Timeline for Solving the geodesic equation for a singularity crossing curve
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2016 at 8:33 | vote | accept | Julian Moore | ||
| Dec 4, 2016 at 5:54 | answer | added | Richard Montgomery | timeline score: 0 | |
| Dec 1, 2016 at 9:51 | answer | added | Matthias Ludewig | timeline score: 0 | |
| Dec 1, 2016 at 9:42 | comment | added | Matthias Ludewig | Sorry, my last sentence was missing a word: "Are you also asking for a way to even interpret it in your situation?". So of course, if you take a curve that intersects your hypersurface transversally at time $t$, it satisfies the geodesic equation on both $[0, t)$ and $(0, t]$, and you want the right and left sided derivatives of $\gamma$ to coincide, then your curve will just be a usual geodesic in Minkowski space (i.e. a straight line), and the discontinuity at the hypersurface plays no role. | |
| Dec 1, 2016 at 6:33 | comment | added | Julian Moore | @matthiasludewig oops, forgot to @ you | |
| Nov 29, 2016 at 20:52 | comment | added | Julian Moore | 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. | |
| Nov 28, 2016 at 19:58 | comment | added | Matthias Ludewig | 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? | |
| Nov 28, 2016 at 17:53 | comment | added | Julian Moore | 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! | |
| Nov 28, 2016 at 2:19 | comment | added | DLIN | Do you mean the initial condition of the geodesics curve, i.e. $\dot X(t)|_{t=0}=V,~ X(0)=U$ ? | |
| Nov 28, 2016 at 0:45 | comment | added | Fan Zheng | Will the metric on $\ | |
| Nov 27, 2016 at 22:03 | review | First posts | |||
| Nov 27, 2016 at 23:11 | |||||
| Nov 27, 2016 at 21:58 | history | asked | Julian Moore | CC BY-SA 3.0 |