Confusion in General Relativity

[klen]

According to equivalence principle, gravity can be treated like acceleration "locally". Based on this principle we can treat a non-inertial frame at rest and explain the fictitious forces (of Newton's Laws) as gravity. From this we can prove that time elapses at different rates at different positions in gravity and that the space is curved.
But Einstein also said that the curved space-time can be treated "locally" as Minkowski space which is flat.
Doesn't these two ideas contradict each other? From the former reason we are saying that space is curved locally because of gravity and latter is saying that it is flat.
Could anyone please explain this reasoning.

 

[PeterDonis]

From this we can prove that time elapses at different rates at different positions

Yes.

and that the space is curved

No. To show that spacetime (not space) is curved, we need to look beyond a single local inertial frame, i.e., beyond the scope of the equivalence principle. The EP alone does not show that spacetime is curved; curvature is a non-local effect, and as you say, the EP is local.

 

[Nugatory]

The equivalence principle says that gravitational effects behave locally like acceleration in flat spacetime. The curvature becomes apparent only when you take a non-local view.

An observer inside a sufficiently small closed room will not be able to tell whether the room is resting in the curved spacetime at the surface of the Earth or being accelerated through flat spacetime at 1g - the spacetime curvature within the room is too small to detect. That's the part about treating curved space-time as locally flat and treating gravity as equivalent to acceleration in flat spacetime.

However, if we make the room large enough, non-local effects that can only be attributed to curvature will start to appear - for example, two objects dropped at the same time will follow very slightly non-parallel paths as their trajectories converge to intersect at the center of the earth.

 

[sweet springs]

Hi. Let me tell you some cases.

1. Suppose you are in the accerelating rocket in the space. You feel gravity. You interprete there is constant gravitational field in the whole universe and it is true in your frame of reference.
Suppose you are on the earth. You feel same gravity as you are in the rocket. Your previous interpretaitin on gravity becomes wrong because the gravity has center at the center of the Earth.
"Locally" the gravitational effect on the Earth is same as that of the accerelating rocket.

2. Suppose your brother floats in space and the accerelatin rokcekt passes him.
You and your brother share the same space time point. You feel gravity but your brother feel no gravity. His spacetime is Minkowsky.
Suppose you are on the Earth and your brother is free falling nearby. You and your brother share the same spacetime point again. You feel gravity but your brother does not. His "Local" spacetime is Minkowsky.

 

[klen]

Hi, I agree if my brother is falling then his local spacetime is Minkowski but when I am standing on Earth will my local spacetime be Minkowski? Because, by above the argument by Einstein, my spacetime appears to be curved.

 

[sweet springs]

Hi, I agree if my brother is falling then his local spacetime is Minkowski .

His spacetime is "Locally" Minkowsky because gravity disappears only where he is.

but when I am standing on Earth will my local spacetime be Minkowski? Because, by above the argument by Einstein, my spacetime appears to be curved.

No. your spacetime is curved thus you get gravity. If you want to straighten the spacetime around you jump off following your brother.

 

[Nugatory]

Hi, I agree if my brother is falling then his local spacetime is Minkowski but when I am standing on Earth will my local spacetime be Minkowski? Because, by above the argument by Einstein, my spacetime appears to be curved.

Either spacetime is curved at a given point or it isn't; it can't be curved for one person and not curved for another. A person standing on the surface of the Earth is accelerating at 1g (we measure this by asking him to hold an accelerometer, or to stand on a spring scale so that we can directly observe the force that is accelerating him) through a locally flat spacetime. The nearby falling person is moving inertially (no force acting on him as long as he remains in free fall) through the same locally flat spacetime.

Of course this local region of spacetime is not exactly flat - it is very slightly curved, and with sufficiently sensitive instruments both observers would be able to measure this very slight curvature. Their measurements would agree, as the curvature is what it is no matter how the person doing the measurement is moving.

No. your spacetime is curved thus you get gravity. If you want to straighten the spacetime around you jump off following your brother.

This is incorrect. It's the same slightly curved locally flat spacetime for both.

 

[sweet springs]

It's the same slightly curved locally flat spacetime for both.

Yes. The free faller should be infinitesimal small dwarf to escape from your criticism.

