R mu nu - 1 over 2g mu nuR plus g mu nu Lambda equals 8 pi G over c4 T mu nu?

Answer 1

Short Answer...

It's an assumption. Check out P. Dirac's book.

Long Answer...

First The Pure Mathematics

Let us denote the Einstein tensor to be constructed from the Ricci curvature tensor and the metric tensor as follows:

(0)... Gμν := Rμν − ½ gμν R

We can fairly easily demonstrate that the follow covariant derivatives are 0.

(1)... gμν:σ = 0

(2)... Gμν:ν = 0

That is, the metric tensor is a "constant" w.r.t. covariant derivatives, and the Einstein tensor is non divergent everywhere.

All of this is done with absolutely NO physics in mind. It's all derived purely mathematically, using differential geometry considering certain manifolds (spacetimes).

Second The Physics

From the mathematics, we note that the "velocity" vector, υμ, of a particle moving through spacetime undergoes no"acceleration" iff it follows the path of a geodesic. This is one of the points (in learning, not in spacetime) where we start to leave the differential geometry, and start incorportaing physical notions. We stop doing math, and start modelling nature with the mathematical tools we've got so far.

How? Well, acceleration brings to mind forces. And forces, if we're only considering gravity, brings to mind mass, velocity, and momentum. We start by consider the scalar field, density, and the velocity vector fields.

We need some expressions to deal with the motion of matter. We only care about momentum. So we want to know how packages of momentum are being transferred through spacetime.

Define:

(3)... T(x)μν = the amount of ν-"momentum"

leaving a surface element (located at x) pointing in the μ-direction

per unit area per unit time.

Think. Take a point x in spacetime. Draw a box around x on the spacetime. Consider the surface element of the box, pointing in the μ-direction, and you notice that the amount of mass leaving the box in that direction, and hence the amount of ν-"momentum" leaving the box through that surface are:

(3a)... dm = ρ υμ . dA dt

(3b)... d ρυν = dm . υν= ρ υμ υν . dA dt

So (3a) & (3b) imply:

(3)... Tμν = ρ υμ υν

This is what we want. We want to describe the evolution of Tμν on the spacetime. If we know this, we'll know how matter is moving about in the universe. Where things are going, etc..

Physical Modelling Assumptions

Physics assumption: Momentum is conserved at all points in spacetime. Thus, Tμν, must be divergent-free.

(Ass 0)... Tμν:ν = 0

We can swallow this assumption. It's perfectly reasonable. (However there's more to think about here, because we're talking about divergence in spacetime, not merely space.)

Now here's where Herr Einstein comes in. (Actually most people don't know by David Hilbert derived this too, independently.) Remember (1) & (2), i.e. that the Einstein and metric tensors are non-divergent too? So obviously any linear combination of them are non-divergent. So Herr Einstein chose:

(Ass 1)... Gμν + Λ gμν = Tμν

or perhaps:

Tμν = Gμν + Λ gμν

For some real, Λ. This is called "the equations of motion" (because there's an equation for each pair {μ,ν}). But I will refer to it as a single equation, as though it were T=G+Λg, where the components of the resulting tensors agree.

Why the Assumption?

Now let's discuss this a bit. I'm sorry if you knew everything above, but it at least helps me to have a bit of ground to work on to explain things.

First I have the option to just go with "Einstein assumed that's all". And I would kind of agree. Though that would be naive. I'm a bit peeved that the culmination of my learning so far in differential geometry has ended in a mere assumption, with only minor mathematical justification. However I want to know why Einstein thought this. So let's discuss this point.

Now, I stated (Ass 1) in two ways, because, although logically equivalent, they in no way are the same expression. One has Tμν as the subject, and the other has a mixture of terms. (People can really frustrate you when they come to writing mathematics. Somehow they think they can drop all their training in language and speak like a two-year old, with fragmented sentences, and grammar that barely reflects their thought processes (or even a logical thought process).) I would say that Tμν is the subject. That is, I would say that Einstein tried to find out what was happening to the momentum, by considering the effect of the geometry on matter.

The next reason for deriving it this way seems to come from Sir Isaac newton. Newton derived his results about gravitation with the tools he had present. And Einstein's, equation are an extension of Newton's. I would go so far as to say that if it weren't for a Newtonian formulation of gravity, then there wouldn't be a leap to the Einstein equations.

I should write more here, but I'm running out of time. Check out P. Dirac's book, for a concise discussion.

Remarks

I need to expand on this, but I don't have time. I want to leave you with one more thing. Einstein's equation are still nevertheless arbitrary. Mere assumptions. And this is clearly illustrated by the fact that there are numerous other formulations for the equations of motion than (Ass 1). However, Einstein's has proven to be the most accurate whilst remaining simple enough to handle computationally. Some people say it's very elegant. My thought's are always of ESA, NASA, etc.. Einstein's equation, just like Newton's is after all just an approximation. But it must be a darn good approximation, for the space agencies to use them!

Finally, I wouldn't be surprised if it will be replaced by something else, as Newton's was by Einsteins. But I would be surprised if they are not an extension of Einstein's.