Shortest Paths

[Johnny5]

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).

 

Yes ok, Rev's question. By tomorrow morning I will know more. I will continue posting the things I learn/recall, so that you can see a bit of what I know just by reading what I do between now and then.

 

Thank you very much.

 

 

 

I was working on expressing a vector, using various different notations. I am going to continue with that for awhile.

[□h=-16πT]

Done.

 

Therefore' date=' 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?

 

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.

[Johnny5]

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.

 

Ok, I need to have a look at this...

 

Definition: The dot product of two vectors is defined as follows:

 

[math] \vec A \bullet \vec B = |\vec A||\vec B| cos(A,B) [/math]

 

Therefore, by definition:

 

[math] \hat i \bullet \hat i = |\hat i | |\hat i | cos (\hat i, \hat i ) [/math]

 

Now, a basis vector must have unit length. Therefore:

 

[math] |\hat i | = 1 [/math]

 

Therefore:

 

[math] \hat i \bullet \hat i = |\hat i | |\hat i | cos (\hat i, \hat i ) = 1*1*cos (\hat i, \hat i ) = cos (\hat i, \hat i ) [/math]

 

Therefore, the dot product of i hat vector with itself is equal to the cosine of the angle between the i hat vector and itself.

 

The i hat vector, in some frame of reference S, is

 

The vector whose tail is the origin of the reference frame, and whose head is located on the positive X axis of the frame, one unit of distance away from the origin. In other words, the tail of the i hat vector is located at the point (0,0,0) in frame S, and the head of the i hat vector is located at the point (1,0,0) in frame S.

 

Now, consider the dot product of the j hat vector of frame S with the i hat vector of frame S.

[□h=-16πT]

Ok' date=' I need to have a look at this...

 

[b']Definition[/b]: The dot product of two vectors is defined as follows:

 

[math] \vec A \bullet \vec B = |\vec A||\vec B| cos(A,B) [/math]

 

Therefore, by definition:

 

[math] \hat i \bullet \hat i = |\hat i | |\hat i | cos (\hat i, \hat i ) [/math]

 

Now, a basis vector must have unit length. Therefore:

 

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

[Johnny5]

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

 

yes I know that.

 

But in the end n=3.

 

But I can remember that anyway...

 

Say it this way...

 

If the fundamental postulate of SR is correct, then the Pythagorean theorem isn't globally true. Perhaps locally, but not globally.

 

The Pythagorean theorem is correct, but there seems to be some confusion on this.

 

What you have said is this...

 

The Pythagorean theorem isn't globally valid on a curved manifold.

 

I don't know what a manifold is, so I am having trouble understanding this, but I do know what the term 'curved' means. It means not straight.

 

Suppose you have a three dimensional rectangular coordinate system.

 

Here are your basis vectors...

 

[math] e_1 [/math]

[math] e_2 [/math]

[math] e_3 [/math]

[□h=-16πT]

yes I know that.

If the fundamental postulate of SR is correct' date=' then the Pythagorean theorem isn't globally true. Perhaps locally, but not globally.

 

The Pythagorean theorem is correct, but there seems to be some confusion on this.

 

What you have said is this...

 

The Pythagorean theorem isn't globally valid on a curved manifold.[/quote']

 

A curved manifold is locally flat (at a point) and so the theorem holds. I was speaking of GR where the metric (generally) isn't flat. In SR space is globally flat, because there is no gravity, and henceforth the pythagorean theorem is true.

[Johnny5]

A curved manifold is locally flat (at a point) and so the theorem holds. I was speaking of GR where the metric (generally) isn't flat. In SR space is globally flat, because there is no gravity, and henceforth the pythagorean theorem is true.

 

I understood that. Good.

 

When the amount of distance away from a point is small, the Pythagorean theorem is true.

 

But apparently, the truth value of statements in GR, vary in space.

 

 

In other words, if you pick a point (x2,y2,z2) very far away from (x1,y1,z1), then

 

If GR is correct, and the space is curved then

 

The distance between the points is not given by the Pythagorean theorem.