 

[A.T.]

According to equivalence principle, gravity can be treated like acceleration "locally". Based on this principle we can treat a non-inertial frame at rest and explain the fictitious forces (of Newton's Laws) as gravity. From this we can prove that time elapses at different rates at different positions in gravity and that the space is curved...From the former reason we are saying that space is curved locally because of gravity

Gravitational acceleration and different clock rates doesn't imply space time curvature. See the intrinsically flat cone the first figure here:

http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

Also this great post:

This is my own non-animated way of looking at it:

• Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[loislane]

curvature is a non-local effect

The curvature becomes apparent only when you take a non-local view.
[...]

non-local effects that can only be attributed to curvature will start to appear

Riemannian curvature is mathematically a local property. So what is non-local about it according to you? If possible give some reference.
The EP must limit arbitrarily to first derivatives of the metric, but how dos this make curvature something non-local?

 

[Nugatory]

Riemannian curvature is mathematically a local property. So what is non-local about it according to you? If possible give some reference.
The EP must limit arbitrarily to first derivatives of the metric, but how does this make curvature something non-local?

When we say that spacetime is locally flat and that curvature is non-local, we're saying that the approximation "it is flat across this region" can be made arbitrarily good by choosing the region in question to be sufficiently small. No matter how small a curvature effect you can detect, there is a region small enough that you can't detect it inside that region.

Conversely, to observe any curvature effect we must consider the trajectories of two test particles that are not colocated, and then we can interpret "non-local" to mean "at least as separated as our two non-colocated particles".

 

[Nugatory]

No. your spacetime is curved thus you get gravity. If you want to straighten the spacetime around you jump off following your brother.

This is incorrect. It's the same slightly curved locally flat spacetime for both.

Yes. The free faller should be infinitesimal small dwarf to escape from your criticism.

I think you may still be missing the point. Yes, you are right that if we were to shrink both observers down to infinitesimal dwarfs, the "it's locally flat" approximation would become exact in the infinitesimally small locality around each observer. But I was objecting to the suggestion that "If you want to straighten the spacetime around you jump off following your brother", because that's just plain wrong. The spacetime is what it is whether you're accelerating through it or in free fall through it.

 

[Dale]

Riemannian curvature is mathematically a local property. So what is non-local about it according to you? If possible give some reference.
The EP must limit arbitrarily to first derivatives of the metric, but how dos this make curvature something non-local?

That is explained in detail in chapters 3 and 4 here
http://www.preposterousuniverse.com/grnotes/

 

[sweet springs]

But I was objecting to the suggestion that "If you want to straighten the spacetime around you jump off following your brother", because that's just plain wrong. The spacetime is what it is whether you're accelerating through it or in free fall through it.

Thanks. Will restating "If you want to change your frame of reference so that it is locally flat where you (an infinitesimal dwarf) are, " satisfy you ?

 

[A.T.]

When we say that spacetime is locally flat and that curvature is non-local, we're saying that the approximation "it is flat across this region" can be made arbitrarily good by choosing the region in question to be sufficiently small. No matter how small a curvature effect you can detect, there is a region small enough that you can't detect it inside that region.

@loislane
As an analogy: The Riemannian curvature of a spherical planet surface is the same at each point. But the effect it has on the accuracy of a flat map depends on the size of the mapped area. Over small areas (locally) the distortion due to curvature is negligible.

 

[loislane]

When we say that spacetime is locally flat and that curvature is non-local, we're saying that the approximation "it is flat across this region" can be made arbitrarily good by choosing the region in question to be sufficiently small. No matter how small a curvature effect you can detect, there is a region small enough that you can't detect it inside that region.

You are restating that the EP arbitrarily restricts itself to the first derivatives of the metric(non-tensorial coordinate-dependent Christoffel coefficients) for that approximation. It might be clarifying to know the reason why a physical principle is restricted to a coordinate approximation when physicists are so concerned about considering physical only what is coordinate-independent(invariants).

Conversely, to observe any curvature effect we must consider the trajectories of two test particles that are not colocated, and then we can interpret "non-local" to mean "at least as separated as our two non-colocated particles".

