# Geodesics and the Lagrangian

Let $$q = (q^1, \cdots, q^n)$$ be a coordinate system on an open subset of some manifold $$M$$ and let $$\dot{q} = (\dot{q}^1, \cdots, \dot{q}^n)$$ be $$n$$ additional variables. We call a function $$L(q, \dot{q}, \tau)$$ of $$q$$, $$\dot{q}$$, and $$\tau$$ a Lagrangian . The action functional $$S$$ is then defined by the integral

S(q) = \int_a^b L(q(\tau), \dot{q}(\tau), \tau)\,d\tau,

where $$q(\tau)$$ is some coordinate curve with $$a \leq \tau \leq b$$ and $$\dot{q}^\mu = dq^\mu/d\tau$$. In classical mechanics, one typically sets $$L = T – U$$, where $$T$$ and $$U$$ are the kinetic and potential energies, respectively. When $$U = 0$$, i.e., when there are no forces present, we say that the motion of a particle is free. Hamilton’s principle states that the actual path of a particle through two given points in a mechanical system is the one that minimizes the action $$S$$. A standard variational argument (see Chapter 1 of [2] Landau and Lifshitz, Volume 1, for example) then produces the Euler-Lagrange equations
\label{eqEulerLagrange}
\frac{d}{d\tau}\left(\frac{\partial L}{\partial \dot{q}^\mu}\right)
– \frac{\partial L}{\partial q^\mu} = 0,\quad 1\leq \mu \leq n,

which determine the dynamics of the system.

We now discuss the relationship between free particle motion on a Lorentzian manifold $$M$$ with metric $$G$$ and its geodesics. Recall that geodesics are the extremal curves of the arc-length functional
\label{eqArcLength}
J(\gamma) = \int_a^b \sqrt{-G(\dot{\gamma}(t), \dot{\gamma}(t))}\,dt,

where $$\gamma$$ is a curve in $$M$$ defined on the interval $$[a, b]$$. (We adopt the sign convention on $$G$$ so that a tangent vector $$v$$ is time-like or null when $$G(v,v) \leq 0$$.) We now take $$S$$ to be the functional
\label{eqEnergy}
S(\gamma)
= – \frac{1}{2}\int_a^b G(\dot{\gamma}(\tau), \dot{\gamma}(\tau))\,d\tau.

We can think of $$-G(\dot{\gamma}(\tau), \dot{\gamma}(\tau)) / 2$$ as the kinetic energy at time $$\tau$$ of a free particle moving along $$\gamma$$ in $$M$$.

It can be shown (see Box 13.3 of [3] Misner, Thorne, and Wheeler) that (\ref{eqArcLength}) and (\ref{eqEnergy}) give rise to the same extremal curves, up to reparametrization. The extremals of (\ref{eqEnergy}) must be parametrized proportional to arc-length, which we denote by $$\tau$$. For a Lorentzian metric, $$\tau$$ is usually called proper time. We now suppose that we have coordinates $$x = (x^0, x^1, x^2, x^3)$$ on an open subset for a $$4$$-dimensional Lorentzian manifold $$M$$, and we define $$L(x, \dot{x}, \tau)$$ by

L(x, \dot{x}, \tau) = – \frac{1}{2}\,G(\dot{x},\dot{x})
= – \frac{1}{2}\,g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu,

where $$g_{\mu \nu}(x)$$ are functions of $$x^\mu$$ alone and $$\dot{x}^\mu = dx^\mu/d\tau$$. The Euler-Lagrange equations (\ref{eqEulerLagrange}) then yield the following $$4$$ equations (indexed by $$\sigma$$)
\label{eqEulerLagrangeMetric}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^\sigma}
\Big( g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu \Big)
\right\}
– \frac{\partial }{\partial x^\sigma}
\Big( g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu \Big)
= 0,\quad 0\leq \mu,\ \nu,\ \sigma \leq 3.

