Ignoring boundary term in the action

[davidge]

We can ignore the boundary contribution, namely the term that arises when we solve
$$\int d^{4}x \sqrt{-g}g^{ab} \delta R_{ab}$$ when varying the action. According to Wikipedia it's appropriate to do so only when the underlying manifold is both without boundary and compact.

That condition seems reasonable to me, because (as I current understand it) a manifold without boundary is a manifold in which every point has an open neighborhood $U$ homeomorphic to an open subset of $\mathbb{R}^4$, such that if any point in $U$ has coordinates $(x,y,z,t)$ in $\mathbb{R}^4$, $t$ is greater than zero.

So the manifold don't having a boundary (or, pheraphs having a boundary at infinity), Stokes' theorem says that the integral vanishes and that integral above vanishes too.

I'd like to know when should we consider closed manifolds in Relativity and when we should not.
What is the physical meaning in terms of the theory of Relativity to have an open/closed manifold?

Also, it seems that the physicists in working on the equations for the action did not realize that we should include the boundary term. Many years later Hawking-Gibbons-York did realize it. Why were the pioneers first considering only closed manifolds, is there a special reason for this choice?

 

[PeterDonis]

According to Wikipedia

Which article? Please give a link.

So the manifold don't having a boundary (or, pheraphs having a boundary at infinity), Stokes' theorem says that the integral vanishes and that integral above vanishes.

That's not quite what the theorem says. You forgot the second condition: that the manifold must be compact.

 

[davidge]

Which article? Please give a link.

Here's the link: https://en.wikipedia.org/wiki/Gibbons–Hawking–York_boundary_term

That's not quite what the theorem says. You forgot the second condition: that the manifold must be compact.

So the Stokes' theorem is valid only if the manifold is compact?

 

[PeterDonis]

So the Stokes' theorem is valid only if the manifold is compact?

No; my language was confusing there, sorry about that. I was referring to what you said in the OP:

According to Wikipedia it's appropriate to do so only when the underlying manifold is both without boundary and compact.

"It" here refers to leaving out the boundary term in the action; the Wikipedia page, as you said in the OP, says that is only appropriate when the manifold is without boundary and compact. In the argument you made in the OP, you only talked about the "without boundary" part; you didn't consider the "compact" part. You need to consider both when determining whether you need to add a boundary term.

The application of Stokes' Theorem is a separate issue; it allows us to convert the volume integral that you give in your OP (which arises from varying the Einstein-Hilbert action) into a surface integral over the boundary of the manifold. Obviously if the manifold has no boundary, this integral vanishes; but that's not a matter of Stokes' Theorem, that's just an obvious property of any integral, that evaluating it over a vanishing range must give a vanishing result. The question is what happens when the manifold does have a boundary: that is the case the Wikipedia article focuses on--to show that the variation of the Gibbons-Hawking boundary term cancels the extra term that comes from varying the Einstein-Hilbert action.

 

[davidge]

Ah, ok. But when should we consider manifolds with boundary?

 

[PeterDonis]

when should we consider manifolds with boundary?

You mean, when should we consider either manifolds with boundary, or manifolds without boundary that are not compact. That was the point of my comment on you leaving out the "compact" condition. AFAIK there are no spacetimes that are useful in GR that are compact manifolds; but many, if not most, of them are manifolds without boundary (more precisely, without a boundary that is not "at infinity"--see below).

The simplest example of a spacetime without boundary that is not compact is Minkowski spacetime. The EFE vanishes on this spacetime (it has zero curvature and zero stress-energy), but AFAIK in order to show this fully, you have to include the Gibbons-Hawking boundary term, heuristically, because there is a "boundary at infinity". Somewhat more precisely, you take the limit of evaluating the surface integrals--the one arising from varying the E-H action and the one arising from the G-H boundary term--at some finite radius, as the radius goes to infinity. AFAIK both of these integrals are nonzero on Minkowski spacetime (they are of opposite sign and equal magnitude), so if you don't include the second, you don't get a vanishing field equation.

 

[PeterDonis]

AFAIK both of these integrals are nonzero on Minkowski spacetime

I'm not completely certain of this for Minkowski spacetime, but I am for Schwarzschild spacetime, which is also an example of a spacetime without boundary (except "at infinity") that is not compact.

