□h=-16πT

Hey people. Great forum, way better than the science forum on totse.com. Love the latex by the way, it's a much appreciated luxury. My favourite branches of maths and physics are geometry (differential geometry, topology etc.) and general/special relativity. Still in secondary school so any chance to discuss science or maths with other people is great.

[math]\sum ^{\infty}_{n=0}\frac{1}{2}\left( \frac{1}{2} \right) ^n=1[/math] 'Tis what the calculus says.

Righto. Also, Johnny5, the components of the SE tensor in the MCRF as previously given are only for those systems in a Minkowski space-time (the flat space of SR). The components of the inverse metric differ for a general space-time. For example the line element in the Schwarzshild metric is (I use this metric again because it's easy to remember) [math] ds^2=-\left( 1-\frac{2M}{r}\right) dt^2+\left( 1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\theta ^2+r^2\sin ^2\theta d\phi ^2 [/math] One usually factors out the r² in the last two terms (as they don't depend on the nature of the space) and calls the remainder [math]d\Omega ^2[/math], i.e. [math]d\Omega ^2=d\theta ^2+\sin ^2\theta d\phi ^2[/math]. The terms before each infinitesimal change on the right hand side are the components of the metric, I couldn't be doing with the hassel of putting them into a matrix. [math] ds^2=-\left( 1-\frac{2M}{r}\right) dt^2+\left( 1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\Omega ^2 [/math] Just thought I'd give you an example of one solution to the field equations. As an aside, the metric above is the only asymptotically flat solution to E's equations (i.e. is flat at r=infinity). If you want to know why, search for Buchdal's theorem as I don't know the proof myself, just the theorem.

Maybe, would proove useful in a unified field theory. I don't think we have any mathematical or experimental evidence at this moment to be able to show this though. Although gravitational waves being the wave part of the duality of light smacks of ether to me. How's it going with your self education by the way?

It is space-time that waves. Ever heard of gravitational waves? My name is the field equation in weak form. The square is the d'Alembertian, or wave operator, and the h is part of the following expression for the metric of a slightly curved space time [math]g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}[/math] Where eta is the Lorentz metric of SR and h is a number that maybe dependant upon position, time etc. The idea of this equation is to demonstrate the small "deviation" from a flat space time (described by the Lorentz metric) by an amount h. So, by the equation I have as my name, the h is a wave and so space-time waves. I tried to get that down ASAP and so the discussion has suffered for brevity. The result is similar for strong fields, but I haven't educated myself in this yet. Remember E=mc²? It is called mass-energy density because the two are equivalent and in SR the units of energy are the same as those of mass because (in normalised units) 1 second=c metres. Before you ask, natural units are those used in SR, where c=1, whereas geometrized units are for GR because in this theory G is going to come up (kg=G/c² metres). Photons don't warp ST as they have no SE tensor (no mass, not a fluid etc.). And of course they don't carry the gravitational force, unless they're the same thing as a graviton...? Yes, volume being frame dependant is a result of SR. In the inertial frame of a fluid element moving in the x direction the volume is [math]\Delta x\Delta y\Delta z[/math]. In the frame of another observer O the length in which the fluid element moves, the x direction, is contracted and so (by the length contraction formula) the volume element of the moving fluid in O's frame is [math]\Delta x\Delta y\Delta z\sqrt ( 1-v^2)[/math] You're really going to have to get used to natural and geometrized units.

That's only in the MCRF (momentarilly comoving reference frame, i.e. a frame that is inertial at t=const. and changes the next, due to acceleration etc.) or inertial frame. The trace in any frame is different.

Could you not take the SE tensor in invariant matrix form, diagonalise it and then see what you get? The components in any frame being: [math](p+\rho)U^{\alpha}U^{\alpha}+g^{\alpha\beta}p[/math] I can't remember the diagonalisation procedure right now. I'll grab my book, flick through the chapter on matrices and give it a go a bit later.

It really doesn't matter, actually. They are what are known as "dummy indices": they all take the values of 0, 1, 2, 3 in some particlular frame, so which letter you use is pretty irrelevant. As long as you use any lower case greek letter it's ok. Although alpha, beta, mu, nu and sigma are usually about as many indices as one may find in a single equation. Was my short "proof" that +ve [math]\rho[/math] gives a +ve-deff. trace adequate, Rev?

I knew you or someone would ask, I couldn't remember at the time. I believe (although my lack of education in QM is great) that it gives the probability of finding a particle in some state [math]\Psi[/math].

