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Null geodesic definition (by extremisation?)

  1. Aug 30, 2014 #1
    Hi,

    How can null geodesics be defined?

    Obviously the concept of parallel-transport, of the tangent to the curve, applies equally well to null curves as to time/space-like curves. Technically this is only the definition for an "auto-parallel", not for a "geodesic". For example in Einstein-Cartan theory, where the connection (e.g. the covariant derivative) is permitted to have nonzero torsion, the autoparallels and the geodesics each follow slightly different trajectories.

    Usually spacelike and timelike geodesics are defined as extremum of the distance between two points. (E.g. using a Euler-Lagrange approach they make the integral of the metric line-element be stationary to first order with respect to any infinitesimal variations of the path coordinates.) Can null geodesics be defined in the same way? GR for physicists seems to think so, but doesn't there seem to be counterexamples in Minkowski space of paths with both timelike and spacelike deviations, that still have the same (zero) total integrated length between the endpoints? I think I've also heard of null geodesics being defined as a limit between a series of timelike and spacelike geodesics, is that unnecessarily cumbersome?
     
    Last edited: Aug 30, 2014
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  3. Aug 30, 2014 #2

    pervect

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    I believe that you can, and that the null geodesics are saddle points rather than maxima or minima (which is true in general when you have a stationary action).

    I'd suggest reading the papers at http://www.eftaylor.com/leastaction.html#actionleast, though it's oriented more towards discusing "least action" and "stationary action" rather than null geodesics. I believe the principals are similar if not identical, though - the action (Lagrangian) for a timelike non-null worldline is just the proper time, for instance.

    [add]About the only difference in solving for a null geodesic rather than a timelike one is that the geodesic curves are parameterized affinely, rather than by proper time (which a photon doesn't have).
     
    Last edited: Aug 30, 2014
  4. Aug 30, 2014 #3

    WannabeNewton

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    pervect is entirely correct. A null geodesic is a local extremum of the action ##S = \int_{\gamma} (\dot{x}^{\mu}\dot{x}_{\mu})^{1/2}d\lambda## where ##\lambda## is an affine parameter along the curve ##\gamma##. An extremum need not be a local maximum or a minimum. It can be a saddle point, which is local by definition, in dimensions greater than 2, and that is exactly what a null geodesic is. A time-like geodesic by contrast is a local maximum.
     
  5. Sep 6, 2014 #4
    If two points of a globally hyberbolic pseudo-Riemannian manifold are connected by a null geodesic, then any curve connecting them is either null-everywhere, or must have a space-like piece. Purely space-like curves (in more than 1 spatial dimensions) are possible. Purely time-like curves are impossible.
     
  6. Sep 6, 2014 #5
    Incnis, I don't follow. [STRIKE]If two events in Minkowski space are connected by a null geodesic, I can't see any way of also connecting them with only space-like curves.[/STRIKE][mistaken] Conversely, at the photon sphere of the Schwarzschild spacetime it is trivial to find a time-like curve between two events that can also be connected by a null geodesic.
     
    Last edited: Sep 6, 2014
  7. Sep 6, 2014 #6

    WannabeNewton

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    Incnis' statement is incorrect and yours is correct.
     
  8. Sep 6, 2014 #7

    PAllen

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    Also worth noting is that while SC geometry is not globally hyperbolic, there is no difficulty constructing a globally hyperbolic manifold that contains a photon sphere. For example, just glue SC just outside of Buchdal radius with a static, stable, matter SET inside the cut.
     
  9. Sep 6, 2014 #8

    PAllen

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    The statement that a smooth spacelike curve, in > 1 spatial direction, can connect events connected by a null geodesic, is true. Consider in Minkowski space the points:

    (t,x,y) = (0,0,0) and (1,1,0). These are connected by a null geodesic. However, they can also be connected by the following spacelike segements:

    (0,0,0) to (.5,.5,.5) to (1,1,0)

    These are both spacelike, and the sharp bend, if smoothed, will be spacelike. This is in contrast to a 1 dimensional attempt:

    (0,0,) to (.5,3) to (1,1)

    If one smooths the sharp bend, it must be timelike.
     
