I am the first in the world

[Reality Check]

I am not surprised by that either, I was kind of surprised though that a coordinate could be defined by its geometry in certain cases.

A coordinate has no geometry.

Why don't you try doing something less trivial, like, I don't know, checking whether the coordinate description I gave works. Oh yeah, that would take work and thought, and it is much easier to go on about trivialities.

Your coordinate description works. That is really obvious since the radial coordiante labels nested concentric spheres and these spheres have an area and Gaussian curvature.
So what?

Your 'coordinate system' is as you stated arbitrary. You have no way of determining what description of the radial coordinate is the better or more valid or whatever other then your personal opinion.

 

[tensordyne]

I notice you haven't responded to my observation that your "solution" isn't a solution (given the information you provided about it).

That is because I did not find it very interesting or sophisticated in thought. I responded to RC because he asked honest questions. I am responding to Vorpal because he said one thing I did find interesting and also because now I would like to get to the bottom of the unresolved issues.

I did not respond to your non-solution comment because you haven't even seen my solution so what is the point? Thanks for your general comments, when I decide to put out the solution, I hope you will have a look.

 

[tensordyne]

A coordinate has no geometry.


Your coordinate description works. That is really obvious since the radial coordiante labels nested concentric spheres and these spheres have an area and Gaussian curvature.
So what?

Your 'coordinate system' is as you stated arbitrary. You have no way of determining what description of the radial coordinate is the better or more valid or whatever other then your personal opinion.

I am done with you for now Reality Check. I have stated what I need to say in response to your questions and you are now just rehashing what you have kind of said before.

Now I return to my previously scheduled conversation with Vorpal, because as far as I can tell, he actually knows quite a bit about this subject. I don't know, maybe you do too RC... whatever, unresolved issues lie in that direction.

All the best to you all!

 

[tensordyne]

maybe, maybe not

If it has g's, you're basically done. Think of a given curve as an immersion of another one-dimensional manifold, and equip it with the pullback metric. The curve is now a manifold equipped with a metric that gives square of its length along it. Voila: the square of the length differential in coordinate-free language.

I think I agree with the general import of the statements above, even though they seem a bit hand-wavy. I do not know how a pullback metric works however. It all sounds plausible, at the very least.

I'm not sure why you're so resistant about the fact that all of the core concepts of differential geometry are definable in a coordinate-free language. If you're not interested in that kind of thing, ok; if you don't find it useful, ok; but this insistence that the concepts need coordinates in order to be meaningful just doesn't mesh with the facts.

My only point was that if I give you a metric expression like the ones used so far for ds^2 for instance, it is not enough to understand what the manifold is without also understanding how the coordinates are measured. I think you agree with this. Imho, some people who work on GR forget this simple fact.

This is nuts. It's a very general statement in the first place, so of course it requires a "generality". What do you want me to do? Characterize a specific geometry in coordinate-free language? You haven't given me one in the first place! Although for any exact solution of GTR, I don't see why that wouldn't be possible in principle, whether or not it's worth doing (and sometimes requires stepping into algebraic geometry). I've seen it done for several nontrivial geometries, including Schwarzschild.

Yes, I wanted you to write down a formula for finding ds^2 for a specific geometry in coordinate-free form. In my own estimation, I thought this could not be done, but it would be great if I was proven wrong on that. I am sorry that I did not make clear which metric formula was to be put in coordinate-free form. It was the ds^2 formula given by you originally that involves the lambda's and nu's.

I think we both agree that there are lots of coordinate-free forms for finding various things like area, and so on, so it is possible to have equations expressing things in coordinate-free form, there you have it.

Why would I you believe I thought this? I've been saying the complete opposite for a long time now. Names are completely and utterly irrelevant mathematically or physically. They have practical value in that having common notational expectations is convenient for communication, but while that's an important concern, it's not a mathematical or physical one.

I think the sentence should be "Why would you..."

