Shortest Paths

[revprez]

pressure = force/area

 

And energy density is what? Force * distance / volume. The components [math]U_{\mu}U_{\nu}[/math] (or [math]U^{\alpha}U^\beta}[/math] if you like that convention) also give you units of [math]\left(\frac{m}{s}\right)^2[/math].

[Johnny5]

And energy density is what? Force * distance / volume. The components [math]U_{\mu}U_{\nu}[/math] (or [math]U^{\alpha}U^\beta}[/math] if you like that convention) also give you units of [math]\left(\frac{m}{s}\right)^2[/math'].

 

 

I'm not following you.

 

You are using the letter U to denote what?

 

Speed apparently.

 

Don't use geometrized units yet.

 

I don't know how to write the stress-tensor in terms of energy density rho yet.

[Johnny5]

[math] G^\alpha^\beta [/math] is called the Einstein tensor.

[math] T^\alpha^\beta [/math] is called either the stress-tensor, or the energy-momentum tensor.

 

The Einstein general relativistic field equation is:

 

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

 

The coefficient of the stress-tensor is just a scalar, a number, and its units are inverse force. G is the Newtonian gravitational constant, and c is the speed 299792458 meters per second. The LHS of the field equation has units of meters. The stress-tensor has units of energy.

 

The Einstein tensor can be written in terms of the Ricci scalar, the Ricci tensor, and the metric tensor as follows:

 

[math] G^\alpha^\beta = R_\mu_\nu - \frac{1}{2} g_\mu_\nu R [/math]

 

[math] R_\mu_\nu [/math] is called the Ricci tensor.

[math] g_\mu_\nu [/math] is called the metric tensor.

R is called the Ricci scalar.

 

[math] R^\mu_\alpha_\mu_\beta [/math] is called the Reimann curvature tensor.

[revprez]

The Einstein tensor can be written in terms of the Ricci scalar' date=' the Ricci tensor, and the metric tensor as follows:

 

[math'] G^\alpha^\beta = R_\mu_\nu - \frac{1}{2} g_\mu_\nu R [/math]

 

Careful here. If you're going to use the upper indice, alpha beta convention, then carry it through to the other side as well. In this case, [math] G^{\alpha\beta} = R^{\alpha\beta} - \frac{1}{2} g^{\alpha\beta} R [/math], and the Riemann tensor should be [math]R_{\lambda}{}^{\alpha\lambda\beta}[/math].

[Johnny5]

Careful here. If you're going to use the upper indice, alpha beta convention, then carry it through to the other side as well. In this case, [math] G^{\alpha\beta} = R^{\alpha\beta} - \frac{1}{2} g^{\alpha\beta} R [/math'].

 

Rev, I would like the notation that I end up using, to be the one the most amount of physicists are already used to, if that tells you anything.

 

 

Which is more common, alpha/beta or mu/nu?

[revprez]

Rev, I would like the notation that I end up using, to be the one the most amount of physicists are already used to, if that tells you anything.

 

I don't know what the break down is, but let's just say that MIT, Wikipedia and Mathworld like lower indices, and Hofstra likes upper indices. You can use either convention, so long as you're consistent. After all, a covariant tensor is just a linear operator on a contravariant one and visa versa; so long as you distinguish between the two in notation it doesn't matter which gets the upper or lower indice. Either way, just look at what you wrote down. Does it seem right to you that the linear combination of two covariant tensors should equal some scalar multiple of a contravariant one?

 

Rev Prez

[Johnny5]

It is very important that you understand that a tensor is a linear mapping.

 

Linear transformation

 

I remember the formula right away

 

If f(ax+by) = af(x)+bf(y) then f is a linear transformation.

 

But to be proper' date=' I need to be clear what my domain of discourse is.

 

And at the wikipedia site, there is reference to

 

fields

vector spaces

 

And i know how complicated that stuff is. I want to keep things as simple as i can, while i learn.

 

 

Rings, fields, modules, submodules, a lot is coming back to me.

 

 

A field is that for which the algebraic axioms are true.

 

 

1. closure under addition.

2. closer under multiplication.

 

3. commutativity of addition

4. associativity of addition.

5. existence of additive identity

6. existence of additive inverses

 

7. commutativity of multiplication

8. associativity of multiplication.

9. existence of multiplicative identity

9. existence of multiplicative inverses

 

10. distributive axiom

11. non triviality not(0=1)

 

and a few other minor ones.

 

For example...

 

Let the domain of discourse be the field know as the real numbers.

 

Let [math'] \mathbb{R} [/math] denote the reals.

 

We postulate two processes on real numbers, addition, and multiplication.

 

The process of addition is denoted by the symbol +

and the process of multiplication is denoted by the symbol [math] \cdot [/math]

Postulate 1 (Closure under addition):

[math] \mathbb{R} [/math] is closed under +

 

What this means is that if we add two real numbers together, the answer is also a real.

 

We can say this using logic:

 

for any symbol x, and any symbol y:

 

[math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x+y \in \mathbb{R} [/math]

 

Translation: If x is an element of the real number system, and y is an element of the real number system, then x+y is an element of the real number system.

