I am the first in the world

[sol invictus]

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?

I'm not sure I understand what you mean by "take the timelike and one of the other components for your tensor, you get dp/dρ".

Let's do an example. Consider a pure cosmological constant. The stress tensor for that has p=-\rho, but there are no sound waves associated with it, because it has no dynamics at all.

Another example. Consider a scalar field f with a potential V(f). Let's say V'(0)=0 is a minimum, and V(0)>0. Then the state with f=0 has p=-rho, but the speed of sound waves will be a function of the wavelength (because in general V''(0)>0, so the perturbations are massive).

 

[tensordyne]

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



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.

Haha, OK, sounds good. Here is a hint, in some geometries (uncountably infinitely many in fact), the Gaussian Curvature is not constant over Schwarzschild coordinates r, so you would not be able to use the Gaussian radius of curvature as one of your 'coordinates' because then the Guassian radius of curvature would not be constant over said surface (r would not equal the GRC over the whole of the surface). The problem, I think, is that you do not understand what geometry means.

Notice that either of the two other people on the thread are not complaining about my statements in this regard? It was discounted as trivial, but not incorrect. Let me ask you this, how do you even define what is a sphere or not a sphere when you have a warped up non-spherically symmetric geometry?

You want examples? OK, not that they would probably mean anything to you, but try this one.

[latex]
ds^2 = dt^2 - dz^2 - (z^{500} \sin \theta)^e d\Omega^2
[/latex]

(The coordinate system is time + cylindrical)

I have not checked, but I am relatively sure that such a geometry will not give K = 1/r^2.

OK, it is all good. Have a nice day RC.

 

[tensordyne]

I'm not sure I understand what you mean by "take the timelike and one of the other components for your tensor, you get dp/dρ".

I wonder if u = (1, 0, 0, 0) type situation and choose one of the mu's in the expression Vorpal gave?

 

[sol invictus]

I wonder if u = (1, 0, 0, 0) type situation and choose one of the mu's in the expression Vorpal gave?

That would simply give you T^0\mu, which is a vector with components (rho, J), where J is the matter current (zero in the dark star solution). It's got nothing much to do with dp/drho.

 

[Vorpal]

I'm not sure I understand what you mean by "take the timelike and one of the other components for your tensor, you get dp/dρ".

For an observer with velocity u, the measured four-momentun density should be dP^μ/dV = T^μνu_ν, ...
Oh, duh. I'm an idiot. I just destroyed why I thought DEC has anything to do with speed of sound.
That's what happens what one accepts claims in textbooks without proving them... methinks I need to brush up on fluids, because they seem to think it obvious.

Let's do an example. Consider a pure cosmological constant. The stress tensor for that has p=-\rho, but there are no sound waves associated with it, because it has no dynamics at all.

Ok, fair point. The stress-energy tensor we're considering here is right on the verge of violating the dominant energy condition; that much seems very clear. What I might be very confused on is the connection between the DEC and the sound waves in perfect fluids, because that's part of the way I thought was the right way to interpret the DEC physically.

Suppose we've restricted the stress-energy tensor to have the perfect fluid form. If the system has no dynamics, then I agree it doesn't make sense to talk about sound waves. But is there any general statement we can make while excluding that case? E.g.: suppose the fluid can support a small adiabatic perturbation. Regardless of whether it's already there, can anything be said about its propagation if it was produced?

 

[alfa1]

Some of you when will die, will meet with Hitler, Stalin, Osama bin Laden…
Anyway when you’ll die please send me a ‘sign’ if my theory is right:
Afterlife in Dark Matter - on my website: http://www.adrianferent.tk

God ‘created’ the small visible Universe and the 5 times bigger, larger the ‘dark matter Universe’.
Buddha, Jesus Christ, Mohammed… are alive, are living in this ‘dark matter Universe’.

 

[Reality Check]

Haha, OK, sounds good. Here is a hint, in some geometries (uncountably infinitely many in fact), the Gaussian Curvature is not constant over Schwarzschild coordinates r, so you would not be able to use the Gaussian radius of curvature as one of your 'coordinates' because then the Guassian radius of curvature would not be constant over said surface (r would not equal the GRC over the whole of the surface). The problem, I think, is that you do not understand what geometry means.