### Christoffel Symbols and the Geodesic Equations

We will now show how the Christoffel symbols and the geodesic equations corresponding to the metric $$G$$ can be derived from (\ref{eqEulerLagrangeMetric}). We first note that, since $$g_{\mu \nu}$$ depends on $$x^\mu$$, but not on $$\dot{x}^\mu$$, we have
\label{eqPartial}
\frac{\partial }{\partial x^\sigma}
\Big( g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu \Big)
= \frac{\partial g_{\mu \nu}}{\partial x^\sigma}\,
\dot{x}^\mu\,\dot{x}^\nu,

and
\label{eqPartialDot}
\begin{aligned}
\frac{\partial}{\partial \dot{x}^\sigma}
\Big( g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu \Big)
&= g_{\mu \nu}\,\frac{\partial}{\partial \dot{x}^\sigma}
\Big(\dot{x}^\mu\,\dot{x}^\nu \Big)\\
&= g_{\mu \nu}\,\left(
\frac{\partial \dot{x}^\mu}{\partial \dot{x}^\sigma}\,\dot{x}^\nu
+ \dot{x}^\mu\,\frac{\partial \dot{x}^\nu }{\partial \dot{x}^\sigma}
\right)\\
&= g_{\mu \nu}\,\left(
\delta_\sigma^\mu\,\dot{x}^\nu + \dot{x}^\mu\,\delta_\sigma^\nu
\right)\\
&= g_{\sigma \nu}\,\dot{x}^\nu + g_{\mu \sigma}\,\dot{x}^\mu.
\end{aligned}

It follows from this and the symmetry $$g_{\mu \nu} = g_{\nu \mu}$$ that (after some reindexing)
\label{eqDdtPartialDot}
\begin{aligned}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^\sigma}
\Big( g_{\mu \nu}\,\dot{x}^\mu\,\dot{x}^\nu \Big)
\right\}
&= \dot{g}_{\sigma \nu}\,\dot{x}^\nu + g_{\sigma \nu}\,\ddot{x}^\nu
+ \dot{g}_{\mu \sigma}\,\dot{x}^\mu + g_{\mu \sigma}\,\ddot{x}^\mu\\
&= 2\,g_{\mu \sigma}\,\ddot{x}^\mu
+ \frac{\partial g_{\sigma \nu}}{\partial x^\rho}\,
\dot{x^\rho}\,\dot{x}^\nu
+ \frac{\partial g_{\mu \sigma}}{\partial x^\rho}\,
\dot{x^\rho}\,\dot{x}^\mu.
\end{aligned}

Plugging (\ref{eqPartial}) and the above into (\ref{eqEulerLagrangeMetric}) then gives us
\label{eqCalc1}
2\,g_{\mu \sigma}\,\ddot{x}^\mu
+ \frac{\partial g_{\sigma \nu}}{\partial x^\rho}\,
\dot{x^\rho}\,\dot{x}^\nu
– \frac{\partial g_{\mu \nu}}{\partial x^\sigma}\,
\dot{x}^\mu\,\dot{x}^\nu
+ \frac{\partial g_{\mu \sigma}}{\partial x^\rho}\,
\dot{x^\rho}\,\dot{x}^\mu
= 0.

Changing $$\rho$$ to $$\mu$$ and $$\nu$$ in the second and fourth terms, respectively, of the left-hand side of (\ref{eqCalc1}) gives us

g_{\mu \sigma}\,\ddot{x}^\mu
+ \frac{1}{2}\,
\left\{ \frac{\partial g_{\sigma \nu}}{\partial x^\mu}
– \frac{\partial g_{\mu \nu}}{\partial x^\sigma}\,
+ \frac{\partial g_{\mu \sigma}}{\partial x^\nu}
\right\}\,
\dot{x^\mu}\,\dot{x}^\nu
= 0.

Multiplying this last equation by the inverse $$g^{\rho \sigma}$$ of the metric tensor $$g_{\sigma \mu}$$ ( $$= g_{\mu \sigma}$$) finally gives us
\label{eqGeodesics}
\ddot{x}^\rho + \Gamma_{\mu \nu }^\rho\,\dot{x^\mu}\,\dot{x}^\nu
= 0,

where the Christoffel symbols (or connection components) $$\Gamma_{\mu \nu }^\sigma$$ are given by
\label{eqChristoffelSymbols}
\Gamma_{\mu \nu }^\rho
= \frac{1}{2}\,g^{\rho \sigma}\,
\left\{ \frac{\partial g_{\nu \sigma}}{\partial x^\mu}
– \frac{\partial g_{\mu \nu}}{\partial x^\sigma}\,
+ \frac{\partial g_{\sigma \mu }}{\partial x^\nu}
\right\}.

