Christoffel symbols in flat spacetime

[homer]

Homework Statement

Consider a particle moving through Minkowski space with worldline $x^\mu(\lambda)$. Here $\lambda$ is a continuous parameter which labels different points on the worldline and $x^\mu = (t,x,y,z)$ denotes the usual Cartesian coordinates. We will denote $\partial/\partial \lambda$ by a dot. In this problem we will assume that the trajectory of the particle obeys the equation of motion $\ddot{x}^\mu = 0$.

(a) Show that this trajectory describes a particle moving at constant velocity.
(b) Show that this trajectory is a local minimum of the action
$$S = \int ds = \int d\lambda\,\sqrt{\eta_{\mu\nu} \dot{x}^\mu \dot{x}^\nu}$$
(c) Consider a new coordinate system $x^{\mu'}$ which differs from the original Cartesian coordinate system; as before, the Cartesian coordinates $x^\mu$ can be written as a function of these new coordinates $x^\mu = x^\mu(x^{\mu'})$. Show that the equation of motion can be written in these new $x^{\mu'}$ coordinates as
$$\ddot{x}^{\mu'} + \Gamma_{\nu'\lambda'}^{\mu'}\dot{x}^{\nu'}\dot{x}^{\lambda'} = 0$$
for some $\Gamma^{\mu'}_{\nu'\lambda'}$ which you must compute; $\Gamma^{\mu'}_{\nu'\lambda'}$ is known as the Christoffel symbol. These extra Christoffel terms in the equation of motion can be thought of as "fictitious" forces that arise in an accelerated reference frame.

(* I only need help with part c *)

Homework Equations

Jacobian matrix:
$$J_{\beta}^{\alpha'} = \frac{\partial x^{\alpha'}}{\partial x^{\beta}}$$

Derivaitves:
$\dot{x}^{\mu'} = J_{\mu}^{\mu'} \dot{x}^\mu$
$\dot{x}^{\mu} = J_{\mu'}^{\mu} \dot{x}^{\mu'}$Notation:
$\partial_\mu = \partial/\partial x^\mu$

The Attempt at a Solution

I feel like I'm just spinning my wheels on this problem, and don't know where to go with it. This is from PHYS 514: General Relativity at McGill. Since I'm not actually taking this class I have no graders nor TA's ask when I get stuck as I learn how to do summation convention calculations. We haven't introduced Christoffel symbols yet in the class videos for the week of this assignment, so I assume we should only find them by deriving the equation of motion in the primed coordinate system. This is what I have come up with so far, but I have no idea if I made an error because this is my first time doing these kind of calculations (in this notation I mean).

Recall we earlier showed that
$$\dot{x}^{\mu'} = J_{\mu}^{\mu'}\dot{x}^\mu\:, \qquad \dot{x}^\nu = J_{\nu'}^{\nu} \dot{x}^{\nu'}.$$

Differentiating the left equation of with respect to $\lambda$ then gives
$$\ddot{x}^{\mu'} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\dot{x}^\mu\big) = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)\,\dot{x}^\mu + J_{\mu}^{\mu'}\ddot{x}^\mu.$$
But since $\ddot{x}^\mu = 0$, this simplifies to
$$\ddot{x}^{\mu'} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)\,\dot{x}^\mu.$$
We can compute the derivative of the Jacobian by swapping the order of derivatives as
$$\frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big) = \frac{d}{d\lambda}\Big(\partial_\mu x^{\mu'}\Big) = \partial_{\mu}\dot{x}^{\mu'}.$$
Thus we have
$$\ddot{x}^{\mu'} = \big(\partial_\mu \dot{x}^{\mu'}\big)\dot{x}^\mu.$$
Since we can write $\dot{x}^{\mu'} = J_{\nu}^{\mu'}\dot{x}^\nu$ and $\dot{x}^\mu = J_{\nu'}^{\mu}\dot{x}^{\nu'}$, we can write the equation above as
$$\ddot{x}^{\mu'} = \partial_\mu\big(J^{\mu'}_{\nu}\dot{x}^\nu\big)J_{\nu'}^{\mu}\dot{x}^{\nu'}.$$
Writing $\dot{x}^\nu = J^{\nu}_{\lambda'}\dot{x}^{\lambda'}$, this equation becomes
$$\ddot{x}^{\mu'} = \partial_\mu\big(J^{\mu'}_{\nu}J^{\nu}_{\lambda'}\dot{x}^{\lambda'}\big)J_{\nu'}^{\mu}\dot{x}^{\nu'}.$$
Applying the product rule for differentiation, we thus find
\begin{align*}
\ddot{x}^{\mu'}
& = \Big(
\partial_\mu\big(J^{\mu'}_{\nu}J^{\nu}_{\lambda'}\big)\dot{x}^{\lambda'} +
J_{\nu}^{\mu'}J^{\nu}_{\lambda'}\partial_\mu \dot{x}^{\lambda'}
\Big)J_{\nu'}^{\mu}\dot{x}^{\nu'} \\
& =
J_{\nu'}^{\mu}\partial_\mu\big(
J^{\mu'}_{\nu}J^{\nu}_{\lambda'}
\big)\dot{x}^{\lambda'}\dot{x}^{\nu'} +
J_{\nu'}^{\mu}J_{\nu}^{\mu'}J^{\nu}_{\lambda'}
\big(\partial_\mu \dot{x}^{\lambda'}\big)\dot{x}^{\nu'}.
\end{align*}

