Levi-Civita connection and pseudoRiemannian metric

[TrickyDicky]

They're not compatible.

That's my impression too.

In fact, while it is possible to impose a metric space structure on the manifold, the distance function will be very much unphysical. It will be completely useless in physics.

I tend to agree, but as I said it is routinely used in relativist literature (I gave the event horizon example) ever since the early works of Hawking, Penrose, Geroch...Of course Hawking just recently changed his mind about black holes so go figure...

So yes, all what matters is the topology and not the metric space structure.

Did you read the mathpages article? If so, do you find it accurate?

 

[WannabeNewton]

Are not null hypersurfaces for instance objects derived from the pseudoriemannian metric pseudometric properties and their induced topology?

It's derived from the pseudo-riemannian metric tensor properties. It has nothing to do with a pseudo-metric. The metric tensor only induces a topology in each tangent space. The topology we work with for space-time (usually) is the manifold topology of the smooth atlas. We never use metrics in GR, period. They're pointless in the GR context.

 

[TrickyDicky]

It's derived from the pseudo-riemannian metric tensor properties.

Agreed.

It has nothing to do with a pseudo-metric. The metric tensor only induces a topology in each tangent space.

Null hypersurfaces topology is not confined to the tangent space, they are submanifolds of M.

The topology we work with for space-time (usually) is the manifold topology of the smooth atlas.

Yes, usually, for instance when applied to anything related to curvature, but usually is not always.

We never use metrics in GR, period. They're pointless in the GR context.

That's a strong statement that doesn't seem right without qualifications. In this context you are saying "we never use distances in GR, they are pointless", are you sure?

 

[WannabeNewton]

Null hypersurfaces topology is not confined to the tangent space, they are submanifolds of M.

Where did I say otherwise? You talked about the topology induced by the metric tensor. This has nothing to do with null hypersurfaces of the subspace topologies of null hypersurfaces. The metric tensor only induces topologies in the tangent spaces.

That's a strong statement that doesn't seem right without qualifications.

Give one example of a metric being used in GR. Metrics are useful in QM not in GR.

In this context you are saying "we never use distances in GR, they are pointless", are you sure?

You're confusing different notions of distance. George already pointed this out.

 

[TrickyDicky]

Where did I say otherwise? You talked about the topology induced by the metric tensor. This has nothing to do with null hypersurfaces of the subspace topologies of null hypersurfaces. The metric tensor only induces topologies in the tangent spaces.

Forget the topology for a minute, I meant the obvious fact that without pseudoRiemannian metric tensor there is no null hypersurface at all, are you disagreeing with this?

Give one example of a metric being used in GR. Metrics are useful in QM not in GR.

Metric tensors can be thouth of as infinitesimal metrics, you just have to integrate them, as is done often in GR to get distances that accommodate or not the requirements of metric spaces. As pointed out in the linked article one can ignore those integrated distances, but that should be consistent with the physical predictions of GR one uses.

You're confusing different notions of distance. George already pointed this out.

Please, enumerate those different notions and tell me how exactly I confuse them so I can get corrected.

 

[micromass]

BTW, there are no metrics in QM, the spaces used thre are seminormed, not normed.

The main used spaces in QM seem to be $\mathbb{R}^n$, $\ell^2$ and $L^2(M)$ for some $M\subseteq \mathbb{R}^n$. How are these not normed.

 

[WannabeNewton]

I'm sorry but you're lacking a lot of analysis knowledge, especially if you think there are no norms or metrics in QM. You need to learn analysis first. This discussion is pointless.

 

[TrickyDicky]

The main used spaces in QM seem to be $\mathbb{R}^n$, $\ell^2$ and $L^2(M)$ for some $M\subseteq \mathbb{R}^n$. How are these not normed.

I'm sorry but you're lacking a lot of analysis knowledge, especially if you think there are no norms or metrics in QM. You need to learn analysis first. This discussion is pointless.

You are right , I wrote that sentence in a rush, I meant that in functional anlysis pseudometrics also arise naturally, thru seminormed spaces. I'll erase it.

Anyway WN this is just a tangential goof that has nothing to do with what we are discussing. Your using it as excuse not to continue discussing is rather odd.

 

[TrickyDicky]

The metric tensor only induces topologies in the tangent spaces.

The structures and objects that I comment below are not confined to the tangent space, they belong to the manifold.

Give one example of a metric being used in GR.

I'll give you examples where distances between events(whatever you want to call those distance functions:metrics, pseudometrics...) based on the pseudoriemannian metric tensor are used: Timelike and null geodesics, and all the objects in GR that are based on any of those geodesics, for instance in the case of null geodesics the notion of null infinity, used to define event horizons or the notion of conformal infinity. In the case of timelike geodesics, the very concept of proper time.

