Is GR superior to Newtonian gravity?

[Ibix]

Possibly worth stating that there is no global energy conservation law in GR in general spacetimes.

 

[PAllen]

You won't be able to get away with this. Newtonian potential is literally a tool for calculating gravitational binding energy. So you are basically saying there is nothing like gravitational binding energy in GR, right?

The Newtonian potential depends position, and this is mathematically incompatible with special relativity because any local change has an instant global effect. Even more generally, a scalar potential would necessarily depend on Lorentz scalars to be consistent with SR. But the only plausible candidate, mass/energy is definitively NOT a Lorentz scalar - it is a component of a tensor (or a component of a 4-vector is you stick to pure kinematics in SR). But no way is it a Lorentz scalar, so you CANNOT have a valid scalar potential in SR (let alone GR). These facts are well known, and derived and universally accepted within a couple years of 1905. As a result, NO ONE bothered looking for a pure scalar potential to make gravity work with SR. If you go to vector potential, you get something more like EM, and the gravito-electromagnetism I mentioned.

 

[PeterDonis]

NO ONE bothered looking for a pure scalar potential to make gravity work with SR. If you go to vector potential, you get something more like EM, and the gravito-electromagnetism I mentioned.

Which also, of course, is not a complete theory (although it is useful as an approximation in the weak field regime), since it also leads to inconsistencies. MTW has a homework problem in one of its fairly early chapters which leads you through building a simple scalar and then vector model of relativistic gravity in flat spacetime and showing how they fail, and then building a simple tensor model and showing how it avoids those failures (but still is inconsistent with spacetime being flat, as in SR).

 

[PeterDonis]

MTW has a homework problem in one of its fairly early chapters

Just looked, it's actually a series of three problems, Exercises 7.1, 7.2, and 7.3.

 

[zonde]

But the only plausible candidate, mass/energy is definitively NOT a Lorentz scalar - it is a component of a tensor (or a component of a 4-vector is you stick to pure kinematics in SR).

Rest mass and binding energy are Lorentz scalars.
You see, you can look at binding energy from two different sides: from one side you can look at it as a radiation (usually EM but in some specific cases gravitational) that is radiated out of the system or from the other side you can look at it as mass defect of the system's rest mass. Considering this and the fact that rest mass is Lorentz scalar, binding energy is Lorentz scalar too.

 

[Dale]

Rest mass and binding energy are Lorentz scalars.

Do you have a reference for the binding energy scalar? I am not familiar with that.

 

[zonde]

Do you have a reference for the binding energy scalar? I am not familiar with that.

No I don't.
Look, mass equivalent of nuclear binding energy is a factor in rest mass of the atom. If rest mass of the atom is Lorentz scalar so is nuclear binding energy. Surely you can see this, right? It's so obvious I'm not sure anybody would bother to make a statement about it.

 

[Dale]

Surely you can see this, right? It's so obvious I'm not sure anybody would bother to make a statement about it.

I don't see it. I don't think it is so obvious, and it is not something that I have heard anyone else claim.

I know that the rest mass is a scalar because I can form it in a manifestly covariant way from the four-momentum. But I don't know how to form the binding energy similarly.

 

[zonde]

I don't see it. I don't think it is so obvious, and it is not something that I have heard anyone else claim.

I know that the rest mass is a scalar because I can form it in a manifestly covariant way from the four-momentum. But I don't know how to form the binding energy similarly.

Look, if $(a+b)$ is scalar and $a$ is scalar how could $b$ be anything else but scalar?

 

[Dale]

What are $a$, $b$, and $a+b$ here?

 

[zonde]

What are $a$, $b$, and $a+b$ here?

Let's say $a$ is rest mass of 2 protons and 2 neutrons, $b$ is negative binding energy of $^4 He$ nucleus and $a+b$ is rest mass of $^4 He$ nucleus.

 

[Dale]

Yes, I agree with that. I thought we were discussing gravity, which is different because of curvature.

Edit: note that $a$, $b$, and $a+b$ in this case are scalars in the sense of numbers not scalars in the sense of scalar fields on spacetime. It is that latter sense in which gravitational potential cannot be a scalar.

