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Solving field equations and the nonphysical nature of the metric. (GR)

  1. Apr 23, 2013 #1
    There seems to be an emphasis in several books on general relativity that the metric (components) in itself does not reflect anything physical, only our choice of coordinates. On the other hand it can seem like the authors, instead of being true to this, treat the metric (components) as containing physical information.

    For example in deriving the Schwarzschild solution one makes some symmetry arguments to restrict the form of the metric (components) and thereby solving the Einstein equations for empty space and thus ending up with this famous solution.

    Since the Einstein equations are now solved for a vacuum one would think that the only reason we did not get a flat-space solution was due to the symmetry assumptions put in at the start of the derivation.

    But is this really a physical assumption at all? Like the authors emphasizing the nonphysical nature of metric (components) I would think that one could reach the assumed form just by choosing a certain basis. But if this is not a physical assumption we have not put any physical information in except that the Einstein equations are to be solved for a vacuum, so then it would be reasonable to expect all solutions consistent with a vacuum (T=0).

    So where does the assumption of a central mass lie in the Schwarzschild solution, and if it actually does lie in the assumption of the form of the metric how is this reconciled with the fact that the metric (only?) reflects choice of coordinates?

    Edit: (components).
     
    Last edited: Apr 23, 2013
  2. jcsd
  3. Apr 23, 2013 #2
    Are you sure you're not just confusing their arguments? The metric is supposed to be physical, but coordinates are not. Maybe it's the components of the metric that are coordinate system dependent (which is true)? Often times physicists consider the components of a tensor as the tensor itself, which isn't true. The tensor doesn't care what coordinates you chose, but the components do.

    If your books are actually telling you that the metric is not physical, then you should get a better book.
     
  4. Apr 23, 2013 #3

    WannabeNewton

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    Coordinates do not have physical / geometrical meaning. Symmetries do have geometrical meaning. The static part of the Schwarzschild space-time tells us that there exists a time-like vector field ##\xi ^{a}##, called a killing vector field, such that ##\mathcal{L}_{\xi}g_{ab} = 0\Leftrightarrow \nabla_{(a}\xi_{b)} = 0## and that ##\xi_{[c}\nabla_{b}\xi_{a]} = 0##. These are geometrical, coordinate independent statements with the first being a property of the space-time and the second being a property of the vector field (hypersurface-orthogonality). Similarly, the spherical symmetry tells us that the isometry group of the space-time ##G## has a subgroup ##H## such that ##H\cong SO(3)## and the orbits of the natural group action are 2-spheres. This is again a purely geometrical property of the space-time.

    All we're doing when writing down the Schwarszchild metric in the usual coordinates is adapting a coordinate system whose coordinate basis takes advantage of these symmetries to give us the simplest possible metric ansatz that we can then plug into Einstein's equations and solve. It is the geometrical properties above, however, that have physical meaning.

    EDIT: What book have you seen that says the metric is non-physical by the way?
     
    Last edited: Apr 23, 2013
  5. Apr 23, 2013 #4
    Yes I ment the metric components which are assumed to have a certain form i solving the central mass problem.

    Sorry of the bad use of words.
     
    Last edited: Apr 23, 2013
  6. Apr 23, 2013 #5

    WannabeNewton

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    Again, we are just choosing coordinates ##(t,r,\theta,\varphi)## that take advantage of the spherical symmetry and the static nature of the space-time so that we can write the metric in the coordinate basis of these coordinates in a particularly simple form. It is akin to taking advantage of the geometrical properties of homogeneity and isotropy of a space-time by choosing coordinates so that the metric takes a simple form in the coordinate basis as we do for the FLRW universe. The physical content is in the geometrical symmetries (which are coordinate independent) described above.
     
  7. Apr 23, 2013 #6
    Ok, as Wannabe newton pointed out, the symmetries act on the abstract tensors, not just the components. Depending on how rigorous the book you are using, you might notice extreme care used to define coordinates, especially the r coordinate. These choices are made to make your choice of coordinates useful for solving the problem. So you have done two things: 1, imposed a physical symmetry on the metric (not it's components), and 2, chosen a coordinate system that allows to take advantage of that symmetry.

    I think the answer to your question is that the symmetries are imposed in a way that is coordinate independent. Otherwise, as you've noticed, the symmetry would be a coordinate symmetry and therefore non physical.
     
  8. Apr 23, 2013 #7
    This solution at wikipedia is fairly similar to the ones I've read before:

    http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution

    To me it seems like the only assumptions being made are statements about metric components (which effects the form of the line element) and also of the role of certain labels; like the statement that invariance under ##\phi \to - \phi## and ##\theta \to - \theta## reflects rotational symmetry. Where exactly does the physical information lie in these assumptions?
     
  9. Apr 23, 2013 #8
    I think I'm currently reading more elementary books than the ones you are referring to. The solution at wikipeda is more like the ones I've read. There the whole thing is solved without ever talking about abstract vectors and tensors. Especially not killing vectors. At the wikipedia solution every assumption seems to be about the metric components.
     
