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I Ignoring boundary term in the action

  1. Apr 28, 2017 #1
    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?
     
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  3. Apr 28, 2017 #2

    PeterDonis

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    Which article? Please give a link.

    That's not quite what the theorem says. You forgot the second condition: that the manifold must be compact.
     
  4. Apr 28, 2017 #3
    Here's the link: https://en.wikipedia.org/wiki/Gibbons–Hawking–York_boundary_term
    So the Stokes' theorem is valid only if the manifold is compact?
     
  5. Apr 28, 2017 #4

    PeterDonis

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

    "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.
     
  6. Apr 28, 2017 #5
    Ah, ok. But when should we consider manifolds with boundary?
     
  7. Apr 28, 2017 #6

    PeterDonis

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    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.
     
  8. Apr 28, 2017 #7

    PeterDonis

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    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.
     
  9. Apr 28, 2017 #8
    The metric in Minkowski space is constant. So how could one vary ##\sqrt{-g}## and ##g_{ab}## itself?
    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?
    I see
     
  10. Apr 28, 2017 #9

    PeterDonis

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    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.
     
  11. Apr 28, 2017 #10

    PeterDonis

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    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?)
     
  12. Apr 29, 2017 #11
    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##.
    Oh, now I see
    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?
     
  13. Apr 29, 2017 #12

    PeterDonis

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    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##.

    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.)
     
  14. Apr 29, 2017 #13
    Thanks
     
  15. Apr 29, 2017 #14
    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?
     
  16. Apr 29, 2017 #15

    PeterDonis

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    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.

    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.

    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").
     
  17. Apr 29, 2017 #16
    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##?
    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?
     
  18. Apr 29, 2017 #17

    PeterDonis

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    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.)

    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.)
     
  19. Apr 29, 2017 #18
    I got it
    :smile:
    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##?
     
  20. Apr 29, 2017 #19

    PeterDonis

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    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.
     
  21. Apr 29, 2017 #20

    PeterDonis

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    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}##.
     
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