We now consider the Schwarzschild metric
\label{eqSchwarzschildMetric}
G = ds^2 = – f(r)\,dt^2 + \frac{1}{f(r)}\,dr^2
+ r^2\,(d\theta^2 + \sin^2\theta\,d\varphi^2),\quad
f(r) = 1 – \frac{2 m}{r},

where $$m$$ is a positive constant and $$2m < r < \infty$$, $$- \infty < t < \infty$$, $$0 < \theta < \pi$$, $$0 < \varphi \leq 2\pi$$. In what follows, we will use the methodology outlined above to compute the geodesic equations corresponding to (\ref{eqSchwarzschildMetric}). Then we will solve these equations in the special case of null geodesics with constant angular coordinates $$\theta = \pi/2$$, $$\varphi = 0$$.

### The Geodesic Equations Corresponding to the Schwarzschild Metric

We have shown that the equations defined by (\ref{eqEulerLagrangeMetric}) give rise to the familiar geodesic equations (\ref{eqGeodesics}, \ref{eqChristoffelSymbols}). We now compute the geodesic equations for the Schwarzschild metric (\ref{eqSchwarzschildMetric}) directly from (\ref{eqEulerLagrangeMetric}).
This will turn out to be easier than computing the Christoffel symbols directly from (\ref{eqChristoffelSymbols}).

First, we establish the notation convention
\label{eqNotation}
\begin{array}{llll}
x^0 = t,
& x^1 = r,
& x^2 = \theta,
& x^3 = \varphi,\\
\\
\dot{x}^0 = \frac{\displaystyle dt}{\displaystyle d\tau},
& \dot{x}^1 = \frac{\displaystyle dr}{\displaystyle d\tau},
& \dot{x}^2 = \frac{\displaystyle d\theta}{\displaystyle d\tau},
& \dot{x}^3 = \frac{\displaystyle d\varphi}{\displaystyle d\tau}.
\end{array}

Since $$G$$ is diagonal (i.e., $$\mu \neq \nu \Rightarrow g_{\mu \nu} = 0$$ ), we have that (\ref{eqEulerLagrangeMetric}) reduces to
\label{eqDiagonalEquations}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^\sigma}
\left( \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2 \right)
\right\}
– \sum_{\rho = 0}^3
\frac{\partial g_{\rho \rho}}{\partial x^\sigma}\,(\dot{x}^\rho)^2
= 0,\quad 0\leq \sigma \leq 3,

where $$g_{00} = -f(r)$$, $$g_{11} = 1/f(r)$$, $$g_{22} = r^2$$, and $$g_{33} = r^2 \sin^2\theta$$.
We can use (\ref{eqNotation}) to obtain
\label{eqDsDt}
\begin{aligned}
\left(\frac{ds}{d\tau}\right)^2
&= \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2\\
&= -f(x^1)\,(\dot{x}^0)^2
+\frac{1}{f(x^1)}\,(\dot{x}^1)^2
+(x^1)^2\,(\dot{x}^2)^2
+(x^1)^2\sin^2(x^2)\,(\dot{x}^3)^2\\
&= -f(r)\left(\frac{dt}{d\tau}\right)^2
+\frac{1}{f(r)}\left(\frac{dr}{d\tau}\right)^2
+r^2\left(\frac{d\theta}{d\tau}\right)^2
+r^2\sin^2\theta\left(\frac{d\varphi}{d\tau}\right)^2.
\end{aligned}

What remains is to use this equation to write out and simplify (\ref{eqDiagonalEquations}) for $$\sigma = 0, 1, 2, 3$$.