 

[PAllen]

Rest mass and binding energy are Lorentz scalars.
You see, you can look at binding energy from two different sides: from one side you can look at it as a radiation (usually EM but in some specific cases gravitational) that is radiated out of the system or from the other side you can look at it as mass defect of the system's rest mass. Considering this and the fact that rest mass is Lorentz scalar, binding energy is Lorentz scalar too.

But total mass/energy is a source in a relativistic gravity theory, and that is definitely NOT a Lorentz scalar. For example, a system of bodies of the same total rest mass, but with differing kinetic energy will attract something else differently. This establishes that you must include kinetic energy. You can't dodge this by saying the total rest energy of the system is a scalar, because that can't account for the mutual motions. The fact is, even in SR, you use a stress energy tensor to capture this, and that completely rules out a gravitational theory based on a scalar potential.

 

[renormalize]

I explain there can’t possibly be anything like a Newtonian potential in a relativistic theory of gravity, and point out that GR is closer to EM in this regard.

Even more generally, a scalar potential would necessarily depend on Lorentz scalars to be consistent with SR. But the only plausible candidate, mass/energy is definitively NOT a Lorentz scalar - it is a component of a tensor (or a component of a 4-vector is you stick to pure kinematics in SR). But no way is it a Lorentz scalar, so you CANNOT have a valid scalar potential in SR (let alone GR).

The fact is, even in SR, you use a stress energy tensor to capture this, and that completely rules out a gravitational theory based on a scalar potential.

How do you reconcile these claims with Nordström scalar gravitation (proposed in 1912-1913)? See, e.g.:
Nordström’s scalar theory of gravity and the equivalence principle

So, while Nordström gravity is indeed excluded in our universe by physical measurements, his theory is nevertheless a mathematically consistent scalar-model for gravity in 4D Minkowski (flat) spacetime. Isn't this an obvious counterexample to your statements?

 

[PeterDonis]

Rest mass and binding energy are Lorentz scalars.

For rest mass, you should really say "invariant mass"--the magnitude of the object's 4-momentum vector.

For binding energy, if you're careful to define it in terms of the appropriate invariants, yes. But note that it's not as simple as "just take the difference in invariant masses". You also have to stipulate that the bound state and the unbound states that compose it (for example, bound state H atom and unbound proton and electron) are all at rest in the same frame. Otherwise your simple summation is not correct. Expressing that in terms of invariants takes some care.

 

[PeterDonis]

How do you reconcile these claims with Nordström scalar gravitation (proposed in 1912-1913)?

How does the "scalar" $\Phi$ in this theory have to transform when you change frames, in order for the tensor $T^{\mu \nu}$ that it contributes to to transform as a tensor, as it must?

Note that the "Newtonian" potential $\Phi = - M / r$ obviously does not transform as a Lorentz scalar.

 

[renormalize]

Note that the "Newtonian" potential $\Phi = - M / r$ obviously does not transform as a Lorentz scalar.

In Nordström gravity the scalar potential $\Phi$ is sourced by the trace $T$ of the energy-momentum tensor $T_{\mu\nu}\,$. For a non-relativistic, massive, pressureless body $T_{\mu\nu}=\text{diag}\left(\mathcal{E},0,0,0\right)$, so there is no difference between the energy-density $\mathcal{E}$ and the trace $T$, meaning Newtonian gravity is recovered in the appropriate limit. But of course this also means that electromagnetic waves, having $T=0$, do not source scalar gravity. Just another reason why Nordström gravity is ruled out physically.

 

[PeterDonis]

@renormalize I deleted the post you just responded to.

 

[zonde]

But total mass/energy is a source in a relativistic gravity theory, and that is definitely NOT a Lorentz scalar.

Of course, it's stress energy tensor.

For example, a system of bodies of the same total rest mass, but with differing kinetic energy will attract something else differently. This establishes that you must include kinetic energy.

Agreed

You can't dodge this by saying the total rest energy of the system is a scalar, because that can't account for the mutual motions.

Sum of the invariant masses of the parts of the system is not the same as invariant mass of the system as a whole.

The fact is, even in SR, you use a stress energy tensor to capture this, and that completely rules out a gravitational theory based on a scalar potential.

Agreed, there is no place for scalar potential in relativistic theory of gravity.