You're confusing different notions of distance.

If you refer to the fact that formally spacetimes are defined as metric spaces with the standard topology, I'm aware of that. I'm just being consequent with it and being strict this definition implies ignoring any distance function incompatible with the definition of metric spaces one could apply to the manifold based on the pseudoriemannian tensor. But I'm highlighting the fact that if one ignores those distance notions one shouldn't be able to use them in physics. The above examples apparently show that is not the case, and I'm just reflecting it.

 

[TrickyDicky]

Quoting from mathpages again: "we can simply take as our basis sets all the finite intersections of Minkowski neighborhoods. Since the contents of an e-neighborhood of a given point are invariant under Lorentz transformations, it follows that the contents of the intersection of the e-neighborhoods of two given points are also invariant. Thus we can define each basis set by specifying a finite collection of events with a specific value of e for each one, and the resulting set of points is invariant under Lorentz transformations.[...] In the case of a spacetime theory, we need to consider whether the temporal and spatial components of intervals have absolute significance, or whether it is only the absolute intervals themselves that are significant."

It would seem this is basically how the issue is solved in general for validly using the physics derived from structures generated by a pseudoRiemannian metric instead of a Riemannian metric, right? considering temporal and spatial components separately?

Is this related to what is called metric identification as explained in Wikipedia http://en.wikipedia.org/wiki/Metric_identification ?

 

[robphy]

Possibly interesting reading:
http://en.wikipedia.org/wiki/Spacetime_topology

 

[TrickyDicky]

Possibly interesting reading:
http://en.wikipedia.org/wiki/Spacetime_topology

Yes, the topology of spacetime must coincide with the manifold topology R4.

I had a pertinent (I think) doubt about how the fact that pseudoriemannian metric tensors can in principle be integrated to distance functions with topology different from the natural manifold topology (that is by definition a metric space) was handled in relativity.
Apparently this question is enough to irritate certain people. And no, I'm not confusing different notions of distance. Anyway it is probably not so hard, but it wasn't obvious to me when I asked. The basic way I think I could answer my own question is the following (wich by the way makes use of the L-C connection, so my initial intuition maybe wasn't so mistaken):
I was concerned with integrated lengths in the case of timelike and mostly null geodesics. I was also stuck at spacetime intervals derived from this integrations, considering the concept as something absolute in itself rather than its spatial and temporal components as it is usually stressed in relativity pedagogy.

But the fact is that the geodesic equation in its canonical form in GR and Riemannian geometry, that is using the unique Levi-Civita affine connection, demands the use of affine parameters, and this simple fact is important here because it only allows geodesics to be expressed in terms of either temporal or spatial lengths that are never zero for distinct temporal or spatial points. And wrt the triangle inequality, one just has to stick to the equality part(test particles). So the manifold topology is preserved.

Ironically it seems to salvage the spacetime topology one has to think in terms of space and time separately.

But it makes one wonder if say explanations based on the reverse triangle inequality for the twin paradox are then valid explanations.

 

[PeterDonis]

pseudoriemannian metric tensors can in principle be integrated to distance functions with topology different from the natural manifold topology

This isn't quite correct as you state it. The correct statement is that the distance function associated with a pseudoriemannian metric does not induce a valid topology on the underlying manifold, because that distance function assigns zero distance to distinct points. Only a riemannian metric can induce a valid topology on the underlying manifold.

But the fact is that the geodesic equation in its canonical form in GR and Riemannian geometry, that is using the unique Levi-Civita affine connection, demands the use of affine parameters, and this simple fact is important here because it only allows geodesics to be expressed in terms of either temporal or spatial lengths that are never zero for distinct temporal or spatial points.

Huh? The geodesic equation applies perfectly well to null geodesics, which have zero length. Null geodesics still have affine parameters; the valid affine parameters for null geodesics just don't include the "distance" along them in terms of the pseudoriemannian metric.

And wrt the triangle inequality, one just has to stick to the equality part(test particles).

I don't understand what you're talking about here.

But it makes one wonder if say explanations based on the reverse triangle inequality for the twin paradox are then valid explanations.

Of course they are. Why wouldn't they be? What does the fact that the pseudoriemannian metric doesn't induce a valid topology have to do with whether the reverse triangle inequality works as an explanation for the twin paradox?

 

[TrickyDicky]

This isn't quite correct as you state it. The correct statement is that the distance function associated with a pseudoriemannian metric does not induce a valid topology on the underlying manifold, because that distance function assigns zero distance to distinct points. Only a riemannian metric can induce a valid topology on the underlying manifold.