If something is convenient for communication in physics and math, then why is that irrelevant? It is irrelevant in an abstract sense, but in terms of communication, it is extremely important and relevant.

This began when I made the statement that conventionally, one would characterize spherical symmetry in terms of the existence of a coordinate chart in which the metric takes a certain form, and further said one can put this criterion into coordinate-independent language. Later, I repeated this so there wouldn't be misunderstanding:

OK, tracking so far.

The condition characterizes spherically symmetric spacetimes. However, what I did mess up on was that (1)'s λ does not correspond to (2)'s λ, and should have made it more general for clarity. Mea culpa; I didn't pay enough attention to the metric because the point was simply that it's possible to characterize spherical symmetry without any reference to coordinates, and as I said before, (1) was simply motivation as to why it would be the case.

Alright.

One can fix this oversight by altering (1) to read (1') [latex]$ds^2 = Fdt^2+2Gdtdr+Hdr^2 + e^{2\lambda}(d\theta^2 + \sin^2d\phi^2)$[/latex] with every coefficient a smooth function of t,r only, as before. It's strictly more general than (1), so everything satisfying (1) automatically satisfies this criterion as well.

OK.

No, it isn't, because there is no r in (2) at all.

I hope it is a function of something then, not just a constant? And if it is a function, then without further definition, it is an arbitrary function, and hence can be whatever you want it to be (could have 50 million variables, singular all over the place, etc. etc.)? But then, how would I know that that function gives a Schwarzschild Geometry assuming even it is a function of one variable?

I do not think this is what you mean though, I think you mean it to be the same as the function lambda used in (1') above, no? But that function is a function of r. Very odd.

To be honest, I am still not sure how to interpret the equations you gave involving g, g_1, etc. The plus sign means matrix addition in the formula for g right? Are you being even more abstract then that? Plus, the
g_2(d\lambda) = 0 equation, what is the motivation behind that, and if possible how do you express that formula in a form I am used to with indices and all, if that is possible?

I ask the above with no malicious intent what-so-ever, I am just trying to understand.

It's not coordinate-free. It's coordinate-independent in just the sense you say here (the coordinates can be whatever), but not coordinate-free (because they're still there). Those meanings are distinct, and represent different approaches to geometry. If you got this idea from wikipedia, then it is simply wrong.

Sounds OK. The only problem I have is that the dx's should be interpreted in terms of math as infinitesimal vectors (contravariant vectors) and in physics as just representing really small vectors. If the dx's are vectors in this sense, and the g's I would hope can be agreed upon to be coordinate-free in whatever sense that that is needed for (it is the metric tensor after all), then the whole expression would then be coordinate-free and coordinate-independent. Maybe that is not the modern approach though.

Here is an article by wikipedia about Abstract Index Notation as a side note.

http://en.wikipedia.org/wiki/Abstract_index_notation

 

[tensordyne]

Oh, I do not mean to imply that the ds^2 formula involving g_mn I gave is in abstract index notation or anything, I just thought that the article was interesting and somewhat related to what is being discussed.

 

[Reality Check]

I am done with you for now Reality Check. I have stated what I need to say in response to your questions and you are now just rehashing what you have kind of said before.

If you cannot explain what you mean coherently then of course I will ask you to explain it better !

Apparently my questions are no longer honest!

I responded to RC because he asked honest questions.

I think that it is obvious that a coordinate has no geometry. Sorry for pointing this out to you in the hope that you would clarify what you mean.

What I have read makes it clear that the radial coordinate in Schwarzschild coordinates labels surfaces with both area and Gaussian curvature. So your A and B scenerios that try to associate r with area or curvature are actually the same. Both A and B describe a set of surfaces.

Thus your concept of a "geordinate system" is in doubt, especially since the two scenerios above refer to the same geometry (same metric).

Perhaps you can give another example?
Or perhaps a mathematical definition of your system?

 

[sol invictus]

That is because I did not find it very interesting or sophisticated in thought.

It may not be "interesting or sophisticated in thought", but it is true, and a serious problem for you.

