Finding geodesic through Killing fields

Question

I am currently reading Wald's General Relativity, but got a bit confused. Given a manifold with a metric, $(M, g_{ab})$, we may find the set of geodesic curves by solving the equation

$$ T^a \nabla_a T^b = \alpha T^b \tag{3.3.2} $$

where $T^a$ is a vector field whose vectors are tangent vectors to the geodesics. Moreover, we assume that the derivative operator $\nabla$ is associated with the metric, i.e. $\nabla_a g_{bc} = 0$.

In Wald, an alternative strategy is suggested when attempting to find the geodesics to the Schwarzschild metric. First and foremost, the (exterior) Schwarzschild metric is defined in the usual sense by

$$ ds^2 = -\left( 1 - \frac{2M}{r}\right)dt^2 + \left( 1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\Omega^2 \tag{6.1.44} $$

with $d\Omega^2 = d\theta^2 + \sin^2\theta d\phi^2$. We assume that the entire geodesic lies in the $\theta = \pi/2$ plane, without any loss of generality, since we can perform a coordinate transformation to achieve this. In order to find the geodesic curves, three equations are obtained. First, it is noted that the norm of the geodesic is either $-1$ or $0$ for timelike and lightlike geodesics (which are to be found). Letting $\kappa$ be either $-1$ or $0$, depending on which type of geodesic we are looking for, the first equation is:

$$ -\kappa = g_{ab} u^a u^b = -(1 - 2M/r)\dot{t}^2 + (1 - 2M/r)^{-1} \dot{r}^2 + r^2 \dot{\phi}^2 \tag{6.3.10} $$

where the dot denotes the derivative with respect to proper time, since the tangent vector in the given coordinate system is, per definition, given by:

$$ u^a = \frac{dt}{d \tau} \left( \frac{\partial}{\partial t} \right)^a + \frac{dr}{d \tau} \left( \frac{\partial}{\partial r} \right)^a + \frac{d \phi}{d \tau} \left( \frac{\partial}{\partial \phi} \right)^a $$

The other two equations are obtained from the following proposition:

PROPOSITION C.3.1. Let $\xi^a$ be a Killing vector field and let $\gamma$ be a geodesic with tangent $u^a$. Then $\xi_a u^a$ is constant along $\gamma$.

Since $(\partial / \partial t )^a$ and $(\partial / \partial \phi )^a$ are Killing fields (corresponding to directions which do not change the metric), two additional equations are given:

$$ E = -g_{ab} \left( \frac{\partial}{\partial t} \right)^a u^b = (1 - 2M/r)\dot{t} $$

where $E$ is constant, and

$$ L = g_{ab} \left( \frac{\partial}{\partial \phi} \right)^a u^b = r^2 \dot{\phi} $$

where $L$ is constant. Obviously, now it is possible to substitute two of the functions back into the first equation, to obtain the final differential equation for the geodesic:

$$ \frac{1}{2} \dot{r}^2 + \frac{1}{2} \left( 1 - \frac{2M}{r}\right) \left( \frac{L^2}{r^2} + \kappa \right) = \frac{1}{2} E^2 \tag{6.3.14} $$

Now, the question I have is, how can we assure that all curves which satisfy equation (6.3.14) are in fact geodesics? More specifically, if we let $C$ be the set of geodesics, corresponding to solutions to (3.3.2), and $C'$ be the set of curves which solve (6.3.10) and the energy- and momentum conservation equations provided above, then necessarily, we must have

$$ C \subseteq C' $$

since all geodesics must satisfy the conservation equations as per the proposition. However, how can we be sure that there is not some further restriction which has to be satisfied in order for the geodesic equation to be fulfilled? More generally, can we always use Killing fields in order to find geodesic curves, and then be sure that this will not include non-geodesics as well?

Any help is greatly appreciated!

Answer 1

One way I can try to answer your question is to invoke the theorem of existence and uniqueness of solutions of differential equations. This is not a complete answer, as there could be issues with the directions of some geodesics, but this would be considered as a rough argument why we would expect an answer to your question in most generic cases. Here is my argument.

I am going to use abbreviated notations. I assume that the metric $g$ is defined on a manifold of dimension $n$, and that $g$ has $n-1$ independent Killing vector fields $K_{(1)}, ..., K_{(n-1)}$. Here, think $n=3$, like in the example you are interested in, or $n=4$. Then, at each point $x$, form the linear operator $K_x : T_xM \,\longrightarrow \, \mathbb{R}^{n-1}$, which can be applied to a tangent vector $v$ at an arbitrary point $x$ as $K_xv = \big(\,g_x(K_{(1), x}, \,v),\, ... ,\, g_x(K_{(n-1), x},\,v)\,\big) \, \in \, \mathbb{R}^{n-1}$.