The problem is that, I think, is that you do not understand what the radial coordinate in the Schwarzschild coordinates is.
Here is a hint: It is not a "Gaussian radius of curvature". It is a label for a spahere with a specific area and Gaussian curvature.

Let me ask you this, how do you even define what is a sphere or not a sphere when you have a warped up non-spherically symmetric geometry?

Let me ask you this: Can you understand the Schwarzschild coordinates article where it shows that 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"?

Your example is dumb. It has no radial coordinate. Thus a relationship between the non-existent r and Gaussian curvature is impossible.
I do not expect that a general metric with a radial coordinate will always result in K = 1/r^2.

OK, it is all rather dumb. Have a nice day tensordyne.

sol invictus and Vorpal: Maybe you can chime in here since tensordyne thinks your silence means he is right.
Hopefully you can explain clearly what he means and that I am wrong.

To me it seems obvious that the radial coordinate is as described in the Schwarzschild coordinates article. It is a label for a specific sphere in "family of nested round spheres". That sphere has an surface area and Gaussian curvature related to the value of r.
Thus r is always a coordinate that descibes the geometry (his 'geordinate').

tensordyne also seems to think that the sphere's Gaussian curvature is more important than it's surface area and that r can somehow be defined to only give the Gaussian curvature.
His general idea seems to be that there are things he calls "geometry dependent coordinates" (like the radial coordinate) in a metric. The problem as I see it is that a metric gives the geometry (lengths and angles) to a manifold. Thus there is only 1 geometry.

Of course if you have a different metric then it will define a different geometry. You can express that metric using the same symbols as another metric. So a coordinate with the same symbol will be dependent on the geometry. An example may be flat space described in spherical coordinates and the Schwarzschild metric. Both have r in them. The r's mean different things.

 

[tensordyne]

thanks for reminding me...

Let me ask you this (yes/no):

1. Should coordinate systems be independent of geometry?
2. Is a 2D sphere defined as a manifold(surface) with the topology of the surface of a globe(you can use common understanding here) and with constant gaussian curvature?
3. Assuming you say yes to 2. above, are the Schw. coordinates defined for the space part as a nested set of spheres (spheres defined as per 2.)?

p.s. The metric I gave is dumb, but not for the reason you listed. It is dumb because it is not actually in time + cylindrical. The problem was the solid angle. The general principle is still the same though.

 

[tensordyne]

Some of you when will die, will meet with Hitler, Stalin, Osama bin Laden…
Anyway when you’ll die please send me a ‘sign’ if my theory is right:
Afterlife in Dark Matter - on my website: http://www.adrianferent.tk

God ‘created’ the small visible Universe and the 5 times bigger, larger the ‘dark matter Universe’.
Buddha, Jesus Christ, Mohammed… are alive, are living in this ‘dark matter Universe’.

I hope you are not serious, because if you are, seek medical attention immediately!

 

[sol invictus]

sol invictus and Vorpal: Maybe you can chime in here since tensordyne thinks your silence means he is right.
Hopefully you can explain clearly what he means and that I am wrong.

To me it seems obvious that the radial coordinate is as described in the Schwarzschild coordinates article. It is a label for a specific sphere in "family of nested round spheres". That sphere has an surface area and Gaussian curvature related to the value of r.

Yes, that's all fine.

tensordyne also seems to think that the sphere's Gaussian curvature is more important than it's surface area and that r can somehow be defined to only give the Gaussian curvature.

For a sphere, the two are essentially the same thing, so one cannot be more important than the other.

Of course if you have a different metric then it will define a different geometry.

Maybe, maybe not. Spherical coordinates and Cartesian coordinates on flat space both describe the same geometry.

In general, it's not at all easy to decide whether two metrics define the same geometry or not.

You can express that metric using the same symbols as another metric. So a coordinate with the same symbol will be dependent on the geometry. An example may be flat space described in spherical coordinates and the Schwarzschild metric. Both have r in them. The r's mean different things.

Yes.

 

[sol invictus]

Let me ask you this (yes/no):

1. Should coordinate systems be independent of geometry?

The coordinates and metric more or less completely define the geometry. So if I understand what you are asking, the answer is no.

