□h=-16πT

The question Rev asked in the first post of this thread, a thread that has now become a lesson. I'll start writing it up ASAP, but it'll take a while so don't expect anything untill the morning. In the mean time I'd buff up on some of the things we've discussed. As I said before, that book is a good starting point. Also by the same author, B. Schutz, is a book on differential geometry "Geometrical Methods of Mathematical Physics" which I'm also reading. This guy is amazing, get those! It assumes your calculus is spot on in both single and many dimensions and that you're linear algebra and group theory is as equally good (although there's a brief chapter on some prerequisites at the beginning of the latter of the afore-mentioned books).

Now, if you want me to, I'll explain the field equations and the problem at hand. You're getting these answers from me because I know how difficult it can be learning maths from websites and that you cannot simply magic a book in front of you.

Yes. Just remember that SR uses greek indices that take the values 0, 1, 2, 3 (time, x, y, z respectively).

Yeah, exactly that. So if one wishes to express a vector as a linear sum one would write [math]\vec{V}=V^i\hat{e}_i[/math] rather than [math]\vec{V}=\sum_iV^i\hat{e}_i[/math]

Ok, first the Einstein summation convention. In an expression that has the same index raised and lowered, a summation is assumed over the index. For example if we take the "dot product" of two vectors [math]\vec{A}[/math] and [math]\vec{B}[/math] in Euclidean space [math]\sum_iA^iB_i=A^iB_i[/math], where i denotes a spacial index (as it does in SR) Here the indices are the same and one is raised and one lowered. These are however not summations over the index [math]A^{\alpha}B^{\alpha}[/math] [math]A_{\alpha}B_{\alpha}[/math] [math] A^{\alpha}B_{i}[/math] etc. Vector componants have their index raised and the i-th basis vectors has its index lowered. One-form componants have a lowered index and the i-th basis one-form has its index raised. I shall digress slightly before discussing the metric to describe some aspects of tensor algebra. The type of tensor is usually demonstrated as a 1x2 matrix, the numbers on the top denote the number of covarient components and the number on the bottom denotes the number of contravarient, the number is the rank. A tensor of type (M N) is a linear mapping of M contravariant vectors (one-forms) and N covariant vectors (vectors) into a scalar. The old-fashioned terms contravariant and covariant vectors are so because they transform with or in the same manner as (co) and oppositely (contra) to the vector basis (that's just a usual vector basis, such as [math]\vec{i}, \vec{j}, \vec{k}[/math]) respectively. So a (2 0) tensor (such as the stress-energy tensor) is a linear mapping of 2 one-forms into a scalar. It is very important that you understand that a tensor is a linear mapping. Now for the metric. The metric you will be familiar with, and yet not familiar with due its nature, will be the Euclidean metric. It's signature is 3 (in E³), signature being its trace, and is denoted by the Kronecker Delta, [math]\delta^{\mu}_{\nu}[/math] which if [math]\mu=\nu[/math] is equal to unity (or as a matrix, equal to the identity matrix). The metric is best defined as a symetric (0 2) tensor, i.e. a linear mapping of two vectors into a scalar, and allows one to define a distance. The linear map is defined as the dot product of the two vectors the tensor operates on and the componants of the metric are the value of the dot product of the basis vectors. Let's give an example. In Cartesian 2-space we have as the basis vectors [math]\hat{i}, \hat{j}[/math], then the componants of the metric are [math]\hat{i}\bullet\hat{j}=0, \hat{j}\bullet\hat{i}=0, \hat{i}\bullet\hat{i}=1[/math] and [math]\hat{j}\bullet\hat{j}=1[/math]. And so the dot product of two vectors [math]\vec{A}[/math] and [math]\vec{B}[/math] is simply [math]\sum_iA^iB^i[/math]. Different spaces have different basis vectors and hence different metrics, I mentioned earlier the Schwarzchild metric. In SR the componants of the (Lorentz) metric are [math]\eta_{\alpha\beta}[/math], which is the same as the metric in ordinary Euclidean 4-space other than the 0 (time) componant is -1 rather than 1. Research metrics for a detailed and much better definition. Mine is slightly weak and lacks detail, the price of brevity I suppose.

