The Covariant Derivative: How Far is a Curve from Being a Geodesic?
I recently discovered a really fun intuition about the covariant derivative. A geodesic is a curve that locally minimizes length, or more conveniently, energy. The covariant derivative of a curve’s time derivative can be thought of as its failure to be a geodesic. I found this to be really neat and wanted to share it with you! :)
Setup
First, notation: the manifold, the metric and its inverse, and the summation convention.
Let $\left(\calM,g\right)$ be an $n$-dimensional Riemannian manifold. For $q \in \calM$, we write the coordinates of $q$ with respect to some coordinate chart as $q \equiv \left(q^1, \dots, q^n\right)$.
- $g_{ij}$ are the components of the metric. They are symmetric and positive definite.
- $g^{ij}$ are the components of the inverse matrix, so $g^{ij} g_{jk} = \delta^i_k$, where $\delta^i_k$ is the Kronecker delta.
- Repeated indices are summed from $1$ to $n$.
The geodesic equation
Recall that geodesics are curves that extremize the length functional on a Riemannian manifold:
\[\Lambda[q] := \int_{t_0}^{t_1} \sqrt{g_{ij}\, \dot q^i \dot q^j} \, dt .\]They also extremize the energy functional
\[E[q] := \int_{t_0}^{t_1} L\left(q, \dot q\right) \, dt, \quad L\left(q, \dot q\right) := g_{ij}\, \dot q^i \dot q^j .\]Here we only look at a small piece of the curve $q$ between $t_0$ and $t_1$, since extremizing length is a local property.
Recall extremal curves of $E$ necessarily satisfies the Euler–Lagrange equations
\[\frac{d}{dt} \frac{\partial L}{\partial \dot q^i} - \frac{\partial L}{\partial q^i} = 0 . \tag{1}\]Since $L$ is quadratic, the two partial derivatives in (1) are nicely computed as
\[\frac{\partial L}{\partial \dot q^i} = 2 g_{ij} \dot q^j, \quad \frac{\partial L}{\partial q^i} = \frac{\partial g_{jk}}{\partial q^i} \dot q^j \dot q^k . \tag{2}\]Using the simply product rule, the time derivative of $\frac{\partial L}{\partial \dot q^i}$ is
\[\frac{d}{dt} \frac{\partial L}{\partial \dot q^i} = 2 g_{ij} \ddot q^j + 2 \frac{\partial g_{ij}}{\partial q^k} \dot q^k \dot q^j .\]The last term is contracted with the symmetric product $\dot q^j \dot q^k$, so only its symmetric part in $j, k$ survives:
\[2 \frac{\partial g_{ij}}{\partial q^k} \dot q^k \dot q^j = \left( \frac{\partial g_{ij}}{\partial q^k} + \frac{\partial g_{ik}}{\partial q^j} \right) \dot q^j \dot q^k .\]Substituting this and $\left(2\right)$ into the left-hand side of $\left(1\right)$ gives
\[\frac{d}{dt} \frac{\partial L}{\partial \dot q^i} - \frac{\partial L}{\partial q^i} = 2 g_{ij} \ddot q^j + \left( \frac{\partial g_{ij}}{\partial q^k} + \frac{\partial g_{ik}}{\partial q^j} - \frac{\partial g_{jk}}{\partial q^i} \right) \dot q^j \dot q^k . \tag{3}\]To clean this up, define the Christoffel symbols
\[\Gamma^l_{jk} := \frac{1}{2} g^{li} \left( \frac{\partial g_{ij}}{\partial q^k} + \frac{\partial g_{ik}}{\partial q^j} - \frac{\partial g_{jk}}{\partial q^i} \right) . \tag{4}\]They are symmetric in $j$ and $k$. Since $g_{il} g^{lm} = \delta_i^m$, the coefficient of $\dot q^j \dot q^k$ in $\left(3\right)$ is exactly $2 g_{il} \Gamma^l_{jk}$ by $\left(4\right)$. Hence
\[\frac{d}{dt} \frac{\partial L}{\partial \dot q^i} - \frac{\partial L}{\partial q^i} = 2 g_{il} \left( \ddot q^l + \Gamma^l_{jk} \dot q^j \dot q^k \right) .\]Setting the left-hand side of $\left(1\right)$ to zero and multiplying by $\frac{1}{2} g^{ki}$ yields the geodesic equation
\[\ddot q^k + \Gamma^k_{ij}\left(q\right)\, \dot q^i \dot q^j = 0 .\]So in conclusion, if we want to compute a geodesic whose position at time $t$ is $q\left(t\right) \equiv \left(q^1\left(t\right), \dots, q^n\left(t\right)\right)$ and whose velocity is $\dot q\left(t\right) \equiv \left(\frac{d q^1}{dt}\left(t\right), \dots, \frac{d q^n}{dt}\left(t\right)\right)$, we solve the second-order ODE
\[\frac{d^2 q^k}{dt^2} + \Gamma^k_{ij}\left(q\right) \frac{d q^i}{dt} \frac{d q^j}{dt} = 0\]with these initial conditions.
Covariant acceleration of an arbitrary curve
Plugging an arbitrary curve into the left-hand side of the geodesic equation gives a vector $\bfa \in T_{q\left(t\right)} \calM$ with components
\[a^k := \ddot q^k + \Gamma^k_{ij}\left(q\right)\, \dot q^i \dot q^j .\]We can call it the covariant acceleration.