2. Is a 2D sphere defined as a manifold(surface) with the topology of the surface of a globe(you can use common understanding here) and with constant gaussian curvature?

That depends a bit on context (for example if you only care about topology, the surface of a cube or egg is equivalent to the surface of a sphere) - but usually, yes.

3. Assuming you say yes to 2. above, are the Schw. coordinates defined for the space part as a nested set of spheres (spheres defined as per 2.)?

Again, I'm not sure what you're asking. The Schw. metric does include a set of spheres, which are "nested" in a certain sense.

 

[Reality Check]

Let me ask you this (yes/no):

1. Should coordinate systems be independent of geometry?
2. Is a 2D sphere defined as a manifold(surface) with the topology of the surface of a globe(you can use common understanding here) and with constant gaussian curvature?
3. Assuming you say yes to 2. above, are the Schw. coordinates defined for the space part as a nested set of spheres (spheres defined as per 2.)?

p.s. The metric I gave is dumb, but not for the reason you listed. It is dumb because it is not actually in time + cylindrical. The problem was the solid angle. The general principle is still the same though.

1. No.
2. Yes.
3. Question not clear enough.
The Schwarzschild coordinates do define a family of nested spheres (nested in the sense that changing r changes which sphere you are referring to).

P.S. Then the metric was doubly dumb (no r, not 'in time + cylindrical')

 

[dafydd]

I hope you are not serious, because if you are, seek medical attention immediately!

Just ignore him,it's the kindest thing to do.

 

[tensordyne]

Hmm, interesting responses and interesting that both you RC and sol agree, it seems.

1. No.

I expected yes, but oh well. So then are you saying that, for instance, if I use 2D Cartesian and in one space the metric is

1.
[latex]
ds^2 = dx^2 + dy^2
[/latex]

and in another space the metric is

2.
[latex]
ds^2 = dx^2 + \sin x dy^2,
[/latex]

that somehow the coordinates in x and y for both depend on which metric is used? In 1. you use 2D Cartesian of style 1. and in 2. 2D Cartesian of style 2., whatever it is that that means? Measurement of x and y will use a different method for the two metrics?

Please elaborate if you can (both of you). I am merely interested in why you think this is the case.

2. Yes.

I concur.

3. Question not clear enough.
The Schwarzschild coordinates do define a family of nested spheres (nested in the sense that changing r changes which sphere you are referring to).

I can take it from your answer that 3. is a yes, given your further elaborations.

P.S. Then the metric was doubly dumb (no r, not 'in time + cylindrical')

Nope, actually, it was careless, but I detected the error (as if you have never made a slip up in an exacting statement here on the forums RC, really!). The statement you made above is dumb in the sense of ignorant, because, why does there have to be an r in the metric?

Have you not heard of coordinate transformations? You know there are lots and lots of coordinate systems out there RC. There is no a priori reason to require the metric to have an r as a coordinate, or for the metric to be expressed in such a coordinate system. Don't believe me? Go ask one of your experts here. I am pretty sure they will concur.

 

[Reality Check]

I expected yes, but oh well. So then are you saying that, for instance, if I use 2D Cartesian...

I would guess that the meaning of the coordinates differs between the 2 metrics (two geometries).
In one you have x and y in a flat geometry, i.e. 2D Cartesian coordinates.
In the other you have x and y in some other kind of geometry, i.e. probably not 2D Cartesian coordinates.

For example compare spherical coordinates to Schwarzschild coordinates. Both have r in them but they are not the same r.

I can take it from your answer that 3. is a yes, given your further elaborations.

No you cannot. The answer is that your question was not clear enough to answer.
The next sentence states a mathematic fact: The Schwarzschild coordinates do define a family of nested spheres (nested in the sense that changing r changes which sphere you are referring to).

The statement you made above is dumb in the sense of ignorant, because, why does there have to be an r in the metric?

That is dumb in the sense of careless, because you stated a metric which did not contain r and then "I have not checked, but I am relatively sure that such a geometry will not give K = 1/r^2".
That is correct - a metric that does not contain r will not give K = 1/r^2.

If you wanted an r in the metric then you should have transformed the coordinates to put an r in it.