  10. Sep 6, 2014 #9

    stevendaryl

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    I don't understand quite how a null geodesic is an extremum of the action [itex]A = \int ds \sqrt{\dot{x^\mu} \dot{x_\mu}}[/itex]

    The usual definition of extremum is this: if we vary [itex]x^\mu(s)[/itex] by replacing it by [itex]x^\mu(s) + \delta x^\mu(s)[/itex], then the first-order variation of [itex]A[/itex] vanishes. But it seems to me that we run into problems applying this definition in the case where [itex]\dot{x^\mu} \dot{x_\mu}=0[/itex]. In that case, if we replace [itex]x^\mu(s)[/itex] by [itex]x^\mu(s) + \delta x^\mu(s)[/itex], then we find:

    [itex]\delta A = \int ds \sqrt{\dot{x^\mu} \dot{x_\mu} + 2\dot{x^\mu}\dot{\delta x_\mu} + \dot{\delta x^\mu}\dot{\delta x_\mu}}[/itex]
    [itex]= \int ds \sqrt{2\dot{x^\mu}\dot{\delta x_\mu} + \dot{\delta x^\mu}\dot{\delta x_\mu}}[/itex]

    It's not obviously the case that the first-order variation in [itex]\delta A[/itex] vanishes.
     
  11. Sep 6, 2014 #10
    Well, ok, but can't you still do that with only one spatial dimension too? Just by taking a large enough detour in space (e.g. (0,0) → (0,10) → (1,1)) then a curve between the two events will be space-like.
     
  12. Sep 7, 2014 #11

    pervect

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    Caroll only does the usual timelike case, stating that the "results are generally appliciable", without going into the details of why or how they are "general" or what needs to be tweaked in the null case you are interested in. But comparing his work to yours, yours appears to be missing some important terms:

    http://preposterousuniverse.com/grnotes/grnotes-three.pdf [Broken] pg 15, eq 3.49

    You are missing terms arising from the second equation.

    Basically, when you change the path, you also change ##g_{\mu\nu}##, which you haven't accounted for.

    An important later step in Caroll's derivation is the choice of proper time for the curve, for a null geodesic this step would be using an affine parameterization of the null curve. But I will wait to address the need for that step unless/until it becomes troubelsome, I think.
     
    Last edited by a moderator: May 6, 2017
  13. Sep 7, 2014 #12

    PAllen

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    Read the rest of my post. I covered that. In one dimension, smoothing the bend introduces a time like segment, while smoothing my 2-d example remains pure space like.
     
  14. Sep 7, 2014 #13

    PeterDonis

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    If you mean the maximally extended Schwarzschild spacetime, yes, it is. The surface ##U = 0## in Kruskal coordinates is a Cauchy surface.

    I think you may be confusing "globally hyperbolic" with "geodesically complete". The maximally extended Schwarzschild geometry is not geodesically complete, because of the singularities. But that doesn't prevent it from being globally hyperbolic.
     
  15. Sep 7, 2014 #14

    PAllen

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    Actually, what I got confused on was that only naked singularities inherently violate the the 'compact intersection of past and future causal sets of events p and q' criterion for global hyperbolicity. Kruskal does not have a nake singularity, so it does not violate this (or any equivalent) definition of global hyperbolicity.
     
  16. Sep 7, 2014 #15

    stevendaryl

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    I was simplifying to the case of flat spacetime. Yes, in general, you need the derivatives of [itex]g_{\mu \nu}[/itex] as well, but let's suppose that those are zero. If they're nonzero, that makes the argument more complicated, but I don't think it resolves the problem.

    Including the derivatives of [itex]g_{\mu \nu}[/itex] gives the more complicated form for [itex]A[/itex]:

    [itex]A = \int ds \sqrt{(\partial_\lambda g_{\mu \nu}) \delta x^\lambda \dot{x^\mu} \dot{x^\nu} + 2 g_{\mu \nu} \dot{x^\mu} \dot{\delta x^\nu}}[/itex]

    In order to prove that you have an extremum, it must be possible to rewrite [itex]\delta A[/itex] in the form:
    [itex]\delta A = \int ds F_\mu(x^\nu, \dot{x^\nu}) \delta x^\mu[/itex]

    Then the vanishing of the first-order variation implies [itex]F_\mu(x^\nu, \dot{x^\nu}) = 0[/itex].

    I don't see how to write [itex]\delta A[/itex] in this form.

