13. Three Basic Premises of General Relativity

Spacetime

General relativity postulates that spacetime (the set of all events) is a smooth 4-dimensional Riemannian manifold M, where points are called events, with the properties A1-A3 listed below.

This means that, if we diagonalize the scalar product on the tangent space at any point, we obtain the matrix

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -1

.

The metric is measurable by clocks and rods.

Before stating the next axiom, we recall some definitions.

Examples 13.2
(a) If a particle moves with constant velocity v in some Lorentz frame, then at time t = x⁴ its position is

x = a + vx⁴.

Using the local coordinate x⁴ as a parameter, we obtain a path in M given by

xi(x4)

=

ai + vix4   if i = 1, 2, 3 x4 if i = 4

||(v¹, v², v³, 1)||² = |v|² - c².

It is timelike at sub-light speeds, lightlike at light speed, and spacelike at faster-than-light speeds.

(b) If u is the proper velocity of some particle in locally Minkowskian spacetime, then we saw (normal condition in Section 10) that u, u = -c² = -1 in our units.

Here, a freely falling particle is one that is effected only by gravity, and recall that a timelike geodesic is a geodesic xⁱ(t) with the property that ||dxⁱ/dt||² < 0 in any paramaterization. (This property is independent of the parameterization -- see the exercise set.)

Note Here are some consequences:

1. No physical laws can use the term "straight line," since that concept has no meaning in M; what's straight in the eyes of one chart is curved in the eyes of another. "Geodesic," on the other hand, does make sense, since it is independent of the choice of coordinates.
2. If we can write down physical laws, such as Maxwell's equations, that work in Minkowski space, then those same laws must work in curved space-time, without the addition of any new terms, such as the curvature tensor. In other words, there can be no form of Maxwell's equations for general curved spacetime that involve the curvature tensor.

An example of such a law is the conservation law, .T = 0, which is thus postulated to hold in all frames.

A Consequence of the Axioms: Forces in Almost Flat Space

Suppose now that the metric in our frame is almost Lorentz, with a slight, not necessarily constant, deviation from the Minkowski metric, as follows.

g**

=

1+2 0 0 0 0 1+2 0 0 0 0 1+2 0 0 0 0 -1+2

...   (I)

or

ds² = (1+2)(dx² + dy² + dz²) - (1-2)dt².

Notes

1. We are not in an inertial frame (modulo scaling) since need not be constant, but we are in a frame that is almost inertial.
2. The metric g_** is obtained from the Minkowski g by adding a small multiple of the identity matrix. We shall see that such a metric does arise, to first order of approximation, as a consequence of Einstein's field equations.

d2xi d2

+

ris

dxr d

dxs d

=

0.

Multiplying both sides by m₀² gives

m0

d2(m0xi) d2

+

ris

d(m0xr) d

d(m0xs) d

=

0.

or

m0

=

dPi d

+

risPrPs

=

0

(since Pi = d(m0xi/d))

dPi d

=

Pi,k

dxk d

so that

Pi,k

dm0xk d

+

risPrPs

=

0,

or

Pⁱ_,kP^k + _rⁱ_sP^rP^s = 0     ...    (I)

Now let us do some estimation for slowly-moving particles v << 1 (the speed of light in our units) where we work in a frame where g has the given form.

Question Why don't we work in an inertial frame (the frame of the particle)?

I give up ... Give me an answer.

First, since the frame is almost inertial (Lorentz), we are close to being in SR, so that

P*		m0U*	=	m0[v1,   v2,  v3,  1]/(1-v2/c2)1/2
				[0,  0,  0,   m0]	(since v << 1)

(in other words, the frame is almost comoving) Thus (I) reduces to

Pⁱ_,4m₀ + ₄ⁱ₄ m₀² = 0    ...   (II)

Let us now look at the spatial coordinates, i = 1, 2, 3. By definition,



4i4

=

1 2

gij (g4j,4 + gj4,4 - g44,j).

We now evaluate this at a specific coordinate i = 1, 2 or 3, where we use the definition of the metric g, recalling that g** = (g_**)^-1, and obtain

1 2

(1+2)-1(0 + 0 - 2,i)

1 2

(1-2)(-2,i)

,i.

(Here and in what follows, we are ignoring terms of order O(²).) Substituting this information in (II), and using the fact that

Pi,4

=

x4

=

(movi),

the time-rate of change of momentum, or the "force" as measured in that frame (see the exercise set), we can rewrite (II) as

m0

x4

(movi) - m02,i

=

0,

or

x4

(movi) - m0,i

=

0.

Thinking of x⁴ as time t, and adopting vector notation for three-dimensional objects, we have, in old fashioned 3-vector notation,

t

(mov)

=

m0,

that is

F = m.

This is the Newtonian force experienced by a particle in a force field potential of . (See the exercise set.) In other words, we have found that we can duplicate, to a good approximation, the physical effects of Newton-like gravitational force from a simple distortion of the metric. In other words -- and this is what Einstein realized -- gravity is nothing more than the geometry of spacetime; it is not a mysterious "force" at all.

Exercise Set 13
1. Show that, if xⁱ = xⁱ(t) has the property that ||dxⁱ/dt||² < 0 for some parameter t, then ||dxⁱ/dts|² < 0 for any other parameter s such that ds/dt 0 along the curve. In other words, the property of being timelike does not depend on the choice of paramaterization.

2. What is wrong with the following (slickly worded) argument based on the Strong Equivalence Principle?

I claim that there can be no physical law of the form A = R in curved spacetime, where A is some physical quantity and R is any quantity derived from the curvature tensor. (Since we shall see that Einstein's Field Equations have this form, it would follow from this argument that he was wrong!) Indeed, if the postulated law A = R was true, then in flat spacetime it would reduce to A = 0. But then we have a physical law in SR, which must, by the Strong Equivalence Principle, generalize to A = 0 in curved spacetime as well. Hence the original law A = R was wrong.

3. Gravity and Antigravity Newton's law of gravity says that a particle of mass M exerts a force on another particle of mass m according to the formula

F

=

-

GMmr r3

,

where r = x, y, z, r = |r|, and G is a constant that depends on the units; if the masses M and m are given in kilograms, then G 6.67 10^-11, and the resulting force is measured in newtons.^* (Note that the magnitude of F is proportional to the inverse square of the distance r. The negative sign makes the force an attractive one.) Show by direct calculation that

F = m,

where

=

GM r

.

Hence write down a metric tensor that would result in an inverse square repelling force ("antigravity").

^*A Newton is the force that will cause a 1-kilogram mass to accelerate at 1 m/sec².