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13. Three Basic Premises of General Relativity
Spacetime
General relativity postulates that spacetime (the set of all events) is a smooth 4dimensional Riemannian manifold M, where points are called events, with the properties A1A3 listed below.
A1. Locally, M is Minkowski spacetime (so that special relativity holds locally). 
This means that, if we diagonalize the scalar product on the tangent space at any point, we obtain the matrix

. 
The metric is measurable by clocks and rods.
Before stating the next axiom, we recall some definitions.
Definitions 13.1 Let M satisfy axiom A1. If V^{i} is a contravariant vector at a point in M, define
(Note that we are not defining V^{i} here.) We say the vector V^{i} is
lightlike if V^{i}^{2}= 0, and spacelike if V^{i}^{2}> 0, 
Examples 13.2
(a) If a particle moves with constant velocity v in some Lorentz frame, then at time t = x^{4} its position is
Using the local coordinate x^{4} as a parameter, we obtain a path in M given by
x^{i}(x^{4})  = 

It is timelike at sublight speeds, lightlike at light speed, and spacelike at fasterthanlight 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 = c^{2} = 1 in our units.
A2. Freely falling particles move on timelike geodesics of M. 
Here, a freely falling particle is one that is effected only by gravity, and recall that a timelike geodesic is a geodesic x^{i}(t) with the property that dx^{i}/dt^{2} < 0 in any paramaterization. (This property is independent of the parameterization  see the exercise set.)
A3 (Strong Equivalence Principle) All physical laws that hold in flat Minkowski space (ie. "special relativity") are expressible in terms of vectors and tensors, and are meaningful in the manifold M, continue to hold in every frame (provided we replace derivatives by covariant derivatives). 
Note Here are some consequences:
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**  = 

... (I) 
or
Notes
d^{2} 
+  _{r}^{i}_{s}  d 
d 
=  0. 
Multiplying both sides by m_{0}^{2} gives
m_{0}  d^{2} 
+  _{r}^{i}_{s}  d 
d 
=  0. 
or
m_{0}  =  d 
+  _{r}^{i}_{s}P^{r}P^{s}  =  0  (since P^{i} = d(m_{0}x^{i}/d)) 
d 
=  P^{i}_{,k}  d 
so that
P^{i}_{,k}  d 
+  _{r}^{i}_{s}P^{r}P^{s}  =  0, 
or
Now let us do some estimation for slowlymoving 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*  m_{0}U*  =  m_{0}[v^{1}, v^{2}, v^{3}, 1]/(1v^{2}/c^{2})^{1/2}  
[0, 0, 0, m_{0}]  (since v << 1) 
(in other words, the frame is almost comoving) Thus (I) reduces to
Let us now look at the spatial coordinates, i = 1, 2, 3. By definition,
_{4}^{i}_{4}  =  1 2 
g^{ij} (g_{4j,4} + g_{j4,4}  g_{44,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 
(12)(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
P^{i}_{,4}  =  x^{4} 
=  (m_{o}v^{i}), 
the timerate of change of momentum, or the "force" as measured in that frame (see the exercise set), we can rewrite (II) as
m_{0}  x^{4} 
(m_{o}v^{i})  m_{0}^{2}_{,i}  =  0, 
or
x^{4} 
(m_{o}v^{i})  m_{0}_{,i}  =  0. 
Thinking of x^{4} as time t, and adopting vector notation for threedimensional objects, we have, in old fashioned 3vector notation,
t 
(m_{o}v)  =  m_{0}, 
that is
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 Newtonlike 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 x^{i} = x^{i}(t) has the property that dx^{i}/dt^{2} < 0 for some parameter t, then dx^{i}/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  =    r^{3} 
, 
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
where
=  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 1kilogram mass to accelerate at 1 m/sec^{2}.
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