Undergrad What's the underlying frame of the Einstein's Field Equation?

[PeterDonis]

I gather that all these coordinates systems assume that the metric is already known?

No. If you would take the time to read the references you have been given, you would see how the coordinates are constructed using assumptions about the symmetries of the spacetime, before solving the EFE to find the full metric. You should realize by now that your current assumptions about how things are done are highly likely to be wrong, so you should stop making them.

For instance in the FRW the curvature is uniform everywhere

The spatial curvature is found to be uniform everywhere, but that is not an assumption, that is a deduction from other assumptions, made in the process of solving the EFE.

for "local" variations of curvature, like in a relative small region with several comparable masses, would it still be accurate?

The FRW metric is treated by cosmologists as an average description of the universe on large distance scales; it is not meant to capture variations on the scale of, say, the solar system, or even a small group of galaxies with empty space between them. The distance scale for the averaging is typically taken to be tens to hundreds of millions of light years.

There are papers in the literature investigating the possibility that variations in density (and hence curvature) on smaller scales might affect the large scale dynamics; so far that research is still open, but I think it's fair to say it has not yet produced any results that have convinced cosmologists in general.

wouldn't the radial coordinates be a little off because of the gravitational lens effect?

"A little off" in what sense? What would they be "off" from? Remember that the radial coordinate is not necessarily the same as radial distance.

I was thinking more of a relatively small region with comparable and meaningful masses, like maybe a neutron stars field, or just two orbiting one around the other.

For this case, the FRW solution is not what is used. There are no known exact solutions for isolated systems containing more than one gravitating mass (such as the binary pulsar systems that have given us evidence for gravitational wave emission); those kinds of cases can only be solved numerically.

 

[Pyter]

No. If you would take the time to read the references you have been given, you would see how the coordinates are constructed using assumptions about the symmetries of the spacetime, before solving the EFE to find the full metric.

I meant: if the FRW is a "metric", that implies it's already a solution of the EFE which assumes a homogeneous distribution of the masses in the area where it's used.
Citing from your reference:

The Friedmann–Lemaître–Robertson–Walker (FLRW; /ˈfriːdmən ləˈmɛtrə ... /) metric is an exact solution of Einstein's field equations of general relativity; it describes a homogeneous, isotropic, expanding (or otherwise, contracting) universe that is path-connected, but not necessarily simply connected.[1][2][3]

 

[vanhees71]

Yes, and as was stressed before in this thread, that's a description of the "large-scale structure" of our universe, i.e., after "coarse graining" (i.e., averaging) over a large distance scale of a few 100 light years. It's based on the assumption of the "cosmological principle", which says that when averaging out the local "fluctuations" of the distribution of energy-momentum-stress of matter and radiation, the universe is homogeneous and isotropic, and this leads to the ansatz of maximally symmetric spacetimes and finally the FLRW solution of the EFE. This implies that on this level of description the energy-momentum-stress tensor is that of an ideal fluid.

 

[Pyter]

"A little off" in what sense? What would they be "off" from?

Excuse me, I meant the "angular" coordinate, off with respect to a situation without dense masses.

 

[PeterDonis]

if the FRW is a "metric" that implies it's already a solution of the EFE

The Wikipedia article uses the term "metric" to mean "general expression for a line element with unknown functions in it still to be determined by solving the EFE". That is clear from the context.

Also, in referencing the article, I clearly said "coordinates", and several different possible coordinate charts used in cosmology are described in the article, including the basics of how they are constructed without knowing the full metric (i.e., without having solved the EFE to find the unknown functions in the line element). The other references you have already been given (Carroll, MTW, Wald) all discuss in greater detail how all this is done.

Please read carefully before posting a knee-jerk reaction that does not correctly reflect what you were told.

 

[PeterDonis]

I meant the "angular" coordinate, off with respect to a situation without dense masses.

The observed angular coordinates can be "off" in this sense with gravitational lensing, yes. That's one of the things cosmologists have to try to correct for when constructing models.