#### The case $$\sigma = 0$$.

For $$\sigma = 0$$, we have $$\partial g_{\rho \rho} / \partial x^0 = \partial g_{\rho \rho} / \partial t = 0$$ for all $$\rho$$. Computing $$\partial g_{\rho \rho} / \partial \dot{x}^0$$ in a similar fashion leads us to the equations

\begin{aligned}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^0}
\left( \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2 \right)
\right\}
&= \frac{d}{d\tau}\Big\{ – 2\,f(x^1)\,\dot{x}^0 \Big\},\\
\sum_{\rho = 0}^3
\frac{\partial g_{\rho \rho}}
{\partial x^0}\,(\dot{x}^\rho)^2
&= 0.
\end{aligned}

Plugging these calculations into (\ref{eqDiagonalEquations}) together with the definition of $$f(r)$$ and the substitutions (\ref{eqNotation}) then
brings us our first equation
\label{eq0}
\boxed{
\frac{d}{d\tau}\left\{ \left(1 -\frac{2m}{r}\right) \,\frac{dt}{d\tau} \right\} = 0.
}

#### The case $$\sigma = 1$$.

This case yields the most complicated equation. As before, referring to the diagonal form of the metric given above, we easily obtain

\begin{aligned}
&\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^1}
\left( \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2 \right)
\right\}
= \frac{d}{d\tau}\left\{\frac{2\,\dot{x}^1}{f(x^1)}\right\},\\
&\sum_{\rho = 0}^3
\frac{\partial g_{\rho \rho}}
{\partial x^1}\,(\dot{x}^\rho)^2
= -f'(x^1)\,(\dot{x}^0)^2
-\frac{f'(x^1)}{f(x^1)^2}\,(\dot{x}^1)^2
+ 2\,x^1\,(\dot{x}^2)^2
+ 2\,x^1\,\sin^2(x^2)\,(\dot{x}^3)^2.
\end{aligned}

Using (\ref{eqNotation}) and (\ref{eqDiagonalEquations}) for $$\sigma = 1$$ then yields the equation
\label{eq1}
\boxed{
\begin{aligned}
2\,\frac{d}{d\tau}\left\{\frac{1}{f(r)}\,\frac{dr}{d\tau} \right\}
+ \frac{df}{dr}\,\left(\frac{dt}{d\tau}\right)^2
&+ \frac{1}{f(r)^2}\,\frac{df}{dr}\,\left(\frac{dr}{d\tau}\right)^2\\
&-2\,r\,\left\{
\left(\frac{d\theta}{d\tau}\right)^2
-\sin^2\theta\left(\frac{d\varphi}{d\tau}\right)^2
\right\}
= 0.
\end{aligned}
}

#### The case $$\sigma = 2$$.

Now let $$\sigma = 2$$. Then $$\partial g_{\rho \rho} / \partial x^2 = \partial g_{\rho \rho} / \partial\theta = 0$$ for $$\rho \neq 3$$, and $$\partial g_{33} / \partial x^2 = 2\,(x^1)^2 \sin(x^2)\,\cos(x^2)$$. Similarly, we find that $$\partial g_{\rho \rho} / \partial \dot{x}^2 = 0$$ for $$\rho \neq 2$$ and that $$\partial g_{22} / \partial \dot{x}^2 = 2\,(x^1)^2\,\dot{x}^2$$. We therefore obtain

\begin{aligned}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^2}
\left( \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2 \right)
\right\}
&= \frac{d}{d\tau}\Big\{2\,(x^1)^2\,\dot{x}^2 \Big\},\\
\sum_{\rho = 0}^3
\frac{\partial g_{\rho \rho}}
{\partial x^2}\,(\dot{x}^\rho)^2
&= 2\,(x^1)^2 \sin(x^2)\,\cos(x^2)\,(\dot{x}^3)^2.
\end{aligned}

Taking the difference of these two equations and making the usual substitutions yields
\label{eq2}
\boxed{
\frac{d}{d\tau}\Big\{r^2\,\frac{d\theta}{d\tau} \Big\}
– r^2 \sin\theta\,\cos\theta\,\left(\frac{d\varphi}{d\tau}\right)^2 = 0.
}