 

[davidge]

The simplest example of a spacetime without boundary that is not compact is Minkowski spacetime. The EFE vanishes on this spacetime (it has zero curvature and zero stress-energy), but AFAIK in order to show this fully, you have to include the Gibbons-Hawking boundary term, heuristically, because there is a "boundary at infinity".

The metric in Minkowski space is constant. So how could one vary $\sqrt{-g}$ and $g_{ab}$ itself?

The simplest example of a spacetime without boundary that is not compact is Minkowski spacetime

Is it not compact because it can have infinite points? -- I mean in the sense that it's not a bounded space --.

Is correct what I wrote in post #1 about what boundary mean?

Somewhat more precisely, you take the limit of evaluating the surface integrals--the one arising from varying the E-H action and the one arising from the G-H boundary term--at some finite radius, as the radius goes to infinity. AFAIK both of these integrals are nonzero on Minkowski spacetime (they are of opposite sign and equal magnitude), so if you don't include the second, you don't get a vanishing field equation.

I see

 

[PeterDonis]

The metric in Minkowski space is constant. So how could one vary $\sqrt{-g}$ and $g_{ab}$ itself?

You're misunderstanding what is being varied. You don't start with a known metric. You start with an action and use the variational principle to find the field equation that extremizes the action. So when I talked about evaluating the variational integral for Minkowski spacetime, what I really meant was: assume that we have a spacetime manifold with global topology $\mathbb{R}^4$ and a metric that extremizes the action derived from the Einstein-Hilbert Lagrangian. Since this manifold is not compact, we have to include the Gibbons-Hawking boundary term in the integral in the general case in order to obtain the Einstein Field Equation. However, this general case includes an infinite number of possible solutions--there are an infinite number of possible spacetimes whose underlying global topology is $\mathbb{R}^4$. Only one of these solutions is Minkowski spacetime; so we can't just say $\sqrt{-g}$ and $g_{ab}$ are constant in advance of going through the procedure. We have to do the variational integral first, and then notice that one of the solutions of the resulting field equation is Minkowski spacetime.

 

[PeterDonis]

Is it not compact because it can have infinite points?

Do you know the definition of a compact manifold? Can you see why Minkowski spacetime--or more generally any spacetime whose underlying topology is $\mathbb{R}^4$--does not satisfy that definition? (We have had a similar discussion in a previous thread, IIRC. Do you remember what was shown there?)

 

[davidge]

Do you know the definition of a compact manifold?[..]
We have had a similar discussion in a previous thread, IIRC. Do you remember what was shown there?

Yes, I do. You have shown there that $\mathbb{R}$ is not compact. If we apply the theorem that says that $\mathbb{R}^n = \mathbb{R}{} \times{}... \times{\ } \mathbb{R}$ we see that $\mathbb{R}^n$ is not compact for any $n > 1$.

You're misunderstanding what is being varied. You don't start with a known metric. You start with an action and use the variational principle to find the field equation that extremizes the action

Oh, now I see

assume that we have a spacetime manifold with global topology $\mathbb{R}^4$ and a metric that extremizes the action derived from the Einstein-Hilbert Lagrangian

What does global topology $\mathbb{R}^4$ means here? Would not it implicitly assume a Minkowski metric before we find the metric to be Minkowskian?

 

[PeterDonis]

What does global topology $\mathbb{R}^4$ means here?

It means the underlying global topological space of the manifold is $\mathbb{R}^4$. That is not true of all spacetimes. Maximally extended Schwarzschild spacetime, for example, has the underlying global topological space $\mathbb{R}^2 \times \mathbb{S}^2$.

Would not it implicitly assume a Minkowski metric

No. $\mathbb{R}^4$ as a topological space has no metric. (And if you look at $\mathbb{R}^4$ the metric space, its metric is Euclidean, not Minkowski.)

 

[davidge]

Thanks

 

[davidge]

So can I conclude from this thread

- We start with an action, choose the underlying topological space, set the variation of the action equal to zero and find the Einstein's equations and these equations will show us what the metric is?

For instance, we let our topological space be homeomorphic to $\mathbb{R}^4$ --i.e. our top space is a manifold-- and solve for the variation of the action. We'll get the Eintein's equations in vaccum. And so we find the metric to be $\eta$ by solving $R_{ab} - \frac{1}{2}g_{ab}R = 0$?