 

 

 

Postulate 2 (Closure under multiplication):

For any symbol x, and any symbol y:

[math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x \cdot y \in \mathbb{R} [/math]

 

Translation: If x is an element of the real number system, and y is an element of the real number system, then [math] x \cdot y [/math] is an element of the real number system.

 

What this means is that if we multiply two reals together, the answer is also a real.

 

Postulate 3(Commutativity of addition):

 

[math] \forall x \forall y [/math]

 

[math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R}) \Rightarrow x+y=y+x[/math]

 

Postulate 4(Associativity of addition):

 

[math] \forall x \forall y \forall z [/math]

 

[math] (x \in \mathbb{R} \ \& \ y \in \mathbb{R} \& \ z \in \mathbb{R}) \Rightarrow x+(y+z) = (x+y)+z [/math]

 

What this postulate says, is intuitive. Suppose we are only permitted to add two numbers at a time, but we wish to add three numbers together. What this says is that it doesnt matter which two we add together first.

 

In other words it tells us this

 

5+(7+8) = 5+(15) = 20

 

(5+7)+8 = (12)+8 = 20

 

In other words it tells us 5+(7+8) = (5+7)+8, but it tells us that we can do this for any three real numbers, not just 5,7, and 8.

 

 

Postulate 5 (existence of additive identity):

 

[math] \exists 0 \in \mathbb{R} \ \forall x \in \mathbb{R} [ 0+x=x ] [/math]

 

Translation: There is at least one real number 0, such that for any real number x, 0+x = x.

 

We leave uniqueness of zero as a theorem.

[Johnny5]

After all' date=' a covariant tensor is just a linear operator on a contravariant one and visa versa

 

Rev Prez[/quote']

 

 

Can you explain this to me quickly?

 

Thanks

[revprez]

Can you explain this to me quickly?

 

Thanks

 

I don't think so, not without writing some lecture notes on linear algebra. You might want to check out Wikipedia. The best I can say in a short amount of time is that if you consider a linear function that takes a column vector and spits out a scalar, it should key in your mind that this function is a row vector matrix multiplied on the argument. Visa versa, too. You can think of contravariant and covariant tensors as generalized linear functions on each other. Does that help?

 

Look at your above reply to h=16, I'm guessing you've got some sense of what a linear map/function/operator is.

 

Rev Prez

[Johnny5]

 

Look at your above reply to h=16' date=' I'm guessing you've got some sense of what a linear map/function/operator is.

 

Rev Prez[/quote']

 

yes.

 

Here is a good discussion of the idea: Linear transformation

 

 

That's the standard notion.

 

f:X-->Y

 

 

function f maps set X into set Y

 

In that article they write:

 

[math] T: X \to Y [/math]

 

They call T a transformation, they call X the domain, and Y the co-domain.

 

Actually, this site seems pretty good.

 

I'm going to go review linear algebra.

 

Is range another name for codomain?

[□h=-16πT]

In the MCRF we have the trace of the SE tensor as follows

 

[math]\rho+p_x+p_y+p_z[/math]

 

You can't have negative pressure in this context as each pressure acts in three different directions and so none of the components are negative. We've already established that the mass-energy distribution is positive, so all terms are positive and so is the trace. I'm no mathematician myself and all my knowledge of this sort of thing is based on books and such. I'm 15 and still in secondary school, I just love maths and physics. How about you Rev?

 

As for the notation, Johnny, I like using lowered indices for covariant rank and visa versa for contravariant, it being the notation used when I taught myself. It really doesn't matter, as long as the author specifies his preferance you'll be fine. In cases like the field equations (where only tensor components are mentioned) you can simply change the indices by a process known as "index raising and lowering", which bares very little on anything until you actually compute the components and you have to use inverse metric components etc. All my posts will use raised for contravariant and visa versa, there could be a different notation for those across the pond.

 

Negative mass distribution...er...

[□h=-16πT]

Prove this symbolically. In other words' date=' 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.

[/quote']

 

If there's no mass-energy distribution in a region of space then there exists no SE tensor and so the Einstein field equation is zero. I believe that the only way that the Einstein tensor be zero would be if the Ricci scalar and tensor were zero and hence the Riemann tensor. If the Riemann curvature tensor is zero, the space is flat. A reasoning follows just from knowing a little about GR: that mass-energy curves space-time, gravitation.

[□h=-16πT]

You are using the letter U to denote what?

 

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

[revprez]

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

 

It looked good to me. Only thing I can add is a clarification. The Euclidean 3-space is described by the 3x3 spatial subset of the 4x4 metric (just take the lower right subset of the 4x4). But I think its clear to anyone who's familiar with the Minkowskian from special relativity ([math]diag(\eta_{\mu\nu}) = (-1,1,1,1)[/math]). Beyond that, I think you're more comfortable with the math and phyiscs--I'll defer to someone more knowledgable.

 

Rev Prez

[revprez]

Is range another name for codomain?