What you write is exactly what I wanted to mean, so if the way you word it is clearer or more correct, I make it mine.

Huh? The geodesic equation applies perfectly well to null geodesics, which have zero length. Null geodesics still have affine parameters; the valid affine parameters for null geodesics just don't include the "distance" along them in terms of the pseudoriemannian metric.

Again, it seems my poor command on the english language must be playing here. What you say here is in perfect agreement with what I'm saying. My point was that it is precisely the fact that the geodesic equation demands that null geodesics are affinely parametrized what makes them perfectly compatible with the manifold topology. This is related to some paragraphs in the mathpages article I linked where towards the end it addresses how to avoid the issues that could arise from the distance function associated with a pseudoriemannian metric. In the end as WN said(although he could have explained it in a less dismissive way (if only for the community spirit) (pseudo)metrics are not used in GR. As simple as that.

I don't understand what you're talking about here.

Yes, I was a bit cryptic here. I meant that for test particles the triangle inequality which is ≤, in a non-euclidean setting the inequality is not strict so d(x,z) can be equal to d(x,y)+d(y,z). Because there are no non-gravitational forces at play.

Of course they are. Why wouldn't they be? What does the fact that the pseudoriemannian metric doesn't induce a valid topology have to do with whether the reverse triangle inequality works as an explanation for the twin paradox?

Well, as I was saying I just meant that the reverse triangle inequality is a property of distance functions that don't induce a valid topology on the manifold, so being rigorous since in the twin paradox are involved accelerations(that is non-gravitational forces) one must use the strict reverse inequality and it feels like an explanation not very appropriate if one is stressing the fact that distances with that property are topologically incompatible with the manifold as a metric space. I don't know maybe my reasoning is too involved.

 

[PeterDonis]

What you write is exactly what I wanted to mean

Ok.

My point was that it is precisely the fact that the geodesic equation demands that null geodesics are affinely parametrized what makes them perfectly compatible with the manifold topology.

Ok.

I meant that for test particles the triangle inequality which is ≤, in a non-euclidean setting the inequality is not strict so d(x,z) can be equal to d(x,y)+d(y,z). Because there are no non-gravitational forces at play.

I still don't understand. Can you give a specific example? That is, can you give three points in Minkowski spacetime (specified by coordinates given in some inertial frame) such that d(x, z) = d(x, y) + d(y, z)? (With d being the Minkowski distance function, of course.)

it feels like an explanation not very appropriate if one is stressing the fact that distances with that property are topologically incompatible with the manifold as a metric space.

First of all, I don't see what this concern has to do with whether or not the reverse triangle inequality must be strict, or whether the equality can hold. (I am unable to think of an example where the equality does hold, which is why I asked you for one above.)

Second, I don't see what the concern is. The pseudo-Riemannian distance function is just a function on the manifold--more precisely, it's a symmetric 2nd-rank tensor field on the manifold, which gets contracted with tangent vector fields and integrated along curves to obtain physical distances, i.e., distances that are used to make physical predictions. There's no requirement that every function or tensor field on a manifold has to induce a valid topology on the manifold; nor is there a requirement that topological "distances" (i.e., distances computed using the metric that corresponds to whatever topology on the manifold we are using) must match "distances" that are used to make physical predictions. The fact that the term "distance" can be used for both concepts does not mean there has to be any relationship between them. It looks to me like you are being confused by terminology issues that have nothing to do with the physics involved.

 

[TrickyDicky]

I still don't understand. Can you give a specific example? That is, can you give three points in Minkowski spacetime (specified by coordinates given in some inertial frame) such that d(x, z) = d(x, y) + d(y, z)? (With d being the Minkowski distance function, of course.)
First of all, I don't see what this concern has to do with whether or not the reverse triangle inequality must be strict, or whether the equality can hold. (I am unable to think of an example where the equality does hold, which is why I asked you for one above.)

My example was trivial I guess, just a test particle following a geodesic path. You cut a segment of that path and call x the first point in the segment, z the last one and y an arbitrary point of the geodesic between x and z.

Second, I don't see what the concern is. The pseudo-Riemannian distance function is just a function on the manifold--more precisely, it's a symmetric 2nd-rank tensor field on the manifold, which gets contracted with tangent vector fields and integrated along curves to obtain physical distances, i.e., distances that are used to make physical predictions. There's no requirement that every function or tensor field on a manifold has to induce a valid topology on the manifold;

I'm not calling the pseudoRiemannian metric tensor field a distance function(it would be in any case an "infinitesimal distance function").

nor is there a requirement that topological "distances" (i.e., distances computed using the metric that corresponds to whatever topology on the manifold we are using) must match "distances" that are used to make physical predictions. The fact that the term "distance" can be used for both concepts does not mean there has to be any relationship between them.

