Shortest Paths

[Dave]

I'd rather talk to rev.

 

First google hit for "matrix trace".

 

For your information, the trace of a matrix is the sum of the diagonal elements.

 

Please also be aware that people don't necessarily want their threads invaded to cover all aspects of GR

 

To Meir Achuz: to typeset LaTeX, you encapsulate with [ math] [/math] (obviously omitting the space).

[□h=-16πT]

What is the definition of the trace of a matrix?

 

The sum of its diagonal componants.

 

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

[Johnny5]

Why so? Can you not just look it up?

 

I guess so...

[Johnny5]

The sum of its diagonal componants.

 

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

 

 

Thank you.

 

 

"The trace of a matrix is the sum of its diagonal components."

 

I wish to understand this question here:

 

Specifically' date=' if T[sub']mn[/sub] is a positive definite matrix, is a geodesic through it always longer than one through empty space-time?

 

Rev Prez

 

Now, a positive matrix is one which has a positive trace. In other words, let A denote a matrix, and let Tr(A) denote the sum of the elements along the main diagonal. That sum is a number. So the question is, "if that sum is greater than zero, does that imply that a geodesic "through it" is always longer than a geodesic through empty space-time"

 

Well, somewhere Rev Prez you need to talk about two points in a frame.

 

Can you please phrase your question using two points?

 

(x1,y1,z1)

(x2,y2,z2)

[□h=-16πT]

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]

[Johnny5]

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' date=' 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]

 

Here's what I know so far...

 

I know what the equation looks like. And don't worry about me getting offended, 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.

 

I also know that the Einstein tensor can be written in terms of the ricci tensor and the ricci scalar, but I don't remember what they look like, Tom showed me.

 

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

 

And I do know a fair amount of electrodynamics, and I remember seeing comments about the Maxwell stress-energy tensor, when I first took the class, but we skimmed over that part.

 

And I also remember that Martin said that sometimes the 8 pi is present in things, and sometimes not.

 

And I know what a positive definite matrix is. It has a positive trace, the trace being equivalent to the sum of the components down the main diagonal.

 

That is:

 

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

 

That's about all I know about it off the top of my head.

[□h=-16πT]

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.

[□h=-16πT]

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.

[□h=-16πT]

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).

[Johnny5]

That's the same field equation that I wrote' date=' it's just that I used geometrized units rather than SI, where G=c=1.

[/quote']

 

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, 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.

[Johnny5]

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]"' date=' if not then I cannot be bothered to explain.

[/quote']

 

Lets see...

 

I just read this.

 

A covariant vector is covariant tensor of rank 1. It is more commonly called a "one form" or "bra".

 

Someplace I just read, "A one form takes a vector as input and outputs a scalar."

 

Now I know bra,ket notation is due to Dirac.

 

Bra...

 

[math] < \psi | [/math]

 

Ket...

 

[math] | \psi > [/math]

 

Ok, here is something intelligible...

 

In Euclidean space the inner product is the familiar dot product.

 

Here is the symbol for Euclidean three dimensional space...

 

[math] \mathbb{R}^3 [/math]

 

Suppose we have two vectors in Euclidean three dimensional space.

 

[math] \vec u [/math]

[math] \vec v [/math]

 

Their dot product is represented by:

 

[math] \vec u \bullet \vec v [/math]

 

Now, suppose that have chosen a frame of reference to view the vectors in.

 

Then we can write the vectors using coordinates, as follows:

 

[math] \vec u = u1 \hat i + u2 \hat j + u3 \hat k [/math]

[math] \vec v = v1 \hat i + v2 \hat j + v3 \hat k [/math]

 

we have:

 

[math] \vec u \bullet \vec v = u1v1+u2v2+u3v3 [/math]

 

The quantity above is a scalar, not a vector. Since scalar multiplication is commutative, we can prove the dot product is commutative:

 

[math] \vec u \bullet \vec v = u1v1+u2v2+u3v3 = v1u1+v2u2+v3u3 = \vec u \bullet \vec v [/math]

 

Now, here is "inner product" notation...

 

[math] <u,v> [/math]

 

If your inner product is on the scalars, i.e. [math] \mathbb R [/math] then it is just elementary multipliction of two numbers, which is commutative.

 

In other words:

 

If U is an element of R, and V is an element of R then

 

<U,V> = UV

 

So for example:

 

<4,7> = (4)(7) = 28

 

But now suppose that your inner product is on Euclidean three dimensional space. In this case you have this:

 

If U is an element of R^3, and V is an element of R^3 then

 

[math] <U,V> = \vec u \bullet \vec v [/math]

 

So lets have a look at an example of this.