 

Not generally. The range is usually just a subset of the codomain. We say that f:A->B where B is the codomain (i.e., reals, complex, some ordered set), and the range are the values in B allowable under the transformation. For something like f = x for x in A, then the codomain and the range are clearly the same thing (in fact, its clear that the codomain and the domain are the same as well).

 

Rev Prez

[revprez]

In the MCRF we have the trace of the SE tensor as follows

 

[math]\rho+p_x+p_y+p_z[/math]

 

You can't have negative pressure in this context as each pressure acts in three different directions and so none of the components are negative. We've already established that the mass-energy distribution is positive' date=' so all terms are positive and so is the trace. I'm no mathematician myself and all my knowledge of this sort of thing is based on books and such. I'm 15 and still in secondary school, I just love maths and physics. How about you Rev?[/quote']

 

25 and an enterprise application developer. I love math and physics, but not sure if I want to do it for a living. However, I'm specifically interested in negative mass-energy distributions and how to demonstrate a class of [math]T_{\mu\nu}[/math] characteristics through a proof.

 

Rev Prez

[□h=-16πT]

Which is more common, alpha/beta or mu/nu?

 

It really doesn't matter, actually. They are what are known as "dummy indices": they all take the values of 0, 1, 2, 3 in some particlular frame, so which letter you use is pretty irrelevant. As long as you use any lower case greek letter it's ok. Although alpha, beta, mu, nu and sigma are usually about as many indices as one may find in a single equation.

 

Was my short "proof" that +ve [math]\rho[/math] gives a +ve-deff. trace adequate, Rev?

[revprez]

Was my short "proof" that +ve [math]\rho[/math'] gives a +ve-deff. trace adequate, Rev?

 

The one problem I have is I can't prove that all positive mass-energy distributions can be represented as a diagonal matrix of the form [math]diag(\rho, p...)[/math]. When I come back later I think I'll kick this over to the appropriate math forum as well. Maybe we can get some feedback from formally trained folk.

 

Rev Prez

[□h=-16πT]

The one problem I have is I can't prove that all positive mass-energy distributions can be represented as a diagonal matrix of the form [math]diag(\rho' date=' p...)[/math']. When I come back later I think I'll kick this over to the appropriate math forum as well. Maybe we can get some feedback from formally trained folk.

 

Rev Prez

 

Could you not take the SE tensor in invariant matrix form, diagonalise it and then see what you get?

 

The components in any frame being:

 

[math](p+\rho)U^{\alpha}U^{\alpha}+g^{\alpha\beta}p[/math]

 

I can't remember the diagonalisation procedure right now. I'll grab my book, flick through the chapter on matrices and give it a go a bit later.

[Johnny5]

Trace of the stress energy tensor.

 

[math]

\rho+p_x+p_y+p_z

[/math]

 

Trace of a matrix is the sum of the components along the main diagonal.

 

The first term corresponds to the entry which is related to the time coordinate, the next three are related to the space coordinates.

[Johnny5]

If there's no mass-energy distribution in a region of space then there exists no SE tensor and so the Einstein field equation is zero. .

 

Wait a moment here....

 

 

Are you saying that in a region of vacuum, the trace of the stress energy tensor is zero, but that if I have a billiard ball in my hand, then within the boundaries of the billiar ball the stress energy tensor is nonzero?

[□h=-16πT]

Trace of the stress energy tensor.

 

[math]

\rho+p_x+p_y+p_z

[/math]

 

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

[Johnny5]

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

 

cool... this is very cool

 

right or wrong it's still cool, I am going to learn this very fast.

 

I will pay close attention.

 

 

In the trace above, what are the four terms called from left to right?

[□h=-16πT]

Wait a moment here....

 

 

Are you saying that in a region of vacuum' date=' the trace of the stress energy tensor is zero, but that if I have a billiard ball in my hand, then within the boundaries of the billiar ball the stress energy tensor is nonzero?[/quote']

 

Yes because a billiard ball is a many bodied system. It has its own inner dynamics, such as mass distribution, energy distribution, momentum flux etc. The SE tensor is zero in a vacuum because there is none of the above, simply because there is no matter to describe dynamically. See one of my posts above for the components of the SE tensor in any frame.

[Johnny5]

Could you not take the SE tensor in invariant matrix form' date=' diagonalise it and then see what you get?

 

The components in any frame being:

 

[math'](p+\rho)U^{\alpha}U^{\alpha}+g^{\alpha\beta}p[/math]

 

I can't remember the diagonalisation procedure right now. I'll grab my book, flick through the chapter on matrices and give it a go a bit later.

 

 

Here it is.

 

what are the terms called from left to right?

 

 

i recognize

 

[math]g^{\alpha\beta}[/math]

as the metric tensor.

 

I am interested in the momentum flux you just mentioned.

 

Is there a momentum wave?

 

And if so, what is the speed?

 

Does inertial mass wave?

 

[math]U^{\alpha}[/math]

 

What is the letter p, what is the letter rho? What do they denote?

 

I am going to do my linear algebra review concurrently. I will learn faster that way. Keep going.

 

I'm not afraid to make mistakes, mistakes are easily corrected.