Well, the thing is we basically don't need this match (fortunately) since we are not using those integrated distances from the pseudoRiemannian metric tensor in practice in GR in the way I was thinking by virtue of the geodesic equation with the affine connection.
I was nitpicking about its use(the reverse triangle equality semimetric property) as a posible explanation of the twin paradox since we don't use (pseudosemi)metrics in relativity.
Anyway, you don't think that the integrated distances that are used to make physical predictions based on a certain mathematical set up that includes a manifold and a topology should be required to "match" the restrictions put forth by that topology on topological distances?

 

[PeterDonis]

My example was trivial I guess, just a test particle following a geodesic path. You cut a segment of that path and call x the first point in the segment, z the last one and y an arbitrary point of the geodesic between x and z.

Ok, so this is just a degenerate "triangle" where all three "sides" are collinear. I agree this satisfies the equality, but I don't see why that creates any sort of problem for cases where the triangle is not degenerate.

Well, the thing is we basically don't need this match (fortunately) since we are not using those integrated distances from the pseudoRiemannian metric tensor in practice in GR in the way I was thinking by virtue of the geodesic equation with the affine connection.

If you mean that we can't use the pseudo-Riemannian distance as an affine parameter, strictly speaking this is only true for null geodesics. For timelike and spacelike geodesics, the pseudo-Riemannian distance does work as an affine parameter, and since affine transformations are linear, any affine parameter that works for a timelike or spacelike geodesic must be a linear function of the pseudo-Riemannian distance.

If you mean that we don't have to use the geodesic equation in order to compute the pseudo-Riemannian distance along a curve, of course this is true; for one thing, non-geodesic curves also have well-defined pseudo-Riemannian distances along them.

I was nitpicking about its use(the reverse triangle equality semimetric property) as a posible explanation of the twin paradox since we don't use (pseudosemi)metrics in relativity.

I don't understand. Are you claiming that the reverse triangle inequality is not valid in Minkowski spacetime? That's obviously false: the Minkowski distance function obeys the reverse inequality as long as at least one side of the triangle is timelike. If you're claiming something else, then I don't understand what you're claiming.

Anyway, you don't think that the integrated distances that are used to make physical predictions based on a certain mathematical set up that includes a manifold and a topology should be required to "match" the restrictions put forth by that topology on topological distances?

Obviously not, since if this were true we could not use Minkowski spacetime, or any spacetime that locally looked like Minkowski spacetime, to make valid physical predictions. Since we can do so, the two "distances" obviously cannot be required to match.

Once again, I think this may be a language confusion. The use of the term "distance" should not mislead you into thinking that topological distances must correspond to physical distances. They don't have to.

 

[TrickyDicky]

Ok, so this is just a degenerate "triangle" where all three "sides" are collinear. I agree this satisfies the equality, but I don't see why that creates any sort of problem for cases where the triangle is not degenerate.

Ok, see below.

If you mean that we don't have to use the geodesic equation in order to compute the pseudo-Riemannian distance along a curve, of course this is true; for one thing, non-geodesic curves also have well-defined pseudo-Riemannian distances along them.

If by pseudoRiemannian distance(not standard term) you mean pseudosemimetric distance, I understood from a previous post(it took me some effort, though) these distances are not used in GR.

I don't understand. Are you claiming that the reverse triangle inequality is not valid in Minkowski spacetime?

Well, yes and no, actually is valid if we consider only test particles, but in the twin paradox one of the twins accelerates

Obviously not, since if this were true we could not use Minkowski spacetime, or any spacetime that locally looked like Minkowski spacetime, to make valid physical predictions. Since we can do so, the quoted statement is false.

Well, we do make physical predictions but apparently we don't need the problematic minkowskian distances in the sense of pseudometrics to make physical predictions. I'm myself still in the process of getting to grips with this.

Once again, I think this may be a language confusion. The use of the term "distance" should not mislead you into thinking that topological distances must correspond to physical distances. They don't have to.

I think it's the other way, physical distances must accommodate to the topology one is using.

 

[PeterDonis]

If by pseudoRiemannian distance(not standard term) you mean pseudosemimetric distance, I understood from a previous post(it took me some effort, though) these distances are not used in GR.