 

[Pyter]

Also, in referencing the article, I clearly said "coordinates", and several different possible coordinate charts used in cosmology are described in the article, including the basics of how they are constructed without knowing the full metric

Even if the full metric is unknown, some assumptions on the mass distribution and thus on the EFE are made beforehand, and they're clearly not fulfilled in the neutron stars field scenario.

or this case, the FRW solution is not what is used. There are no known exact solutions for isolated systems containing more than one gravitating mass (such as the binary pulsar systems that have given us evidence for gravitational wave emission)

I thought so, the FRW, including the Cartesian coordinates, can't be used in this "extreme" case.

 

[PeterDonis]

Even if the full metric is unknown, some assumptions on the mass distribution and thus on the EFE are made beforehand, and they're clearly not fulfilled in the neutron stars field scenario.

Of course not. I explicitly said that the FRW assumptions apply to the universe as a whole.

the FRW, including the Cartesian coordinates, can't be used in this "extreme" case.

This isn't an "extreme" case, it's just a different case from modeling the universe as a whole. It's different because the assumptions are obviously different.

(Btw, there is nothing stopping you from using Cartesian coordinates in this case. The final metric obtained by numerically solving the EFE just won't look like an FRW metric in those coordinates.)

 

[Pyter]

This isn't an "extreme" case, it's just a different case from modeling the universe as a whole.

"Extreme" compared to the Schwa. metric, which can be solved in closed form without recurring to numerical methods.
Speaking of which, I think that what I wrote in post #46 of this thread wasn't so far fetched, at least in the Schwa. case.

Now I'm aware that both sides of the EFE contribute to ten differential equations whose unkown are the components of , and thus also the RHS depends on the linear coordinates through the unknown functions .

And as we saw the RHS does depends on g**, through T**.

The physical observation of your masses only set the boundary conditions of the EFE, and they contain the unknown in a parametrized form, for instance in the measurement of your distance from objects of known mass.

Maybe they don't set the boundary conditions, but at least an additional set of constraints helping solve the differential equations.
In Waner's op.cit., pag. 121, we can see that the observable quantity "total mass" helps solve for $\Phi$:

I wonder it the above also applies to the neutron stars field or, to further simplify, the binary neutron star scenario? I'm not talking about actually solving the EFE, just setting it up.

 

[PeterDonis]

I think that what I wrote in post #46 of this thread wasn't so far fetched, at least in the Schwa. case.

Unfortunately, still no. See below.

Maybe they don't set the boundary conditions, but at least an additional set of constraints helping solve the differential equations.

The additional constraints in the Schwarzschild case are the symmetries of the spacetime, which drastically reduce the number of independent equations, plus the assumption of vacuum outside the star, which makes the differential equations homogeneous (zero on the RHS) and thereby further simplifies the solution. They have nothing to do with applying any boundary conditions or other conditions based on observation of the total mass.

In Waner's op.cit., pag. 121, we can see that the observable quantity "total mass" helps solve for $\Phi$

No, it doesn't. The form of the solution comes from the drastic reduction in the number of independent equations due to symmetry. In the Schwarzschild case, this is what results in one of the EFE components becoming the separable first order equation with solution as described. The total mass $M$ is an undetermined constant in that solution, and it's known to be constant outside the star because outside the star is vacuum, not because knowing its actual value helps us in any way in solving the equations. (There is a boundary condition for $\Phi$, that it should vanish as $r \to \infty$, but this is true regardless of the actual value of the total mass $M$, so knowing the total mass is irrelevant to applying it.)

I wonder it the above also applies to the neutron stars field

The general form given applies to any static, spherically symmetric mass distribution. One of my Insights articles that I referenced earlier in this thread discusses this (and also discusses how the symmetries simplify the solution).

or, to further simplify, the binary neutron star scenario?