I guess you are referring here to geodesic deviation that uses the Riemannian tensor in its formulation and therefore a local notion o curvature. Riemannian curvature is preserved by local isometries.

 

[PeterDonis]

It might be clarifying to know the reason why a physical principle is restricted to a coordinate approximation when physicists are so concerned about considering physical only what is coordinate-independent(invariants).

The curvature tensor is covariant, so scalars derived from it that measure "the amount of spacetime curvature" (the Ricci scalar is the simplest one) are coordinate-independent. The EP can just as easily be stated as applying only over length and time scales that are small compared to the radius of curvature derived from such scalars. That is a coordinate-independent statement.

 

[loislane]

The curvature tensor is covariant, so scalars derived from it that measure "the amount of spacetime curvature" (the Ricci scalar is the simplest one) are coordinate-independent.

Sure.

The EP can just as easily be stated as applying only over length and time scales that are small compared to the radius of curvature derived from such scalars. That is a coordinate-independent statement.

But here you are introducing the second derivatives of the metric(derivatives of the Christoffels) since you are introducing curvature . These are explicitly left out in the EP, that is only concerned with the first derivatives of the metric, that vnish at the point by using Riemann normal coordinates(and you need this local vanishing or you don't have any equivalence principle). So you can't use a local measure of curvature here to state a principle about equivalence to local flatness by arguing that curvature is coordinate-independent. Curvature can never be equivalent to flatness, that's precisely the reason the EP must be stated in terms of only first derivtives and Riemann normal coordinates.

 

[loislane]

That is explained in detail in chapters 3 and 4 here
http://www.preposterousuniverse.com/grnotes/

page 108 of chapter 4 indeed explains what I'm saying about the EP starting by:"
In fact, let us be honest about the principle of equivalence: it serves as a useful guideline,
but it does not deserve to be treated as a fundamental principle of nature. From the modern
point of view, we do not expect the EEP to be rigorously true".
But it doesn't give any alternative other than wait and see if the principle is empiracally ruled out in practice(explaining why this would be quite hard in fact).

 

[Dale]

page 108 of chapter 4 indeed explains what I'm saying about the EP starting by:"
In fact, let us be honest about the principle of equivalence: it serves as a useful guideline,
but it does not deserve to be treated as a fundamental principle of nature. From the modern
point of view, we do not expect the EEP to be rigorously true".
But it doesn't give any alternative other than wait and see if the principle is empiracally ruled out in practice(explaining why this would be quite hard in fact).

If what you are saying is consistent with what Carroll is saying then I don't think that there is any real disagreement here.

I am not sure why you think some alternative is needed. We have a theory which makes testable predictions. The theory reflects the principle meaning that it follows the "useful guideline". So what more is needed?

 

[PeterDonis]

you can't use a local measure of curvature here to state a principle about equivalence to local flatness by arguing that curvature is coordinate-independent.

Sure you can. I gave you an explicit criterion for deciding what "local" means: it means "over length and time scales that are small compared to the radius of curvature of spacetime". That's all the EP requires.

let us be honest about the principle of equivalence: it serves as a useful guideline,
but it does not deserve to be treated as a fundamental principle of nature

I agree with Carroll on this point; but note that in that case, since the EP is not supposed to be a fundamental principle of nature, the sort of practical criterion I gave above is perfectly adequate.

 

[loislane]

If what you are saying is consistent with what Carroll is saying then I don't think that there is any real disagreement here.

I am not sure why you think some alternative is needed. We have a theory which makes testable predictions. The theory reflects the principle meaning that it follows the "useful guideline". So what more is needed?

The disagrement here was about teaching a "useful guideline" as if was a fundamental principle which I think it was what PeterDonis and Nugatory and others are doing, that in the text you recomnended is clearly said it doesn't deserve to be treated as, because it is not even strictly true, in Carroll's words.
Now GR as theory is formulated in terms in which the EP(mathematically minimal coupling) must be exactly true, not just a guide. So it is no wonder that researchers working on quantum gravity consider GR to be just a useful approximation but ultimately wrong. I guess you should be asking them why some alternative is needed and what more is needed. I am just a perplexed mathematician and don't care for alternatives. It does bother me to read certain things.