When you speak of hbar do you actually mean [math]\hbar[/math] (planks constant over [math]2\pi[/math]) or was it [math]\bar{h}[/math], used to represent the field equations in weak form? If the former I'd like to know what the problem was as I've not seen it used in the field equations (although I am new to GR).

I think I see what you're getting at. If we have [math]T^{\mu\nu}_{;\nu}=0[/math] for the fluid then you're asking that in order for energy and momentum to be conserved through a fluid the geodesics are altered in such a manner as to make them of greater length. If so I would have thought the answer would be a yes.

That's the same field equation that I wrote, it's just that I used geometrized units rather than SI, where G=c=1. The equation already given, that involves Ricci scalars/tensors, is simply the Einstein Tensor, but Einstein realised that this particular tensor was so important in GR that he gave it its own name. Also the position and symbols given to the indices are pretty irrelevant, as long as [math]\alpha\beta[/math] is in the same position on both sides of the equality and that the greek indices (which denote 0, 1, 2, 3; or time, x ,y, z) are the same on both sides then it's fine. The special relativistic definition of the stress-energy tensor, taken from "A First Course in General Relativity": [math]\mathbf{T}(\tilde{d}x^{\alpha}, \tilde{d}x^{\beta})=T^{\alpha\beta}[/math] is the [math]\alpha[/math] componant of four-momentum across a surface of constant [math]x^{\beta}[/math] You may wonder where the "energy" part of its name comes from when we are speaking of momentum: in SR the four-momentum vector has its 0 (time) componant as its energy. If you know anything about one-forms ([math]\left( \begin{array}{ccc} 1\\ 0 \\ \end{array} \right)\][/math] tensors or covariant vectors as they used to be called) then you may understand the "across a surface of constant [math]x^{\beta}[/math]", if not then I cannot be bothered to explain. The componants of the tensor are: [math](\rho+p)U^{\alpha}U^{\alpha} + g^{\alpha\beta}p[/math] or in tensor form [math]\mathbf{T}=(\rho+p)\vec{U}\otimes\vec{U} + \mathbf{g}^{-1}p[/math] Where [math]\rho[/math] is the mass density (or energy density, the two being equivalant), [math]U^{\alpha}[/math] is the [math]\alpha[/math] componant of four-velocity and [math]g^{\alpha\beta}[/math] is the [math]\alpha\beta[/math] componant of the inverse metric. The above is an invariant expression, the componants in an MCRF or inertial frame have already been given by Rev. For what each componant actually means physically use Google or find a decent book on it, such as the one I mentioned above.

Well the question it self would imply some points: if the length were between two points of infinite separation the length would be infinite in a space with no or positive-defininte stress-energy tensor, if this were so the question would be trivial. I'm new to here so forgive me if I cause insult, but do you know what the stress energy tensor is and that it is present in the field equations? [math]G^{\alpha\beta}=8 \pi T^{\alpha\beta}[/math] If there were no mass-energy distribution in the space we would have a flat, Euclidean space, yes? By the field equations the distribution causes curvature in the space leading to a metric differing from that of Special Relativity ([math]\eta_{\alpha\beta}[/math]) and giving a different line element. If we use the exterior Schwarzchild metric (which uses a static star of mass M with positive mass-energy distribution) the proper radial length between two points is: [math]ds=\int^{r_2}_{r_1}(1-\tfrac{2M}{r})^{-\tfrac{1}{2}}dr[/math] The integrand of which is always positive-definite (that is unless we're dealing with a Schwarzchild black hole). In flat space the radial length is: [math]ds=\int^{r_2}_{r_1}dr=r_2-r_1[/math]

The sum of its diagonal componants. [math]Tr(A) =\sum A_{ii}[/math]