Four-velocity. Rev, have you read my short discussion of the field equations for Johnny? Would like to know how accurate and coherant it is, and I doubt Johnny can really be a detractor of my accuracy due to his lack of knowledge in the field.

In the MCRF we have the trace of the SE tensor as follows [math]\rho+p_x+p_y+p_z[/math] You can't have negative pressure in this context as each pressure acts in three different directions and so none of the components are negative. We've already established that the mass-energy distribution is positive, so all terms are positive and so is the trace. I'm no mathematician myself and all my knowledge of this sort of thing is based on books and such. I'm 15 and still in secondary school, I just love maths and physics. How about you Rev? As for the notation, Johnny, I like using lowered indices for covariant rank and visa versa for contravariant, it being the notation used when I taught myself. It really doesn't matter, as long as the author specifies his preferance you'll be fine. In cases like the field equations (where only tensor components are mentioned) you can simply change the indices by a process known as "index raising and lowering", which bares very little on anything until you actually compute the components and you have to use inverse metric components etc. All my posts will use raised for contravariant and visa versa, there could be a different notation for those across the pond. Negative mass distribution...er...

The field equations are (in geometrized units) [math]G^{\alpha\beta}=8 \pi T^{\alpha\beta}[/math] Where G is the Einstein tensor defined as: [math] R^{\mu\nu} - \frac{1}{2} g^{\mu\nu}R [/math] The Ricci tensor has the following definition [math]R_{\alpha\beta}=R_{\alpha\mu\beta} ^{\mu}=R_{\beta\alpha}[/math] It is the contraction of the Riemann Curvature Tensor [math]R_{\alpha\nu\beta} ^{\mu}[/math] on the first and third indices (the sum basically, the first and third index being the same). The Ricci Scalar is another contraction on the Riemann tensor, as follows [math]R=g^{\mu\nu}R_{\mu\nu}[/math] The Riemann curvature tensor (which gives a mathematical definition of the curvature of a manifold) is, in terms of metric components, [math]R^{\alpha}_{\beta\mu\nu}=\tfrac{1}{2} g^{\alpha\sigma}\left (g_{\sigma\nu,\beta\mu}- g_{\sigma\mu,\beta\nu}+ g_{\beta\mu,\sigma\nu}- g_{\beta\nu,\sigma\mu}\right)[/math] Here a comma in the index denotes partial differentiation with respect to the subsequent indices. The above definition is a second order differential equation in [math]g_{\alpha\beta}[/math], a fact we shall use later in the construction of the field equations. In addition, raised indices on the metric are the components of the inverse metric and, conversely, lowered indices are the components of the usual metric. It would be a good idea for you to study and research the interpretations and derivations of the above definitions, rather than skimming over them. Having decided upon a description of gravity based on the idea of the curvature of a manifold with a metric (a Riemannian manifold), Einstein set out to devise an equation that could describe his idea in mathematical form. First, we give the Newtonian field equation a mention, in which the mass density is the source of the field, [math]\nabla^2\phi=4\pi\rho[/math], where [math]\phi[/math] is the Newtonian gravitational potential and [math]\rho[/math] is the mass density. The above is a second order differential equation in [math]\phi[/math]. We aim to find a field equation that is analogous to the Newtonian field equation and that, in the “Newtonian Limit”, reduces to Newton’s equation. Before we can formulate our relativistic field equation, we must discuss the generation of the field, which was the mass density in Newtonian gravitation. However mass density (or energy density) is a frame dependant quantity and other observers will measure the source’s energy density to be different, this would not give a very satisfactory or invariant equation. Here one needs to have some education in the stress-energy tensor. The energy density is the 00 component ([math]T^{00}[/math]) of this tensor, but , as we said before, if we used this as the source of the field we would be saying that one class of observers is preferred over others (namely those that for whom [math]\rho[/math] is the observed mass density). To avoid this variance, this coordinate and frame dependency, we can use the whole of the stress energy tensor (tensors being coordinate and frame invariant entities). Therefore, we can come to some conclusions as to the form of our field equation: it must involve some second order operator on the metric; reduce to Newton’s field equations in non-relativistic cases; be invariant; and include the stress-energy tensor. If we write it in a manner analogous to the field equation above [math]\mathbf{O}(\mathbf{g})=k\mathbf{T}[/math] O is our second order differential operator acting on g (the metric, which we have decided is the