In order to determine the geodesics, one must solve the following set of differential equations \begin{align} \frac{d^2 x^j}{ds^2} + {j\brace h\,\,k}\frac{dx^h}{ds}\frac{dx^k}{ds} = 0, \end{align} where ${j\brace h\,\,k}$ is the Christoffel symbol of second kind, which is defined as \begin{align*} {j\brace h\,\,k} = \frac{1}{2} g^{jk}\left[\frac{\partial g_{hk}}{\partial x^l} + \frac{\partial g_{kl}}{\partial x^h} - \frac{\partial g_{hl}}{\partial x^k}\right]. \end{align*} Often, these equations can not be solved analytically, and we can only solve them numerically.

When solving the geodesic differential equations, one of the main concerns seem to be the conjugate metric $g^{jk}$, and it's existence. If the metric $g_{jk}$ is singular, i.e $det(g) = 0$, then the conjugate metric does not exist, as \begin{align*} g_{jk}g^{jk} = \delta_j^{\phantom{j}k}. \end{align*}

My questions is as following :

Given an initial condition, does there still exist a geodesic that is a unique solution, or is it simply not possible to solve the equation?

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    $\begingroup$ 1. How is this a physics question? In which physical situation does $\det(g) = 0$ arise? 2. The notation ${j\brace h\,\,k}$ is horribly antiquated, why do you not use the more modern ${\Gamma^j}_{hk}$? $\endgroup$ – ACuriousMind Nov 24 '15 at 14:03
  • $\begingroup$ I like the curly braces, they remind me that the Christoffel symbols are not a tensorial quantity. Here is one author at least who still uses them : J. H. Heinbockel. Introduction to Tensor Calculus and Continuum Mechanics (2001). Thats actually the 21 century...oh gosh $\endgroup$ – imranal Nov 24 '15 at 14:08
  • $\begingroup$ Well, I'm not starting a war over this (i.e. I'm not editing your post, and I won't debate this further than this comment), but if you conceive of $\Gamma$ as an abstract connection form on the tangent bundle (explained e.g. here), then ${\Gamma^j}_{hk}$ are indeed its components. The curly braces just intensify the idea that these objects are somehow "unnatural" and "don't transform right", when you just have to understand what geometrical object they actually belong to. $\endgroup$ – ACuriousMind Nov 24 '15 at 14:24
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    $\begingroup$ I do not understand the question. If $det(g)=0$, then $g^{ij}$ does not exist and the Christoffel symbol is not defined. In this sitiation there is no geodesical equation at all. Existence and uniqueness of what, if you do not have an equation? $\endgroup$ – Valter Moretti Nov 24 '15 at 15:04
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    $\begingroup$ $g^{ij}$ is the inverse matrix of $g_{ij}$. The inverse exists if and only if $det(g)\neq 0$. So, in the degenerate case, you cannot define the Christoffel symbol. Yes, the equation itself is not valid in the singular case. $\endgroup$ – Valter Moretti Nov 24 '15 at 16:02

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