Tests of EFE: Assessing General Relativity's Validity Beyond Cosmology

[CycoFin]

There are several tests of general relativity:

- gravitational redshift
- deflaction of light
- perihelion precession of Mercury
- Shapiro delay
- Lense-Thirring precession
- binary pulsars

Now comes the part that problems me. These all are based on vacuum solution of Einstein field equations.
With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.

So is there any real tests of EFE, other than Cosmology? (which have it's problems as we all know)

Also looking for good explanations why left side of EFE must equal right side.

 

[Mentz114]

There are several tests of general relativity:

- gravitational redshift
- deflaction of light
- perihelion precession of Mercury
- Shapiro delay
- Lense-Thirring precession
- binary pulsars

Now comes the part that problems me. These all are based on vacuum solution of Einstein field equations.
With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.

So is there any real tests of EFE, other than Cosmology? (which have it's problems as we all know)

Also looking for good explanations why left side of EFE must equal right side.

The field equations follow from extrememizing an action as in any field theory.

Your concerns about vacuum solutions are groundless. The vacuum solutions have zero Ricci tensor but still possesses non-zero Weyl curvature which
describes the field. It is very technical and if you study hard for a few years you will understand.

 

[Dale]

With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.

You can always simplify any equation to that form if you choose. This is a complete non issue.

All of the tests you mention are non-trivial solutions to the EFE.

 

[CycoFin]

The field equations follow from extrememizing an action as in any field theory.

Your concerns about vacuum solutions are groundless. The vacuum solutions have zero Ricci tensor but still possesses non-zero Weyl curvature which
describes the field. It is very technical and if you study hard for a few years you will understand.

Can you then write EFE using only Weyl tensors? Is it so that basically Riemann tensor = Ricci tensor (trace part) + Weyl tensor(everything else)?

You can always simplify any equation to that form if you choose. This is a complete non issue.

All of the tests you mention are non-trivial solutions to the EFE.

I disagree, the vacuum solution is the Schwarzschild metric which have 0 = 0

 

[Mentz114]

Can you then write EFE using only Weyl tensors? Is it so that basically Riemann tensor = Ricci tensor (trace part) + Weyl tensor(everything else)?

The only way to change the EFE is to change the action being extremized.

A vacuum solution is created by solving the differential equations
$G(g)^{\mu\nu} =0$
where $g$ is a function of the coordinates. We thus find a valid metric, which gives us $0=0$, which is what we wanted.

See here http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution

 

[CycoFin]

Yes, deriving the schwarzschild solution we assume that rigth side must be zero because vacuum and then find metric that gives also left side zero (R = 0).

But this 0 = 0 does not prove that EFE is correct hypothesis. One can put anything one wants to the right side, just make sure it is zero (e.g. number of magnetic monopoles).

I ask again, is there any experimental tests that have rigth side of EFE not zero?

 

[Mentz114]

Yes, deriving the schwarzschild solution we assume that rigth side must be zero because vacuum and then find metric that gives also left side zero (R = 0).

But this 0 = 0 does not prove that EFE is correct hypothesis. One can put anything one wants to the right side, just make sure it is zero (e.g. number of magnetic monopoles).

I ask again, is there any experimental tests that have rigth side of EFE not zero?

That is rubbish. You have to put the correct metric in. The metric is the solution. Your understanding of logic and mathematics is flawed.

The Schwarzschild metric satisfies all the experimental tests that have been tried - so your assumption is flawed in any case.

The equation $x^2-2ax+a^2=0$ has a solution $x=a$ which gives $0=0$. What is wrong with that logic ?

 

[CycoFin]

Nothing is wrong with that logic.

You have proved that x²-2ax+a² = 0 is true (have solution) but you have not proved that x²-2ax+a² = b when b is not 0.

I agree that the Schwarzschild metric satisfies all the experimental tests that have been tried and those I listed on first post.

I just don't understand how Schwardschild metric can "prove" EFE if equation reads 0 = 0.

 

[Mentz114]

Nothing is wrong with that logic.

You have proved that x²-2ax+a² = 0 is true (have solution) but you have not proved that x²-2ax+a² = b when b is not 0.

I agree that the Schwarzschild metric satisfies all the experimental tests that have been tried and those I listed on first post.

I just don't understand how Schwarzschild metric can "prove" EFE if equation reads 0 = 0.

Well, you need to understand what solving an equation means. It is like a set of scales. The scales still work when they have nothing to weight. They are saying $0=0$.

Take the quadratic equation I wrote. I could have written $x^2-2ax = -a^2$. Now put in the solution $x=a$. Now the equation reads $-a^2=-a^2$.

We know the solution is correct when both sides are the same.

I don't think I can help any further with this.