The binary neutron star scenario is not simpler than a single neutron star; it's more complicated, and, as I have already said, no exact solution is known for it. Certainly you can't use a single Schwarzschild solution for it, since the scenario is neither static nor spherically symmetric. Not only that, but in the binary neutron star case, the field is too strong for nonlinearities to be ignored, so even the technique of simply adding together multiple Schwarzschild solutions centered on multiple massive objects does not work, since that technique assumes linearity.

 

[Pyter]

I wonder it the above also applies to the neutron stars field

@PeterDonis by the "above" I meant: 1) that the RHS depends on g**; 2) that the physical observations help reducing the degrees of freedom.

The first point for the Schwa. case is verified, the second maybe not because, as you said, due to the many symmetries, the equations are so simplified that the physical observations aren't even needed.

For the neutron stars field scenario — by that I mean a region with many neutron stars, and I called it sometimes the "general case" exactly because you can't rely on particular symmetries or uniform mass distribution hypothesis like in the "special" Schwa. case — I wondered if point 1) still holds. And if you can't simplify through symmetries, you're somewhat forced to use the physical observations to reduce the degrees of freedom, so I don't see why point 2) should not apply.

 

[PeterDonis]

@PeterDonis by the "above" I meant: 1) that the RHS depends on g**;

That's always true to the extent that the SET involves the metric.

2) that the physical observations help reducing the degrees of freedom.

As I have already said, what helps in reducing the degrees of freedom is assumptions about symmetries of the spacetime.

One could add to that assumptions restricting the form of the SET, which can reduce the number of unknown functions that must be solved for. The simplest of these is, of course, assuming vacuum; that means all SET components are zero. But we could also, for example, assume a perfect fluid with some particular equation of state (which is another common assumption for modeling, say, the interiors of stars and planets). In a sense assumptions like the latter are based on "physical observations", in that scientists have spend considerable time and effort figuring out useful equations of state for various phases of matter, but they're not observations of the particular system you're trying to model; they're just general observations that are useful background knowledge.

The first point for the Schwa. case is verified

It's always true for any case whatever.

the second maybe not because, as you said, due to the many symmetries, the equations are so simplified that the physical observations aren't even needed.

Correct. See above.

For the neutron stars field scenario — by that I mean a region with many neutron stars, and I called it sometimes the "general case" exactly because you can't rely on particular symmetries or uniform mass distribution hypothesis like in the "special" Schwa. case — I wondered if point 1) still holds.

See answers to this above.

And if you can't simplify through symmetries, you're somewhat forced to use the physical observations to reduce the degrees of freedom

But they don't. See above.

 

[Pyter]

But they don't. See above.

You mean that in the "general" (neutron stars) scenario, the physical observations (radial and angular positions, estimated (?) masses) are useless in setting up the EFE, let alone solve it?
Does it mean that in that case the EFE is not even solvable in theory?

 

[vanhees71]

Of course it is solvable, at least numerically. The really interesting question is about the correct equation of state of the strongly interacting matter the neutron star is made of. There is the hope to learn much more about it in the near future using gravitational wave observations as well as new experiments with heavy ions.

 

[Orodruin]

That's interesting, maybe topic for another thread. I haven't delved into it so far, but I figured that the homogeneous EFE only admitted the identically flat space solution or the planar wave solution, for analogy with the homogeneous Maxwell equations.

I don’t think it has been pointed out so...

The homogeneous Maxwell equations does not allow only one solution. The solution becomes unique only if you are also given appropriate boundary conditions.

 

[Pyter]

Of course it is solvable, at least numerically.

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique. In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs, in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.
Or are there supplementary constraints I'm not aware of?

 

[PeterDonis]

You mean that in the "general" (neutron stars) scenario, the physical observations (radial and angular positions, estimated (?) masses) are useless in setting up the EFE, let alone solve it?

Yes. More precisely, they don't help in either reducing the number of equations you have to solve or the number of unknown functions you have to solve for. All those physical observations do is give you particular parameters to plug into the solution after you've obtained it. (For example, knowing the mass of some spherically symmetric body does not help you obtain the Schwarzschild solution for the spacetime geometry outside the body; all it does is give you the number $M$ to plug into the solution once you've got it.)