By "pseudo-Riemannian distance" I mean you have a curve in spacetime with tangent vector field $u^{\mu}$, you have a metric $g_{\mu \nu}$, which is a tensor field, and you integrate $ds = \sqrt{g_{\mu \nu} u^{\mu} u^{\nu}}$ along the curve to get a path length. I certainly hope you are not claiming that this is not used in GR; I don't think anyone in previous posts was claiming it was. I think they were only saying that the "topological distance" (the "distance" you would get if you tried to use the distance function induced by the standard topology on $\mathbb{R}^4$) was not used in GR to model actual physical distances.

Well, yes and no, actually is valid if we consider only test particles, but in the twin paradox one of the twins accelerates

Technically, yes, but in the usual formulation the path length of the accelerating portion of the traveling twin's worldline is negligible, so the small curved section at the "corner" where the traveling twin turns around is simply approximated as a vertex of the triangle. The reversed triangle inequality applies just fine to this approximate model.

That said, even if you refuse to use the inequality, you can still obviously compute the path lengths for each twin directly, using the method I gave above. So I don't see a big issue here in any case.

Well, we do make physical predictions but apparently we don't need the problematic minkowskian distances in the sense of pseudometrics to make physical predictions.

I have no idea what you mean here. I think you're getting confused by terminology again. As I asked above, are you saying the process I described of integrating $ds = \sqrt{g_{\mu \nu} u^{\mu} u^{\nu}}$ along a curve to get a path length is not used to make physical predictions? If you are, that's just wrong. If you aren't, what are you saying?

I think it's the other way, physical distances must accommodate to the topology one is using.

Why?

 

[TrickyDicky]

As I asked above, are you saying the process I described of integrating $ds = \sqrt{g_{\mu \nu} u^{\mu} u^{\nu}}$ along a curve to get a path length is not used to make physical predictions? If you are, that's just wrong. If you aren't, what are you saying?

Yes, it is used but those curves are parametrized by an affine parameter and that makes them compatible with the manifold topology.

Why?

Because the physics whenever possible obeys the math.

 

[PeterDonis]

Yes, it is used but those curves are parametrized by an affine parameter and that makes them compatible with the manifold topology.

I don't understand. It's trivially true that every curve can be affinely parametrized; but it is not true that every affine parametrization on every curve must have a direct physical meaning in terms of physical path length along the curve. Certainly that's not true for null curves in spacetime; yet the equation I gave for path length applies perfectly well to null curves. [Edit: With corrections--see following post.] So I don't understand what point you're trying to make here.

Because the physics whenever possible obeys the math.

It obeys the math that appropriately models the physics, yes. The distance function induced by the topology does not appropriately model the physics.

In every physical theory, there is math involved that does not directly model the physics; it's "extra" math that is there because without it we can't give the theory a complete formulation. The distance function induced by the topology on spacetime is an example of such "extra" math; it's there because we need to have a topology to define a manifold structure on spacetime, and if you have a topology, you can always use it to define a distance function. But that distance function has no physical meaning. There is no requirement that every single piece of math in every theory has a physical meaning.

 

[PeterDonis]

By "pseudo-Riemannian distance" I mean you have a curve in spacetime with tangent vector field $u^{\mu}$, you have a metric $g_{\mu \nu}$, which is a tensor field, and you integrate $ds = \sqrt{g_{\mu \nu} u^{\mu} u^{\nu}}$ along the curve to get a path length.

I should have checked myself before posting. The correct equation for $ds$ is

$$
ds = \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}
$$

The tangent vector field comes into play when you want to integrate this and you need a relationship between different coordinate differentials $dx^{\mu}$ and $dx^{\nu}$ in order to reduce the integral to one variable; in general, we use relationships of the form $dx^{\nu} = dx^{\mu} \left( u^{\nu} / u^{\mu} \right)$.

 

[TrickyDicky]

I don't understand. It's trivially true that every curve can be affinely parametrized; but it is not true that every affine parametrization on every curve must have a direct physical meaning in terms of physical path length along the curve. Certainly that's not true for null curves in spacetime; yet the equation I gave for path length applies perfectly well to null curves. [Edit: With corrections--see following post.] So I don't understand what point you're trying to make here.

That equation for null curves is ds=0, in order to integrate this to a valid distance function we can't parametrize with arc length s, like we do with spacelike curves, nor with proper time tau like we do with timelike geodesics, we must use an affine parameter different from tau and s that will give a finite distance between two different events.

It obeys the math that appropriately models the physics, yes. The distance function induced by the topology does not appropriately model the physics.

In every physical theory, there is math involved that does not directly model the physics; it's "extra" math that is there because without it we can't give the theory a complete formulation. The distance function induced by the topology on spacetime is an example of such "extra" math; it's there because we need to have a topology to define a manifold structure on spacetime, and if you have a topology, you can always use it to define a distance function. But that distance function has no physical meaning. There is no requirement that every single piece of math in every theory has a physical meaning.

