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Metric tensor in spherical coordinates

  1. Jul 4, 2012 #1
    Hi all,

    In flat space-time the metric is

    [itex]ds^2[/itex][itex]=[/itex][itex]-dt^2[/itex][itex]+dr^2+r^2[/itex][itex]\Omega^2[/itex]

    The Schwarzschild metric is

    [itex]ds^2[/itex][itex]=-[/itex][itex](1-\frac{2MG}{r})[/itex][itex]dt^2[/itex][itex]+[/itex][itex]\frac{dr^2}{(1-\frac{2MG}{r})}[/itex][itex]+[/itex][itex]r^2d[/itex][itex]\Omega^2[/itex]

    Very far from the planet, assuming it is symmetrical and non-spinning, the Schwarzschild metric reduces to the flat space-time metric (as [itex]r[/itex] goes to infinity)

    Now, from this equation it can be concluded that [itex]g_{00}{}^{}[/itex][itex]=[/itex][itex]1-2GM/R[/itex]

    Then, [itex]g_{rr}{}^{}[/itex][itex]=[/itex][itex]\frac{1}{(1-2MG/R)}[/itex]

    If I want to get a matrix for this, what are other non-zero components of the metric [itex]g[/itex]?

    I also set c=1.

    Also, I would like to get components of the Stress-Energy tensor:

    [itex]\mathrm T_{}{}^{00}[/itex][itex]=[/itex][itex]\rho[/itex][itex]c^2[/itex]

    [itex]\mathrm T_{}{}^{10}[/itex][itex]=[/itex][itex]\rho[/itex][itex]v_{x}{}^{}[/itex]

    [itex]\mathrm T_{}{}^{20}[/itex][itex]=[/itex][itex]\rho[/itex][itex]v_{y}{}^{}[/itex]

    [itex]\mathrm T_{}{}^{30}[/itex][itex]=[/itex][itex]\rho[/itex][itex]v_{z}{}^{}[/itex]

    Please check those which I wrote and add those which I haven't mentioned.

    Thanks!
     
  2. jcsd
  3. Jul 4, 2012 #2
    Here are some more: (I am looking for comments!)

    [itex]g_{θθ}{}^{}[/itex][itex]=[/itex][itex]r^2[/itex]

    [itex]g_{\phi\phi}{}^{}[/itex][itex]=[/itex][itex]r^2sin^2θ[/itex]

    Also,

    is [itex]R^a_{mab}[/itex][itex]=[/itex][itex]R_{mb}[/itex]?
     
    Last edited: Jul 4, 2012
  4. Jul 4, 2012 #3

    WannabeNewton

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    The solution you're talking about is a solution to the vacuum field equations where the energy - momentum tensor is identically zero.
     
  5. Jul 4, 2012 #4
    For the Schwarzschild metric in SC coordinates and the Minkowski metric in spherical coordinates, those are all of the components. They are both diagonal metrics, so all other components are zero:

    [tex]\eta _{\mu \nu}=\left [ \begin{matrix}
    -1 & 0 & 0 & 0\\
    0 & 1 & 0 & 0\\
    0 & 0 & r^2 & 0\\
    0 & 0 & 0 & r^2 sin^2\theta
    \end{matrix} \right ][/tex]

    [tex]g_{\mu \nu}=\left [ \begin{matrix}
    -(1-\frac{r_s}{r}) & 0 & 0 & 0\\
    0 & (1-\frac{r_s}{r})^{-1} & 0 & 0\\
    0 & 0 & r^2 & 0\\
    0 & 0 & 0 & r^2 sin^2\theta
    \end{matrix} \right ][/tex]


    The stress-energy tensor you give is incorrect. All of its components should be zero for both flat spacetime and Schwarzschild spacetime. What you gave looks like the stress-energy tensor of dust or a fluid of some kind.

    As for the Ricci curvature, yes, that is the definition of the Ricci curvature tensor. Note that all of its components should be zero for both metrics.
     
  6. Jul 4, 2012 #5
    So Schwarzschild metric is another way of representing the flat space-time metric? I read that Schwarzschild metric describes the gravitational field outside the spherical non-rotating uncharged body (Ex. a planet). So if the [itex]G^{\mu\nu}_{}{}[/itex][itex]=0[/itex] then Stress Energy is also zero, right? But we cannot have Stress Energy to be zero because the planet, in this case, has density, has mass, etc. So what point am I missing?