 

[AdMan]

Some of you when will die, will meet with Hitler, Stalin, Osama bin Laden…
Anyway when you’ll die please send me a ‘sign’ if my theory is right:
Afterlife in Dark Matter - on my website: http://www.adrianferent.tk

God ‘created’ the small visible Universe and the 5 times bigger, larger the ‘dark matter Universe’.
Buddha, Jesus Christ, Mohammed… are alive, are living in this ‘dark matter Universe’.

I think it's best if you stay out of this thread; there's actual physics being discussed.

 

[sol invictus]

I expected yes, but oh well. So then are you saying that, for instance, if I use 2D Cartesian and in one space the metric is

1.
[latex]
ds^2 = dx^2 + dy^2
[/latex]

and in another space the metric is

2.
[latex]
ds^2 = dx^2 + \sin x dy^2,
[/latex]

that somehow the coordinates in x and y for both depend on which metric is used?

Of course - why wouldn't they? For example, if that sin x was squared, the second metric is at least locally that of a sphere with unit radius, and "x" and "y" are coordinates that people usually denote \theta and \psi.

In 1. you use 2D Cartesian of style 1. and in 2. 2D Cartesian of style 2., whatever it is that that means? Measurement of x and y will use a different method for the two metrics?

Please elaborate if you can (both of you). I am merely interested in why you think this is the case.

2. is not a Cartesian coordinate system. Did you think that if you decided to label the coordinates with the letters "x" and "y", that would make them Cartesian?

 

[tensordyne]

message, in a metric bottle...

I would guess that the meaning of the coordinates differs between the 2 metrics (two geometries).
In one you have x and y in a flat geometry, i.e. 2D Cartesian coordinates.
In the other you have x and y in some other kind of geometry, i.e. probably not 2D Cartesian coordinates.

I hope you will not take too much offence, but as far as I know, the above shows that you do not understand how metrics and coordinate systems work. You did say it was your guess after all.

Let's flesh out things some. First off, how does one measure the metric? Unless you believe the metric is some a priori thing, it should be measurable right? Before how metrics can be measured can be understood though, one needs to understand local metrics. I will stay with GR as my context.

In a small enough region of space-time (locally, that is), the metric should be Lorentzian (free fall frame). In other words, the metric in such a situation will be

[latex]
(1) ds^2 = c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2,
[/latex]

where tau is the propre-time. (please, no comments about how on the inside of a blackhole the signature of the space changes, I am well aware of those ideas and I am just talking about normal, plain jane situations here)

Now, how does this help in measuring a metric? Well, if we have another observer who is not as close to things but measures in their own coordinate system (of which type does not change, even if measuring other geometries in GR) measures their own differences

[latex]
(2) dx^{\mu},
[/latex]

then conceivably I know of two ways to measure the metric. Perhaps there are more, but these are the ones I know of. A note, we are trying to find the metric at the point P.

I. Use the Differences.

In physics we do not measure dx's but small differences. One could therefore for instance say the measured metric at P is

[latex]
(3) g_{\mu\nu} =
(\Delta ct / \Delta x^{\mu})(\Delta ct / \Delta x^{\nu}) +
\mbox{"same for x, y, z"}.
[/latex]

Experimentally one would want several differences measured from a given point where the metric is to be measured. Then the metric at that point would be averaged.

II. Use the Square Lengths.

The ds^2 from equation (1) should be the same as the ds^2 from the following:

[latex]
(3) ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}.
[/latex]

(understand that experimentally though we would be using small differences and not differentials, also, there is no requirement in GR that coordinates not measured locally have to have a Lorentzian signature RC, that is part of the point behind GR before you go complaining how dumb I am again).

So you get the ds^2 from (1) because the differences in both coordinate systems are known, and that can give you ds^2 immediately.

Now you could either measure a bunch of differences from P to other points close by and use (3) in one of two ways to find g at P.

1. Measure from P to 16 other points and use matrix algebra (the differences are constants and the metric components are the unknowns of which there are 16 in number corresponding to the metric tensor components).

2. Over determine the metric in (3) by measuring more then 16 points from P and use the method of least squares (I think you can do least squares on linear regressions right?).