generalisation of [math]\phi[/math]), k is a constant and T is the SE tensor. We also have the requisite that O be a tensor of covariant rank 2, because T is a tensor of this type and, of course, we must have this requirement for the equality to be true. In other words [math]{O^{\alpha\beta}}[/math] must be the components of a (2 0) tensor and must be combinations of [math] g_{\beta\mu,\sigma\nu}[/math], [math] g_{\beta\mu,\sigma}[/math] and [math]g_{\beta\mu}[/math]. It is clear that the Ricci tensor satisfies these conditions (as it is a contraction on the Riemann tensor, a tensor that is second order in [math] g_{\alpha\beta}[/math]) and so any tensor of the form [math]O^{\alpha\beta}=R^{\alpha\beta}+\mu g^{\alpha\beta}R+\Lambda g^{\alpha\beta}[/math] will suffice. R is the Ricci scalar, [math]R^{\alpha\beta}[/math] the Ricci tensor, [math]g^{\alpha\beta}[/math] the inverse metric and mu and lambda are constants. To determine mu and lambda, we use a property of the SE tensor: local conservation of energy and momentum. Mathematically this conservation law takes the form of [math]T^{\alpha\beta}_{;\beta}=0[/math] The conservation law usually takes the form of a partial derivative and uses a comma in place of the semi-colon, but here we use the “Strong Equivalence Principle” or “comma goes to semi-colon law”. The semi-colon denotes covariant differentiation; similar to the partial derivative, the covariant derivative is more general and, in some sense, extends the PD to curved and non-Cartesian surfaces (you may not have seen it before because the covariant derivative in Cartesian coordinates is the same as the partial derivative, as the Christoffel symbols vanish). So if we apply the covariant derivative to both sides of our, as of yet incomplete, field equations (in component form) we get [math](O^{\alpha\beta})_{;\beta}=(R^{\alpha\beta}+\mu g^{\alpha\beta}R+\Gamma g^{\alpha\beta})_{;\beta}= kT^{\alpha\beta}_{;\beta}=0 [/math] So the components of O must satisfy [math](R^{\alpha\beta}+\mu g^{\alpha\beta}R+\Gamma g^{\alpha\beta})_{;\beta}=0[/math] If we set gamma to zero and mu to -1/2, the term inside parentheses is the Einstein tensor, defined above, and the covariant derivative of the Einstein tensor is zero (a property that is derived from the Bianchi identities) and so we now have the form and name of our differential operator. Therefore, our field equations take the form [math]G^{\alpha\beta}=kT^{\alpha\beta}[/math] We find that, in order that the equations reduce to that of Newton’s, k must be equal to [math]8\pi[/math] and so we complete the field equation [math]G^{\alpha\beta}=8 \pi T^{\alpha\beta}[/math] Therefore, when you speak of “solving for the metric” we are solving the second order differential equation. Some solutions include the Schwarzschild metric for spherically symmetric, static (non-rotating) stars and the Kerr metric for stationary (rotating), axially symmetric stars. Although the derivation given is far from rigorous or elegant, its emphasis is on the qualitative reasoning behind it and gives a much better understanding to those new to GR. There exists a more elegant derivation of the field equation, discovered by Hilbert, but I have not looked into it. There have been many other field equations developed since 1915, but Einstein’s is by far the most successful. There, done, didn’t take me too long. If you there any errors or problems anyone has, tell me.

If by that you mean "are all events simultaneous?", then the answer is no according to SR: time is frame-dependant. Oh, one thing you said earlier that I didn't see, this is not the same stress-energy as the Maxwell one. This tensor encorporates relativistic effects, such as momentum being frame dependant and, of course, uses four-vectors. The way I have learnt is that the notation for tensor type has the covariant rank on the top. It doesn't matter all too much as long as the notation is explained.

"Manifold" is a mathematically precise substitution for the word "space". A manifold is an (intrinsically) N dimensional surface that is differentiable and continuous everywhere. Intrinsically means the geometry of the surface with regard to its self, where as extrinsic geometry is the geometry when the surface is regarded as existing in a manifold of higher dimension. For example A 2D ant on a cylinder regards the surface of the cylinder to be a flat 2D surface, whilst a 3D observer observing the cylinder says it is a curved 3D shape.

That definition is true only in Euclidean space, because of the Pythagorean theorem (which isn't globally valid on a curved manifold).

The time basis vector is just another orthonormal basis vector, as the others are, because in SR we treat time as another dimension. It just differs in that its dot product with itself is -1 rather than 1. Read up on four-vectors. To understand the definitions I give whilst explaining the field equations you'll need to understand transformation of componants etc. and index raising and lowering.