)e-2) + 2'(r)pe-2 = dpdr e-2.

Now for the next term:

a = 1, b = 2: T12|2 =  T12x2 + 212T22 + 221T11
=  0 + 12 gl1(g2l,2+gl2,2-g22,l)T22 + 12 gl2(g1l,2+gl2,1-g21,l)T11
=  12 g11(-g22,1) T22 + 12 g22(g22,1)T11
=  12 e-2(-2r) pr2 + 12 1r2sin 2rsinpe-2
=  0

Similarly (exercise set)

T13|3 = 0.

Finally,

a = 1, b = 4: T14|4 =  T14x4 + 414T44 + 441T11
=  12 g11(-g44,1) T44+ g44(g44,1)T11
=  12 e-2(2'(r)e2)e-2 12 (-e-2)(-2'(r)e2)pe-2
=  e-2'(r)[ + p].

Hence, the conservation equation becomes

T1a|a = 0  dpdr + ddr ( + p) e-2 = 0
 dpdr = -( + p) ddr

This gives the pressure gradient required to keep the plasma static in a star.

Note In classical mechanics, the term on the right has rather than +p. Thus, the pressure gradient is larger in relativistic theory than in classical theory. This increased pressure gradient corresponds to greater values for p, and hence bigger values for all the components of T. By Einstein's field equations, this now leads to even greater values of (manifested as gravitational force) thereby causing even larger values of the pressure gradient. If p is large to begin with (big stars) this vicious cycle diverges, ending in the gravitational collapse of a star, leading to neutron stars or, in extreme cases, black holes. You can go directly to Lecture 15 on stellar collapse to find out more.

(b) Einstein Field Equations Gab = 8Tab

Looking at the (4, 4) component first, and substituting from the expressions for G and T, we find

 1r2 e-2 ddr [r(1-e-2] = 8e-2.

If we define

 12 r(1-e-2) = m(r),

then the equation becomes

 1r2 e-2 dm(r)dr = 4e-2,

or

 dm(r)dr = 4r2 ...   (I)

This looks like an equation for classical mass, since classically,

 M(R) = R0 4r2(r)  dr ,

where the integrand is the mass of a shell whose thickness is dr. Thus,

 dM(R)dr = 42(r).

Here, is energy density, and by our choice of units, energy is equal to rest mass, so we interpret m(r) as the total mass of the star enclosed by a sphere of radius r.

Now look at the (1, 1) component:

 2r 'e-4 - 1r2 (1-e-2) = 8pe-2
 2r ' - e2r2 (1-e-2) = 8pe2
 2r' - e2 (1-e-2) = 8r2pe2
 ' = e2 (1-e-2) + 8r2p2r

In the expression for m, solve for e2 to get

 e2 = 11-2m/r ,

giving

 ddr = 8r2p + 2m/r2r(1-2m/r)

or

 ddr = 4r3p + m r(r-2m) ...   (II)

It can be checked using the Bianchi identities that we in fact get no additional information from the (2,2) and (3,3) components, so we ignore them.

Consequences of the Field Equations: Outside the Star

Outside the star we take p = 0, and m(r) = M, the total stellar mass, getting

 (I): dmdr = 0 (nothing new, since m = M = constant) (II): ddr = Mr(r-2M) ,

which is a separable first order differential equation with solution

 e2 = 1 - 2Mr ,

if we impose the boundary condition 0 as r+. (See the exercise Set).

Recalling from the definition of m that

 e2 = 11-2M/r ,

we can now express the metric outside a star as follows:

Schwarzschild Metric
g** =
 11-2M/r 0 0 0 0 r2 0 0 0 0 r2sin2 0 0 0 0 -(1-2M/r)

In the exercise set, you will see how this leads to Newton's Law of Gravity.

Exercise Set 13

1. Use 2 = 4G and the divergence theorem to deduce Newton's law = GM/r2 for a spherical mass of uniform density .

2. Calculate the Einstein tensor for the metric g = diag(e2, r2, r2sin, -e2), and verify that it agrees with that in the notes.

3. Referring to the notes above, show that T13|3 = 0.

4. Show that Ti4|4 = 0 for i = 2, 3, 4.

5. If we impose the condition that, far from the star, spacetime is flat, show that this is equivalent to saying that r+(r) = r+(r) = 0. Hence obtain the formula e2 = 1 -2M/r.

6. A Derivation of Newton's Law of Gravity
(a) Show that, at a large distance R from a static stable star, the Schwarzschild metric can be approximated as

g**  1 +2M/R 0 0 0 0 R2 0 0 0 0 R2sin2 0 0 0 0 -(1-2M/R)

(b) (Schutz, p. 272 #9) Define a new coordinate by R = (1+M/)2, and deduce that, in terms of the new coordinates (ignoring terms of order 1/R2)

g**  1 + 2M/ 0 0 0 0 2(1+2M/)2 0 0 0 0 2(1+2M/)2sin2 0 0 0 0 -(1-2M/)

(c) Now convert to Cartesian coordinates, (x, y, z, t) to obtain

g**  1 +2M/ 0 0 0 0 1+2M/ 0 0 0 0 1+2M/ 0 0 0 0 -(1-2M/)

(d) Now refer to the last formula in Lecture 13, and obtain Newton's Law of Gravity. To how many kilograms does one unit of M correspond?

Last Updated: January, 2002