    If, on the other hand, [itex]g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}[/itex] is nonzero, then we can do it:

    [itex]A = \int ds \sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu} + (\partial_\lambda g_{\mu \nu}) \delta x^\lambda \dot{x^\mu} \dot{x^\nu} + 2 g_{\mu \nu} \dot{x^\mu} \dot{\delta x^\nu}}[/itex]
    [itex]= \int ds (\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}} + \frac{1}{2}\dfrac{(\partial_\lambda g_{\mu \nu}) \delta x^\lambda \dot{x^\mu} \dot{x^\nu} + 2 g_{\mu \nu} \dot{x^\mu} \dot{\delta x^\nu}}{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}})[/itex]

    So
    [itex]\delta A = \int ds (\frac{1}{2}\dfrac{(\partial_\lambda g_{\mu \nu}) \delta x^\lambda \dot{x^\mu} \dot{x^\nu} + 2 g_{\mu \nu} \dot{x^\mu} \dot{\delta x^\nu}}{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}})[/itex]

    We can integrate by parts to get rid of the derivative with respect to [itex]s[/itex] on [itex]\delta x^\mu[/itex]:

    [itex]\delta A = \int ds (\frac{1}{2}\delta x^\lambda \dfrac{(\partial_\lambda g_{\mu \nu}) \dot{x^\mu} \dot{x^\nu}}{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}} - \delta x^\nu \dfrac{d}{ds}(\dfrac{g_{\mu \nu} \dot{x^\mu} }{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}}))[/itex]

    Finally, we relabel dummy indices to get:

    [itex]\delta A = \int ds (\frac{1}{2} \dfrac{(\partial_\lambda g_{\mu \nu}) \dot{x^\mu} \dot{x^\nu}}{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}} - \dfrac{d}{ds}(\dfrac{g_{\mu \lambda} \dot{x^\mu} }{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}})) \delta x^\lambda[/itex]

    So the vanishing of first-order variation gives:
    [itex]\frac{1}{2} \dfrac{(\partial_\lambda g_{\mu \nu}) \dot{x^\mu} \dot{x^\nu}}{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}} - \dfrac{d}{ds}(\dfrac{g_{\mu \lambda} \dot{x^\mu} }{\sqrt{g_{\mu \nu} \dot{x^\mu}\dot{x^\nu}}}) = 0[/itex]

    Which is the geodesic equation with an arbitrary (not necessarily affine) parameter [itex]s[/itex].

    But it clearly doesn't work if [itex]g_{\mu \nu} \dot{x^\mu} \dot{x^\nu} = 0[/itex], because this expression appears in the denominator.
     
  17. Sep 8, 2014 #16

    pervect

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    At the moment, I don't quite see how to resolve the problem. However, since we know the formula works for timelike geodesics in the limit as they approach lightspeed, and for spacelike geodesics in the limit as they approach a lightlike geodesic, I don't find it surprising that it actually works for null geodesics, even though I don't quite see how to prove it at the moment.

    If we let g = diag(-1,1) , a flat minkowskii space with one time dimension ##x_0## and one space dimension ##x_1##, and let the affine parameter be s then we expect

    ##x_0 = ks \quad x_1=ks \quad \dot{x_0}=k \quad \dot{x_1}=k## to be a null geodesic.

    More generally, in a higher dimension flat space the geodesic equation should just tell us that the components of ##\dot{x}## are constant, where the dot represents differentaion with respect to the affine parameter s. Thus our geodesic is just:

    ##x_\mu = k_\mu s## where each of ##k_0##, ##k_1##, etc are constant.

    This condition works for timelike, spacelike, and null geodesics in Minkowskii spacetime, what separates out null geodesics is the normalization condition ##x^\mu x_\mu = 0##

    If we assume that ##g_{\mu\nu}## is constant, then we need to go from

    to the above conclusion, which would be trivial (we've assumed ##g_{\mu\nu} is constant) if the denominator was constant and nonzero, but unfortunately the denominator is constant (with an affine parameterization), but as you point out for null geodesics that constant is zero :(.
     
  18. Sep 8, 2014 #17
    It was silly, about impossibility of any time-like curve. Even gravitational lensing provides a counterexample. But for a null geodesic in general position, can a curve arbitrarily close to the geodesic be timelike (everywhere)? This has nothing to do with global properties.
     
  19. Sep 8, 2014 #18

    PAllen

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    I suspect not. Trivially not if you consider a region of sufficiently limited extent (as defined by deviation of metric from Minkowski).

    However, concerning the OP idea of approaching a null geodesic with time like paths, this seems possible if, while fixing one end, you have family of timelike paths whose other ends approach the the null geodesic ever closer.
     
