Symmetries and Conserved Quantities along Geodesics

This is a short note just to recall the calculation of an often useful fact in computing geodesics. First, some notation: let $(M,g)$ be a Riemannian / Lorentzian manifold, with a one-parameter group of isometries $u\mapsto ({\mathrm{\Phi }}_u:M\to M)$. The derivative of this 1 parameter family determines a vector field $K(p)=\frac{d}{du}{|}_{u=0}{\mathrm{\Phi }}_u(p)$, known as the Killing Field corresponding to these isometries. These fields provide a direct link between the symmetries of $M$ and conserved quantities¹ along its geodesics. Precisely,

To show this is constant, we wish to show the derivative with respect to the geodesic parameter $t$ vanishes. Differentiating via the product rule,

$$\frac{d}{dt}g(\dot{\gamma },K)=g(\frac{D}{dt}\dot{\gamma },K)+g(\dot{\gamma },\frac{D}{dt}K)$$

Where $D/dt$ is covariant differentiation pulled back to the real line. As $\gamma$ is a geodesic, $\frac{D}{dt}\dot{\gamma }=0$ by definition. And, as $\frac{D}{dt}$ coincides with ${\mathrm{\nabla }}_{\dot{\gamma }}$ for vector fields along a curve which are induced by smooth vector fields on $M$, this becomes $\frac{d}{dt}g(\dot{\gamma },K)=0+g(\dot{\gamma },{\mathrm{\nabla }}_{\dot{\gamma }}K)$. Thus, it suffices to show

$$g(\dot{\gamma },{\mathrm{\nabla }}_{\dot{\gamma }}K)=0$$

Here’s the big picture. The Lie derivative of the metric along a killing field is zero (we prove this in an appendix below): $${\mathcal{L}}_{K}g\equiv 0$$

This directly implies via a calculation (which we also do in an appendix below) that if $X,Y$ are arbitrary vector fields on $M$ and $K$ is a killing field, $$g({\mathrm{\nabla }}_{X}K,Y)+g(X,{\mathrm{\nabla }}_{Y}K)=0$$

Now set² $X=Y=\dot{\gamma }$ and use the symmetry of the metric tensor: $$0=g({\mathrm{\nabla }}_{\dot{\gamma }}K,\dot{\gamma })+g(\dot{\gamma },\mathrm{\nabla }\dot{\gamma }K)=2g(\dot{\gamma },{\mathrm{\nabla }}_{\dot{\gamma }}K)$$

That’s it! $g(\dot{\gamma },{\mathrm{\nabla }}_{\dot{\gamma }}K)$ vanishes, and so $g(\dot{\gamma },K)$ is constant along geodesics as claimed.

Appendix: Lie Derivative of the Metric

Here we prove the fact that underlies our whole calculation:

The intution for this result is clear from the definitions involved:

• The Lie derivative of an object measures how it changes when flowing along a vector field
• Flowing along a Killing Field generates a 1-parameter group of isometries
• Isometries do not change the metric
• Thus, you should expect to notice no changes in the metric tensor, the Lie derivative should be zero.

Here we turn this intuition into a short proof, using the geometric, or coordinate-invariant description of the Lie derivative. Setting up notation, let $K$ be a killing field and ${\mathrm{\Phi }}_t$ the associated flow by isometries. The Lie derivative of $g$ at a point $p$ is calculated by comparing $g_p$ to $g$ a bit further along the flow $g_{{\mathrm{\Phi }}_t(p)}$. As these two tensors are based at different points of $M$ we cannot compare them directly, but instead must *pull back to p*. That is, the well-defined difference is

${[({\mathrm{\Phi }}_t{)}^{\ast }{g}_{{\mathrm{\Phi }}_t(p)}]}_p-g_p$. This difference tends to zero as $t\to 0$, and the derivative is defined as usual, as the rate this quantity vanishes in comparison to $t$:

$$(\mathcal{L}kg)p = \lim{t\to 0}\frac{\left[(\Phi_t)^\ast g{\Phi_t(p)}\right] - g_p}{t}$$

Alright, with the formal definition out of the way, we can turn to the case of interest, where $K$ is a killing field so ${\mathrm{\Phi }}_t$ is a flow by isometries. By the very definition of isometry³ being a map which leaves the Riemannian metric invariant, we have $({\mathrm{\Phi }}_t{)}^{\ast }{g}_{{\mathrm{\Phi }}_t(p)}=g_p$, so the numerator of our difference quotient is *constant, and equal to zero*. Thus, the limit is zero, and the Lie derivative of $g$ is zero, as claimed.

Appendix: Killing Fields and the Metric

Here we prove that if $K$ is a killing field and $X,Y$ are arbitrary vector fields on $M$ then $g({\mathrm{\nabla }}_{X}K,Y)+g(X,{\mathrm{\nabla }}_{Y}K)=0$. In fact, we give a computation of the metric’s Lie derivative along an arbitrary vector field, and then specify to the required case using that this quantity vanishes along Killing fields.

First, a note on remembering how operators (like the Lie derivative, or covariant derivative) extend to general tensors. Say that $T$ is some tensor that takes in vector fields $X,Y$ and $\star$ is some operator. If you think of applying the tensor $T$ as a kind of multiplication (for instance, when $T$ is written like a generalized matrix), then the Leibniz rule for differentiating a product implies a formula for $\star$ applied to the smooth function $T(X,Y)$:

$$\star\left(T(X,Y)\right)=\star T+T(\star X, Y)+T(X,\star Y)$$

If we already know how to define $\star$ on smooth functions and on (co-)Vector fields, we can view such a rule as implicitly defining $\star$ on the tensor $T$: $$\star T=\star(T(X,Y))-T(\star X,Y)-T(X,\star Y)$$

We use this to explicitly compute the derivative ${\mathcal{L}}_V$ of the metric tensor $g$ below.