I did not respond to your non-solution comment because you haven't even seen my solution so what is the point?

The point is, you told me enough for me to know you are wrong. More specifically, two things you said are in contradiction (the stress tensor you gave is inconsistent with your statement that the metric derivatives are discontinuous at the shell).

Here is "your" "solution" (which is quite obvious to anyone with any experience in GR):

[latex]$ds^2 = -f(r)dt^2+dr^2/f(r) + r^2 d\Omega^2$[/latex], where [latex]$f(r) = 1-r^2/R^2$[/latex] for r<R₀, and [latex]$f(r) = 1-2Gm/r$[/latex] for r>R₀, and there is a simple relation between Gm, R, and R₀. Surprised?

As I told you, this does not have the Einstein tensor you asserted it does, because there is a delta function at r=R₀ that you missed.

And FYI, such solutions have been considered (with considerable care) by various other people before in the past. They are of no physical relevance.

 

[tensordyne]

RC, read post #71. It is all in there. I do not like having to repeat myself. Your questions are honest, they just have already been answered is all. Thanks for asking though.

 

[tensordyne]

It may not be "interesting or sophisticated in thought", but it is true, and a serious problem for you.



The point is, you told me enough for me to know you are wrong. More specifically, two things you said are in contradiction (the stress tensor you gave is inconsistent with your statement that the metric derivatives are discontinuous at the shell).

Here is "your" "solution" (which is quite obvious to anyone with any experience in GR):

[latex]$ds^2 = -f(r)dt^2+dr^2/f(r) + r^2 d\Omega^2$[/latex], where [latex]$f(r) = 1-r^2/R^2$[/latex] for r<R₀, and [latex]$f(r) = 1-2Gm/r$[/latex] for r>R₀, and there is a simple relation between Gm, R, and R₀. Surprised?

As I told you, this does not have the Einstein tensor you asserted it does, because there is a delta function at r=R₀ that you missed.

And FYI, such solutions have been considered (with considerable care) by various other people before in the past. They are of no physical relevance.

Well, all that I can say is let me get back to you, I wrote the metric in a file and it is on another computer I do not have access to currently and I do not remember it well enough to reproduce it right now (I would not like to give out the wrong metric).

I am curious though, the metric you give, I would have to do the calculations and all but are you sure the Einstein Tensor for it does not have a discontinuity at r = R_0 and not a delta function there? Just because you have a discontinuity does not mean you have a delta-function (although the derivative of a function at a discontinuity is a delta-function).

Until then.

 

[sol invictus]

Well, all that I can say is let me get back to you, I wrote the metric in a file and it is on another computer I do not have access to currently and I do not remember it well enough to reproduce it right now (I would not like to give out the wrong metric).

The one I posted is the unique metric with the properties you described.

I am curious though, the metric you give, I would have to do the calculations and all but are you sure the Einstein Tensor for it does not have a discontinuity at r = R_0 and not a delta function there?

Yes.

Just because you have a discontinuity does not mean you have a delta-function (although the derivative of a function at a discontinuity is a delta-function).

The Einstein tensor is second order in derivatives. Anyway it's trivial to compute, and it does have a divergence.

 

[Reality Check]

RC, read post #71. It is all in there. I do not like having to repeat myself. Your questions are honest, they just have already been answered is all. Thanks for asking though.

I read post #71 when you posted it. There is nothing there.

But if you want then I will take your answers as written. That means that your 'geordinate systems' do not exist as I pointed out in my previous post. They are just coordinate systems.

You seem to think that the radial coordinate labels different things

• spheres with an area.
• spheres with a Gaussian curvature.

The fact is that the radial coordinate label spheres. These sphers have

• an area and
• a Gaussian curvature.

 

[tensordyne]

The one I posted is the unique metric with the properties you described.

Yes.

The Einstein tensor is second order in derivatives. Anyway it's trivial to compute, and it does have a divergence.

