# How can one define ordinary derivative of a vector field along a curve?

Definition 1: A curve $C$ on a manifold $M$ is a smooth function $C:(a,b) \rightarrow M$.

Definition 2: A vector field $v^a=v^a(t)$ along a curve $C$ is an assignment of vectors on the tangent space at each point $C(t)$ on the curve.

Definition 3: Let $v^b$ be a vector field on $M$. The derivative operator $\partial_a v^b$ is defined by taking partial derivative at each component of $v^b$, given that a fixed coordinate system is chosen.

Definition 4: $v^a$ is said to be parallelly transported along the curve $C$ if $t^a \nabla_a v^b=0$.

In Wald's General Relativity, the derivative operator $\nabla_a$ is defined on vector field on the manifold, but still the above is valid by defining similar thing for vector fields along a curve.

Theorem: Suppose a coordinate system is fixed. The following are equivalent.

(1) A vector $v^a$ is parallelly transported along a curve $C$.

(2) $t^a\partial_a v_b+t^a {\Gamma^c}_{ab} v_c=0$ [Tensor form]

(3) $\displaystyle\frac{dv^\nu}{dt}+\sum_{\mu,\lambda=1}^n t^\mu {\Gamma^\lambda}_{\mu\nu}v_\lambda=0$ [Component form]

This confuses me a lot. In particular, we are unable to define the derivative operator on vector field along a curve in the manner in definition 3. To make sense of taking partial derivatives at each component, formally speaking, it is defined as partial derivatives of the composition of the components and the parametrization of the manifold (which makes sense because the components themselves are formally functions from $M$ to $\mathbb{R}$).

However, the curve may have self-intersection and two different tangent vectors at the same point might be defined for a vector field along a curve, which means the components now cannot be a function from $M$ to $\mathbb{R}$. So we cannot make composition with parametrizations now in order to allow us to define partial derivatives. Therefore, I am unable to understand the meaning of statement (2) in the above theorem.

Can anyone give me some clue?

• Would Mathematics be a better home for this question? – Qmechanic Aug 26 '18 at 21:12
• @Qmechanic: I don’t think so: is about math tools used in general relativity, which is physics. – Frédéric Grosshans Aug 29 '18 at 16:26

You can formalize it as such:

Given a "derivative operator" $\nabla$ (I will refer to it as a connection), and a curve $\gamma:\mathbb R\rightarrow M,\ \lambda\mapsto \gamma(\lambda)$, there exists a unique map $$\frac{D}{d\lambda}:\Gamma(TM,\gamma)\rightarrow\Gamma(TM,\gamma),\ v\mapsto \frac{Dv}{d\lambda},$$ mapping vector fields along $\gamma$ to vector fields along $\gamma$ (the notation $\Gamma(TM,\gamma)$ refers to sections of $TM$ along $\gamma$, smooth maps of the form $v:\mathbb R\rightarrow TM$ such that $\pi\circ v=\gamma$ with $\pi$ being the projection to the base - vector fields along a curve essentially) satisfying the following.

• Linearity: $D/d\lambda$ is $\mathbb R$-linear.

• Product rule: For any $\alpha\in C^\infty(\mathbb R)$, we have $$\frac{D}{d\lambda}(\alpha v)=\frac{d\alpha}{d\lambda}v+\alpha\frac{Dv}{d\lambda}.$$

• Consistency: Let $V\in\Gamma(TM)$ be any smooth vector field, let $t=\frac{d}{d\lambda}$ be the tangent vector field of the curve $\gamma$. We then define $v=V\circ\gamma$, with $v$ being a vector field along $\gamma$ now. Then $$\frac{Dv}{d\lambda}=(\nabla_tV)\circ\gamma,$$ where $\nabla_t$ is in Wald's notation $t^a\nabla_a$ (sorry for breaking the index notation, but it is quite counterintuitive here).

This may be proven quite easily, and is often proven in differential geometry books.

Physics books often (and as far as I remember, Wald's is not really an exception to this most of the time) omit any discussions on locality, and even when not, things are often not proven explicitly.