Does it mean that in that case the EFE is not even solvable in theory?

Of course not. See above.

if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique.

You are misstating this. See below.

In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs

None of those are physical observations. See my comment above about the Schwarzschild solution for the vacuum region outside a spherically symmetric massive body. You use spherical symmetry plus vacuum to obtain a mathematical expression for the metric. But this expression does not describe just one metric; it describes a family of them, all having the same form, but with different numbers for the constant $M$ that appears in the solution. Your knowledge of the actual mass of the particular body you are interested in tells you what number to plug in for $M$ in the metric.

In other words: even after you have used symmetries, etc., to reduce the number of degrees of freedom, you still don't have a "unique" solution; you will have an expression for the metric that will have some undetermined parameters in it (like $M$ in the Schwarzschild case). If you want a unique solution, you need to fill in values for those parameters. Physical observations of the particular system you are interested in can help you to do that; but they can't help you to get to that point, where you have an expression for the metric that only needs some parameters filled into be unique.

in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.

I have no idea what you are talking about here. Perhaps rethinking your understanding in the light of the above will help.

 

[vanhees71]

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique. In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs, in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.
Or are there supplementary constraints I'm not aware of?

As any gauge theory the equations of motion for GR do not lead to unique solutions. That's the same for electromagnetism, where Maxwell's equations do not lead to unique solutions for the four-potential, which is a gauge field and thus contains redundant degrees of freedom. This does, however, not mean that you don't have a unique description of the physical situation. It's just that different solutions, connected by a gauge transformation, describe the same physical situation. In GR the gauge transformations are the (local) diffeomorphisms between different coordinates, but the physical situation doesn't depend on it. That's what's behind the general covariance of GR, and as in any gauge theory you can fix the gauge in any way you like to facilitate your calculation.

 

[PeterDonis]

different solutions, connected by a gauge transformation, describe the same physical situation.

Note that while this is of course true in GR for arbitrary diffeomorphisms, it is also the case that solving the field equation in a given case can give you a family of solutions even after the gauge freedom is taken into account. For example, as I said in a previous post, solving the EFE for the case of spherical symmetry and vacuum gives the Schwarzschild solution, but "the" Schwarzschild solution is really an infinite family of solutions, one for each possible value of $M$. Those solutions are physically different; you can't transform a Schwarzschild solution with $M_1$ into a Schwarzschild solution with $M_2 \neq M_1$ by a gauge transformation. But they are both solutions of the EFE under the same set of assumptions.

 

[vanhees71]

Of course not, because there you have physically different situations, i.e., vacuum Schwarzschild solutions for "matter"/"black holes" of different "mass", $M$. If there is vacuum in the region $r>R$, then there can be anything in $r<R$, as long as it's spherically symmetric. In the extreme singular case, you have a (non-rotating) black hole with a "point mass" $M$ at rest in the origin of the symmetry center.

It's always good to consider the electromagnetic analogy. In the case of electromagnetism, when you assume a spherically symmetric vacuum solution of the Maxwell equations (in Minkowski space) you get a Coulomb field (assuming that there's vacuum outside some sphere of radius $R$). The only relevant parameter is the electric charge in the "interiour", $r<R$, which can take any value and can be due to completely different charge distributions. The singular case is that of a point charge sitting at rest in the symmetry center.

 

[PeterDonis]

[Edit: Removed statement based on misreading of quoted post.]

In the extreme singular case, you have a (non-rotating) black hole with a "point mass" $M$ at rest in the origin of the symmetry center.

This is not a correct description of a black hole. The "symmetry center" is not a point in space, it's a moment of time (it's a spacelike curve, not a timelike curve), and there is no "point mass" there, it's vacuum all the way down to the singularity (which is itself not part of the manifold).