AND HERE IS WHERE I'M STUCK

Any help would be greatly appreciated, as this is a somewhat daunting subject to go it alone.

 

[homer]

The rest of the problem is simple enough, but I'm stuck on part (c).

 

[homer]

Forgot to mention $\eta_{\mu\nu}$ is the metric of flat spacetime in cartesian coordinates, with signature - + + +.

 

[Orodruin]

Hint: The chain rule
$$
\frac d{d\lambda} = \frac{d x^\mu}{d\lambda} \partial_\mu
$$

 

[homer]

Hint: The chain rule
$$
\frac d{d\lambda} = \frac{d x^\mu}{d\lambda} \partial_\mu
$$

ARGGGGHHHHH! Flubbing the chain rule! Thank you so much for seeing through my BS argument! OK, so the answer should be:

Recall we earlier showed that
\begin{equation}
\dot{x}^{\mu'} = J_{\mu}^{\mu'}\dot{x}^\mu.
\end{equation}
Differentiating again, we get
\begin{align*}
\ddot{x}^{\mu'}
& = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)\dot{x}^\mu + J_{\mu}^{\mu'}\ddot{x}^\mu.
\end{align*}
Since $\ddot{x}^\mu = 0$, this equation becomes
\begin{equation}
\ddot{x}^{\mu'} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)\dot{x}^\mu.
\end{equation}
From the chain rule we know that
\begin{equation}
\frac{d}{d\lambda}
= \frac{d x^\nu}{d\lambda}\frac{\partial}{\partial x^\nu} = \dot{x}^\nu \partial_\nu,
\end{equation}
so our equation of motion simplifies to
\begin{equation}
\ddot{x}^{\mu'} = \dot{x}^\nu\big(\partial_\nu J_{\mu}^{\mu'}\big)\dot{x}^\mu.
\end{equation}
But we also know that
\begin{equation}
\dot{x}^\nu = J_{\nu'}^{\nu}\dot{x}^{\nu'}, \qquad \dot{x}^mu = J_{\lambda'}^{\mu}\dot{x}^{\lambda'},
\end{equation}
so our equation of motion becomes
\begin{equation}
\ddot{x}^{\mu'} =
J_{\nu'}^{\nu}\dot{x}^{\nu'}\big(\partial_\nu J_{\mu}^{\mu'}\big)J_{\lambda'}^{\mu}\dot{x}^{\lambda'} =
J_{\nu'}^{\nu}J_{\lambda'}^{\mu}\big(\partial_\nu J_{\mu}^{\mu'}\big)\dot{x}^{\nu'}\dot{x}^{\lambda'}.
\end{equation}
Moving everything to the left hand side, our equation of motion is then
\begin{equation}
\ddot{x}^{\mu'} -
J_{\nu'}^{\nu}J_{\lambda'}^{\mu}\big(\partial_\nu J_{\mu}^{\mu'}\big)\dot{x}^{\nu'}\dot{x}^{\lambda'} = 0.
\end{equation}
Thus by inspection we find the Christoffel symbols are given by
\begin{equation}
\Gamma_{\nu'\lambda'}^{\mu'} = -J_{\nu'}^{\nu}J_{\lambda'}^{\mu}\partial_\nu J_{\mu}^{\mu'}.
\end{equation}