The point is that one can show that if the $D/d\lambda$ operator exists, then it is local in the sense that let $\lambda_0$ be a point in the domain of $\gamma$, and let $v$ and $\tilde{v}$ be two vector fields along $\gamma$ that agree on an open interval around $\lambda_0$ (say, $(\lambda_0-\epsilon,\lambda_0+\epsilon)$), then we have $$\frac{Dv}{d\lambda}(\lambda_0)=\frac{D\tilde v}{d\lambda}(\lambda_0).$$

However any smooth curve is going to be non-selfintersecting when restricted to near a point. Then, by locality, you can construct $Dv/dt(\lambda_0)$ as $(\nabla_t V)(\gamma(\lambda_0))$ and extend it to all points $\lambda$.

Wald was being somewhat handwave-y here, and he just skipped this construction.

EDIT: Rereading my post, I realize I haven't been the most clear, especially near the end, so I'll say some additional things.

1)

The validity of the expression in the (3) statement of your theorem is a direct consequence of my $D/d\lambda$. We can use locality to calculate things in basis expansions. Let $\partial_\mu$ denote the coordinate basis fields, and let $e_\mu=\partial_\mu\circ\gamma$ be the corresponding vector fields along $\gamma$. Due to locality and the above definition, we have for any $v$ along $\gamma$: $$\frac{D}{d\lambda}v=\frac{D}{d\lambda}(v^\mu e_\mu)=\frac{dv^\mu}{d\lambda}e_\mu+v^\mu\frac{D}{d\lambda}e_\mu=\frac{d v^\mu}{d\lambda}e_\mu+v^\mu(\nabla_t\partial_\mu)\circ\gamma=\frac{dv^\mu}{d\lambda}{e_\mu}+v^\mu(\Gamma^\nu_{\sigma\mu}\circ\gamma)t^\sigma(\partial_\nu\circ\gamma)=\left(\frac{dv^\mu}{d\lambda}+t^\nu(\Gamma^\mu_{\nu\sigma}\circ\gamma)v^\sigma\right)e_\mu.$$

2)

Strictly speaking, the expression (2) $$\partial_a v^b+\Gamma^b_{ac}v^c$$ is not valid for arbitrary $v$ along $\gamma$, due to the above mentioned self-intersections. However, the equation for parallel transport is a local differential equation, hence, you can consider it restricted to a sufficiently small open interval in the curve's domain.

Then the curve will no longer be self-intersecting, and one may show that a (non-unique) open extension of $v$ exists (however the result will be extension-independent). For that open extension it is valid that $$\frac{Dv^a}{d\lambda}=t^b\partial_b v^a+t^b\Gamma^a_{bc}v^c.$$

Statement (2) is the definition of the directional covariant derivative, i.e. the covariant derivative of a vector along a curve, which vanishes in case of parallel transport. The vector in (2) is represented by its covariant components (dual vector). But a minus sign is missing before the second term (because of the covariant components).

Statement (3) in the intentions would be exactly the same, but there is an inconsistency in the index $\nu$ that in the first term is up, while in the second term is down. If we follow (2), you should lower the $\nu$ in the first term and post a minus sign before the second term. The $\sum$ symbol is not necessary as the Einstein convention on the indices is in place.

Note that the first term in (2) and the first term in (3) are the same, in fact:
$\dot x^\mu \partial_\mu A^\nu = dA^\nu / d\lambda$
where:
$x^\mu = x^\mu (\lambda)$ curve parameterized by $\lambda$
$\dot x^\mu = dx^\mu / d\lambda$ tangent to the curve
$\lambda$ affine parameter

I believe you are misreading statement (2) of the theorem. $v^b$ are the components of a vector field on $M$, and are therefore maps from $M$ to $\mathbb{R}$. In particular, we are not talking about a vector field along a curve, so your objection does not apply and we can differentiate freely.

It is $t^a$ which are the components of a vector field along a curve - presumably the tangent vectors to the curve. These are not being differentiated with respect to the coordinates, so we have no problem.

• But there is something like parallel tranporting a vector along a closed curve will not return to the same vector, so if you do not allow two different vectors being defined at the same place then the discussion as I just mention do not make any sense. – Jerry Aug 27 '18 at 11:35
• @Jerry A vector field $X$ is said to be parallel to a curve $C:(a,b)\rightarrow M$ if the covariant derivative of $X$ along the tangent vector to $C$ vanishes for all $t\in (a,b)$. There's no reason to think that, given an arbitrary curve $C$, a global vector field which is parallel to $C$ exists (and in general, it does not, as you point out). However, one can always consider locally defined vector fields which don't cover the whole manifold. – J. Murray Aug 27 '18 at 15:59
• Uldreth's answer goes into this in more detail. – J. Murray Aug 27 '18 at 16:05