 

[PeterDonis]

The disagrement here was about teaching a "useful guideline" as if was a fundamental principle which I think it was what PeterDonis and Nugatory and others are doing

If you're teaching GR, you teach GR. In GR, the minimal coupling principle is exactly true, so that's what we teach.

If we're talking about whether or not GR is an exactly correct theory of everything, that's a different question. Of course it isn't. But we don't have a more fundamental theory that covers the same domain, so we don't have anything else to teach.

it is not even strictly true, in Carroll's words.

None of our physical theories are "strictly true". They all break down at some point at the edges of their domains of validity. If you're looking for a physical theory that is strictly true in all respects, you're going to be looking a long, long time. In the meantime, physicists use (and teach) actual theories that exist now and give answers that are correct to within our accuracy of measurement, within their domains of validity.

 

[loislane]

Sure you can. I gave you an explicit criterion for deciding what "local" means: it means "over length and time scales that are small compared to the radius of curvature of spacetime". That's all the EP requires.

I agree with Carroll on this point; but note that in that case, since the EP is not supposed to be a fundamental principle of nature, the sort of practical criterion I gave above is perfectly adequate.

And again that "criterion" for what "local" means in mathematics is just a physical heuristic "guideline"(if not just something you just thought up now basically). That is not what local means in differential geometry. Your criterions(since you introduced a second one) as I explained are simply either "let's arbitrarily forget about derivatives higher that the first ones, even if anything concerning coordinate independence in general requires at least the second derivatives", or "let's use something added to the EP which is curvature itself at a point and let's define the EP with that addition included"..

 

[PeterDonis]

That is not what local means in differential geometry.

Ok, what does "local" mean in differential geometry?

 

[loislane]

Ok, what does "local" mean in differential geometry?

Obviously properties defined in the neigbourhood of a point. Clearly that is not restricted to the first derivatives of a tensor field at a point.
More broadly, those properties that when are fulfilled in open sets of a manifold are also true globally.

 

[PeterDonis]

Obviously properties defined in the neigbourhood of a point. Clearly that is not restricted to the first derivatives of a tensor field at a point.

Agreed; if it were, it would be impossible to define tensors involving higher derivatives, such as the Riemann tensor.

So in your view, can the minimal coupling principle be formulated as a "local" principle?

 

[loislane]

Agreed; if it were, it would be impossible to define tensors involving higher derivatives, such as the Riemann tensor.

So in your view, can the minimal coupling principle be formulated as a "local" principle?

I wouldn't say in "my view". Minimal coupling cannot be formulated as a procedure that respects local properties as defined in Riemannian geometry(since we are concerned about Riemann curvature) because it basically let's you get away ignoring terms that are necessary to take into account in Riemannian differential geometry.
But I have the feeling(I don't know enoug about seudo-riemannian manifolds to be sure) that in the pseudo-Riemannian generalization used in GR the distinction between local and global properties is not so clear as in classical differential geometry.

 

[PeterDonis]

Minimal coupling cannot be formulated as a procedure that respects local properties as defined in Riemannian geometry(since we are concerned about Riemann curvature) because it basically let's you get away ignoring terms that are necessary to take into account in Riemannian differential geometry.

I'm not sure I understand this. Consider the two alternative field equations for EM that Carroll gives (equations 4.24 and 4.32):

$$
\nabla_{\mu} F^{\mu \nu} = 4 \pi J^{\nu}
$$

vs.

$$
\nabla_{\mu} \left[ \left( 1 + \alpha R \right) F^{\mu \nu} \right] = 4 \pi J^{\nu}
$$

Both of these are perfectly valid equations from the point of view of differential geometry. The $\alpha R$ term is certainly allowed by differential geometry, but I don't see why it is made necessary by differential geometry. Adopting the minimal coupling principle in order to make $\alpha = 0$ in standard GR is not "ignoring" anything about differential geometry; it's making a physical judgment about what terms to include in a physical theory.

Note, also, that if we're really going to insist on including all possible terms involving curvature, there's no reason to stop at the $\alpha$ term; we could have higher order terms as well, so we would have something like:

$$
\nabla_{\mu} \left[ \left( 1 + \alpha R + \beta R^2 + \gamma R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} + ... \right) F^{\mu \nu} \right] = 4 \pi J^{\nu}
$$

I don't see why the above wouldn't be just as much "necessary" as the version with only $\alpha$ in it, by the logic you are using.