 

[PeterDonis]

With vacuum solution EFE is 0 = 0

No, it isn't. The EFE is a second order partial differential equation (more precisely, it's a system of 10 of them in the general case). So the RHS being zero just means it's a homogeneous second order partial differential equation, which certainly has solutions other than the trivial one. See here for an overview:

http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)

 

[John F. Gogo]

No, it isn't. The EFE is a second order partial differential equation (more precisely, it's a system of 10 of them in the general case). So the RHS being zero just means it's a homogeneous second order partial differential equation, which certainly has solutions other than the trivial one. See here for an overview:

http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)

So, you wrote this for wiki? Cool.

 

[PeterDonis]

you wrote this for wiki?

If you mean, did I write the Wikipedia page I linked to, no, I didn't. I just linked to it as a reference.

 

[Dale]

I disagree, the vacuum solution is the Schwarzschild metric which have 0 = 0

So what? This is true of any solution of any equation. You can always simplify it to 0=0:

Given A=B
A-B=0 by subtraction
0=0 by substitution

It is a completely useless complaint because it is always true for any equation. If you over-simplify you can always get 0=0. Your query about experiments or observations with a non-vacuum metric is a valid question, but this bit about 0=0 is not.

 

[atyy]

There are several tests of general relativity:

- gravitational redshift
- deflaction of light
- perihelion precession of Mercury
- Shapiro delay
- Lense-Thirring precession
- binary pulsars

Now comes the part that problems me. These all are based on vacuum solution of Einstein field equations.
With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.

Conceptually, these are not tests of the pure vacuum solutions, because in addition to the vacuum solution, geodesic motion is assumed. The geodesic motion is derived by assuming the presence of matter: http://arxiv.org/abs/0806.3293.

So is there any real tests of EFE, other than Cosmology? (which have it's problems as we all know)

What problems does cosmology have? Are you thinking about dark matter?

For general relativity with matter, there seems to still be research going in these areas;
http://www.einstein-online.info/spotlights/hydrodynamics_realm
http://arxiv.org/abs/1210.4921
http://arxiv.org/abs/1206.2503

 

[PeterDonis]

geodesic motion is derived by assuming the presence of matter

Just to clarify, this is referring to geodesic motion of bodies that are not "test bodies", because they have significant self-gravity, correct? For example, geodesic motion of the Earth in the gravitational field of the Sun.

This "presence of matter" is still very different from the "matter" that is present in cosmological solutions. In the latter, the matter significantly affects the spacetime geometry everywhere. In the former, the matter only significantly affects the spacetime geometry (in the sense of requiring a nonzero stress-energy tensor in the local solution of the EFE) in isolated regions corresponding to the central gravitating body producing the overall geometry (e.g., the Sun), and the self-gravitating bodies undergoing geodesic motion in this geometry (e.g., the Earth). Everywhere else, the solution is a vacuum solution, and that vacuum solution is what is used to predict the results of the classic solar system test experiments. (For example, nobody uses a nonzero stress-energy tensor for the Sun to predict the bending of light by the Sun; they just use the Schwarzschild geometry of the vacuum region exterior to the Sun.)

 

[atyy]

Just to clarify, this is referring to geodesic motion of bodies that are not "test bodies", because they have significant self-gravity, correct? For example, geodesic motion of the Earth in the gravitational field of the Sun.

I only meant that there are no test bodies, so although we treat the Earth in the gravitational field of the sun as a test body, we do it by assuming a solution in which matter is present, and derive the treatment of the Earth as a test body as a very good approximation.

 

[CycoFin]

So what? This is true of any solution of any equation. You can always simplify it to 0=0:

Given A=B
A-B=0 by subtraction
0=0 by substitution

It is a completely useless complaint because it is always true for any equation. If you over-simplify you can always get 0=0. Your query about experiments or observations with a non-vacuum metric is a valid question, but this bit about 0=0 is not.

First I like to thank to get my thread back. I know my questions are very edge of the this forum rules. I will try not to go over.

I will try to make my point by some simple examles.

One can write equation:

apples = oranges

This is true if set both apples and oranges to zero. Equation reads 0 = 0. But surely it is not true if there are nonzero items of either.

One can write equation:

polar bears = constant * pink elephants.

Assume we make some experimetal tests on Afrika. Because there is no polar bears and neither pink elephants on Afrika, we are happy that test results agree with our equation which reads 0 = 0. We know there is polar bears on north pole but we have not yet made any tests because north pole is so far away from Afrika.

Can we now say from our test results that our polar bear/pink elephant equation is correct? No we don't. We have only tested one point (0=0).
This is the reason I asked is there any experimental test results where we have EFE something else than 0 = 0.