 

Suppose that space is not curved, and instead 'flat', then no matter how far away (x2,y2,z2) is from (x1,y1,z1), the Pythagorean theorem is true. That would mean that the distance D between the points is given by the following formula:

 

[math] D = \sqrt{(x2-x1)^2 +(y2-y1)^2+(z2-z1)^2} [/math]

 

Now I am saying that is true, but relativists entertain an impossibility, so ok whatever.

 

How did you reach the conclusion that in SR, space is globally flat?

[□h=-16πT]

"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.

[Johnny5]

"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.

 

 

Strange but ok.

 

Manifold mathematically precise substitution for word 'space'.

 

Here is what I meant by strange... the ant thing

 

 

You have two types of ways to look at what is only one thing in reality.

 

 

Extrinsically

Intrinsically

 

Extrinsic geometry

Intrinsic geometry

[□h=-16πT]

How did you reach the conclusion that in SR' date=' space is globally flat?[/quote']

 

That is one of the assumptions that SR is based upon: that space is globally Euclidean. It is certainly not curved because it assumes no gravity. It's only when one progresses to GR that space becomes curved.

 

It's not really strange, it's the analogy that is. Look for a proper definition if you don't quite get it. And I know that's what you found strange. One could regard our universe as intrinsically 4-D (including time that is) and extrinsically 5-D from the point of view of a 5 dimensional observer. Although this isn't considered at all in the sciences because it's totally irrelevant.

[Johnny5]

That is one of the assumptions that SR is based upon: that space is globally Euclidean. It is certainly not curved because it assumes no gravity. It's only when one progresses to GR that space becomes curved.

 

Let me ask you this, is simultaneity absolute, if SR is correct?

[Johnny5]

Let me think about this ant thing.

 

You say the ant is a two dimensional being.... mmm well ok.

 

And that he is constrained to move along the surface of a three dimensional cylinder. Thats a little better but...

 

He isn't really a two dimensional being, if his body is bent.

 

in other words, any tiny patch of surface area, on the surface of a cylinder, is not flat, even locally.

 

You can take your limit, and it shouldnt matter.

 

Now matter how close you are to a point on the surface of the cylinde, if...

 

You let R denote the vector from that point to wherever you are... then

 

There is a point on the vector, which isn't on the cylinder.

 

I know how to say what I mean better...

[□h=-16πT]

Let me think about this ant thing.

 

You say the ant is a two dimensional being.... mmm well ok.

 

And that he is constrained to move along the surface of a three dimensional cylinder. Thats a little better but...

 

He isn't really a two dimensional being' date=' if his body is bent.

 

in other words, any tiny patch of surface area, on the surface of a cylinder, is not flat, even locally.

 

You can take your limit, and it shouldnt matter.

 

Now matter how close you are to a point on the surface of the cylinde, if...

 

You let R denote the vector from that point to wherever you are... then[/quote']

 

It's just an analogy, but if you're going to be picky replace the ant with a point on the cylinder that has the ability to perceive.

 

The vector thing you've just mentioned is well known and the set of all tangent vectors, of arbitrary length and direction, to a surface defines a space known as a tangent vector space (or when we draw the tangents in a different way, a fibre bundel) and which has a dimension higher than the manifold it is tangent to.

[□h=-16πT]

Let me ask you this, is simultaneity absolute, if SR is correct?

 

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.

[□h=-16πT]

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.

[Johnny5]

I need to catch up to you. I am going learn this as fast as I can. The rate you are going is good. I am going to post things as I learn them, and possibly questions. Oh and I didn't think it was the same as the Maxwell stress tensor.

 

Thank you

[Johnny5]

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.

 

[math] \vec V = V^ie_i = \sum_i V^ie_i [/math]

 

For non SR:

 

[math] \vec V = V^ie_i = \sum_i V^ie_i = \sum_{i=1}^{i=3} V^ie_i = v1\hat e_1 + v2\hat e_2 + v3\hat e_3 [/math]

 

Where:

 

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

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

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

 

i hat is a unit vector along the positive x axis, with its tail at the origin, j hat is a unit vector along the positive y axis, with its tail at the origin, and k hat is a unit vector along the positive z axis, with its tail at the origin.

 

 

For SR:

 

[math] \vec V = V^ie_i = \sum_i V^ie_i = \sum_{i=0}^{i=3} V^ie_i = v0 \hat e_0 + v1\hat e_1 + v2\hat e_2 + v3\hat e_3 [/math]

 

I could subscript v0,v1,v2, etcetera, but I don't want to.