 

[loislane]

I'm not sure I understand this. Consider the two alternative field equations for EM that Carroll gives (equations 4.24 and 4.32):

$$
\nabla_{\mu} F^{\mu \nu} = 4 \pi J^{\nu}
$$

vs.

$$
\nabla_{\mu} \left[ \left( 1 + \alpha R \right) F^{\mu \nu} \right] = 4 \pi J^{\nu}
$$

Both of these are perfectly valid equations from the point of view of differential geometry. The $\alpha R$ term is certainly allowed by differential geometry, but I don't see why it is made necessary by differential geometry. Adopting the minimal coupling principle in order to make $\alpha = 0$ in standard GR is not "ignoring" anything about differential geometry; it's making a physical judgment about what terms to include in a physical theory.

I was talking about something a bit less general than differential geometry, indeed it can be said that both are valid in differential geometry without more qualifications.
The distinction I was referring to was between Riemannian and pseudo-riemannian geometry.

Note, also, that if we're really going to insist on including all possible terms involving curvature, there's no reason to stop at the $\alpha$ term; we could have higher order terms as well, so we would have something like:

$$
\nabla_{\mu} \left[ \left( 1 + \alpha R + \beta R^2 + \gamma R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} + ... \right) F^{\mu \nu} \right] = 4 \pi J^{\nu}
$$

I don't see why the above wouldn't be just as much "necessary" as the version with only $\alpha$ in it, by the logic you are using.

You can see this explained here: https://www.physicsforums.com/threads/einstein-hilbert-action-origin.838487/ in posts #15 and #17

 

[PeterDonis]

The distinction I was referring to was between Riemannian and pseudo-riemannian geometry.

I don't see any real distinction between the two with regard to the equations I wrote down; both of them are valid regardless of the metric signature.

You can see this explained here: https://www.physicsforums.com/threads/einstein-hilbert-action-origin.838487/ in posts #15 and #17

The point being made in those posts is that, if we look at the Einstein-Hilbert action for gravity alone, it can only be linear in second derivatives of the metric (i.e., in $R$, the Ricci scalar); otherwise we would get a third or higher order PDE for the field equation for gravity.

But we haven't been looking at the field equation for gravity; we've been looking at the field equation for electromagnetism, which is derived by varying the Lagrangian with respect to the vector potential $A_{\mu}$, not the metric. Terms of higher order in $R$ aren't functions of the vector potential, so, from the standpoint of Maxwell's Equations, if we're going to include any curvature terms, we should include them all, regardless of order.

We could, however, look at how adding those terms to the Lagrangian would affect the Einstein Field Equation; what we will find is that even the $\alpha$ term would add extra unwanted terms to that equation. The usual EFE is derived from the action (leaving out irrelevant factors in the terms and including the EM Lagrangian--we assume there is no other matter present):

$$
S = \int d^4 x \sqrt{-g} \left( R - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right)
$$

Varying this with respect to the metric gives the usual EFE with the EM stress-energy tensor on the RHS (which I won't bother writing out explicitly since its form won't matter here):

$$
R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = T_{\mu \nu}
$$

But now suppose we include the extra $\alpha$ term in the Lagrangian, so that varying with respect to the vector potential puts the extra $\alpha$ term in Maxwell's Equations:

$$
S = \int d^4 x \sqrt{-g} \left( R - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \alpha R \right)
$$

Now, when we vary this with respect to the metric, we get:

$$
\left( R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R \right) \left( 1 - \alpha \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right) = T_{\mu \nu} \left( 1 + \alpha R \right)
$$

This mess is clearly not a clean field equation for the metric (i.e., an equation where the LHS has no matter terms and the RHS has no curvature terms. (And if we try to make it into one by algebraically moving terms and factors around, what we end up with will not be a clean second-order PDE anyway, since it will have to divide the Einstein tensor by the factor $\left( 1 + \alpha R \right)$ on the LHS.)