  20. Dec 14, 2015 #19
    Dear All.
    I've read your discussion with interest and I think there is something that solves your conundrum with the missing quadratic term in the Least Action principle in the null world-line case.
    Namely, for a null geodesic you should change the Lagrangian form in the Least Action principle, because it is then a constrained functional with the condition ##\dot{x}.\dot{x}=0##, and even more
    importantly, because this condition is not analytically compatible with the usual Lagrangian form ##\sqrt{\dot{x}.\dot{x}}## on account that the momentum ##\frac{\dot{x}}{\sqrt{\dot{x}.\dot{x}}}## is then
    ill defined. For a null case, the Lagrangian is simply ##\lambda\, \dot{x}.\dot{x}## with a Lagrange multiplier ##\lambda## (which is an additional variable along with ##x##'es; ##\lambda## must
    appropriately transform with re-parameterization in order the Lagrangian be of the first degree in the velocities, as required by the relativity theory). Then, in addition to the Euler-Lagrange
    equation for ##x## (which assumes a geodesic-like form) we get another equation ##\dot{x}.\dot{x}=0## by varying w.r.t. ##\lambda##.

    Now, we may get rid off the term linear in ##\dot{x}## in the geodesic-like form by introducing an appropriate parameterization (this is the affine parameterization) in which case we obtain the proper geodesic form and the condition ##\dot{x}.\dot{x}=0## (the latter unaffected by this special choice of parameterization).

    It is interesting to note, that the time-like and space-like geodesics can also be derived in a similar manner with the condition ##\dot{x}.\dot{x}=\pm 1##, in which case the Lagrangian is:

    A)
    ##L=\frac{1}{2} \lambda\, \dot{x}.\dot{x} + \frac{1}{2} a \lambda^{-1}## in general (with a dimensional constant ##a##) if we want to have re-parameterization invariance, or

    B)
    ##L=\sqrt{\pm\dot{x}.\dot{x}}+\lambda(\dot{x}.\dot{x}\mp1)## if we want a special affine parameterization.

    In the above cases, you obtain from A) the geodesic-like form (by varying w.r.t. ##x##), while by varying w.r.t. ##\lambda## you obtain a second equation: ##\dot{x}.\dot{x}- \lambda^{-2}a=0##
    which gives ##\lambda^{-1}=\sqrt{a^{-1}\,\dot{x}.\dot{x}}## /we see that the sign of ##a## determines the time- or space- like case/).

    By plugging this ##\lambda## into A) you obtain ##L=\sqrt{a\,\dot{x}.\dot{x}}## -- which reduces to the ordinary case without the ##\lambda## variable (for ##a>0## we have ##a={\rm mass}^2##). In the B) case you get the condition ##\dot{x}.\dot{x}=\pm 1## which again reduces the geodesic-like form
    to one with affine parametrization without the linear term in ##\dot{x}##.

    The A) Lagrangian form is universal both for normalizable ##\dot{x}.\dot{x}## and for ##\dot{x}.\dot{x}=0##. In the latter case you require that the action is extremal for any ##\lambda##, that
    is, ##\lambda## is NOT a function of the velocities and this is possible only for ##a=0## (massless case); in the former case when ##\lambda## does is a function of the velocities, you get ##a\neq0##
    and a condition for function ##\lambda## which you may plug into the Lagrangian (which then attains the ordinary form). So, from this standpoint we see, that the geodesic-like forms are obtained in the null and the non-null case in the same general scheme, while the qualitative difference lies in that ##\lambda## is or is not a function of ##\dot{x}##.

    In any case, the above reasoning shows that for null curves you also get a Lagrangian quadratic in ##\dot{x}##.

    Note also the qualitative difference, that for ##a\neq 0## the condition ##\dot{x}.\dot{x}=\pm1## sets a parameterization, while for ##a=0## the condition ##\dot{x}.\dot{x}=0## does NOT set any
    parameterization and it is still re-parameterization invariant!

    This explains the presence of an additional degree of freedom in the null-like case represented by the independence of ##\lambda## on ##\dot{x}##. This also shows a fundamental difference which
    distinguishes null curves as such. In fact null curves are distinguished geometrically by the invariant null-cone structure of the empty spacetime -- there is nothing similar in the Euclidean space!. Physically it is thus justified to consider always two Lagrangians in every mechanical case (one for timelike
    curves, the other for null curves): eg. for a structureless particle you consider ##m\sqrt{\dot{x}.\dot{x}}## for a massive particle or ##\lambda \dot{x}.\dot{x}## for a massless particle with independent additional intrinsic degree of freedom ##\lambda##.

    .B.
     

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