 

[atyy]

@CycoFin, as DaleSpam and others have said, your objection makes no sense. Every equation can be written in the form U = 0, including the full Maxwell's equations with charges and currents, and Newton's laws etc. It is true that there are regimes of GR that remain untested. However, you can consider the vacuum solutions of GR analogous to Maxwell's equations without charges. Even in the absence of charge, Maxwell's equations predict interesting phenomena such as electrotromagnetic waves that travel over great distances.

 

[atyy]

apples = oranges

This is true if set both apples and oranges to zero. Equation reads 0 = 0. But surely it is not true if there are nonzero items of either.

In fact, the equation can be rewritten as

apples - oranges = 0.

Also, the equation can be true if there are nonzero items of either - it means that the number of apples and oranges is the same.

 

[CycoFin]

No, it isn't. The EFE is a second order partial differential equation (more precisely, it's a system of 10 of them in the general case). So the RHS being zero just means it's a homogeneous second order partial differential equation, which certainly has solutions other than the trivial one. See here for an overview:

http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)

Now I am little bit lost. You say that even we have RHS stress-energy T zero, LHS Einstein tensor G is not?

I know there are several diffrent vacuum solutions with different metrics but it does not change that in tensor point of view EFE reads 0 = 0 (these are zero tensors, not zeros)

 

[CycoFin]

In fact, the equation can be rewritten as

apples - oranges = 0.

Also, the equation can be true if there are nonzero items of either - it means that the number of apples and oranges is the same.

Yes but if you make experimental tests when you never have neither apples nor oranges, you can not say if this equation is explainng nature.

All you can say that there is no apples or oranges on universe.

 

[atyy]

Yes but if you make experimental tests when you never have neither apples nor oranges, you can not say if this equation is explainng nature.

All you can say that there is no apples or oranges on universe.

But the vacuum solutions correspond to the case where where apples and oranges are not each zero.

 

[CycoFin]

Vacuum solutions makes our RHS zero. There is no energy or stress on vacuum (not talking about cosmology here now). So our oranges are zero.

Then we assume that our LHS side must be zero also. We assume apples zero. We make measurements that agrees our assumption that apples are zero.

Can we really then say that apples = oranges because they both are zero? What about changing that energy-stress tensor to pink elephant tensor. That is also zero in vacuum.

So why we don't just use apples = 0 and be happy. And add those oranges once we have tests that contain non-zero numbers of oranges (of course if our apples = oranges is still valid)

 

[Nugatory]

Now I am little bit lost. You say that even we have RHS stress-energy T zero, LHS Einstein tensor G is not?

Of course not.
It's the metric tensor that we're solving for. The EFE supplies constraints on what this unknown metric tensor might be; one of these constraints is that the Einstein tensor calculated from the unknown metric tensor must be zero at points where the stress-energy tensor is zero. Given the complexity of the relationship between the metric and the Einstein tensors, $G=0$ is a seriously non-trivial constraint.

 

[CycoFin]

Yes, we have G = 0, I don't disagree this and metric explanation of gravity makes sense and this have been very successfully tested.

But surely there must be some experimental test results so we can write G = constant * T, where T is nonzero energy-stress tensor.

If not, R = 0, where R is ricci tensor is enough to descripe gravity effects on vacuum.

 

[atyy]

Vacuum solutions makes our RHS zero. There is no energy or stress on vacuum (not talking about cosmology here now). So our oranges are zero.

Then we assume that our LHS side must be zero also. We assume apples zero. We make measurements that agrees our assumption that apples are zero.

Can we really then say that apples = oranges because they both are zero? What about changing that energy-stress tensor to pink elephant tensor. That is also zero in vacuum.

So why we don't just use apples = 0 and be happy. And add those oranges once we have tests that contain non-zero numbers of oranges (of course if our apples = oranges is still valid)

No. The analogy is more like

apples - oranges = 0.

There the RHS is zero, but clearly, apples and oranges can both be nonzero, as long as they are equal.

 

[PeterDonis]

You say that even we have RHS stress-energy T zero, LHS Einstein tensor G is not?

No; I am saying that the Einstein tensor being zero does not mean "zero apples" or "zero oranges". It means something more like "the rate of change of the rate of changes of apples plus the rate of change of the rate of change of oranges is zero", and nine more equations of that sort. The fact that a bunch of derivatives of something add up to zero does not require that something itself to be zero. And gravity is the "something" here, so this is just saying that the Einstein tensor being zero is not the same as gravity being zero.

we have G = 0, I don't disagree this and metric explanation of gravity makes sense and this have been very successfully tested.