In other words, you can't add any curvature terms to the field equations for matter (such as Maxwell's Equations) without also adding unwanted terms to the field equation for gravity (as above). That's a powerful argument in favor of the minimal coupling principle.

 

[loislane]

I don't see any real distinction between the two with regard to the equations I wrote down; both of them are valid regardless of the metric signature.

They both use the formulation of EM in pseudo-riemannian space, so how would you know? Are you aware of any formulation of the electromagnetic field tensor in Riemannian geometry to compare?

The point being made in those posts is that, if we look at the Einstein-Hilbert action for gravity alone, it can only be linear in second derivatives of the metric (i.e., in $R$, the Ricci scalar); otherwise we would get a third or higher order PDE for the field equation for gravity.

But we haven't been looking at the field equation for gravity; we've been looking at the field equation for electromagnetism, [...]

Well, this thread is about GR and the EP and so is the reference the equations are taken from, not about electromagnetism. The last equation is an example of what one equation of EM in the Lorentzian space formulation would look like in the curved spacetime of GR if the minimal coupling was not used. I agree they should have used the equation you wrote in your post instead if the point was about electromagnetism, but Carroll's point was about GR and the EP and I guess that is why he used 4.31 instead of your version. And that's why the posts I linked that are about GR and the EP in relation to the absence of higher derivatives are pertinent

 

[Dale]

The disagrement here was about teaching a "useful guideline" as if was a fundamental principle

So you dislike that the Equivalence Principle is called a principle? OK. But that is what it is called. Weren't you earlier complaining that other people were focusing on semantics?

Do you have a substantive objection or question about GR , or just want to complain about the "window dressings"?

 

[PeterDonis]

They both use the formulation of EM in pseudo-riemannian space, so how would you know?

Are we talking about mathematics, or about physics? If we're talking about physics, then the objections you're raising don't seem relevant, since they're mathematical, not physical. You can't argue that a particular term "should" be in a physical equation because "differential geometry says so". What belongs in a physical equation depends on the physics.

If we're talking about mathematics, then my statement means just what it says: you can't tell, just by looking at either of the equations under discussion, what metric signature they apply to. They are valid equations, mathematically, in a Riemannian metric just as they are in a pseudo-Riemannian metric. Of course they say different things physically if the metric is Riemannian vs. pseudo-Riemannian, but that's a question of physics, not differential geometry.

the posts I linked that are about GR and the EP in relation to the absence of higher derivatives are pertinent

Do you have any comment on the rest of my post, where I argue that if those arguments are pertinent, they mean that the $\alpha$ term also doesn't belong?

 

[loislane]

If we're talking about mathematics, then my statement means just what it says: you can't tell, just by looking at either of the equations under discussion, what metric signature they apply to. They are valid equations, mathematically, in a Riemannian metric just as they are in a pseudo-Riemannian metric. Of course they say different things physically if the metric is Riemannian vs. pseudo-Riemannian, but that's a question of physics, not differential geometry.

I guess I'm missing something here, or you are missing my point. What I'm saying is that the electromagnetic field tensor in those equations is formulated in Lorentzian space, again is there a formulation of the electromagnetic tensor that doesn't have indefinite signature, i.e. that uses a Riemannian metric?

Do you have any comment on the rest of my post, where I argue that if those arguments are pertinent, they mean that the $\alpha$ term also doesn't belong?

No. It doesn't sem obviously wrong to me but my knowledge of physics is limited.

 

[martinbn]

I am a bit confused what the discussion is about? I always thought that the equivalence principle is quite clear and any non-physicist would understand it (that includes mathematicians). There are a couple of threads discussing it and it seems that there are different opinions, but I am not sure what they are. Loislane, can you state what your position is as clearly as possible? Now it seems to me that you are arguing for the sake of arguing and I am sure I am wrong.

 

[PeterDonis]

What I'm saying is that the electromagnetic field tensor in those equations is formulated in Lorentzian space

Physically, yes, we only use the version of those equations that is formulated with a Lorentzian metric signature. But that's a physical restriction, not a mathematical restriction. Mathematically, the equations themselves are perfectly valid equations with a Riemannian metric.