 

Now, in SR, e0 hat is the "unit time vector," that's what I'm calling it for now.

 

Since it's special I am going to represent it by t hat.

 

[math] \hat e_0 = \hat t [/math]

 

Hence:

 

[math] v0 \hat e_0 + v1\hat e_1 + v2\hat e_2 + v3\hat e_3 = v0 \hat t + v1\hat i + v2\hat j + v3\hat k [/math]

 

So now we can go back and forth from one notation to the other. Right now, the coefficient of the unit time vector is meaningless.

 

The vector above, is an example of a four vector. .

[revprez]

If there were no mass-energy distribution in the space we would have a flat, Euclidean space, yes?

 

Yes, I am familiar with the energy momentum tensor. My question is with a negative mass-energy distribution, what is the proof that [math]T_{\mu\nu}[/math] (or [math]T^{\alpha\beta}[/math] is positive definite, negative definite, semidefinite either way, or indefinite? Bear in mind I'm not a mathematician. I intuitively grasp that a negative mass-energy distribution is either indefinite, negative semidefinite, or negative definite, and in the case of the Schwarzchild metric I can see clearly that the energy momentum tensor is positive definite. I want to understand this generally in terms of the trace.

[Johnny5]

 

[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' date=' Euclidean space, yes?

[/quote']

 

Prove this symbolically. In other words, prove that if there were no mass-energy distribution in the space, that space would be "flat" ie Euclidean. Someone, anyone, I don't care who proves it.

 

That is one way to write one of the general relativistic field equations, and

 

the units are "geometrized."

 

Another way to write the same statement is this...

 

[math] G^\alpha^\beta = \frac{8 \pi G}{c^4} T^\alpha^\beta [/math]

 

But the equation above uses SI units. G is the Newtonian gravitational constant, which is 6.672 x 10^-11 m^3/kgs^2, and c is the speed 299792458 meters per second.

 

The coefficient of the stress-tensor, also called the energy-stress tensor, has units of force, and is apparently also called the energy-momentum tensor.

 

In either of the expressions above, the term

 

[math] G^\alpha^\beta [/math]

 

is called the Einstein tensor, and is equivalent to:

 

[math] R_\mu_\nu - \frac{1}{2} g_\mu_\nu R [/math]

 

In the formula above, R_mu_nu is called the Ricci curvature tensor.

In the formula above, R is called the Ricci scalar.

In the formula above, g_mu_nu is called the metric tensor.

[revprez]

I think I see what you're getting at. If we have [math]T^{\mu\nu}_{;\nu}[/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.

 

Thank you. That matches my intuition. But my problem remains. I can't even prove that a positive mass-energy distribution implies a positive trace; I just don't know how. I've never seen the proof and I am just a little lazy. On the other hand, I do have a question as to what happens to geodesics when flat space-time is introduced to a negative mass-energy distribution. Follow me so far?

 

Rev Prez

[revprez]

The coefficient of the stress-tensor, also called the energy-stress tensor, has units of force, and apparently also called the energy-momentum tensor.

 

It has units of inverse force. Now you know the left side of the equation is in units of distance. What does that tell you about the right side of the equation? (ignore the volumetric components of [math]T_{\mu\nu}[/math].

 

Rev Prez

[Johnny5]

It has units of inverse force. Now you know the left side of the equation is in units of distance. What does that tell you about the right side of the equation? (ignore the volumetric components of [math]T_{\mu\nu}[/math].

 

Rev Prez

 

Yes inverse force, the gravitational constant G is in the numerator, and that has units of kilograms in the denominator, therefore inverse force. Yes, thank you.

 

LHS units of distance...

 

for that to be true RHS must have units of distance also. Therefore...

 

the stress-tensor must have units of...

 

Force times distance.... i.e. units of "energy"

 

What does "volumetric component" mean?

[Johnny5]

A vector in Euclidean space can be written, using the coordinates of a frame.

 

Suppose we have frame S. That means we have three axes, which are mutually orthogonal.

 

One is always called the X axis of the frame, and the others are the Y axis, and the Z axis of the frame.

