□h=-16πT

I can see this simple thing getting in the way here. So lets forget about it.

Well if you can unify the two or come up with a theory that describes gravity as some form of electromagnetism then yes. Why do you think this? Is it just some wild assertion?

They're just units! Physicists change them because they're lazy and one can easily change the units back into SI. If you'd rather not convert them because it seems convoluted then so be it. To me it seems perfectly rational.

The mass term is in the SE tensor, and the extra 1/c² bit comes from other units that need changing back into SI. Ok, geometrized units. c=3x10^8 m/s if we take one second to be equal to 3x10^8 metres then we can normalise c and make it dimensionless. These are just units we're playing with, it doesn't matter if we start changing them, as long as we change them back when we compare to experiment etc. and besides we're taking time to be a dimension so why not give it dimensions of length...? Next: normalising G G=6x10^-11m³/kg/s² in SI If we slap in s=c m we have G=6x10^-11m/kg/c² (the metres cancel) So if we make one Kg equal to G/c² metres we can normalise G and make it dimensionless as well. Thats all ther is to it. Don't start debating the validity of this, they're just units and can easily be converted back into SI. It makes sense to have such fundamental constants unified and at the end of the day it's a labour saving device not a method of confusing students. To your last question: no there isn't. Just accept this principle and read up on the Michelson-Morely experiment. Whether you believe it or not is irrelevant if you want to learn a theory in which it is true.

I'll be leaving them out. If you see the mention of mass anywhere in any equations I use, just place a G/c² in front of it. In the more complex cases you'll have to convert the units back to SI for yourself or just take it as it is (it changes nothing) because I'm lazy; if there's a method for making my work easier I'll use it. In deriving the Riemann tensor there'll be no need for the use of geometrized units, as it's simply pure maths and no physics.

I knew you'd say something in reply to my mention of the speed of light being constant. The very thing you're learning here is based upon this principle, so if you're going to reject it you may as well stop learning SR/GR right now. Why do you believe simultanaety is absolute: that all observers observe the same event at the same time? Don't you dare give it up though, I've spent ages teaching you this! If you do I'll punch you. Thanks for complementing my teaching ability, it's not as good as you say in person sadly .

The Riemann curvature tensor, it's what I said I'd explain and what I've started to discuss above. It'll be finished by the weekend. Physicists use geometrized units in GR because it saves their wrists a little bit of strain from writing the same two constants out over and over. The physical interpretation of equations is not just in the units, if one becomes accustomed to geometrized form then it's just as simple as SI. The important point about the field equations is that space time is warped by a body described dynamically by the SE tensor, not what units we're using. They're called geometrized units because dealing with the geometry of space-time etc. Just as we may call the units of SR, where just c=1, "natural" because it seems natural to have such a fundamental constant, of which the entire theory is based on, just blend into the background, so to speak.

You can leave out your terms with c and G in them if you like. Geometrized units are of no difficulty, and they don't alter the interpretation of an expression at all (other than the dimensions, obviously). I'd also ignore the field equations for now until you know about the Riemann tensor, Christoffel symbols (of the second kind), covariant derivative etc. Also it's the speed of light in a vacuum, theres is no need to define a frame to which the light is travelling relative to. This is because the speed of light is constant in all inertial frames.

I turned the comlumn matrix 90 degrees anti-clockwise so that I wouldn't have to bother with letexing the maths for it, so one-forms are on the left and vectors on the right. A letter with a tilde over it denotes a one-form and so omega alpha is the alpha-th one form. However, I have been slightly sloppy with my notation as omega is usually reserved for basis one-forms (just as e is for basis vectors), a notation which I shall use in my discussion of the Riemann tensor. In the example of linearity above just take the omega to be any one-form rather than part of a basis. Yes, tensors are linear operators.

Yes, I was wondering why you have called the stress-energy tensor a pseudo-tensor.