In other words, it isn't differential geometry that tells us to use a Lorentzian metric signature when we apply the equations; it's physics, picking a particular formulation out of all the possible ones that are valid mathematically.

 

[loislane]

Physically, yes, we only use the version of those equations that is formulated with a Lorentzian metric signature. But that's a physical restriction, not a mathematical restriction. Mathematically, the equations themselves are perfectly valid equations with a Riemannian metric.

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric? Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean), it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients? You are assuming they are zero as if it could only be a Euclidean tensor.

 

[PeterDonis]

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric?

Um, because you can take the covariant divergence of a tensor field, and validly equate it to some vector field, just as easily in Riemannian geometry as in pseudo-Riemannian geometry? I don't understand what the problem is. Remember we are talking math, not physics, in this particular case; we're not putting any physical interpretation on $F^{\mu \nu}$ or $J^{\nu}$, we're just taking the covariant divergence of the first and equating it to the second. Mathematically, that can obviously be done in Riemannian geometry.

If you want a concrete example, suppose $F^{\mu \nu}$ and $J^{\nu}$ are tensor and vector fields on a 2-sphere, a curved Riemannian manifold. Are you saying we can't take the covariant divergence of $F$, using the 2-sphere's metric, and equate it to $J$?

Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean)

What is a "Euclidean" tensor field? I understand what a Euclidean metric is, but we're not assuming the metric (in the Riemannian case you are talking about) is Euclidean. Tensor fields other than the metric aren't Euclidean or non-Euclidean (or, in the pseudo-Riemannian case, Minkowskian or non-Minkowskian). They're just tensor fields on the manifold, which has whatever metric we are assuming for the particular problem under discussion.

it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients?

Um, the same thing you do with them in a curved pseudo-Riemannian metric? Are you saying that Christoffel symbols somehow don't work in Riemannian geometry?

You are assuming they are zero as if it could only be a Euclidean tensor.

I'm assuming no such thing. The equation under discussion has a covariant divergence in it, not an ordinary divergence; the whole point of using $\nabla$ instead of $\partial$ is to emphasize that we are in a curved manifold so you have to include the connection coefficients when you take derivatives.

 

[martinbn]

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric? Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean), it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients? You are assuming they are zero as if it could only be a Euclidean tensor.

This makes no sense! For tensors you only need a manifold, so you can have a tensor $F^{\mu\nu}$ (antisymetric or not) on a given manifold. If on top of that you have a connection, say from a Riemannian metric, then you can differentiate the tensor.

 

[loislane]

If you want a concrete example, suppose $F^{\mu \nu}$ and $J^{\nu}$ are tensor and vector fields on a 2-sphere, a curved Riemannian manifold. Are you saying we can't take the covariant divergence of $F$, using the 2-sphere's metric, and equate it to $J$?
Um, the same thing you do with them in a curved pseudo-Riemannian metric? Are you saying that Christoffel symbols somehow don't work in Riemannian geometry?

No, I was asking you where are the Christoffel symbols in 4.24.

I'm assuming no such thing. The equation under discussion has a covariant divergence in it, not an ordinary divergence; the whole point of using $\nabla$ instead of $\partial$ is to emphasize that we are in a curved manifold so you have to include the connection coefficients when you take derivatives.

Exactly, I gues what you are saying is that they are implicit in the equation in vector field in the rhs.
I think we are basically talking at cross purposes here, so I'm leaving the thread, (apparently there are some attempts at trolling too so it is better to leave it alone).

 

[PeterDonis]

I was asking you where are the Christoffel symbols in 4.24.

Um, by definition, the $\nabla$ operator includes them. That's why I said the whole point of writing $\nabla$ instead of $\partial$ in a curved manifold is to ensure that the Christoffel symbols are taken into account.

I gues what you are saying is that they are implicit in the equation in vector field in the rhs.

No, they are implicit in the $\nabla$ operator on the LHS.

 

[loislane]

So in your own words, since you agreed with Caroll's quote, why doesn't the EP deserve to be treated as a fundamental principle?

And subsequently why should the practical consequence of something that is not a fundamental principle, i.e. the minimal coupling, should be taught as something exact in GR?

 

[PeterDonis]

why doesn't the EP deserve to be treated as a fundamental principle?