 

And there are three unit vectors, called orthonormal basis vectors.

Now, a point in a frame, is represented by a three tuple.

 

Instead of an ordered pair, its an ordered triple.

 

So here is how to specify a unique location in the frame:

 

(3,2,4)

 

But order matters in the above notation of that location, it is a three tuple. The first number you see, 3, is the point which is three units of distance away from the origin of the frame, in the positive i hat direction.

 

Do not forget that, i^ is a unit vector, whose tail is at the origin of the frame, and whose head is one say meter away from the origin an on the positive X axis of the frame, where we have chosen the unit of distance to be the meter.

 

The other numbers 2, 4 are interpreted similarly.

 

Now, the theory of special relativity introduced a fourth orthonormal basis vector t^.

 

Thus, the coordinates of a point in four dimensional Minkowski space need to be represented by a four tuple, instead of the more logical three tuple used for points in Euclidean space.

 

I'm reading this now.

 

 

A point in Euclidean three space is represented as follows:

 

(x,y,z)

 

A point in Minkowski space is represented as follows:

 

(t,x,y,z)

 

The previous point is a four tuple.

 

Here is a general example of a four tuple:

 

 

 

[math] x^\mu = (x^0,x^1,x^2,x^3) [/math]

 

Now, here is a general example of a four-vector:

 

[math]

\left[

\begin{array}{c}

x^0\\

x^1\\

x^2\\

x^3\\

 

\end{array} \right]

[/math]

 

In the theory of relativity, the first element in the four tuple is the time component of the "position four vector of SR", the second element in the four tuple is the x coordinate of the ...., the third element in the four tuple is the y coordinate, and the fourth element is the z coordinate.

 

At wolfram, the position four-vector of SR theory, is written as follows:

 

[math]

\left[

\begin{array}{c}

ct\\

x\\

y\\

z\\

 

\end{array} \right]

[/math]

 

It's a vector that can be represented with only one indice.

 

Let's choose mu for the indice, and write the equation at wolfram.

 

[math]

x^\mu =

\left[

\begin{array}{c}

x^0\\

x^1\\

x^2\\

x^3\\

\end{array} \right]

 

[/math]

 

 

So for SR theory, the first component of the four vector is ct. So when we see ct, we know we are dealing with SR just by looking at the symbolism.

 

So merely by writing the following

 

[math]

x^\mu =

\left[

\begin{array}{c}

x^0\\

x^1\\

x^2\\

x^3\\

\end{array} \right]=\left[

\begin{array}{c}

ct\\

x\\

y\\

z\\

\end{array} \right]

 

[/math]

 

We know the symbol on the RHS, is the position-four vector of SR theory.

 

The earlier notation should be connected to this one.

 

Let us have chosen the following symbol for a vector in m dimensional-space.

 

[math] \vec V [/math]

 

If we choose to leave summation explicit, then we can write:

 

[math] \vec V = \sum_{i=1}^{i=m} v^i \hat e_i [/math]

 

If we want to have summation implicit, we can use the Einstein summation convention, and write the arbitrary m dimensional vector as follows:

 

[math] \vec V = v^ie_i [/math]

 

Notice, we have lost information about the dimension of the space, because we lost the symbol m. At any rate, here is how we know there is a summation of the dummy indice i, one of the indices is superscripted, the other subscripted, and the indice symbols are identical.

 

 

Now, when dealing with SR theory... the sum is started from 0, and goes up to 3, so that there are four coordinates for a point in the space. And specifically, when dealing with SR theory, the zeroth component corresponds to the time coordinate.

 

To be quite honest, here is the notation I like the best for this nightmare:

 

[math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i [/math]

 

In other words, leave the summation explicit.

 

Then expand the series...

 

[math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i = v0 \hat e_0 +v1\hat e_1 +v2 \hat e_2 +v3\hat e_3 [/math]

 

Then in order to switch to SR theory, the e0 unit vector must represent the unit time vector, whose dot product with itself is negative one, unlike the orthonormal basis vectors of ordinary Euclidean space, which when dotted with themselves, have their dot product equal to 1.