    Also, we are saying (and I used Mathematica to prove it) that Ricci tensor here is zero. But the Riemann tensor is not zero. So one of the components of the Riemann tensor here is [itex]R^\Theta_{r\Theta r}{}[/itex][itex]=[/itex][itex]\frac{m}{(2m-r)r^2}[/itex]

    By [itex]R^\Theta_{r\Theta r}{}[/itex][itex]=[/itex][itex]R^{}_{rr}{}[/itex] so basically the r-r component of the Ricci tensor is not zero yet we have here the statement that the Ricci tensor is zero (Mathematica says that Ricci tensor components are zero too).

    Can anyone shed some light on this confusion?
     
    Last edited: Jul 4, 2012
  7. Jul 4, 2012 #6
    The stress energy tensor is zero outside the object, which is what that metric describes. The inside of the object will necessarily be described by some other metric.
     
  8. Jul 4, 2012 #7
    ^
    If we assume that [itex]T^{\mu\nu}[/itex] is zero outside the object this would mean that all objects have the same degree of curvature outside them. Or do (1-2MG/R) terms take care of that? That would make some sense. Also, please look up above this to edited version of my recent post. Thank you.
     
  9. Jul 4, 2012 #8

    WannabeNewton

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    It is not another way to represent flat space - time. The solution describes space - time in the exterior of said massive body so there is no energy - momentum where the solution is valid. You use [itex]R_{\mu\nu} = 0[/itex], with certain assumptions made about the metric (that it should describe a static, spherically symmetric space -time), and solve for the unknown components of the metric. You aren't assuming the metric describes flat space -time though.
     
  10. Jul 4, 2012 #9
    The components of the Riemann tensor are not zero, using the Schwarzschild metric.
     
  11. Jul 4, 2012 #10
    ^^
    Yes, but I would still QUOTE my question because I didn't get a clear answer to this

     
  12. Jul 4, 2012 #11

    WannabeNewton

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    What you have said isn't correct. [itex]R_{rr} = R^{\sigma }_{r\sigma r} = R^{t}_{rtr} + R^{r}_{rrr} + R^{\theta }_{r\theta r} + R^{\phi }_{r\phi r} [/itex] in that coordinate basis. It is a repeated index so you have to sum over it: http://en.wikipedia.org/wiki/Einstein_summation
     
  13. Jul 4, 2012 #12
    ^
    Ok. So now it is clear to me why the Ricci tensor, Ricci scalar, and Einstein tensor are zero. However, if [itex]G^{\mu\nu}{}[/itex][itex]=0[/itex], then it would mean that the einstein tensor--the average curvature of the space-time--is zero. But if we have a planet with some density and some mass, it would mean that it actually curves the geometry around it by some extend. This point is not so clear.
     
  14. Jul 4, 2012 #13
    I was trying to write down the components of the Stress-Energy tensor of a planet.

    So we have that


    [itex]T^{\mu \nu}=\left [ \begin{matrix}
    T^{00} & T^{01} & T^{02} & T^{03}\\
    T^{10} & T^{11} & T^{12} & T^{13}\\
    T^{20} & T^{21} & T^{22} & T^{23}\\
    T^{30} & T^{31} & T^{32} & T^{33}
    \end{matrix} \right ] [/itex]


    [itex]T^{\mu \nu}=\left [ \begin{matrix}
    \rho c^2 & T^{01} & T^{02} & T^{03}\\
    \rho v_{x} & T^{11} & T^{12} & T^{13}\\
    \rho v_{y} & T^{21} & T^{22} & T^{23}\\
    \rho v_{z} & T^{31} & T^{32} & T^{33}
    \end{matrix} \right ] [/itex]

    What I meant was that I was asking for expressing other components of Stress-Energy tensor in terms of [itex]\rho, v, c, etc[/itex]. For example, I had [itex]T^{00}[/itex], then I had [itex]\rho c^2[/itex]. So I was asking for a substitution.

    Also [itex] T^{10}+T^{20}+T^{30}[/itex] is a momentum density according to Wikipedia, so I broke down it into 3 components. I would like to have other components evaluated as well.

    Not for a dust, not for a liquid, but for a medium size spherically symmetrical uncharged planet.
     
  15. Jul 4, 2012 #14

    pervect

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    Those are called interior solutions. You'll find the solution for a perfect fluid interior in most textbooks (for instance Wald, General Relativity, pg 125). It's usually called the Schwarzschild interior solution or some such.

    You say you don't want a perfect fluid solution (at least I think you are trying to say that), but it's not clear to me why.

    A perfect fluid has the property that the pressure is isotropic, and that the density is some function of the pressure. The relation between pressure and density is called the "equation of state".