If I were an experimentalist I would use method II.2. since it will give the tightest fit. It does not matter though because the point is that the coordinate system x^{\mu} does not change, and nor should it. A coordinate system is like an algorithm for finding points and that is bijective between how the points are labeled and the points themselves. That is all it is.

Incidentally, this is part of the reason why in math they talk about Tangent Vector spaces and all that (for sol invictus and Vorpal), to make the idea of metrics precise. Vorpal, you have been kind of quiet recently.

Another note is that method I. will always give you symmetric metrics as per

[latex]
g_{\mu\nu} = g_{\nu\mu}.
[/latex]

It is therefore perhaps better to always use II. methods in measuring metrics because it can allow for non-symmetric metrics, as far as I can tell. We do not want to make any assumptions when measuring things after all.

For example compare spherical coordinates to Schwarzschild coordinates. Both have r in them but they are not the same r.

Haha, I agree that the two r's are not the same, but that is getting ahead of the story as it were.

No you cannot. The answer is that your question was not clear enough to answer.
The next sentence states a mathematic fact: The Schwarzschild coordinates do define a family of nested spheres (nested in the sense that changing r changes which sphere you are referring to).

Fair enough, the logical import of your statements was that, but it is your prerogative whether to say yes or no, even if it disagrees with logic. I can just hold you to what you have said already.

That is dumb in the sense of careless, because you stated a metric which did not contain r and then "I have not checked, but I am relatively sure that such a geometry will not give K = 1/r^2".
That is correct - a metric that does not contain r will not give K = 1/r^2.

If you wanted an r in the metric then you should have transformed the coordinates to put an r in it.

Now I get to use a word that you much love to abuse, trivial. Transforming from a coordinate system that is in (time + cylindrical) to one containing r as per (time + polar) is trivial. The point was more abstract then that anyways, and is, sadly, lost on you, because you, in my estimation, do not have the maths currently.

Calculating K can be a pain. Let's see how you do with the above stuff on how to measure metrics before I give a metric with a K I have worked out completely. My guess is you will not get the ideas about how metrics and coordinate systems work, but perhaps sol invictus might.

 

[Vorpal]

The metric is Lorentzian in any frame; you probably meant Minkowksi.

Anyway, if I now understand you correctly, then I was right in guessing that frame fields/tetrads already do what you want to do, and you're simply calling 'coordinate system' what mathematicians, and at least some physicists, would call 'frame'. An observer can some axes operationally in some appropriate way to measure things in the immediate vicinity and thus create a frame. Repeating this over a region gives at least part of a local frame field. A local frame field determines the metric in a region.

Neglecting any concerns about the mathematical formalism of frames, I don't understand is why you didn't just say "yep, a coordinate system is a frame, rather than a chart" back then, because I wouldn't have gotten the impression that you were looking you meant something different from both charts and frames, and have it make a lot less sense.

Or are you?

Vorpal, you have been kind of quiet recently.

Several consecutive days of insomnia have left me a bit pooped. I'm going to go get some beer soon and see if it makes me sleepy enough.

 

[Reality Check]

...snipped things I already knew...
Haha, I agree that the two r's are not the same, but that is getting ahead of the story as it were.

This seems to be the entire point of the story.
You are assuming that just because you use x and y for 2D Cartesian coordinates in flat space that a metric for non-flat space that has x and y in it is using 2D Cartesian coordinates.

Fair enough, the logical import of your statements was that, but it is your prerogative whether to say yes or no, even if it disagrees with logic. I can just hold you to what you have said already.

You do not seem to understand what I stated so here is another try:
Your question was not clear enough for any answer . Thus I did not answer it. I did not say yes. I did not say no.
There was no disagreement with logic because there was no understandable question to be answered.

What I have already sid was:

Originally Posted by tensordyne
Let me ask you this (yes/no):

1. Should coordinate systems be independent of geometry?
2. Is a 2D sphere defined as a manifold(surface) with the topology of the surface of a globe(you can use common understanding here) and with constant gaussian curvature?
3. Assuming you say yes to 2. above, are the Schw. coordinates defined for the space part as a nested set of spheres (spheres defined as per 2.)?

1. No.
2. Yes.
3. Question not clear enough.
The Schwarzschild coordinates do define a family of nested spheres (nested in the sense that changing r changes which sphere you are referring to).