 

Now, in SR theory we have:

 

[math] \hat e_0 = \hat t [/math]

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

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

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

 

And the space is called Minkowski space, and a point in Minkowski space is represented as follows:

 

(ct,x,y,z)

 

So in SR theory we have:

 

[math] \vec V = \sum_{i=0}^{i=3} v^i \hat e_i = v0 \hat e_0 +v1\hat e_1 +v2 \hat e_2 +v3\hat e_3 = ct \hat t +x\hat i +y \hat j + z\hat k [/math]

 

And we can write that as a column vector, as follows:

 

 

[math] \vec V =

\left[

\begin{array}{c}

ct\\

x\\

y\\

z\\

\end{array} \right]

[/math]

 

 

I guess there is a small problem, in that the above notation doesn't lead to t^ dot t^ = -1.

 

A simple way to cure this, is to demand that the unit time vector be imaginary.

 

In other words we have this kind of thing going on here...

 

[math] \hat i \bullet \hat i \equiv |\hat i| |\hat i| cos(\hat i,\hat i) [/math]

 

And since the magnitude of unit vector is equal to one, we must have:

 

[math] \hat i \bullet \hat i = cos(\hat i,\hat i) [/math]

 

And the angle between ihat and itself is zero, and the cosine of zero is equal to one so that we have:

 

[math] \hat i \bullet \hat i = 1 [/math]

 

And the angle between i^ and j^ is 90 degrees, and cos(90)=cos(pi/2) = 0, therefore i^ dot j^ =0, and for the same reason, i^ dot k^ = 0, and for the same reason, j^ dot k^ = 0, so that we must have:

 

[math] \hat i \bullet \hat i = \hat j \bullet \hat j = \hat k \bullet \hat k= 1 [/math]

 

and

 

[math] \hat i \bullet \hat j = \hat i \bullet \hat k = \hat j \bullet \hat k = 0 [/math]

 

 

Now, consider the unit time vector t^.

 

Using the definition of dot product we must have this:

 

[math] \hat t \bullet \hat t \equiv |\hat t| |\hat t| cos(\hat t,\hat t) [/math]

 

Now, if we stipulate that the magnitude of the unit time vector is equal to the square root of negative one, and that the angle between the unit time vector and itself is 0, then we get this:

 

[math] \hat t \bullet \hat t \equiv |\hat t| |\hat t| cos(\hat t,\hat t) = \sqrt {-1} \sqrt{-1} cos 0 = i^2 = -1[/math]

 

But now if we do that, then the magnitude of a unit vector no longer has length equal to 1. And I believe the magnitude of a unit vector is stipulated to be one. But apparently not so, for the "unit time vector" of SR theory.

 

On the other hand, suppose that we stipulate that the magnitude of the unit time vector must equal one.

 

Then the only way to have the dot product come out to be negative one, is to have cosine (t^,t^) = -1

 

cos(0)=1

cos(90) = 0

cos(180) = -1

cos(270)=0

cos(360)=1

cos(450)=0

cos(540)=-1

cos(630)=0

cos(720)=1

 

and so on... the values of cosine which yeild negative 1, occur when the angle is 180,540, 900, 1260, and so on...

 

and also

 

-180, -540, -900, and so on...

 

and we can write this in terms of pi.

 

[math] \pi radians = 180 degrees [/math]

 

So use the following symbol for the integers...

 

[math] \mathbb{Z} [/math]

 

Now, let n denote an arbitrary element of [math] \mathbb{Z} [/math].

 

In other words:

 

[math] n \in \mathbb{Z} [/math]

 

So, |t^|=1 and cosine (t^,t^) = -1

 

Provided

 

[math] cos(\pi + 2n\pi) [/math]

 

for some n an element of [math] \mathbb{Z} [/math].

 

And there are no other ways to make the dot product of the unit time vector with itself come out to equal negative one. Therefore, let us close the scope of the assumption right here.

 

If [math] \hat t \bullet \hat t [/math] = -1

then ( [math] |\hat t | = \sqrt{-1} [/math] or [math] cos(\hat t,\hat t) = cos(\pi + 2n\pi) [/math]

 

 

Let's clean the previous statement up, using logical symbols.

