In general relativity, we can write the geodesic equation as a contraction $v^\alpha \nabla _\alpha v^\mu = f(\lambda)v^\mu$ along a path defined by coordinates $x^\mu(\lambda)$, and where $v^\mu = \frac{dx^\mu}{dt}$.

I understand how we can interpret the covariant derivative as the invariant version of partial derivative. But what I don't understand is the contraction with the velocity vector.

Nice thing about this formula is that if we think of it as sort of giving the "rate of change" along the curve. Then if we take lambda as proportional to the proper time measured by the particle from its starting point, we get $f(\lambda) = 0$ which is nice since the tangent velocity vector is "not changing" in a geodesic. This allows us to interpret the geodesic as sort of a straight path.

How can we interpret $v^\alpha \nabla _\alpha$ analogous to differentiation?

  • $\begingroup$ I've found an answer here that seems to answer my question but I will still leave the question open if one desires to answer in a different way physics.stackexchange.com/q/61640 $\endgroup$
    – Ron
    Commented Jul 1, 2016 at 12:57
  • $\begingroup$ Do you know what a directional derivative is? $\endgroup$
    – Ryan Unger
    Commented Jul 1, 2016 at 14:12
  • $\begingroup$ This is probably easy if you use more covariant notation. The covariant way to write this is $$ v^a \nabla_a v^\mu \partial_\mu = \nabla_vv $$ the RHS is clearly a directional derivative. $\endgroup$
    – zzz
    Commented Jul 3, 2016 at 17:01

1 Answer 1


It's just like a directional derivative in standard Cartesian coordinates.

In $\mathbb{R}^3$, if I have a function $f : \mathbb{R}^3 \rightarrow \mathbb{R}$, the gradient $\nabla$ gives me a map $\nabla f : \mathbb{R}^3 \rightarrow \mathbb{R}^3.$ For any vector $\vec{v} \in \mathbb{R}^3$, I can define a map $\vec{v} \cdot \nabla f : \mathbb{R}^3 \rightarrow \mathbb{R},$ which is called the directional derivative of $f$ with respect to the vector $\vec{v}.$ That is, it tells me how $f$ is changing as I move along the vector $\vec{f}.$

In curved space, there's no dot product. Instead, we have the metric, which gives us a way of contracting tensors. In general, moving from flat to curved spacetime you can often replace the dot product with an inner product with respect to the metric. That is, $\vec{v} \cdot \vec{w} \mapsto v^a w_a.$

So the natural way to generalized $\vec{v} \cdot \nabla f$ is $v^a \nabla_a f.$

Using the language of tensors, we can generalize to cases where $f$ has a higher rank than zero. If I replace $f$ by a vector $v^{a}$ for example, then the directional derivative of $v^a$ with respect to itself is given by $v^b \nabla_b v^a.$

So the geodesic equation $0 = v^b \nabla_b v^a$ indicates, as you suspected, that the tangent vector to a path does not change with respect to itself as you move along the path.


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