Question about the the velocity and acceleration in tensor notation

Question

When computing the volicty of a particle moving along a curve parametrized by $Z^i(t)$ for each component i, the components of the velocity $V^i$ are given by $$V^i = (d/dt)Z^i$$ and the components fo the acceleration are given by $$A^i=(d/dt)V^i + \Gamma^i_{jk} V^j V^k.$$

My question is: why the derivative of the basis vectors doesn't appear in the expression for the velocity? Because the for the Christoffel symbol to appear in the acceleration expression there has to be a derivative in respect to the basis vectors. What am I missing here? Any help will be appreciated.

Answer 1

Definition of directional covariant derivative
The directional covariant derivative of a vector along a curve in a manifold is defined as
$D/d\sigma P^\mu$
where:
$D/d$ directional covariant derivative
$\gamma (\sigma)$ curve
$\sigma$ proper time (massive particle) or affine parameter (massless particle, e.g. photon)
$\mu = 0, 1, 2, 3$
$P^\mu$ vector
If $x^\mu(\sigma)$ describes the curve, it may be written as
$D/d\sigma P^\mu = \dot x^\nu \nabla_\nu P^\mu$ (1)
where:
$\dot x^\mu = dx^\mu/d\sigma$
$\nabla_\mu$ covariant derivative
In case of a vector, you have
$\nabla_\nu P^\mu = \partial_\nu P^\mu + \Gamma^\mu_{\nu \lambda} P^\lambda$

Velocity
If the vector represents the position, that is $P^\mu = x^\mu$, (1) measures the velocity $V^\mu$ of the particle
$V^\mu = D/d\sigma x^\mu = \dot x^\nu \nabla_\nu x^\mu = \dot x^\nu \partial_\nu x^\mu + \dot x^\nu \Gamma^\mu_{\nu \lambda} x^\lambda$ (2)

Acceleration
As for the acceleration $A^\mu$ you apply the (1) once again to the velocity $V^\mu$ in (2)
$A^\mu = D/d\sigma V^\mu = \dot x^\nu \nabla_\nu V^\mu$

Note
The covariant derivative already embeds the change of the basis vectors along a curve in a manifold to describe correctly the change of the geometric object. This is accounted for by the connection $\Gamma$.
The formulas in the question do not seem correct.

Answer 2

This answer is based on Pavel Grinfeld's Youtube lecture.

The trajectory is parameterized with respect to time as a given. That is, the contravariant bases are expressed as a function of time $Z^i \equiv Z^i(t)$. The velocity and acceleration vectors are the first and second derivatives of the position vector ($\mathbf{R}$) as a function of time. Let $$\mathbf{R}(t) \equiv R(Z^i(t))$$ To find the velocity vector, we differentiate $\mathbf{R}$ once with respect to time: $$\mathbf{V} = \frac{d\mathbf{R}}{dt} = \frac{dR(Z^i(t))}{dt}$$ Applying the chain rule, we have $$\mathbf{V} = \frac{\partial \mathbf{R}}{\partial Z^i}\frac{d Z^i}{d t}$$ By definition, $\frac{\partial \mathbf{R}}{\partial Z^i}$ is the covariant basis $Z_i$. This makes $\frac{dZ^i}{dt}$ the contravariant component of the velocity vector $\mathbf{V}$. $$\mathbf{V} = V^i Z_i$$ $$V^i(t) = \frac{dZ^i}{dt}$$

To find the acceleration, we need to differentiate the velocity vector once with respect to time. Using the product rule, we have $$\mathbf{A}(t) = \frac{dV^i}{dt} Z_i + V^i \frac{dZ_i(Z^i(t))}{dt}$$ Notice that we do not have the covariant bases expressed directly as a function of time, so we need to express them as a function of the contravariant bases which are a function of time (given). The derivative in the second term on the right side can be evaluated using the chain rule. $$\frac{dZ_i(Z^i(t))}{dt} = \frac{\partial Z_i}{\partial Z^j}\frac{dZ^j}{dt}$$ The derivative $\frac{\partial Z_i}{\partial Z^j}$ is $\Gamma^k_{ij}Z_k$, by definition. While the derivative$\frac{dZ^j}{dt}$ is the contravariant velocity component $V^j$ as shown in the first step. Plugging these back into the expression for the acceleration, we get $$\mathbf{A}(t) = \frac{dV^i}{dt} Z_i + V^i \Gamma^k_{ij}Z_k V^j$$ To factor out the covariant basis $Z_i$, indices are renamed in the second term to get $$\mathbf{A}(t) = \frac{dV^i}{dt} Z_i + V^k \Gamma^i_{kj}Z_i V^j$$ Finally, we can factor our $Z_i$ as follows $$\mathbf{A}(t) = \left(\frac{dV^i}{dt} + V^k \Gamma^i_{kj} V^j \right) Z_i$$ This yields the contravariant component of the acceleration vector $$A^i(t) = \frac{dV^i}{dt} + V^k \Gamma^i_{kj} V^j$$

Answer 3

“When computing the velocity of a particle moving along a curve“ I think perhaps the confusion arises from the ambiguity in the question: is the velocity of the particle itself is moving along the given curve or is the curve represents particle’s velocity (representing the change in its position ) ; does Z represent a scalar field or a vector field?