And these are tests of the EFE, because, once again, the LHS of the EFE is not just a simple expression whose value we test to be zero; it's a combination of a bunch of derivatives of different things, and we don't measure the combination, we measure the derivatives separately, so we can test whether they add up to zero as the EFE predicts. That's a valid test.

surely there must be some experimental test results so we can write G = constant * T, where T is nonzero energy-stress tensor.

There are, but they're indirect. The sorts of experimental tests you need here are tests of the structure of matter, but it's difficult to find pieces of matter where relativistic effects are significant in determining its structure--i.e., where the difference between the predictions of GR and the predictions of Newtonian gravity are significant. Nothing in our solar system even comes close: Newtonian gravity is sufficient to explain the internal structure of all ordinary objects, and of the Sun and all the planets and other bodies in the solar system.

The kinds of matter (other than what we model in cosmology) for which relativistic effects are significant are, basically, white dwarfs, neutron stars, black holes (these aren't exactly "matter", but they are formed from collapsing matter), and supermassive stars. There aren't any such objects nearby, so we can only observe them indirectly, from far away, hence our observational tests of GR regarding the internal structure of such objects are still very rough.

However, I don't understand why you leave out cosmology here; the cosmological models we use to accurately match many observations, such as the properties of the CMBR and the curvature in the Hubble diagram, certainly require a nonzero stress-energy tensor and are valid tests of GR with a nonzero RHS of the EFE.

The analogy is more like

apples - oranges = 0.

I'm not sure even this conveys it, because "apples" and "oranges" here correspond to different metric coefficients, and the EFE contains derivatives of those. See above.

 

[Mentz114]

Is this thread what mathematicians call 'reducio ad absurdum' ?

 

[Vanadium 50]

Is this thread what mathematicians call 'reducio ad absurdum' ?

In the words of Monty Python, "That's not argument! It's just contradiction!"

 

[CycoFin]

And these are tests of the EFE, because, once again, the LHS of the EFE is not just a simple expression whose value we test to be zero; it's a combination of a bunch of derivatives of different things, and we don't measure the combination, we measure the derivatives separately, so we can test whether they add up to zero as the EFE predicts. That's a valid test.

These are tests of very specific form of EFE where we have RHS zero tensor. This specific vacuum EFE can also be written as Ricci tensor = 0, R = 0.
Surely Ricci tensor is made by combination of 2nd derivates of metric but that is not my point here.

We cannot use these test results to make any claim about general EFE (where RHS is not zero tensor), except obivious that general EFE is true when T = 0.
If we do that, it is like testing equation x² = x³ only on point x = 0 (0 = 0) and then using those results to make claim that our equation is correct with every x.

However, I don't understand why you leave out cosmology here; the cosmological models we use to accurately match many observations, such as the properties of the CMBR and the curvature in the Hubble diagram, certainly require a nonzero stress-energy tensor and are valid tests of GR with a nonzero RHS of the EFE.

I leaved cosmology out because problems we have there (dark energy, inflation, flatness problem, also dark matter but is not that much because FLRW). Because these one can start to think if we are using correct equation there.

I have some ideas how to tackle those if we accept only the vacuum EFE and dispose the general EFE but this is against forum rules and I am not going to discuss these here. Private conversation if interested to hear.

 

[Mentz114]

In the words of Monty Python, "That's not argument! It's just contradiction!"

No it isn't !

It looks like we have a crackpot so this thread is doomed. Phew.

 

[PeterDonis]

These are tests of very specific form of EFE where we have RHS zero tensor.

You're shifting your ground. Before, you were saying vacuum tests weren't tests of the EFE at all. Now you're saying they're tests, just tests of "a very specific form" of the EFE. (See below for more on that.)

We cannot use these test results to make any claim about general EFE

By this argument, you can't use any test results to make a claim about the general EFE, only about the "specific form" that you tested. You do realize that saying the RHS of the EFE is not zero does not pin down one specific "value" for the RHS, right? The RHS is the stress-energy tensor; there are many, many different forms that that tensor can take, depending on what kinds of matter and energy are present. Any test of the EFE is only going to test one "specific form" of the stress-energy tensor, corresponding to the particular kinds of matter and energy that are present during the test. A vacuum (no matter or energy present at all) is just one particular case among many.

I leaved cosmology out because problems we have there

So what? Cosmology still gives tests of the EFE, using the particular kinds of matter and energy present, on average, in the universe. And GR passes those tests for a large portion of the universe's history (basically back to times early enough that we're not sure what stress-energy tensor to use).

 

[PeterDonis]

I have some ideas how to tackle those if we accept only the vacuum EFE and dispose the general EFE

You appear to have basic misunderstandings about how the EFE is tested in the first place. We've done our best to address them in this thread, but I don't think there's much point in further discussion. This thread is closed.