 

[math] (\hat t \bullet \hat t = -1) \Rightarrow (|\hat t | = i \vee cos(\hat t,\hat t) = cos(\pi + 2n\pi) ) [/math]

 

Where:

[math] n \in \mathbb{Z} [/math]

 

Translation: If the dot product of the unit time vector with itself is equal to negative one, then either the magnitude of the unit time vector is equal to the square root of negative one or the cosine of the spatial angle between the unit time vector and itself is equal to the cosine of pi radians plus a multiple of 2pi radians.

 

So, for the sake of argument, suppose that:

 

1. The dot product of the unit time vector with itself is equal to -1.

2. The magnitude of the unit time vector isn't root -1.

 

Then it must be the case that the angle between the unit time vector and itself is equal to:

 

[math] \pi + 2n\pi [/math]

 

For at least one integer n.

 

 

In other words, the following is a fact:

 

If the dot product is defined in the conventional way, and the dot product of the unit time vector with itself is equal to -1, and it is not the case that the magnitude of the unit time vector is equal to the square root of negative one, then it must be the case that there is at least one integer n, for which

the spatial angle between the unit time vector and itself is equal to

 

[math] \pi (2n+1) [/math]

 

And the symbols above, are just a fancy way of saying, "an odd multiple of pi."

 

Definition: The dot product of two vectors [math] \vec A, \vec B [/math] is defined as follows:

 

[math] \vec A \bullet \vec B \equiv |\vec A| |\vec B| cos(\vec A,\vec B) [/math]

 

Let [math] \theta [/math] denote the angle between the unit time vector and itself.

 

Therefore, theta is an element of the following set:

 

[math] \mathcal{f} ...,-5\pi, -3\pi, -\pi, \pi, 3\pi, 5\pi, ... \mathcal{g} [/math]

 

where I have used the roster method of set denotation.

 

For a discussion on set theory, see here set theory .

 

 

Now, the natural question to be raised, would be how in the world can a temporal vector, have a nonzero spatial angle between it and itself? The correct answer is that it cannot. I will come up with a proof for this later.

 

So, it is not the case that there is at least one integer n, for which the angle between the unit time vector and itself is equal to:

 

[math] \pi(2n+1) [/math]

 

We can use the existential quantifier to help us say this symbolically...

 

[math] \neg \exists n \in \mathbb{Z} [(\hat t, \hat t) = \pi(2n+1)] [/math]

Translation: It is not the case that there is at least one integer n, such that the angle between the "unit time vector" and itself is equal to pi times (2n+1) times.

 

Recap

 

After a bit of work, the following statement can be proven:

 

If the dot product of the unit time vector with itself is equal to -1, and the magnitude of the unit time vector isn't [math] \sqrt{-1} [/math], then it must be the case that the angle between the unit time vector and itself is equal to:

 

[math] \pi + 2n\pi [/math]

 

For at least one integer n.

 

And later, it will be proven that it is impossible for the consequent of the conditional above to be true.

 

So the negation of the antecedant of the conditional above is false.

 

[math] \therefore [/math]

 

Either the dot product of the unit time vector with itself isn't equal to negative one, or the magnitude of the unit time vector is equal to the square root of negative one.

 

Right now, it has been stipulated that the dot product of the unit time vector with itself is equal to negative one. Therefore, we are forced to conclude that the magnitude of the unit time vector is equal to the square root of negative one.

 

The strange thing about this, is that a unit vector is supposed to have a magnitude of 1, not root -1.

[revprez]

What does "volumetric component" mean?

 

We usually write [math]T_{\mu\nu}[/math] in terms of energy density [math]\rho[/math] (where c = 1 and [math]\rho [/math] has units of pressure) and pressures p.

 

Rev Prez

[Johnny5]

We usually write [math]T_{\mu\nu}[/math] in terms of energy density [math]\rho[/math] (where c = 1 and [math]\rho c^2[/math] has units of pressure) and pressures p.

 

Rev Prez

 

pressure = force/area

 

 

Let p denote pressure.

 

energy density = energy/volume

 

Let r denote energy density.

 

Now [math] T_\mu_\nu [/math] (the stress tensor) has to have units of energy' date=' so that the LHS of the general relativistic field equation has units of meters, units of distance.

 

p= pressure = force/area = (force)(distance)/(volume) = (energy)/(volume)

 

I am reading this right now.

 

And also this.