- Depending on our choice of the manifold --with/out boundary, compact or not-- we'll get some extra terms on varying the action. So we introduce the Gibbons-Hawking-York term in order to the vanishing of the action, as you said in #6?

By the way, is it relatively easy to work on General Relativity because of the principle of equivalence? To put it in another way, if there were no principle of equivalence, we would have to work with topological spaces that aren't manifolds?

 

[PeterDonis]

We start with an action, choose the underlying topological space, set the variation of the action equal to zero and find the Einstein's equations and these equations will show us what the metric is?

They will give you equations that the metric (more precisely the Einstein tensor derived from the metric) must satisfy. But there won't be just one metric that satisfies them. There will be an infinite number of them.

Also, since the field equation is local, the same solution can actually be obtained with different assumptions for the underlying topological space. That is why I specified "global" when I was describing the underlying topological spaces of different solutions in my earlier post. The global topological space is the one that matches the maximal analytic extension of a solution. So, for example, $\mathbb{R}^2 \times \mathbb{S}^2$ is the underlying topological space of the maximal analytic extension of Schwarzschild spacetime. But of course any solution of the EFE can be restricted to an open neighborhood of some event (this is one way of saying what it means for the field equation to be "local"), and the underlying topological space of any open neighborhood is $\mathbb{R}^4$. So you can also obtain the Schwarzschild metric as a solution of the EFE if you assume $\mathbb{R}^4$ as the underlying topological space. You just won't be able to obtain the maximal analytic extension that way.

For instance, we let our topological space be homeomorphic to R4\mathbb{R}^4 --i.e. our top space is a manifold-- and solve for the variation of the action. We'll get the Eintein's equations in vaccum

Not just in vacuum. You can have a non-vacuum solution whose underlying topological space is $\mathbb{R}^4$. For example, the flat and open FRW spacetimes have this underlying topology (global topology, in the sense I gave above).

Whether or not you get the vacuum EFE or the non-vacuum EFE depends on whether you include "matter" (i.e., any non-gravitational fields) in the Lagrangian before you vary it. If you do, you get the non-vacuum EFE. If you don't, you get the vacuum EFE. But that's a separate question from the questions of underlying topology, boundary vs. no boundary, etc.

Depending on our choice of the manifold --with/out boundary, compact or not-- we'll get some extra terms on varying the action. So we introduce the Gibbons-Hawking-York term in order to the vanishing of the action, as you said in #6?

You have to include the GBH term if the manifold is not compact. A non-compact manifold might not have a boundary (except "at infinity", but mathematically, such manifolds are called "manifolds without boundary").

 

[davidge]

They will give you equations that the metric (more precisely the Einstein tensor derived from the metric) must satisfy. But there won't be just one metric that satisfies them. There will be an infinite number of them

So it remain for us to choose which of them is most convenient? How do we know that the one most convenient is the $\eta$?

So, for example, $\mathbb{R}^2 \times \mathbb{S}^2$ is the underlying topological space of the maximal analytic extension of Schwarzschild spacetime. But of course any solution of the EFE can be restricted to an open neighborhood of some event (this is one way of saying what it means for the field equation to be "local"), and the underlying topological space of any open neighborhood is $\mathbb{R}^4$. So you can also obtain the Schwarzschild metric as a solution of the EFE if you assume $\mathbb{R}^4$ as the underlying topological space. You just won't be able to obtain the maximal analytic extension that way.

Oh, ok. I think I see the point.
What is specifically the meaning of $\mathbb{R}^2 \times \mathbb{S}^2$? Would it be that we are dealing with a space that has the topology of both the 2-sphere and the Euclidean plane?

 

[PeterDonis]

So it remain for us to choose which of them is most convenient?

Not which one is most convenient. Which one makes correct predictions for the actual physical situation we are trying to model. If we're trying to model empty space far away from all gravitating bodies, we choose Minkowski spacetime. If we're trying to model the vacuum region around a spherically symmetric mass, we choose Schwarzschild spacetime. And so on. (I'm a bit surprised I need to say this; I think you are getting too focused on technical points and losing sight of the ultimate purpose of all this.)

