□h=-16πT

Yes please. What is the problem at hand?

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).

Done.

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.

Right.

 

So if your vector is entirely in the XY plane of some reference frame' date=' then the vector will only have two components. So that the indice i is going to go from 1 to 2.

 

So suppose that the symbol [math'] \vec V [/math] was used to denote the vector.

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.

Can you briefly explain geometrized units?

 

It doesn't make any sense to equate the gravitational constant with a speed. Speed has units of distance' date=' divided by units of time. The gravitational constant has oddball units, in SI cubic meters per kilogram, divided by seconds squared. So... that formula mixes up apples and oranges.

 

What that does amounts to this...

 

(mmm)/(Kg)(ss) = m/s

 

So that we get...

 

mm/(Kg)(s) = unitless

 

so that we get...

 

mm = Kg s

 

It's totally strange I cannot make any sense whatsoever out of it, so can you explain it? I've seen that before G=c=1.[/quote']

 

In geometrized units G and c are equal to unity and dimensionless. First of all: making c dimensionless.

 

c=299,792,458 m/s

 

If we measure time in metres and take one second to be equal to299,792,458 m then we have normalized c. This now means that the lazy physicists out there no longer have to write c any more, as it comes up everywhere in SR and GR.

 

Now for doing the same to G. G is equal to 6.673x10^-11 m³/kg/s² in SI units. If we introduce our definition of a second as given above the units become m/kg and if we now measure mass in metres as well we can normalise and make dimensionless our gravitational constant, kg=(G/c²)m.

 

So when you came up with m²=kg s, you were actually correct.

 

Although we now have to change all our units of SI into their geometrized form (energy, pressure etc.) it gives one's wrist slightly less strain in the long run.

 

You're right about the definition of a one-form you gave. More specifically a one form is a linear mapping (as with all tensors the operation is linear) of 1 vector into a scalar. Only upon supliment with a vector does the one form map into a number though. Usual notation, for what is known as contraction, for the mapping is

 

[math]<\tilde{\omega}, \vec{V}>=\tilde{\omega}(\vec{V})[/math]

 

Where we have [math]\tilde{\omega}[/math] as a one form and [math]\vec{V}[/math] as a vector. The geometrical definition is too tedius for me to want to explain here. Also one never encounters one-forms in Euclidean geometry because their componants are identical to that of vectors, again I cannot be bothered to explain. Find a good book on tensor algebra or differential geometry and you'll be fine. Or to encorporate both of these and GR/SR get "A First Course in GR", the book I am reading at the moment.

 

Also, you owe me one for answering quite a few of your questions. I am very much like yourself, in the sense of asking many questions and wanting to acquire as much knowledge as possible. How old are you?

 

If you're defining componants of vectors in a particular frame it would make sense to make them four-vectors, as you are speaking of SR.

Do you realize how that looks to someone who doesn't know GR? If i had to sift through that to find an error' date=' it could take me years.

 

Let me ask you this. You say "we are solving for a metric." Explain that to me if you can. Maybe I'm not so blind after all.[/quote']

 

You know what a metric is, yes? Also are you familiar with the Einstein summation convention and what a tensor actually is? If you want the answer to your question then post your answers to mine. However if the answer is no to the last question then I shalln't bother, as even I lack the time and patience to sit here and type up an article on tensor algebra for the use of one person, so I suggest you find out if you don't know. If no for the 1st and 2nd then I'll explain, as they're pretty simple.

Here's what I know so far...

And I also remember that Martin was eager to mix hbar in somehow. He gave me a series of problems to solve.

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).

This is a conservation question, nothing more, nothing less.

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.

Here's what I know so far...

 

I know what the equation looks like. And don't worry about me getting offended' date=' it wont happen. I know the future, well a very tiny portion of it anyways.

 

Kind regards

 

Actually, I know a tiny bit about the stress energy tensor. I do know that it is present in what everyone is calling the field equations. Also, that isn't the formula that Martin, or Tom Mattson showed me (theirs had three terms), and I know there is one that involves the ricci tensor. I don't know what the stress tensor represents though.

 

Let me see if i remember what Martin showed me...

 

 

[math'] G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} [/math]

 

G/c^4 is supposed to have units of inverse force I believe, and...

 

G mu nu is called the umm

 

The Einstein tensor.

 

And T mu nu is called the stress-energy tensor, but as I said I don't know what that is supposed to symbolize.

 

I also know that G is the Newtonian gravitational constant, and c is the speed 299792458 meters per second. [/math]

 

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]

Wow' date=' solutions on the web, first hit. Yeah that's the one. Hmm.

 

Well let me ask you something related then.

 

What is psi*psi?[/quote']

 

The probability. The asterix denotes complex conjungation.

What is the definition of the trace of a matrix?

The sum of its diagonal componants.

[math]Tr(A) =\sum A_{ii}[/math]