It's always good to consider the electromagnetic analogy.

Not in cases where it is misleading, as in the case of a "point charge", which does not have an analogue of a "point mass" in GR (although, ironically, it does in Newtonian mechanics).

 

[Orodruin]

Sure there can, it just can't be static.

I believe he is intending R to be arbitrary, not the Schwarzschild radius.

 

[PeterDonis]

I believe he is intending R to be arbitrary, not the Schwarzschild radius.

Ah, yes, I see I misread "can" for "can't" in his post. I'll edit my previous post to correct.

 

[Dale]

In both cases you have an invertible map that biunivocally connects a point on the curved N-dimensional manifold to a point in the N-dimensional Cartesian space.

Not quite. There is indeed a smooth invertible map from events in the manifold to points in R4. But the points in R4 are not Cartesian. The “distance” between them is given by the metric, not the Pythagorean theorem.

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique.

That isn’t really a problem. Many different mathematical solutions to many different differential equations can describe the same physics.

 

[vanhees71]

[Edit: Removed statement based on misreading of quoted post.]This is not a correct description of a black hole. The "symmetry center" is not a point in space, it's a moment of time (it's a spacelike curve, not a timelike curve), and there is no "point mass" there, it's vacuum all the way down to the singularity (which is itself not part of the manifold).Not in cases where it is misleading, as in the case of a "point charge", which does not have an analogue of a "point mass" in GR (although, ironically, it does in Newtonian mechanics).

You are right. I forgot that in standard Schwarzschild coordinates $r$ becomes the time-like coordinate for $r<r_{\text{S}}$.

 

[Pyter]

you still don't have a "unique" solution; you will have an expression for the metric that will have some undetermined parameters in it (like in the Schwarzschild case). If you want a unique solution, you need to fill in values for those parameters. Physical observations of the particular system you are interested in can help you to do that;

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too: If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun. Do you find this value by solving other differential equations, this time involving the observations?

 

[Pyter]

Not quite. There is indeed a smooth invertible map from events in the manifold to points in R4. But the points in R4 are not Cartesian. The “distance” between them is given by the metric, not the Pythagorean theorem.

According to the manifold's definition I had posted, which I repeat here for convenience:

all the "local coordinates" $\in E_1$, in the sense that they're rectlinear.
Example: latitude and longitude for a $S_2$ sphere. If you want to organize them as Cartesian coordinates (i.e. consider the $(\phi, \theta)$ plane, or not), it's up to you. Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes. In that case you only use the metric to map the points in the manifold to their (latitude, longitude).
If you instead need the length of an arbitrary path connecting two points on the manifold and lying entirely on the manifold, you will integrate the $ds^2$, which as you say is a function of the metric, along this path.

 

[Dale]

Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes.

Since this distance has no physical meaning I would not call that space “Cartesian”.

 

[martinbn]

According to the manifold's definition I had posted, which I repeat here for convenience:

View attachment 295032
all the "local coordinates" $\in E_1$, in the sense that they're rectlinear.

No, they are in $E_1$ in the sense that they have values in the real numbers. It has nothing to do with rectlinear, which doesn't make sense in genereal.

Example: latitude and longitude for a $S_2$ sphere. If you want to organize them as Cartesian coordinates (i.e. consider the $(\phi, \theta)$ plane, or not), it's up to you. Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes.

You could do that if you had a chart that covers the whole manifold otherwise you cannot.

In that case you only use the metric to map the points in the manifold to their (latitude, longitude).

That is not what a metric is!

If you instead need the length of an arbitrary path connecting two points on the manifold and lying entirely on the manifold, you will integrate the $ds^2$, which as you say is a function of the metric, along this path.

This is also not true, even if you ignore the imprecisions you want $ds$.

 

[Orodruin]

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too: If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun. Do you find this value by solving other differential equations, this time involving the observations?

No, r is not a parameter of the solution. It is one of the coordinates. For the Sun-Earth system, it is to very good approximation given by the distance from the Sun.