What is specifically the meaning of $\mathbb{R}^2 \times \mathbb{S}^2$?

It's the Cartesian product of $\mathbb{R}^2$ and the 2-sphere $\mathbb{S}^2$, with the product topology.

https://en.wikipedia.org/wiki/Product_topology

If you are not familiar with this sort of notation, I strongly recommend spending some time with a textbook on the basics of topology. (This is the sort of background knowledge you would normally be assumed to have for an "I" level discussion.)

 

[davidge]

Not which one is most convenient. Which one makes correct predictions for the actual physical situation we are trying to model. If we're trying to model empty space far away from all gravitating bodies, we choose Minkowski spacetime. If we're trying to model the vacuum region around a spherically symmetric mass, we choose Schwarzschild spacetime. And so on.

I got it

(I'm a bit surprised I need to say this; I think you are getting too focused on technical points and losing sight of the ultimate purpose of all this.)

If you are not familiar with this sort of notation, I strongly recommend spending some time with a textbook on the basics of topology. (This is the sort of background knowledge you would normally be assumed to have for an "I" level discussion.)

I'm sorry. It's not easy to find material that covers this subject. Just to know, would that Cartesian product just mean that $\mathbb{R}^2{} \times{} \mathbb{S}^2 = (t,x,y,z) | x²+y²+z² = r²$?

 

[PeterDonis]

Just to know, would that Cartesian product just mean that $\mathbb{R}^2{} \times{} \mathbb{S}^2 = (t,x,y,z) | x²+y²+z² = r²?$

No. Please take the time to learn how this works from a textbook. You are approaching it the wrong way: you should not be starting out by trying to view a topological space as an embedding into $\mathbb{R}^n$. (For one thing, not all topological spaces can be so embedded.) You need to understand from first principles what a Cartesian product is and how to construct one.

 

[PeterDonis]

You need to understand from first principles what a Cartesian product is and how to construct one.

If you want a simple example to start on, try this: $\mathbb{R}^2$ is the Cartesian product $\mathbb{R} \times \mathbb{R}$, with the natural product topology. See if you can figure out why this is from the definition of a Cartesian product and your knowledge of $\mathbb{R}$.

 

[DrGreg]

Just to know, would that Cartesian product just mean that $\mathbb{R}^2{} \times{} \mathbb{S}^2 = (t,x,y,z) | x²+y²+z² = r²$?

In the context used in post #15, you would denote the coordinates of $\mathbb{R}^2{} \times{} \mathbb{S}^2$ by $(t, r, \theta, \phi)$, that is, $(t, r) \in \mathbb{R}^2$ and $(\theta, \phi) \in \mathbb{S}^2$

 

[davidge]

No. Please take the time to learn how this works from a textbook. You are approaching it the wrong way: you should not be starting out by trying to view a topological space as an embedding into $\mathbb{R}^n$. (For one thing, not all topological spaces can be so embedded.) You need to understand from first principles what a Cartesian product is and how to construct one.

Okay

If you want a simple example to start on, try this: $\mathbb{R}^2$ is the Cartesian product $\mathbb{R} \times \mathbb{R}$, with the natural product topology. See if you can figure out why this is from the definition of a Cartesian product and your knowledge of $\mathbb{R}$.

Would this be all the ordered pairs $(a,b)$ with $a \in \mathbb{R}$ and $b \in \mathbb{R}$?

In the context used in post #15, you would denote the coordinates of $\mathbb{R}^2{} \times{} \mathbb{S}^2$ by $(t, r, \theta, \phi)$, that is, $(t, r) \in \mathbb{R}^2$ and $(\theta, \phi) \in \mathbb{S}^2$

And would $(\theta, \phi) \in \mathbb{S}^2$ be so because $\theta, \phi$ are the only intrinsic coordinates of the sphere?

 

[PeterDonis]

Would this be all the ordered pairs $(a,b)$ with $a \in \mathbb{R}$ and $b \in \mathbb{R}$?

Yes. And the construction DrGreg described for $\mathbb{R}^2 \times \mathbb{S}^2$ is similar; the only nuance is that, strictly speaking, he should have written the ordered pair $((t, r), (\theta, \phi))$.

would $(\theta, \phi) \in \mathbb{S}^2$ be so because $\theta$, $\phi$ are the only intrinsic coordinates of the sphere?

