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[Apologies if this question is not considered research level, but it received no substantive comments and no answers at math.SE; I thought it was straightforward, but maybe it isn't.]

Part I - What are the formal conditions for the solubility of the geodesic equation

$$\frac{d^2 x^\mu}{ds^2} =- \Gamma^{\mu}_{\alpha \beta} \frac{dx^\alpha}{ds} \frac{dx^\beta}{ds} ?$$

Part II - Consider (see figure) the particular case of a pseudo-Riemannian manifold whose metric is everywhere constant except on the surface $\Sigma$ (and the metric on $\Sigma$ may vary smoothly from place to place) such that every causal curve from $p$ to $q$ intersects the surface $\Sigma$ at a single point (except possibly in the null case), which surface - appearances in the figure not withstanding - might be or have null regions, and across which the metric is discontinuous.

Are there any conditions on the metric on $\Sigma$ (or otherwise) that make the geodesic equation soluble? (Possibly piecewise?).

Notes

The usual approach to solving the geodesic equation will not work, I believe, because of the curvature discontinuity introduced by $\Sigma$ that entails undefined derivatives across $\Sigma$, but maybe if one took as an assumption that $\Sigma$ is nowhere null, one could consider the curves to $\Sigma$ on each side separately and seek identity of their limit points.

If there are regions of $\Sigma$ that are null it seems to me it might be possible for the limit points of the two segments of $\gamma$ to be separated in $\Sigma$.

PS I'm not a mathematician, so in attempting to be suitably precise for those who are, I may in fact have had the opposite effect; I hope others can read between the lines.

Figure

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  • $\begingroup$ Will the metric on $\ $\endgroup$ – Fan Zheng Nov 28 '16 at 0:45
  • $\begingroup$ Do you mean the initial condition of the geodesics curve, i.e. $\dot X(t)|_{t=0}=V,~ X(0)=U$ ? $\endgroup$ – DLIN Nov 28 '16 at 2:19
  • $\begingroup$ Thanks for the question but no, I meant the differentiability conditions; hence part II where I wondered whether single sided derivatives, and limit points of geodesic sections meeting at a point (etc.) might succeed in producing a valid solution despite the lack of suitable differentiability everywhere (hence the curve crossing constraints). My thinking: if the metric is "greater" on $\Sigma$ there could be a solution, as moving even infinitesimally along $\Sigma$ would produce a longer path; but... I'm a physicist and not a mathematician hence the request for an assist! $\endgroup$ – Julian Moore Nov 28 '16 at 17:53
  • $\begingroup$ What do you mean with the metric being constant? Do you mean in local coordinates? Because if you meant parallel (which is the usual replacement for "constant" in the covariant setting), then a metric is always parallel with respect to its own Levi-Civita connection. Also, if your metric is not even continuous, it not at all clear what your differential equation even means. Are you also for a way to even interpret it in your situation? $\endgroup$ – Matthias Ludewig Nov 28 '16 at 19:58
  • $\begingroup$ Re constant metric I meant that, except at the discontinuity, the space is Minkowski space; the metric is continuous everywhere except at the discontinuity, so the d.e. would, I thought be meaningful by default except there, and the question asks whether the natural insolubility of the equation across the surface of discontinuity could be overcome by a piecewise approach, taking one side derivatives up to the surface on either side. I'm sorry I don't know how to express it more precisely. NB I'm afraid I didn't understand the last sentence. $\endgroup$ – Julian Moore Nov 29 '16 at 20:52
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This is probably a longer comment, but let me make this into an answer.

Suppose $\gamma$ is a curve in your manifold and we are asking if it satisfies the geodesic equation in some sense. Then either

(a) it hits $\Sigma$ at a discrete set of points $t_k$. Then the natural requirement that $\gamma$ satisfies the geodesic equation on the intervals $(t_k, t_{k+1}), \dots$ plus asking that the left and right sided derivatives at the points $t_k$ coincide already makes it into a geodesic of usual Minkowski space, i.e. a straight line, and the modification of the metric on $\Sigma$ does not play a role.

(b) The set of times $t$ such that $\gamma(t) \in \Sigma$ is not discrete. Then one would need to make sense of the geodesic equation at $\Sigma$, which seems kind of hard, because there is no way to define the Christoffel symbols etc. Also, a weak formulation (using Sobolev spaces etc.) does not really make sense, because $\Sigma$ is a zero set, hence the metric modified on $\Sigma$ as your describe defines the same element in $L^2$ as the original Minkowski metric.

So it seems that either we are in a trivial situation, and your modification just doesn't play a role, or we are in a situation where it is impossible to say what the geodesic equation even means.

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  • $\begingroup$ Thank you. (a) was my intuition, but what may be "trivial" is absolutely dependent on knowing you know the correct approach - thanks for being definitive ;) (b) The Christoffel issue generally I had considered, but not from the $\gamma(t) \in \Sigma$ not discretem geodesic equation at $\Sigma$ case - except for being distracted by null sections. I don't follow the weak formulation concept, but insofar as the answer remains -ve I'm content. $\endgroup$ – Julian Moore Dec 4 '16 at 8:33
  • $\begingroup$ I also note that your points should carry through to more general spacetimes where the metric is smooth except for a surface of discontinuity as described. $\endgroup$ – Julian Moore Dec 4 '16 at 8:38
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Perhaps you mean to say that your metric behaves rather like those arising in Snell's law from geometric optics where to model the refraction of light as it passes from one medium to another you take an index of refraction $n$ constant along slabs with a jumps at the interface? The optical metric is $(1/n) (dx^2 + dy^2 + dz^2)$ where the scalar function $n$ is piecewise constant, suffering a jump at the interface. Do you perhaps mean your pseudo-Riemannian metric to have the form $A_- (dx^2 + dy^2 + dz^2 - c^2 dt ^2 )$ on one side of your hypersurface $\Sigma$, and $A_+ (dx^2 + dy^2 + dz^2 - c^2 dt ^2 )$ on the other side, where $A_+, A_-$ are positive constants? If so, a kind of `Snell law'' will hold. To compute this law for the angle of ``refraction' as your geodesic crosses $\Sigma$ you extremize its arclength relative to the constraint that the curve is piecewise linear with a jump at $\Sigma$. This will yield a calculus problem in 4-space.

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  • $\begingroup$ Thanks, but I really did mean a delta function rather than a step function and @MatthiasLudewig has resolved the question. $\endgroup$ – Julian Moore Dec 4 '16 at 8:40
  • $\begingroup$ Notice however, that a delta function is a completely different thing than what you describe. If you meant your metric to be given as the sum of the Minkowski metric and a singular metric with distributional support on a hypersurface, the answer to your question will be more complicated and also completely different! $\endgroup$ – Matthias Ludewig Dec 5 '16 at 10:33
  • $\begingroup$ @matthiasludewig Yes, I understand about the delta function; physicist's sloppiness :) Sorry for muddying the waters. $\endgroup$ – Julian Moore Dec 5 '16 at 12:55

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