That's fine, I'm not concerned with this. My concern was with the distance functions derived from the pseudoRiemannian metric being incompatible with that "extra math" but I see now that is not the case due to the structure of GR.

 

[PeterDonis]

That equation for null curves is ds=0, in order to integrate this to a valid distance function we can't parametrize with arc length s, like we do with spacelike curves, nor with proper time tau like we do with timelike geodesics, we must use an affine parameter different from tau and s that will give a finite distance between two different events.

Basically what you're saying here is that arc length $s$ is not a valid affine parameter along null curves. This is true. So what? The fact that $ds = 0$ along null curves is still used to make physical predictions: the "affine parameter distance" along null curves is not.

 

[TrickyDicky]

Basically what you're saying here is that arc length $s$ is not a valid affine parameter along null curves. This is true. So what? The fact that $ds = 0$ along null curves is still used to make physical predictions: the "affine parameter distance" along null curves is not.

It is used along null geodesics, since they must obey the geodesic equation, which is all I needed physically, in the classical theory of GR light rays are null geodesics. Note I didn't have a problem with the equation ds=0, but with its integration. Of course ds=0 is valid to make physical predictions(think of light deflection or gravitational redshift), but for physical light rays distances it is always integrated using the affine parameter, just like test particles, which are the only ones in GR(no nongravitational forces for particles) have their distances parametrized with proper time.

Of course that's GR, I'm afraid when applying Minkowski space to QM in QFT, where EM self-interactions are taken into account(no gravitational test particles anymore), you get in trouble with nonsensical infinites.

 

[PeterDonis]

It is used along null geodesics, since they must obey the geodesic equation

Ok, yes, this is a valid point; to show that a given null curve is a geodesic you need some affine parametrization in order to apply the geodesic equation. But you don't need to integrate anything; see below.

Note I didn't have a problem with the equation ds=0, but with its integration.

What's the problem? If $ds = 0$, then $\int ds = 0$ as well. You don't need an affine parametrization to do that.

Of course ds=0 is valid to make physical predictions (think of light deflection or gravitational redshift), but for physical light rays distances it is always integrated using the affine parameter

No, it isn't. You don't need to do an integral to apply the geodesic equation; you just need to verify that, as a differential equation, it holds at each event on the curve. And the number you get if you do integrate an affine parameter along a null curve is physically meaningless. The correct integral to get the actual, physical path length along a null curve is, as I noted above, $\int ds = 0$, integrating the physical interval $ds$, which is zero.

 

[TrickyDicky]

But you don't need to integrate anything

That's the thing, actually those integrations are not usually done in GR

What's the problem? If $ds = 0$, then $\int ds = 0$ as well. You don't need an affine parametrization to do that.

No, but 0 is not a valid distance between 2 distinct points according to the manifold topology. A trivial example, when it is informally said that it takes for a light ray ten light-years, to reach some point which is actually a distance, we don't get that length for the geodesic light ray by integrating ds=0 to 0, right? we are using some arbitrary affine parameter so that we get some distance in arbitrary units.

 

[PeterDonis]

That's the thing, actually those integrations are not usually done in GR

Yes, because they aren't necessary.

No, but 0 is not a valid distance between 2 distinct points according to the manifold topology.

Which has nothing to do with anything, physically speaking. See below.

A trivial example, when it is informally said that it takes for a light ray ten light-years, to reach some point which is actually a distance, we don't get that length for the geodesic light ray by integrating ds=0 to 0, right? we are using some arbitrary affine parameter so that we get some distance in arbitrary units.

No, we are using some coordinate chart in which the distance is ten light-years. This "distance" is not a spacetime interval between two events on the null geodesic; it is a spacetime interval along a spacelike geodesic between two events which have the same time coordinate in the chosen coordinate chart, and which have the same space coordinates as the two endpoints of the null geodesic. Nowhere in this process do we have to assign any affine parametrization to the null geodesic.

 

[TrickyDicky]

Yes, because they aren't necessary.

Exactly, it took me some time to realice this.

No, we are using some coordinate chart in which the distance is ten light-years. This "distance" is not a spacetime interval between two events on the null geodesic; it is a spacetime interval along a spacelike geodesic between two events which have the same time coordinate in the chosen coordinate chart, and which have the same space coordinates as the two endpoints of the null geodesic. Nowhere in this process do we have to assign any affine parametrization to the null geodesic.

My example maybe is not very good. I was trying to talk in terms of invariants, distances as geometric objects so there is no need of coordinate charts, but it is not that easy as it is not really utilized in GR.