    Planets may not quite be perfect fluids, but it's an excellent approximation. You could not approximate the gravitational field of , say, a large cube with a perfect fluid, the pressure would be a tensor rather than a scalar. But the materials that planets are made of are not strong enough to make planets that are large cubes, the structural strength to hold a cubical shape against gravity just isn't there. This lack of strength to hold anything but a spherical shape is what makes the perfect fluid approximation a good one.

    If you really want the gravitational field of, say, a cube, you could most easily study it with the Newtonian approximation or weak field gravity. The effects of GR aren't really very important for objects as small as planets anyway. They become even less important for objects small enough to have a non-spherical shape.

    I hope this brief discussion helps. If you want more detail of the perfect fluid solution and can't find it in a textbook, let us know. Offhand I don't know of any solution for a non-perfect fluid. Assuming you have a perfect fluid the math a LOT simpler (and it's still not terribly easy), and it's also a very realistic assumption for objects large enough that GR would actually make a difference. So I rather doubt anyone has worked it out, but I haven't really tried to look for one.
     
  16. Jul 4, 2012 #15

    Matterwave

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    Maybe it'd be easier if I mentioned that as the EFE's are a set of differential equations, certain "boundary conditions" are required to specify them completely, and so even in a vacuum, you can get solutions which are not the flat space solution if your boundary condition (in this case, the central mass) is not the same.
     
  17. Jul 5, 2012 #16
    Seems like no one understood what I was trying to do. Probably this can shed some light (because I did some search on wikipedia)

    [itex]T^{\mu \nu}=\left [ \begin{matrix}
    \dfrac{E}{V} & \dfrac{\Delta E}{A^x \Delta t} & \dfrac{\Delta E}{A^y \Delta t} & \dfrac{\Delta E}{A^z \Delta t}\\
    \rho v_{x} & \dfrac{\Delta mv_{x}}{A^x \Delta t} & \dfrac{\Delta mv_{x}}{A^y \Delta t} & \dfrac{\Delta mv_{x}}{A^z \Delta t}\\
    \rho v_{y} & \dfrac{\Delta mv_{y}}{A^x \Delta t} & \dfrac{\Delta mv_{y}}{A^y \Delta t} & \dfrac{\Delta mv_{y}}{A^z \Delta t}\\
    \rho v_{z} & \dfrac{\Delta mv_{z}}{A^x \Delta t} & \dfrac{\Delta mv_{z}}{A^y \Delta t} & \dfrac{\Delta mv_{z}}{A^z \Delta t}
    \end{matrix} \right ][/itex]

    Please comment on this tensor--whether it is a general definition of a Stress-Energy tensor (the one which I wrote above!)


    Look up here: http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg
     
    Last edited: Jul 5, 2012
  18. Jul 5, 2012 #17

    pervect

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    The wiki definition makes some sense. Offhand, I can't imagine a sensible set of definitions for your collection of undefined symbols (you write E - does E represent E energy, or electric field? What about the A's?) that will make what you wrote equivalent to what wiki wrote. WHich doesn' mean that one doesn't exist, but it certainly doesn't look like any standard textbook result that I"ve seen. I'm not even sure if it's a tensor or not.

    Do you have an actual GR textbook? It seems like you're trying to learn GR from reading Wiki.

    The basic idea is simple. As wiki says, it's just a tensor that represents the amount of energy and momentum contained in a unit volume. The problem is, that the defintion of a unit volume varies from observer to observer - the "unit volume" of an observer at rest in frame A is not the same as the "unit volume" of an observer at rest in frame B who is moving with respect to frame A.

    T_ab * u^b, where u^b is the four-velocity of an observer, will give you a vector valued quantity which is the energy / unit volume and the momentum / unit volume for the observer moving with the four-velocity u^b.
     
  19. Jul 5, 2012 #18
    I denoted [itex]E[/itex] as energy, [itex]A[/itex] as area cross-section. The [itex]\Delta E[/itex][itex]=[/itex][itex]E_{2}-E{1}[/itex] (the finite change in the Energy).
     
    Last edited: Jul 5, 2012
  20. Jul 5, 2012 #19
    I don't understand why you are not understanding the matrix that I wrote. It perfectly, in my own view, corresponds to that in Wikipedia. For example, [itex]T^{10}[/itex] is a density of the momentum in the [itex]x[/itex] direction, so I write in [itex]\rho v_{x}[/itex] for it.
     
  21. Jul 5, 2012 #20
    Also, how can you intuitively distinguish the Riemann from Ricci tensor?
     
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