  10. Apr 23, 2013 #9

    WannabeNewton

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    I suggest you get a better book then. Try Wald "General Relativity". The physical content of the symmetries are exactly as I stated in post #3. Wald goes into some detail on how these symmetries are used to then place the metric in the usual simple form by choosing a coordinate system adapted to said symmetries. Spherical symmetry is a purely coordinate-independent, geometrical statement about the space-time; that Wikipedia trivializes it by talking about it in terms of parity transformations in a specific coordinate system is reason enough for you to not use it as a source regarding this matter.
     
  11. Apr 23, 2013 #10
    Basically, this is the problem with teaching GR at a lower level of rigor: you don't really know exactly what's going on. What the wikipedia definition is technically doing is making the coordinate independent statements "behind the scenes", then using the results to motivate their coordinate dependent (and explicit) statements. Notice, they don't really tell you how r, t, phi or theta are defined.

    Books that are more explicit are Sean Carroll's book, and Wald's book, but they may be a bit tough, depending on your background.

    Edit: I keep saying the same things as Wannabe Newton, so I'm just going to back away from the thread...
     
  12. Apr 23, 2013 #11
    I felt that Wald was a bit heavy for an introduction to GR. Would you then claim that the more elementary solution at Wikipedia is not satisfactory (wrong)?
     
  13. Apr 23, 2013 #12

    WannabeNewton

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    Actually, it would probably be better if you stayed because I have class in a few minutes:smile:!
     
  14. Apr 23, 2013 #13

    WannabeNewton

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    I would not say it's wrong, I would say it is giving the coordinate dependent properties of the metric in that particular coordinate basis which is not how you want to view symmetries of space-time. For example, wiki says the stationary part implies ##\partial _{t}g_{\mu\nu} = 0## but this itself is a purely coordinate dependent statement and in fact comes right out of the killing vector definition of being stationary stated above.

    Indeed, given a stationary space-time we know there exists a time-like killing vector field ##\xi^{a}## satisfying ##\nabla_{(a}\xi_{b)} = 0##. It can be shown that there exist coordinates ##(t,x^1,x^2,x^3)## in which ##\xi^{a} = (\frac{\partial }{\partial t})^{a}##. In the coordinate basis, this is just ##\xi^{\mu} = \delta^{\mu}_{t}##. We see that ##\nabla_{\mu}\xi_{\nu} = \nabla_{\mu}(g_{\alpha\nu}\xi^{\alpha}) = \nabla_{\mu}g_{t\nu} = \partial _{\mu}g_{t\nu} - \Gamma _{t\mu\nu} = \partial _{\mu}g_{t\nu} - \frac{1}{2}(\partial_{\nu}g_{t\mu} + \partial_{\mu}g_{t\nu} - \partial_{t}g_{\mu\nu})## therefore, using the killing vector field condition above, ##\nabla_{(\mu}\xi_{\nu)} = \partial _{\mu}g_{t\nu} - \frac{1}{2}(\partial_{\nu}g_{t\mu} + \partial_{\mu}g_{t\nu} - \partial_{t}g_{\mu\nu}) + \partial_{\nu}g_{t\mu} - \frac{1}{2}(\partial_{\mu}g_{t\nu} + \partial_{\nu}g_{t\mu} - \partial_{t}g_{\mu\nu}) = \partial _{t}g_{\mu\nu} = 0##. As you can see, this condition is a coordinate dependent condition (it is quite obviously not covariant) that comes out of the physical / geometrical definition of a stationary space-time.
     
  15. Apr 23, 2013 #14
    We impose the restriction that there is spatial spherical symmetry about (at least) one point, and together with the other restrictions (see below) this leads to a family of solutions characterized by a single free parameter. If that parameter is set to zero we get flat spacetime (which is actually spherically symmetrical about every point), but if we set that parameter to a positive value we get a solution that is spherically symmetrical only about the central point, and the parameter corresponds to the central mass.

    The reason this solution does not have a unique form is that we have freedom in our choice of coordinates, and this affects the form of the metric components. Since we are assuming spatial spherical symmetry, it's convenient to use polar space coordinates, but this doesn't fully constrain the form of the solution, because we are still free to define the radial spatial coordinate in different ways. For example, if we stipulate that the circumference of a circle of radius r about the central point is 2 pi r (and also assign the "r=0" point suitably) we arrive at the usual Schwarzschild coordinates. On the other hand, if we want the speed of light to be isotropic in terms of our coordinates, we need to define the radial parameter differently (so the radius of a central circle is not 2 pi r). Naturally in terms of these "isotropic coordinates" the components of the metric have a different form, even though it represents physically the same family of solutions. We could also choose infinitely many other systems of coordinates, and for each system the metric components would have a different form, even though it is the same solution. This is why your books tell you that the metric components are coordinate-dependent.

    By the way, in view of Birkhoff's theorem, it isn't actually necessary to assume a static solution. All you need is spherical symmetry, but most elementary derivations make use of the static assumption to make things easy.
     
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