Because GR is not a theory of everything, and on quantum field theoretical grounds, we expect the additional terms that GR does not include, because of the minimal coupling principle, to be present, but suppressed by inverse powers of something like the Planck mass, which makes them completely unobservable in any experimental regime we are likely to be able to reach any time soon (or even not so soon). So including them in our models would just clutter up the models to no purpose, since our calculations would crank out a bunch of additional terms that are harder to deal with mathematically and predict effects that are way too small to measure.

why should the practical consequence of something that is not a fundamental principle, i.e. the minimal coupling, should be taught as something exact in GR?

Because, as I said before, if you're going to teach GR, you teach GR. In GR, the minimal coupling principle is exact, so that's how we teach it. Just as, when we teach Newtonian physics, we treat the principle of conservation of mass as exact, without cluttering students' heads with all the complications brought in by mass-energy equivalence in relativity. Or, when we teach special relativity, we treat flat Minkowski spacetime as exact, without cluttering students' heads with all the complications brought in by the fact that there is no such thing as true flat Minkowski spacetime in the real world, since there is nonzero stress-energy present.

In other words, physics is not taught as the theory of everything, with all its complications, all in one lump. It is taught in stages, and at each stage things are taught as exact which turn out to be not quite exact in later stages.

 

[loislane]

Because GR is not a theory of everything, and on quantum field theoretical grounds, we expect the additional terms that GR does not include, because of the minimal coupling principle, to be present, but suppressed by inverse powers of something like the Planck mass, which makes them completely unobservable in any experimental regime we are likely to be able to reach any time soon (or even not so soon).

You mean like GR is some sort of classical "effective field theory"?

 

[PeterDonis]

You mean like GR is some sort of classical "effective field theory"?

That is pretty much the current mainstream view, yes. The question we don't yet know the answer to is what underlying quantum theory it is the classical effective field theory of.

 

[loislane]

That is pretty much the current mainstream view, yes. The question we don't yet know the answer to is what underlying quantum theory it is the classical effective field theory of.

I see. I must say I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context. In the absence of a particular absolute invariant group, like say the Lorentz group in the Minkowski space of SR, the general symmetry of GR is reduced to its coordinate covariance(invariance under general local coordinate transformations), in other words a trivial property of any theory is raised to the category of physical symmetry(dynamical metric-gravitational fields).

 

[PeterDonis]

I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context.

It has nothing to do with the indefinite signature. A Riemannian manifold is locally Euclidean in the same way that a pseudo-Riemannian manifold of the type used in GR is locally Minkowskian.

 

[martinbn]

I see. I must say I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context.

As PeterDonis said the signature is irrelevant. To me the intuitive picture is that there are no sharp point or edges, the space-time has a tangent space at each point and in a small enough neighborhood the space-time and the tangent space are almost the same (including the metric).

In the absence of a particular absolute invariant group, like say the Lorentz group in the Minkowski space of SR, the general symmetry of GR is reduced to its coordinate covariance(invariance under general local coordinate transformations), in other words a trivial property of any theory is raised to the category of physical symmetry(dynamical metric-gravitational fields).

I don't understand this and how it is related to the above.

 

[loislane]

It has nothing to do with the indefinite signature. A Riemannian manifold is locally Euclidean in the same way that a pseudo-Riemannian manifold of the type used in GR is locally Minkowskian.

Not in the same way. That's an important misconception you should look into.
-Being locally euclidean has nothing to do with the Euclidean metrics or signature types, it is a topological or differentiable structure equivalence with ℝn locally depending on the kind of manifold(topological or smooth) one is talking about. It has nothing to do with having a Euclidean metric no matter how small a region you want to consider.
-Being locally Minkowskian obviously involves having locally a specific Minkowskian metric tensor with vanishing curvature and a specific signature.
This is basically the EP content and one could infer logically it is the reason why mathematically it must be limited to the first derivatives of the metric, otherwise it would collide with the differential geometry concept of curvature at a point.
I think that is what is behind the split between the "spacetime tells mtter how to move" and "matter tells spacetime how to curve" separate parts put forward by Wheeler and also determines the peculiar implementation of "general covariance" in GR that I refer to in my last post.