What does $\bfa$ measure?
For one, $\bfa = 0$ if and only if the curve $q$ is a geodesic. So, in a sense, $\bfa$ measures the failure of $q$ to be a geodesic. Being a geodesic is a local property: at every instant, the curve is extremizing length (or energy). So $\bfa$ can also be interpreted as the failure of $q$ to extremize length. The curve is traveling a bit more distance than it needs to.
More importantly, as the title of this entry suggests, $\bfa$ is precisely the covariant derivative of $\dot q$:
\[\frac{D}{dt} \dot q = \nabla_{\dot q} \dot q = \bfa .\]Isn’t that neat? The covariant derivative is usually introduced axiomatically, via the Fundamental Theorem of Riemannian Geometry: it is the unique operator satisfying a short list of properties. I like to call this the algebraic way of doing things, as opposed to the constructive, analytic definition of the derivative as a limit that we learn in calculus. I understand why it is done this way, and I enjoy it: in my opinion, a derivative operator is more of an algebraic object than an analytic one. Still, it is nice that there is an analytic route to the covariant derivative as well, and we arrived at it without using that theorem.
Variations
To make “failure to extremize” precise, we perturb the curve and see how the energy responds. Fix a curve $q$. A variation of $q$ is a smooth vector field $\delta q^i\left(t\right)$ along $q$ with $\delta q^i\left(t_0\right) = \delta q^i\left(t_1\right) = 0$.
For a small real parameter $s$, define the perturbed curve $q_s$ whose coordinates are
\[q_s^i\left(t\right) := q^i\left(t\right) + s\, \delta q^i\left(t\right) .\]The first variation of $E$ at $q$ in the direction $\delta q$ is a scalar defined as
\[\delta E_q [ \delta q ] := \frac{d}{ds} E [ q_s ] \Big|_{s = 0} = \lim_{s \to 0} \int_{t_0}^{t_1} \frac{L\left(q_s, \dot q_s\right) - L\left(q, \dot q\right)}{s} \, dt = \int_{t_0}^{t_1} \lim_{s \to 0} \frac{L\left(q_s, \dot q_s\right) - L\left(q, \dot q\right)}{s} \, dt .\]It is linear in $\delta q$. The curve $q_s$ depends on the coordinate chart, but the derivative at $s = 0$ does not. A curve $q$ is extremal for $E$ exactly when $\delta E_q [ \delta q ] = 0$ for every variation $\delta q$. Next, we compute $\delta E_q$ explicitly, which is where the Euler–Lagrange equations come from.
The Euler–Lagrange equations
We now compute $\delta E_q$ explicitly and see the Euler–Lagrange equations fall out. Since $\dot q_s = \dot q + s\, \delta \dot q$, differentiating under the integral and using the chain rule gives
\[\delta E_q [ \delta q ] = \int_{t_0}^{t_1} \left( \frac{\partial L}{\partial q^i} \delta q^i + \frac{\partial L}{\partial \dot q^i} \delta \dot q^i \right) dt . \tag{5}\]Integrate the second term of $\left(5\right)$ by parts. The boundary term vanishes because $\delta q$ does, and moving the time derivative off $\delta q$ produces a minus sign:
\[\int_{t_0}^{t_1} \frac{\partial L}{\partial \dot q^i} \delta \dot q^i \, dt = \left[ \frac{\partial L}{\partial \dot q^i} \delta q^i \right]_{t_0}^{t_1} - \int_{t_0}^{t_1} \left( \frac{d}{dt} \frac{\partial L}{\partial \dot q^i} \right) \delta q^i \, dt = - \int_{t_0}^{t_1} \left( \frac{d}{dt} \frac{\partial L}{\partial \dot q^i} \right) \delta q^i \, dt .\]Substituting this into $\left(5\right)$ gives
\[\delta E_q [ \delta q ] = - \int_{t_0}^{t_1} \left( \frac{d}{dt} \frac{\partial L}{\partial \dot q^i} - \frac{\partial L}{\partial q^i} \right) \delta q^i \, dt . \tag{6}\]Now suppose $\delta E_q [ \delta q ] = 0$ for every variation $\delta q$. By the fundamental lemma of the calculus of variations, the bracket in $\left(6\right)$ must vanish identically. These are exactly the Euler–Lagrange equations $\left(1\right)$.
Earlier, we showed that the bracket in $\left(6\right)$ equals $2 g_{ik} \left( \ddot q^k + \Gamma^k_{ij} \dot q^i \dot q^j \right)$, which is $2 g_{ik} a^k$:
\[\frac{d}{dt} \frac{\partial L}{\partial \dot q^i} - \frac{\partial L}{\partial q^i} = 2 g_{ik}\, a^k .\]Substituting into $\left(6\right)$,
\[\delta E_q [ \delta q ] = - 2 \int_{t_0}^{t_1} g_{ik}\, a^k \delta q^i \, dt = -2 \int_{t_0}^{t_1} g(\bfa, \delta \bfq) \ dt .\]The integrand is the Riemannian metric applied to $-2 \bfa$ and $\delta q$. In particular, $\delta E_q [ \delta q ] = 0$ for every $\delta q$ exactly when $\bfa = 0$. More than that, $-2 \bfa$ is the gradient of $E$ at $q$, so $\bfa$ points in the direction of steepest energy descent, and its size measures how far $q$ is from being a critical point of $E$. This gives $\bfa$ a variational meaning.