Shortest Paths

[revprez]

Quick question. Does positive definite energy momentum always imply a lengthened geodesic (compared to the vacuum state)? Also, is there an example of macroscopic indefinite energy momentum state in nature?

 

Rev Prez

[Johnny5]

Quick question. Does positive definite energy momentum always imply a lengthened geodesic (compared to the vacuum state)? Also' date=' is there an example of macroscopic indefinite energy momentum state in nature?

 

Rev Prez[/quote']

 

What is energy momentum?

 

Energy and momentum are different.

[Meir Achuz]

revprez:

I assume your question refers to an energy-momentum four-vector with

E>p (with c=1). The velocity, v=p/E, will always be less than 1, implying a lengthened geodesic. By "indefinite energy momentum state", I assume you mean one with E<p. This would have v>1, implying a tachyon. Some speculative theories do postulate tachyons, but they are not in the "standard model", and have never been observed.

[revprez]

My question has to do with the energy momentum tensor. If [math]T_{\mu\nu}[/math] is positive definite, is a geodesic across the corresponding topology always longer than its projection across a Minkowski spacetime? This is obvious for M>>0 in the Schwarzchild metric--the interval increases timelike. I'm asking if it is a general condition of all positive definite [math]T_{\mu\nu}[/math].

 

Rev Prez

[Meir Achuz]

Positive definite for a matrix means that if the matrix is diagonalized the trace will be positive. This trace is just the length squared of the E-p four-vector, so I think my previous answer applies.

By the way, I see you used Latex. I know latex, but how do you get this forum to

typset it? Thanks.

[Johnny5]

I see now that you were referring to GR.

 

So let me ask you this, what does

 

[math] T_{\mu \nu} [/math]

 

have to do with "shortest path"

 

Thank you

[revprez]

Positive definite for a matrix means that if the matrix is diagonalized the trace will be positive. This trace is just the length squared of the E-p four-vector' date=' so I think my previous answer applies.

By the way, I see you used Latex. I know latex, but how do you get this forum to

typset it? Thanks.[/quote']

 

You're only considering positive energy states. If E were negative to begin with, you could still have a timelike path even though the trace is negative.

 

To typeset latex, refer here.

 

Rev Prez

[revprez]

I see now that you were referring to GR.

 

So let me ask you this' date=' what does

 

[math'] T_{\mu \nu} [/math]

 

have to do with "shortest path"

 

Thank you

 

Um, what?

 

Rev Prez

[Johnny5]

Um' date=' what?

 

Rev Prez[/quote']

 

Well I know your question is about GR, and I don't know GR, therefore I cannot answer it. But I am not telling you something, I am asking you something.

 

You asked about a geodesic. Now this i do know something about. It is the concept of the shortest path through a space.

 

So my question was a very simple one, at least I think so...

 

What does the symbol

 

[math] T_{\mu \nu} [/math]

 

Have to do with a geodesic?

 

What is the sequence of logical relations leading from the one to the other?

You ask in post #4

 

whether or not if the energy-momentum tensor T_mn is positive definite, is the geodesic lengthened?

 

I was hoping to learn something from you. Perhaps even assist you in answering your own question.

[Meir Achuz]

Revprez,

The trace is E^2-p^2 so the sign of E is not involved.

[revprez]

Revprez, the trace is E^2-p^2 so the sign of E is not involved.

 

Uh, that's not the trace.

 

Rev Prez

[revprez]

What does the symbol

 

[math] T_{\mu \nu} [/math]

 

Have to do with a geodesic?

 

[math]T_{\mu \nu} [/math] is the energy momentum tensor in [math]R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8 \pi G T_{\mu\nu}[/math]' date=' where [math']R_{\mu\nu}[/math] is the Ricci tensor, R is the Ricci scalar, G is the gravitational constant, and [math]g_{\mu\nu}[/math] is the metric we're solving for.

 

Rev Prez

[Johnny5]

[math]T_{\mu \nu} [/math] is the energy momentum tensor in [math]R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8 \pi G T_{\mu\nu}[/math]' date=' where [math']R_{\mu\nu}[/math] is the Ricci tensor, R is the Ricci scalar, G is the gravitational constant, and [math]g_{\mu\nu}[/math] is the metric we're solving for.

 

Rev Prez

 

Do you realize how that looks to someone who doesn't know GR? If i had to sift through that to find an error, 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.

[Tom Mattson]

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.

[/quote']

 

Have you considered the possibility that he is looking for discussion with someone who knows what he is talking about, and not from a nagging kid who asks a million questions?

[revprez]

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.

 

I'm looking for the shortest path through a region of space-time containing some distribution of matter described dynamically by [math]T_{\mu\nu}[/math]. Specifically, if [math]T_{\mu\nu}[/math] is a positive definite matrix, is a geodesic through it always longer than one through empty space-time?

 

Rev Prez

[Johnny5]

Have you considered the possibility that he is looking for discussion with someone who knows what he is talking about, and not from a nagging kid who asks a million questions?

 

Yes, I considered that. But you know, questions lead to answers.

[Johnny5]

I'm looking for the shortest path through a region of space-time containing some distribution of matter described dynamically by [math]T_{\mu\nu}[/math]. Specifically' date=' if [math']T_{\mu\nu}[/math] is a positive definite matrix, is a geodesic through it always longer than one through empty space-time?

 

Rev Prez

 

This question almost makes sense to me, and I know nothing of GR. Why is that?

 

Revprez, if you don't mind my asking, how many dimensions is space?

 

Also, what is a positive definite matrix?

 

Kind regards

[revprez]

Yes, I considered that. But you know, questions lead to answers.

 

They haven't here.

 

Rev Prez

[revprez]

This question almost makes sense to me, and I know nothing of GR. Why is that?

 

Because you haven't taken the time to pick it up?

 

Revprez, if you don't mind my asking, how many dimensions is space?

 

I don't know. N.

 

Also, what is a positive definite matrix?

 

Any matrix A where Tr(A) > 0.

 

Rev Prez

[Meir Achuz]

E's eq. is irrelevant to T. T depends only on E and p. The problem (usually insoluble) posed by E's eq. is finding g. Writing E's eq. does not change the fact that the trace of T, if diagonalized is E^2-p^2. The short answer to your original question is "Yes."

[Johnny5]

 

Any matrix A where Tr(A) > 0.

 

 

What is the definition of the trace of a matrix?

[Bob182]

Try searching on the net for it, or any maths text book.

[revprez]

E's eq. is irrelevant to T. T depends only on E and p.

 

This is wrong and of no consequence. Yes, the components of [math]T_{\mu\nu}[/math] are [math]\rho[/math] and [math]p[/math]. I am asking if geodesics are always lengthened by positive definite [math]T_{\mu\nu}[/math]? This is a conservation question, nothing more, nothing less.

 

The problem (usually insoluble) posed by E's eq. is finding g. Writing E's eq. does not change the fact that the trace of T, if diagonalized is E^2-p^2. The short answer to your original question is "Yes."

 

I have no idea where you're getting squared terms in the tensor. The components are [math]\rho[/math] and [math]p[/math] for a perfect fluid. The trace is simply [math]\rho + p_x + p_y + p_z[/math].

 

Rev Prez

[Johnny5]

Try searching on the net for it, or any maths text book.

 

I'd rather talk to rev.

[Bob182]

Why so? Can you not just look it up?