 

You have chosen some reference frame, and are now able to express two vectors in the frame, in terms of the coordinates of the frame.

 

Pick two random points in the frame...

 

P1 = (x1,y1,z1)

P2 = (x2,y2,z2)

 

Now, we can write the vector from either point to the other in a simple manner.

 

Suppose we want the vector which points from P2 towards P1, then we have this...

 

[math] \vec {P2P1} = (x1-x2) \hat i + (y1-y2) \hat j + (z1-z2) \hat k [/math]

 

And the vector which points from P1 towards P2 is similarly given by:

 

[math] \vec {P1P2} = (x2-x1) \hat i + (y2-y1) \hat j + (z2-z1) \hat k [/math]

 

 

So, suppose we have two specific vectors in the frame...

 

[math] \vec u = \vec {P1P2} = u1 \hat i + u2 \hat j + u3 \hat k [/math]

[math] \vec v = \vec {P3P4} = v1 \hat i + v2 \hat j + v3 \hat k [/math]

 

The dot product of them is equal to:

 

u1v1+u2v2+u3v3

 

As our specific example, suppose that:

 

[math] \vec u = \vec {P1P2} = 2 \hat i + 3 \hat j + 4 \hat k [/math]

[math] \vec v = \vec {P3P4} = 6 \hat i + 7 \hat j + 8 \hat k [/math]

 

then

 

[math] \vec u \bullet \vec v = 12 + 21 + 32 = 65 [/math]

 

And since the dot product is commutative, v dot u = 65 also.

 

We have to be careful about the domains of discourse being used, so that we can use first order logic to remember things.

 

Two domains of discourse have appeared so far, they are:

 

[math] \mathbb R [/math]

[math] \mathbb R^3 [/math]

 

An element of the first domain of discourse is a scalar... a number, a real number, like the number 3, or the square root of two, or pi.

 

An element of the second domain of discourse, is usually called a vector.

 

A vector quantity is different from a scalar quantity, in that it has both magnitude and direction.

 

So the inner product on the reals is just scalar multiplication, and the inner product on Euclidean three space is the dot product.

 

If x,y denote elements of R then <x,y> = xy

If x,y denote elements of R^3 then <x,y> = x dot y

 

Ok, so the bra ket notation, is related to inner product notation.

 

The names bra and ket, are related to the symbols used to write them, namely brackets... actually angle brackets

 

parentheses ( )

braces { }

square brackets [ ]

 

angle brackets < >

 

To recapitulate...

 

A covariant vector, also called a "one form" or "bra" can be written as follows:

 

[math] < \psi | [/math]

 

A contravariant vector, also called a "ket" can be written as follows:

 

[math] | \psi > [/math]

 

And the inner product of a bra with a ket can be written as follows:

 

[math] <\psi | \psi > [/math]

 

 

Now you also used the following notation for a one form...

 

[math]

\left( \begin{array}{ccc}

1\\

0 \\

\end{array} \right)\]

[/math]

 

I still don't understand this notation. Here is the closest thing I could find about it:

 

Notation for general relativity and other advanced topics

 

But as you can see, that guy says that a one form is expressed as:

 

[math]

 

\left (

\begin{array}{c}

0\\

1\\

\end{array}

\right)

[/math]

 

So which of you is right?

[Bob182]

For G=C=1 google on Natural Units - e.g.

 

http://quantumrelativity.calsci.com/Relativity/Appendix2.html

 

It's really just a way of simplifying calculations cause people that work in fields with lots of G's, C's and h bars can't be bothered to write them over and over.

[□h=-16πT]

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.

[□h=-16πT]

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.

[Johnny5]

Also' date=' 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.

[/quote']

 

Yeah I know... thank you.

 

 

And im in my thirties.

[Johnny5]

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.

 

Metric has something to do with distance. I am not familiar with the Einstein summation convention, however I do remember reading that he thought it was his greatest contribution to physics.

 

No, I don't have a perfect definition of "tensor" yet.

 

Let me go get one.

 

Here is mathworld's definition of tensor.

 

It will take me a few seconds to make sense out of that. I will learn what a tensor is as fast as possible.

 

Here is the first sentence:

 

An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and mⁿ components and obeys certain transformation rules.

 

Let me fix m at 3.

 

 

An nth-rank tensor in 3-dimensional space is a mathematical object that has n indices and 3ⁿ components and obeys certain transformation rules.

 

Suppose that n=0. Then a zero rank tensor in three dimensional space is a mathematical object that has 0 indices and 1 component, and obeys certain transformation rules. This mathematical object is a scalar.

 

Suppose that n=1. Then a first rank tensor in three dimensional space is a mathematical object that has one indice and 3^1=3 components, and obeys certain transformation rules. This mathematical object is a vector.