If you have the geodesic of a light ray from event A to a different event B, and you want to compute that distance by integration, you can either take ∫ds=0, which doesn't agree with the manifold topology so it's not valid mathematically, or you can use an affine parameter and interpret it as a distance different from zero which is compatible with the natural topology. This is independent of using ds=0 to derive physical predictions.
In the Schwarzschild geometry for instance one can use the radial coordinate as affine parameter(Bill_k showed this in a past thread).

 

[PeterDonis]

My example maybe is not very good. I was trying to talk in terms of invariants, distances as geometric objects so there is no need of coordinate charts, but it is not that easy as it is not really utilized in GR.

Um, what? Invariants such as "distances as geometric objects" are *fundamental* to GR. You can't possibly mean what the above quote says. If by "distance" you actually mean "topological distance", i.e., the "distance" induced by the standard topology on $\mathbb{R}^4$, then, as I've said before, you appear to be confused by the term "distance"; the topological distance is not a "geometric object". The "distance as a geometric object" is the distance obtained using the physical metric tensor, $g_{\mu \nu}$, by the procedure I described before. If you want to describe this distance without using coordinates, that can be done, but such a description will still tell you that the "distance as a geometric object" along a null curve is zero.

If you have the geodesic of a light ray from event A to a different event B, and you want to compute that distance by integration, you can either take ∫ds=0, which doesn't agree with the manifold topology so it's not valid mathematically, or you can use an affine parameter and interpret it as a distance different from zero which is compatible with the natural topology. This is independent of using ds=0 to derive physical predictions.

You appear to be confused, as I said above, by the term "distance". If I receive a light signal from a distant object, and I want to know "how far" the light has traveled, I don't compute any integral along the null geodesic that the light followed. I compute an integral along a spacelike geodesic that lies in some surface of simultaneity for me, i.e., some spacelike hypersurface that is orthogonal to my worldline. Both $\int ds = 0$ along the light's null geodesic, and affine parametrization along that geodesic (including any topological "distance" computed using it), are simply irrelevant to this.

In the Schwarzschild geometry for instance one can use the radial coordinate as affine parameter(Bill_k showed this in a past thread).

Sure, but the $r$ coordinate isn't the same as radial "distance". If I want to know "how far" a radial light signal travels in the Schwarzschild geometry, I don't integrate $dr$ by itself, and I don't integrate anything along the null geodesic the light followed. I integrate $dr / \sqrt{1 - 2M / r}$ along a spacelike geodesic from the $r$ where the light is emitted to the $r$ where the light is detected. In other words, I integrate along a spacelike geodesic that lies in a surface of constant Schwarzschild coordinate time. (Since the spacetime is static, it doesn't matter which coordinate time I pick.)

 

[TrickyDicky]

Um, what? Invariants such as "distances as geometric objects" are *fundamental* to GR. You can't possibly mean what the above quote says.

I'm obviously referring to what we have been stressing, that these integrations in the sense of metrics are useless in GR.

If by "distance" you actually mean "topological distance", i.e., the "distance" induced by the standard topology on $\mathbb{R}^4$, then, as I've said before, you appear to be confused by the term "distance"; the topological distance is not a "geometric object". The "distance as a geometric object" is the distance obtained using the physical metric tensor, $g_{\mu \nu}$, by the procedure I described before. If you want to describe this distance without using coordinates, that can be done, but such a description will still tell you that the "distance as a geometric object" along a null curve is zero.

The standard topology doesn't induce any distance, topology has nothing to do with distances, metrics induce topologies, that should be compatible with the manifold topoogy.

"the "distance as a geometric object" along a null geodesic is zero" is a description that is not used in GR and for good reason. It would make it inconsistent. "Infinitesimal distance(ds) as a geometric object along a null geodesic is zero" is a fine description and used all the time.

If I receive a light signal from a distant object, and I want to know "how far" the light has traveled, I don't compute any integral along the null geodesic that the light followed. I compute an integral along a spacelike geodesic that lies in some surface of simultaneity for me, i.e., some spacelike hypersurface that is orthogonal to my worldline. Both $\int ds = 0$ along the light's null geodesic, and affine parametrization along that geodesic (including any topological "distance" computed using it), are simply irrelevant to this.

Agreed. The important here is that you actually end up using spacelike geodesics.