With that out of the way I would like to point out what was my point.
Let's say you know mass/energy distribution within the system. So you plug it in stress energy tensor and do the calculations.
But then let's say you have different question: what would be a reasonable mass/energy distribution within the system? Can you somehow distinguish between reasonable and unreasonable distributions of mass/energy with the system with the help of GR. My hunch is that Newtonian gravity (within it's limited domain of applicability!) is much better at that than GR.
Obviously Newtonian gravity just breaks down in strong field limit and in case of gravitomagnetism (not sure how much the two overlap).

So my question was how wrong or right I am? Can we tell apart reasonable vs unreasonable distributions of mass/energy in strong field limit?

 

[Dale]

Can you somehow distinguish between reasonable and unreasonable distributions of mass/energy with the system with the help of GR

Yes. That is formally what the energy conditions do.

My hunch is that Newtonian gravity (within it's limited domain of applicability!) is much better at that than GR.

To my knowledge the opposite is true. I am not aware of anything similar to the energy conditions for Newtonian gravity.

 

[Ibix]

Can we tell apart reasonable vs unreasonable distributions of mass/energy in strong field limit?

I think that depends on what you mean by "reasonable", and whether Newtonian theory is better at it probably depends what you mean by "better". And it probably depends on what circumstances you're thinking about analysing and what your given values are.

For example, you might ask what is the maximum spin rate of a rocky planet before it disintegrates. That has no answer without a material model describing how much tensile stress rock can handle, so you'll need to posit that. Once you have it, determining when the planet disintegrates is just a matter of cranking the handle on the maths. Doing that with Newton's theory is probably easier, which might be one measure of better, but it won't warn you when it's invalid (it'll be perfectly comfortable with an equatorial speed exceeding $c$ for example), so you may need to do something post-Newtonian to check your result. And in all of this, the "reasonable" question mostly lies in the material model, not the gravitational theory. I can certainly posit a planet made of scrith and crank the maths in either theory - and whether that's a reasonable thing to do or not depends entirely on whether I know how to make a science fiction material, not on the theory of gravity I choose.

So I think working with Newtonian gravity is easier, but whether that equates to better is not clear. And I think the choice of gravitational theory has little bearing on reasonable.

 

[PAllen]

How do you reconcile these claims with Nordström scalar gravitation (proposed in 1912-1913)? See, e.g.:
Nordström’s scalar theory of gravity and the equivalence principle
View attachment 375139
View attachment 375140
So, while Nordström gravity is indeed excluded in our universe by physical measurements, his theory is nevertheless a mathematically consistent scalar-model for gravity in 4D Minkowski (flat) spacetime. Isn't this an obvious counterexample to your statements?

Yes, to some extent it gets around my criticism - the trace of T is obviously a Lorentz scalar, and it avoids instant changes at a distance from local changes. However, by having EM radiation not source gravity, it does not meet the requirements that all energy contributes to gravity. Because it avoids instant action at a distance, it does include gravitational radiation - though of a weird type.

 

[PAllen]

Just as an aside - one of the ways to address binding energy directly in GR is to assume, for a large local system, e.g, a galaxy, you can approximately assume it is embedded in asymptotically flat space time. Then, as a system evolves from a widely separated state to a bound state, the decline in Bondi energy of the system is the binding energy.

 

[renormalize]

Yes, to some extent it gets around my criticism - the trace of T is obviously a Lorentz scalar, and it avoids instant changes at a distance from local changes.

However, by having EM radiation not source gravity, it does not meet the requirements that all energy contributes to gravity.

That's true, but was the requirement that all energy (including that of EM radiation) contributes to gravity generally accepted by physicists prior to the development of GR in 1916, wherein such a contribution is predicted?

 

[PAllen]

That's true, but was the requirement that all energy (including that of EM radiation) contributes to gravity generally accepted by physicists prior to the development of GR in 1916, wherein such a contribution is predicted?

it was certainly suspected from SR because light has momentum in SR, and if absorbed contributes to rest mass. So this non inclusion would be suspect.

 

[PeterDonis]

My hunch is that Newtonian gravity (within it's limited domain of applicability!) is much better at that than GR.

I have no idea what you are basing that on. Newtonian gravity, the theory, says nothing whatever about what mass distributions are "reasonable". It doesn't even have things like the energy conditions, which as @Dale pointed out are conditions in the GR context that are used to judge whether a particular stress-energy tensor describes a "reasonable" distribution of matter and energy. So Newtonian gravity, if anything, is worse than GR at this.