Well then I am kippered it seems. Thanks for pointing this out. Is there a name to this metric? It seems then one should check to make sure the metric is continuous to 2nd derivs or are first good enough generally speaking?

I was happy I solved a nonlinear 2nd order differential equation too.
Oh well. Turns out sol you had the most important thing to say. Thanks again.

[P.S. RC, think whatever you want, I am already bored with going into it with you]

 

[sol invictus]

Well then I am kippered it seems. Thanks for pointing this out. Is there a name to this metric?

There was some nonsense about "dark energy stars" a few years ago. I think those are like the metric I posted, but with the infinitely thin shell replaced by a thick shell with some (probably unphysical) energy density inside.

None of these solutions have any bearing on the existence of black holes. To show BHs don't exist, you'd have to prove that stars cannot collapse into them, not that there are some other random vaguely star-like solutions. And you'd need to deal with the very strong observational evidence for BHs too.

It seems then one should check to make sure the metric is continuous to 2nd derivs or are first good enough generally speaking?

You should check Einstein's equations, first of all. If there are discontinuities in anything you have to do that carefully. Then you should compute as many curvature invariants as possible, or find an argument that they are all finite. Even then, other things can still go wrong (closed timelike curves, violations of energy conditions, etc.).

GR is hard.

 

[tensordyne]

Hmmm, I hate to have to ask this question then, but what is the unphysicalness of the metric we have been discussing thus far (besides at the boundary of the sphere)? I only ask because I intended to look up whatever name you gave but I am not sure if I will be able to find the specific papers you mention. Is it Vorpal's note that it would be of a light-like fluid (not completely sure why that would be true either), or something else?

 

[Vorpal]

BTW, I meant that the perflect fluid has a speed of sound equal to speed of light: mass-energy flow is lightlike. That was inept wording on my part.

 

[tensordyne]

BTW, I meant that the perflect fluid has a speed of sound equal to speed of light: mass-energy flow is lightlike. That was inept wording on my part.

How can you tell that? I am going to call the solution the Dark Star Sol. for the moment (one given by sol invictus). The Dark Star Solution is static so I am not sure what would be flowing is all.

 

[tensordyne]

Oops, forgot, it is not supposed to be a solution, call it the Dark Star metric maybe then.

 

[sol invictus]

BTW, I meant that the perflect fluid has a speed of sound equal to speed of light: mass-energy flow is lightlike. That was inept wording on my part.

Actually, that isn't quite right. First of all, to determine the speed of sound requires dp/drho, not p/rho. Second, the equation is c_s²=dp/drho. If we assume dp/drho=p/rho, that gives c_s²=-c², not +c².

What that means is that dark energy of that form does not propagate sound waves. Of course there might still be waves, but their speed isn't determined by p/rho.

 

[Vorpal]

How can you tell that? I am going to call the solution the Dark Star Sol. for the moment (one given by sol invictus). The Dark Star Solution is static so I am not sure what would be flowing is all.

Given an observer with four-velocity u, the density of four-momentum is T^μ_νu^ν as measured by that observer. If you take the timelike and one of the other components for your tensor, you get dp/dρ, which is the square of the speed of adiabatic sound waves in that direction.

Or am I crazy?

 

[Reality Check]

[P.S. RC, think whatever you want, I am already bored with going into it with you]

P.S. tensordyne, if you cannot be bothered giving supporting evidence for your assertion then I (at least) will conclude that you have no such evidence.

As already noted, your 'geordinate system' is just a coordinate system.
The definition of the radial coordinate in Schwarzschild coordinates makes this clear

The defining characteristic of Schwarzschild chart is that the radial coordinate possesses a natural geometric interpretation in terms of the surface area and Gaussian curvature of each sphere. However, radial distances and angles are not accurately represented.

A value of r (the radial coordinate) is a label for a specific sphere. That sphere has values for

• surface area and
• Gaussian curvature

It does not matter if you like area or curvature: r still has the geometric meaning of labeling the sphere. r is alway a 'geordinate' and so we should call it what it is - a coordinate.