Sure, but the $r$ coordinate isn't the same as radial "distance". If I want to know "how far" a radial light signal travels in the Schwarzschild geometry, I don't integrate $dr$ by itself, and I don't integrate anything along the null geodesic the light followed. I integrate $dr / \sqrt{1 - 2M / r}$ along a spacelike geodesic from the $r$ where the light is emitted to the $r$ where the light is detected. In other words, I integrate along a spacelike geodesic that lies in a surface of constant Schwarzschild coordinate time. (Since the spacetime is static, it doesn't matter which coordinate time I pick.)

Agreed.

So neither of the two integrations(the vanishing one and the affine parametrized one) is actually used in GR, but I was highlighting the fact that the latter was compatible with the manifold topology. But yes nobody computes it except maybe use the radial coordinate as parameter for null geodesics in Schwarzschild spacetime for whatever reason I can't think of now.

 

[PeterDonis]

I'm obviously referring to what we have been stressing, that these integrations in the sense of metrics are useless in GR.

Since the term "metric" is ambiguous, I'm not sure which integration you are referring to here, unless it's the same as what I think you're talking about further on, when you say that neither of the two integrations is used in GR. See my comment on that below.

The standard topology doesn't induce any distance, topology has nothing to do with distances, metrics induce topologies, that should be compatible with the manifold topology.

Yes, I should have said "the distance function that induces a topology that is the same as the manifold topology".

"the "distance as a geometric object" along a null curve is zero" is a description that is not used in GRand for good reason. It would make it inconsistent. "Infinitesimal distance(ds) as a geometric object along a null curve is zero" is a fine description and used all the time.

So you're saying that GR says that the arc length along an infinitesimal null curve is zero, but GR somehow doesn't say that the arc length along a *finite* null curve is zero, because that would be inconsistent? That makes no sense. If $ds = 0$, then $\int ds = 0$ follows trivially. And if $ds$ is the arc length along an infinitesimal curve, then $\int ds$ must be the arc length along a finite curve.

There is a substantive comment you could have made here; you could have said that in order to define $\int ds$, I have to define the range of integration, and how do I do that without an affine parameter? Or, to put it another way, how do I make sense of the concept of a "finite" curve without an affine parameter? But there's an answer to that, too: a "finite" curve is just a curve whose points form a closed, compact set. Closure and compactness can be defined without an affine parametrization; you need a topology (the underlying manifold topology), but that's all.

As for the range of integration, I can define that using any convenient coordinate chart. Technically, I can always construct a chart for which one coordinate can serve as an affine parameter along the curve I'm interested in (that's true whether the curve is null, timelike, or spacelike); but even if I'm using such a chart, the fact that one coordinate can serve as an affine parameter along a curve is not necessary for any computations.

So neither of the two integrations(the vanishing one and the affine parametrized one) is actually used in GR

Not quite; the vanishing integration *is* used to describe null curves. See above.

 

[TrickyDicky]

So you're saying that GR says that the arc length along an infinitesimal null curve is zero, but GR somehow doesn't say that the arc length along a *finite* null curve is zero, because that would be inconsistent? That makes no sense. If $ds = 0$, then $\int ds = 0$ follows trivially. And if $ds$ is the arc length along an infinitesimal curve, then $\int ds$ must be the arc length along a finite curve.

I've corrected my post to make clear I was referring to null geodesics rather than arbitrary null curves.

Arc length is not defined along an infinitesimal segment of curve, ds is simply a differential, an infinitesimal segment of a curve. Generally ds=0 refers to null vectors in the tangent space at a point in the manifold.
Null geodesics can't be parametrized by arc length. By definition.

 

[PeterDonis]

Arc length is not defined along an infinitesimal segment of curve, ds is simply a differential, an infinitesimal segment of a curve.

Do you have any mainstream references to support this unusual position? The usual position is that $ds$ is the differential change in arc length along the curve; $ds = 0$ for a null curve just means the differential change in arc length along the curve is zero.

Null geodesics can't be parametrized by arc length. By definition.

They can't be affinely parametrized by arc length, yes. That doesn't mean they don't *have* arc length. Arc length being zero is not the same as arc length not being defined.

 

[TrickyDicky]

Do you have any mainstream references to support this unusual position? The usual position is that $ds$ is the differential change in arc length along the curve; $ds = 0$ for a null curve just means the differential change in arc length along the curve is zero.



They can't be affinely parametrized by arc length, yes. That doesn't mean they don't *have* arc length. Arc length being zero is not the same as arc length not being defined.

My position is simply that arc length is not defined in "infinitesimal curves" whatever that means(infinitesimally they are linearized) but for curves. ds is not an arc length, s is. There is no need for references here.
ds=0 refers in GR to null vectors at a point's tangent space. It doesn't mean that arc length is zero.
Null geodesics for instance don't have arc length because since they can't be parametrized by arc length, they don't integrate to an arc length.