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Solving for spacetime

  1. May 29, 2006 #1
    Hi all,

    I would like to write down the arc-length parametrization of a curve in 3-dimensional Euclidean space, [tex]\gamma(\lambda)[/tex], then specify that in a certain spacetime this path is a null geodesic [tex]z^\mu[/tex] and solve for the metric of that spacetime.

    My first question is, does this even make sense as something to attempt, or does the initial reliance on coordinates to write down the curve (and the assumption that it's done in Euclidean space) mean that this will fail? If it does make sense, then my second question is does there exist some sort of framework in which to work that you could point me to?

    Also, when I convert the parametrized spatial curve into a null geodesic I think I will have something like [tex]z^\mu\stackrel{?}{=}(f(\lambda),\gamma(\lambda))\stackrel{?}{=}z_\mu[/tex]. Which = is correct? Should the curve gamma be the spatial component of the covariant or contravariant form of the geodesic?

    My plan right now is to impose symmetry and invertibility on an otherwise arbitrary metric tensor [tex]g_{\mu\nu}[/tex] and solve for [tex]g^{\mu\nu}[/tex]. I'll then impose [tex]g_{\mu\nu}z^\mu z^\nu=0[/tex] and the geodesic equation to obtain DE's for the components of g.

    I have attempted this for the simple case of trying to recover 2D Minkowski space [g=diag(1,-1)]. With [tex]z^\mu=(\lambda,\pm\lambda)[/tex] I found that [tex]g_{11}=g_{00}+f(x^0+x^1)[/tex] and [tex]g_{01}=g_{10}=\frac{1}{2}(g_{00}+g_{11})=g_{00}+\frac{1}{2}f(x^0+x^1)[/tex]. By choosing [tex]g_{00}\equiv 1[/tex] and [tex]f(x^0+x^1)\equiv -2[/tex] we see that the 2D Minkowski metric satisfies the equations, so that's reassuring. I used the plus or minus lambda in the geodesic because I wasn't sure whether the spatial curve [tex]\gamma(\lambda)=\lambda[/tex] should be covariant or contravariant (and I cheated a little by knowing what the resulting metric would do to it if it should be covariant) -- it turned out not to matter in this case, which didn't help me reach a conclusion :).

    Thanks for any help!

  2. jcsd
  3. May 30, 2006 #2


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    You can parameterize a curve on an arbitrary manifold, I don't see why you think there should be any problem with that.

    For a curve [itex]z^i(\lambda)[/itex], the 4-velocity of the curve at any point, u would be a tangent vector. (I forget whether that is co or contra, sorry, though I could look it up if it's really important).

    [tex]u^i = \frac{d z^i(\lambda)}{d \lambda}[/itex]

    is a tangent vector to the curve.

    For u^i to be null, [itex]u^i u_i = 0[/itex], thus

    [tex]g_{ij} u^i u^j = 0[/tex]

    I'm not quite sure how to solve this offhand, I don't think the solution is necessarily unique.
    Last edited: May 30, 2006
  4. May 30, 2006 #3


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    You want [itex]z^{\mu}=(f(\lambda),\gamma^{i}(\lambda))[/itex]. Coordinate functions are always written with the indices up.

    If you then try to solve
    [tex]g_{\mu\nu}(z(\lambda)) \dot{z}^{\mu}(\lambda) \dot{z}^{\nu}(\lambda)=0[/tex]
    with a given z, realize that you obtain only one (algebraic) equation for the metric at each point of the worldline. There are 10 algebraically independent components of [itex]g_{\mu\nu}[/itex], so the solutions aren't unique. Does this answer your question?
  5. May 30, 2006 #4


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    I suppose that if one requires that g{1..3,1..3} be diagonal(1,1,1) that leaves one with 4 unknows, g_00, g_01, g_02, and g_03 - and one equation.
  6. May 30, 2006 #5


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    But what would be the motivation for that? It would seem more natural to me to let the metric be diagonal. There are of course the same number of unknowns either way, though.
  7. May 30, 2006 #6


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    I probably wasn't very clear. (Unless I made a boo-boo that I didn't notice?)

    I used 0 as the subscript for time, so be requiring g{1..3,1..3} to be diagonal, I am saying that the hypersurfaces of constant time have a diagonal metric.

    IIRC the usual 3+1 approach uses a lapse function and shift vector, which I think is equivalent to the above, though I don't recall the details.
  8. May 30, 2006 #7


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    There's something bothering me about the OP: he says he is starting with a curve in 3-space, not a curve in 4-space.