This is what I have so far, Johnny5. I'll work on the rest and you can read through what I have at the moment. Throughout this discussion raised indices denote covariant components; lowered contravariant; and [math]g_{\alpha\beta}[/math] and [math]g^{\alpha\beta}[/math] are the components of the metric and inverse metric respectively. A comma in the index denotes differentiation with respect to the subsequent indices and a semi-colon followed by an index is the something-th component of the covariant derivative. I shall first set down some properties that will prove useful in the construction of the Riemann tensor. The first being that at a point P of any Riemannian manifold [math]g_{\alpha\beta}(P)=\eta_{\alpha\beta}[/math] (1) where [math]\eta[/math] is the metric of SR (obviously we don't have the SR metric in discussions of general manifolds but we still have a flat metric), i.e. that at a point on any Riemannian manifold the space is flat (for a proof see the local flatness theorem). Since at any point the space is flat, we have [math]g_{\alpha\beta}_{,\mu}(P)=0[/math] and [math]g_{\alpha\beta}_{,\mu\nu}(P)\neq 0[/math] for any index. The first equality is obvious from (1). The second is such for curved manifolds, as the derivatives change about a point P and so a second derivative exists in general. If you have looked up the Christoffel symbols, a definition you may have found is that they are the coefficients of the basis vectors upon differentiation of the basis, i.e. [math]\frac{\partial\vec{e_{\alpha}}}{\partial x^{\beta}}=\Gamma^{\mu}_{\alpha\beta}\vec{e_{\mu}}[/math] In terms of metric components the symbols are [math] \Gamma^{\gamma}_{\beta\mu}=\tfrac{1}{2}g^{\alpha \gamma}(g_{\alpha\beta,\mu}+ g_{\alpha\mu,\beta}- g_{\beta\mu,\alpha})[/math] (2) You should also have accustomed yourself with the manipulation of indices by now, as proficiency in this will prove useful (such a dummy, free and changing of indices).

If T is a (1 0) tensor then (the sum of two one-forms being a one-form) [math]\mathbf{T}(a\tilde{\omega^{\alpha}}+b\tilde{\omega^{\beta}})=a\mathbf{T}(\tilde{\omega^{\alpha}})+b \mathbf{T} (\tilde{\omega^{\beta}})[/math] This is very important. Here I've just used the example of a (1 0) tensor for simplicity, but it's the same for any (N M) tensor. Oh, when one writes the arguements of a general (N M) tensor in a notation similar to the above, the vectors go on the left of the brackets and the one-forms on the right and the two are separated by a semi-colon. For example a (1 1) tensor [math]\mathbf{T}(\vec{V^{\alpha}}; \tilde{\omega^{\beta}})[/math]

You also must be aware that a tensor is linear in its arguements. I will set about writing something on the curvature tensor on thursday or later this evening, the intermediate time being occupied with an art exam I'm doing (God only knows why I chose to do an art GCSE )

Could you give me the definition of a tensor? I want to know if you actually know what one is.

To understand the Ricci tensor you really need to understand the Riemann curvature tensor, geodesics, parrallel transport etc. If I get time in the next week or so I'll write something up on it for you. It's nothing of great difficulty, but it's incredibly elegant. I may add something on Geodesic deviation as well. Oh, read up on covariant differentiation and the Christoffel symbols. The column matrix one-form thing is only notation for the type of tensor. If one were to represent a one-form as a matrix it would be a row matrix. The notation for a general tensor with N one-forms and M vectors as its arguements is [math] \left ( \begin{array}{c} N\\ M\\ \end{array} \right ) [/math] I think before you go onto general tensors you need to educate yourself fully in one-forms and their geometrical interpretation and vector spaces (as you're doing now). As I mentioned above, look up the covariant derivative (this is a one-form).

And...?

Yes. I'm still a novice in GR' date=' so I don't know how to answer your question on expansion of ST. As to the other, no the Euclidean metric does not give an accurate description of space-time, obviously as it contains no negative time component, and hence fails to predict such phenomenon as time dilation, length contraction etc. which [b']do[/b] occur in "reality". If you didn't know, the space-time of SR is called Minkowskian ST. I also must appologise as I realised an error in my notation last night whilst browsing my posts on this thread. That is that I had inadvertantly written a one-form as a column matrix with a 1 in the top slot; this is not true and it should be in the bottom, as you discovered upon research. This is, of course, because a one-form is a linear mapping whose arguement is one contravarient vector (a vector) and so the notation requires a 1 in the bottom slot (as the bottom slot denotes the number of vectors supplied to the mapping).

*two thumbs up* Aww...this thread has kinda dried up; I was enjoying this discussion. Want to know anything at all or require any help, Johnny?

Hey everyone. Was just wondering if any of you had read "Gravitation"-Wheeler et al for self or university study. I'm thinking about getting it, so it would be useful if I had other peoples' opinions so as to convince me whether to buy it or not. I'm pretty much skint at the moment and it's an expensive book, so I need to know if it's worth it.

And...?