#### The case $$\sigma = 3$$.

Finally, let $$\sigma = 3$$. Then $$\partial g_{\rho \rho} / \partial x^3 = \partial g_{\rho \rho} / \partial\varphi = 0$$ for all $$\rho$$. Similarly, $$\partial g_{\rho \rho} / \partial \dot{x}^3 = 0$$, except for $$\rho = 3$$, which yields $$\partial g_{\rho \rho} / \partial \dot{x}^3 = 2\,(x^1)^2\sin^2(x^2)\,\dot{x}^3$$. We therefore obtain

\begin{aligned}
\frac{d}{d\tau}
\left\{\frac{\partial}{\partial \dot{x}^3}
\left( \sum_{\rho = 0}^3 g_{\rho \rho}\,(\dot{x}^\rho)^2 \right)
\right\}
&= \frac{d}{d\tau}\Big\{2\,(x^1)^2\sin^2(x^2)\,\dot{x}^3 \Big\},\\
\sum_{\rho = 0}^3
\frac{\partial g_{\rho \rho}}
{\partial x^3}\,(\dot{x}^\rho)^2
&= 0.
\end{aligned}

Plugging these calculations into (\ref{eqDiagonalEquations}) and simplifying leads to our fourth equation
\label{eq3}
\boxed{
\frac{d}{d\tau}\left\{ r^2 \,\sin^2\,\theta\frac{d\varphi}{d\tau} \right\} = 0.
}

To summarize, the geodesic equations (or equations of motion) are given by (\ref{eq0}, \ref{eq1}, \ref{eq2}, \ref{eq3}). Note that the first and fourth equations (\ref{eq0}, \ref{eq3}) derived above give us two conserved quantities, or \emph{integrals of motion}. These are obtained by integrating with respect to $$\tau$$.

### Null Geodesics in the Schwarzschild Metric

Recall that a curve with tangent vector $$\dot{x}^\mu(\tau)$$ is a \emph{null curve} if $$g_{\mu \nu}\,\dot{x}^\mu \dot{x}^\nu = 0$$. Given a curve that satisfies $$\theta = \pi/2$$ and $$\varphi = 0$$ for all $$\tau$$, we must also have $$\dot{\theta} = 0$$ and $$\dot{\varphi} = 0$$.
The form of $$(ds/d\tau)^2$$ then simplifies to
\label{eqDsDtPlane}
\left(\frac{ds}{d\tau}\right)^2
= -f(r)\left(\frac{dt}{d\tau}\right)^2
+\frac{1}{f(r)}\left(\frac{dr}{d\tau}\right)^2.

It follows that the solution curves lie in the $$(r, t)$$-plane. For null geodesics, we must also have $$ds/ d\tau = 0$$, and hence that
\label{eqDsDtPlaneSimplified}
f(r)\left(\frac{dt}{d\tau}\right)^2
-\frac{1}{f(r)}\left(\frac{dr}{d\tau}\right)^2 = 0.

Since $$f(r) > 0$$ for $$r > 2m$$, this simplifies to

\frac{dt}{d\tau} = \pm\frac{1}{f(r)}\,\frac{dr}{d\tau}.

From this we obtain the first order differential equation

dt = \pm\frac{dr}{1 – 2m/r},

which has the solution

\begin{aligned}
t &= \pm\int\frac{dr}{1 – 2m/r}\\
&= \pm\left\{r + 2m\,\log(r – 2m)\right\} + c.
\end{aligned}

We now define \emph{null coordinates} $$(v, w)$$ by

v = t + r + 2m\,\log(r – 2m),\quad
w = t – r – 2m\,\log(r – 2m).

The curves $$v = c_1$$ and $$w = c_2$$ for constants $$c_1$$ and $$c_2$$ define null geodesics. These curves are shown in Figure 1. See Section 5.5 of [1] Hawking and Ellis (pp. 149–61) for a discussion and further development of this calculation.

### References

1. S.W. Hawking and G.F.R. Ellis. The Large Scale Structure of Space-Time. Cambridge Monographs in Mathematical Physics. Cambridge University Press. 1973.
2. L.D. Landau and E.M. Lifshitz. Mechanics, 3rd Edition. Course of Theoretical Physics. Pergamon Press. 1976.
3. Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler. Gravitation. W.H. Freeman and Company. 1973.