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
<br /> S = \int ds = \int d\lambda\,\sqrt{\eta_{\mu\nu} \dot{x}^\mu \dot{x}^\nu}<br />
(c) Consider a new coordinate system x^{\mu&#039;} 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&#039;}). Show that the equation of motion can be written in these new x^{\mu&#039;} coordinates as
<br /> \ddot{x}^{\mu&#039;} + \Gamma_{\nu&#039;\lambda&#039;}^{\mu&#039;}\dot{x}^{\nu&#039;}\dot{x}^{\lambda&#039;} = 0<br />
for some \Gamma^{\mu&#039;}_{\nu&#039;\lambda&#039;} which you must compute; \Gamma^{\mu&#039;}_{\nu&#039;\lambda&#039;} 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:
<br /> J_{\beta}^{\alpha&#039;} = \frac{\partial x^{\alpha&#039;}}{\partial x^{\beta}}<br />

Derivaitves:
\dot{x}^{\mu&#039;} = J_{\mu}^{\mu&#039;} \dot{x}^\mu
\dot{x}^{\mu} = J_{\mu&#039;}^{\mu} \dot{x}^{\mu&#039;}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
<br /> \dot{x}^{\mu&#039;} = J_{\mu}^{\mu&#039;}\dot{x}^\mu\:,<br /> \qquad<br /> \dot{x}^\nu = J_{\nu&#039;}^{\nu} \dot{x}^{\nu&#039;}.<br />

Differentiating the left equation of with respect to \lambda then gives
<br /> \ddot{x}^{\mu&#039;} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu&#039;}\dot{x}^\mu\big)<br /> = \frac{d}{d\lambda}\big(J_{\mu}^{\mu&#039;}\big)\,\dot{x}^\mu + J_{\mu}^{\mu&#039;}\ddot{x}^\mu.<br />
But since \ddot{x}^\mu = 0, this simplifies to
<br /> \ddot{x}^{\mu&#039;} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu&#039;}\big)\,\dot{x}^\mu.<br />
We can compute the derivative of the Jacobian by swapping the order of derivatives as
<br /> \frac{d}{d\lambda}\big(J_{\mu}^{\mu&#039;}\big)<br /> = <br /> \frac{d}{d\lambda}\Big(\partial_\mu x^{\mu&#039;}\Big)<br /> = \partial_{\mu}\dot{x}^{\mu&#039;}.<br />
Thus we have
<br /> \ddot{x}^{\mu&#039;} = \big(\partial_\mu \dot{x}^{\mu&#039;}\big)\dot{x}^\mu.<br />
Since we can write \dot{x}^{\mu&#039;} = J_{\nu}^{\mu&#039;}\dot{x}^\nu and \dot{x}^\mu = J_{\nu&#039;}^{\mu}\dot{x}^{\nu&#039;}, we can write the equation above as
<br /> \ddot{x}^{\mu&#039;} = \partial_\mu\big(J^{\mu&#039;}_{\nu}\dot{x}^\nu\big)J_{\nu&#039;}^{\mu}\dot{x}^{\nu&#039;}.<br />
Writing \dot{x}^\nu = J^{\nu}_{\lambda&#039;}\dot{x}^{\lambda&#039;}, this equation becomes
<br /> \ddot{x}^{\mu&#039;} = \partial_\mu\big(J^{\mu&#039;}_{\nu}J^{\nu}_{\lambda&#039;}\dot{x}^{\lambda&#039;}\big)J_{\nu&#039;}^{\mu}\dot{x}^{\nu&#039;}.<br />
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}