    Anyways, for the problem y'all are discussing, I think you're making it too hard. :smile:

    (1) Chose a moving frame on R^4. That is, choose 4 vector fields that are linearly independent at each point.

    (2) Choose a vector field V that is tangent to your curve at each point on your curve.

    (3) Do some linear algebra to produce a new moving frame where V is the sum of the zeroeth and first vector field.

    (4) Choose your metric so your moving frame is orthonormal, with the zeroeth timelike and other three spacelike.

    The only problem is in (3) where you have to take some care so that your linear algebra is valid at all points simultaneously.
  9. May 31, 2006 #8
    Pervect I agree with you in general, but what is bothering Hurkyl was what was also bothering me as I posted :smile: .

    My original idea was to find a way to say, for example, that light was to follow a parabola. I would then have a curve in 3D like [tex]\gamma(\lambda)=(\lambda^2,0,0)[/tex] and want to find a 4D Lorentzian metric such that the prescribed path was a null geodesic.

    I realized later that attempting the construction in 2D space was pointless since all 2D Lorentzian metrics are conformally related to Minkowski space.

    Yes, I think it does. Thanks all...
  10. May 31, 2006 #9


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    As you're stating it, it's either easy or impossible: if you have an embedding of R^3 in R^4, then you have a natural way to view your 3D curve as a 4D curve. If you don't have such an embedding, then you don't have a reasonable way to view your curve as being in 4D space.

    But I don't quite think that's what you're asking (which is why it was bothering me) -- I was wondering if you meant, maybe, that your 3D curve is the spatial projection of your particles actual 4D trajectory... that it's really tracing out some 4D curve like [itex](\lambda, \lambda^2, 0, 0)[/itex], where the first coordinate is coordinate time,
  11. May 31, 2006 #10
    That's the original reason for my post - I first thought it would be easy, then impossible, then wasn't sure so came here :) .

    That is essentially what I want to do. Parametrizing a curve in 4D isn't a problem, but requiring the arbitrary curve to be a geodesic is the part at which I got a little stuck. Perhaps though it is easy and I'm just worrying about it too much...

    I'll reformulate this with a more concrete example to try to clarify what I was trying to do. Given the Schwarzschild metric we can calculate the path that a photon will take, and find a hyperbola. Could we instead, given that hyperbola and the knowledge that it's a null geodesic, recover the Schwarzschild metric? (Among others; as I and others previously pointed out it's not going to be unique.)
  12. May 31, 2006 #11
    This is a rather puzzeling statement of a problem. You refer to Euclidean space which means that a certain spatial metric is used, the components of which are coordinate dependant. Then you go on to inquire about spacetime and metric. A particular parameterization of a certain curve in Euclidean space is insufficient information to determine a spacetime metric. E.g. consider the curve that a particle of light travels on in flat spacetime. The 3-curve can be a straight line while the 4-curve would not. There is also a difference between curves and geodesics. Also, you didn't say if this curve was a geodesic.

  13. May 31, 2006 #12
    I have restated a more specific version of the problem in post #10... Does it make the problem I little more clear?

    I'm a bit confused by this - don't the paths of photons in a flat spacetime have to be straight lines (in the Euclidean sense) in both 3 and 4 dimensions?

    This is the entire point of my question, and I did say - I want to find the spacetime in which the specified curve is a geodesic, even though it's probably not a geodesic in a flat spacetime.
  14. Jun 1, 2006 #13
    Yes. Thanks.
    It depends on the spacetime coordinates that you've chosen. E.g. consider the curve in a coordinate system that corresponds to an inertial frame in flat spacetime. First we'll use Lorentz coordinates (ct, x, y, z). In such coordinates plot the worldline of a free particle. It will be a straight line if the coordinate axes are mutually perpendicular to each other. Now change coordinates to those corresponding to an accelerating frame of reference (coordinate axes mutually perpendicular to each other). Then the worldline is no longer a straight curve but a hyperbola.
    Given one geodesic then this is insufficient information to determine the metric tensor. Consider as an example a geodesic on a sphere. All geodesics are great circles passing through each pole. If you say to me "The curve is a circle. What is the metric?" then I cannot answer that since a geodesics on a cylinder include circles (The intersection of a plane whose normal is parallel to the cylinder's axis with the cylinder). In each case the curve is a circle and each is a geodesic. Yet there is insufficient information to determine the metric since the metric of a sphere is different than the metric of a cylinder.

    Hope that helps! :smile:

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