They aren't the only possible coordinates on a 2-sphere. But they are a way of labeling the points on a 2-sphere, yes.

 

[davidge]

Yes. And the construction DrGreg described for $\mathbb{R}^2 \times \mathbb{S}^2$ is similar; the only nuance is that, strictly speaking, he should have written the ordered pair $((t, r), (\theta, \phi))$.

In this case, what restriction would $((t, r), (\theta, \phi))$ obey?

 

[PeterDonis]

what restriction would $((t, r), (\theta, \phi))$ obey?

You should be able to work this out from the fact that $(t, r)$ are coordinates on $\mathbb{R}^2$ and $(\theta, \phi)$ are coordinates on $\mathbb{S}^2$. Remember here that "coordinates" just means "sets of numbers that label points"; there is no metric since we're just talking about topological spaces. You already worked out how ordered pairs label points in $\mathbb{R}^2$; for $\mathbb{S}^2$, think of the two coordinates as latitude and longitude.

 

[davidge]

Remember here that "coordinates" just means "sets of numbers that label points"; there is no metric since we're just talking about topological spaces.

So could we say that points of $\mathbb{R}^2 \times \mathbb{S}^2$, seen as a independently topological space, will have rectangular coordinates $(t, r, \theta, \phi)$. Now if we embed this topological space into $\mathbb{R}^4$ and let the metric be that of $\mathbb{R}^4$ and if the points on $\mathbb{R}^4$ are labeled $(t,x,y,z)$, we have
$$x = x(r, \theta, \phi) \\ y = y (r, \theta, \phi) \\ z = z(r, \theta, \phi)$$ with the usual expressions on the RHS of these functions.

However, as a manifold is locally Euclidean, it is on each point like $\mathbb{R}^4$ and so we would not need to make that change of coordinates, we would just continue with our natural coordinates $(t,r,\theta,\phi)$. Of course, as you said in previous post, we would not get the maximal extension of a solution if we do that way. So the conclusion would be that the most smart choice is to make that transformations above for $x, y, z$?

 

[PeterDonis]

could we say that points of $\mathbb{R}^2 \times \mathbb{S}^2$, seen as a independently topological space, will have rectangular coordinates $(t, r, \theta, \phi)$.

You can't say the coordinates are "rectangular" because you don't have a metric. (Note that those coordinates aren't usually called "rectangular" for the standard metrics they are usually used for.)

The rest of your post just builds on this mistake.

 

[PeterDonis]

Of course, as you said in previous post, we would not get the maximal extension of a solution if we do that way.

What I said in that previous post only applies if you have a metric; the concept of a maximal analytic extension only has meaning if you have a metric. You don't have a metric in this example; we're just talking about a topological space.

 

[davidge]

You can't say the coordinates are "rectangular" because you don't have a metric.

The rest of your post just builds on this mistake.

So even if I were not considering a metric in my post, those things about the coordinates are wrong?

 

[PeterDonis]

even if I were not considering a metric in my post, those things about the coordinates are wrong?

If there is no metric, the things you are saying about the coordinates make no sense. Read your own post again:

rectangular coordinates

let the metric be that of $\mathbb{R}^4$

 

[davidge]

What I mean is, "let the natural coordinates --just the label of the points of $\mathbb{R}^2 \times \mathbb{S}^2$-- be $(t,r, \theta, \phi)$. Now, let's consider a metric space, namely $\mathbb{R}^4$ with its usual metric."

 

[PeterDonis]

What I mean is, let the natural coordinates --just label of the points of $\mathbb{R}^2 \times \mathbb{S}^2$-- be $(t,r, \theta, \phi)$. Now, let's consider a metric space, namely $\mathbb{R}^4$ with its usual metric.

Why do you want to do this? It makes no sense.

Also, you still haven't understood: you need to understand the properties of a topological space without using any embedding of that space in another space.

I'm sorry, but I don't see the point of continuing to repeat the same advice and have a discussion if you continue to ignore that advice.

 

[PeterDonis]

The OP question in this thread has been answered, and the thread is closed. @davidge if you want to ask questions about the topological concept of a manifold with boundary, you can do that in the appropriate math forum; but please take the time to work through a textbook first.