The geodesic equations can be written as $\nabla_{\dot{\gamma}}\dot{\gamma} = 0$, where $\nabla$ is the Levi-Civita connection of $g$ and every solution $\gamma(\tau)$ is a proper time, proper distance, or affinely (for light-like) parametrized geodesic. These equations can be even taken as a definition of a geodesic. According to Noether's theorem, which is the proposition you have quoted in the OP, each quantity $g(K_{(i)}, \dot{\gamma})$ is constant along any geodesic $\gamma(\tau)$. This translates into the fact that any geodesic $\gamma(\tau)$, for which $\gamma(0) = x_0$ and $\dot{\gamma}(0) = v_0$, satisfies the constraints $g_{\gamma(\tau)}\big(\,K_{(i), \gamma(\tau)},\, \dot{\gamma}(\tau)\,\big) = c_i$, where $c_i = g_{x_0}\big(\,K_{(i), x_0},\, v_0\,\big)$, and the special parametrization of $\gamma(\tau)$ is equivalent to $g_{\gamma(\tau)}\big(\,\dot{\gamma}(\tau),\, \dot{\gamma}(\tau)\,\big) = \epsilon$, where $\epsilon = -1, 0, 1.$ We can write the $n$ equations of constraints as follows:

$$K_{\gamma}\dot{\gamma} = c \,\,\text{ and }\,\, g_{\gamma}(\dot{\gamma}, \dot{\gamma}) = \epsilon$$ where $c = (c_1, ..., c_{n-1})$, and let us further abbreviate the latter system as follows: $$K(\gamma, \dot{\gamma}) = c$$

where now $c = (c_1, ..., c_{n-1}, \epsilon)$. Hence, if you take a geodesic $\gamma(\tau)$, which by definition solves $\nabla_{\dot{\gamma}}\dot{\gamma}=0$, then it definitely satisfies the system $K(\gamma, \dot{\gamma}) = c$, because the latter is formed by the $n$ independent conserved quantities.

Now, after this notational introduction, let us take a solution $\sigma(\tau)$ of the equations $$K(\sigma, \dot{\sigma}) = c$$ Then let $\sigma(0)=x_0$ and $\dot{\sigma}(0)=v_0$, where we can see that $v_0$ is a solution to $$K(x_0, v_0) = c$$

Observe that by construction (Noether's theorem), a geodesic $\gamma(\tau)$ with $\gamma(0) = x_0$ and $\dot{\gamma}(0) = v_0$ is a solution to $K(\sigma, \dot{\sigma}) = c$. Since, due to the independence of the Killing vector fields, $\partial_{v}K(x_0, v_0)$ is an $n \times n$ operator of rank at least $n-1$, and whenever the direction $v_0$ is not in the span of the Killing vectors $K_{(1),x_0}, ..., K_{(n-1),x_0}$, the $n \times n$ operator $\partial_{v}K(x_0, v_0)$ becomes rank $n$, i.e. invertible. Consequently, we can apply the implicit function theorem to solve uniquely the system of equations $K(x, v) = c$ as $v = f(x, c)$ in a neighbourhood of $x_0, v_0$, i.e. $$K\big(x, \,f(x, c)\big) = c$$ Hence, by the implicit function theorem, the solutions of $$K(\sigma, \dot{\sigma})=c$$ coincide with the solutions of $$\dot{\sigma} = f(\sigma, c)$$ in an open neighbourhood of $x_0, v_0$.

Let us take a solution of the initial value problem $K(\sigma, \,\dot{\sigma}) = c$ where $\sigma(0) = x_0$. Then, as discussed above, $\sigma(\tau)$ is a solution to $\dot{\sigma} = f(\sigma, c)$ with $\sigma(0)=x_0$ for $\tau$ in an open interval of $0$. By the existence and uniqueness theorem for systems of differential equations, it is the unique solution to the latter initial value problem. Moreover, observe that $\dot{\sigma}(0) = v_0$.

Next, let us take a geodesic $\gamma(\tau)$, which must solve $\nabla_{\dot{\gamma}}\dot{\gamma} = 0$ by virtue of being a geodesic, so that $\gamma(0)=x_0$ and $\dot{\gamma}(0) = v_0$. Then, $\gamma$ satisfies the system $K(\gamma, \dot{\gamma}) = c$ with $\gamma(0)=x_0$, which means that it is a solution of the initial value problem $\dot{\gamma} = f(\gamma, c)$ with $\gamma(0)=x_0$ in an open interval of $\tau=0$. However, since there can be only one solution to the initial value problem $\dot{\sigma}=f(\sigma, c)$ where $\sigma(0) = x_0$, the only conclusion left is that $\sigma(\tau) = \gamma(\tau)$ in an open interval of $\tau=0$.

Hence, a solution of $K(\sigma, \dot{\sigma})=c$ is locally a geodesic when its tangent vector at $x_0$ is not in the span of the Killing vectors at $x_0$.