 

Suppose that n=2. Then a second rank tensor in a three dimensional space is a mathematical object that has two indices, and 9 components, and obeys certain transformation rules.

 

Here is what it says about notation...

The notation for a tensor is similar to that of a matrix (i.e., [math] (A_i_j) [/math], except that a tensor may have an arbitrary number of indices.

 

The article makes it clear that whether an indice is a superscript or subscript matters in the symbol's interpretation.

 

contravariant... upper indice

covariant... lower indice

 

 

The author of the article then says something very important...

While the distinction between covariant and contravariant indices must be made for general tensors, the two are equivalent for tensors in three-dimensional Euclidean space, and such tensors are known as Cartesian tensors.

 

Tensor notation

 

 

Consider our vector u from earlier.

 

[math] \vec u = u1 \hat i + u2 \hat j + u3 \hat k [/math]

 

 

In the mathworld article at Wolfram, we are told the following:

An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and components and obeys certain transformation rules.

 

Fixing the dimension of the space at 3, leads to:

 

An nth-rank tensor in 3-dimensional space is a mathematical object that has n indices and 3^m components and obeys certain transformation rules.

 

Now, keeping in mind that a vector is a first rank tensor, it follows that:

 

An first-rank tensor in 3-dimensional space is a mathematical object that has 1 indice and 3 components and obeys certain transformation rules.

 

So if we use lower indices, the notation for our vector u is:

 

[math] \vec u = u_1 \hat i + u_2 \hat j + u_3 \hat k [/math]

 

instead.

 

The author then says this

 

Tensor notation can provide a very concise way of writing vector and more general identities. For example, in tensor notation, the dot product is simply written

 

[math] \vec v \bullet \vec u = u_i v^i [/math]

 

So far the author has been consistent.

 

A vector in three space will have one indice (apparently either upper or lower, the choice is yours), and it will have three components.

 

The logic of tensor notation seems quite convoluted.

 

 

Actually, what I can do is this...

 

[math] \vec u = \sum_{i=1}^{i=n} u_i [/math]

 

Now I just need to get the basis vectors into the notation somehow.

 

Here is the notation for an arbitrary point in three dimensional space, in a frame of reference with an x axis, y axis, and z axis

 

P = (x,y,z)

 

Note that this is a three tuple, and what that means is this... that means that order matters.

 

In other words...

 

not( (2,-4,3) = (3,2,-4) )

[□h=-16πT]

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.

[Johnny5]

Ok' date=' 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.

 

Ok I got you. When you see the same index raised, and lowered, you don't have to write the giant Sigma symbol, you are to know that you are summing over the index.

 

In other words, there is a semantical equivalence of exactly what you wrote, namely...

 

[math] \sum_{i} A_iB^i = A_iB^i [/math]

 

In other words, when Einstein summation convention is used, the meaning of the left hand side of the expression above is equivalent to the meaning of the RHS.

 

A shorthand notation.

 

So that is for when you see the same index.

 

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]

 

etc. [\quote]

[□h=-16πT]

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]

[Johnny5]

Yeah' date=' 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]

 

Right.

 

So if your vector is entirely in the XY plane of some reference frame, 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.

 

Let it be the case that:

 

[math] \vec V = x \hat i + y \hat j [/math]

 

 

So as a time saving technique, we are going to write that using the summation convention as:

 

[math] \vec V = V^i e_i [/math]

 

Where we have used e_i to represent the i'th basis vector.

 

So if we let i range from 1 to 3 we get this:

 

[math] \vec V = V^i e_i = v1 \hat i +v2 \hat j + v3 \hat k = v1 \hat e_1+v2\hat e_2+v3\hat e_3 [/math]

 

Just trying to work out some equivalences right now.

 

Now, change to SR theory..., which means go from zero...

 

[math] \vec V = V^i e_i = v0\hat e_0 + v1 \hat e_1+v2\hat e_2+v3\hat e_3 [/math]

[□h=-16πT]

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).

[Johnny5]

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

 

Done.

 

Therefore, in SR:

 

[math] e_1 = \hat i [/math]

[math] e_2 = \hat j [/math]

[math] e_3 = \hat k [/math]

 

and i suppose that:

 

[math] e_0 = [/math] direction of time?

 

Is there a way to use the notation to express the fact that the time coordinate t of a frame changes unidirectionally?

[□h=-16πT]

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.

[Johnny5]

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 please. What is the problem at hand?

 

I know this field equation here:

 

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

 

The quantity on the LHS is called the Einstein tensor, and the quantity on the RHS is a scalar quantity whose units are that of force (Kg m/s^2), times the Maxwell stress-energy tensor.

[□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).