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Christoffel symbols in flat spacetime

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  1. Aug 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a particle moving through Minkowski space with worldline [itex]x^\mu(\lambda)[/itex]. Here [itex]\lambda[/itex] is a continuous parameter which labels different points on the worldline and [itex]x^\mu = (t,x,y,z)[/itex] denotes the usual Cartesian coordinates. We will denote [itex]\partial/\partial \lambda[/itex] by a dot. In this problem we will assume that the trajectory of the particle obeys the equation of motion [itex]\ddot{x}^\mu = 0[/itex].

    (a) Show that this trajectory describes a particle moving at constant velocity.
    (b) Show that this trajectory is a local minimum of the action
    [tex]
    S = \int ds = \int d\lambda\,\sqrt{\eta_{\mu\nu} \dot{x}^\mu \dot{x}^\nu}
    [/tex]
    (c) Consider a new coordinate system [itex]x^{\mu'}[/itex] which differs from the original Cartesian coordinate system; as before, the Cartesian coordinates [itex]x^\mu[/itex] can be written as a function of these new coordinates [itex]x^\mu = x^\mu(x^{\mu'})[/itex]. Show that the equation of motion can be written in these new [itex]x^{\mu'}[/itex] coordinates as
    [tex]
    \ddot{x}^{\mu'} + \Gamma_{\nu'\lambda'}^{\mu'}\dot{x}^{\nu'}\dot{x}^{\lambda'} = 0
    [/tex]
    for some [itex]\Gamma^{\mu'}_{\nu'\lambda'}[/itex] which you must compute; [itex]\Gamma^{\mu'}_{\nu'\lambda'}[/itex] 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 *)


    2. Relevant equations
    Jacobian matrix:
    [tex]
    J_{\beta}^{\alpha'} = \frac{\partial x^{\alpha'}}{\partial x^{\beta}}
    [/tex]

    Derivaitves:
    [itex]\dot{x}^{\mu'} = J_{\mu}^{\mu'} \dot{x}^\mu[/itex]
    [itex]\dot{x}^{\mu} = J_{\mu'}^{\mu} \dot{x}^{\mu'}[/itex]


    Notation:
    [itex]\partial_\mu = \partial/\partial x^\mu[/itex]


    3. 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
    [tex]
    \dot{x}^{\mu'} = J_{\mu}^{\mu'}\dot{x}^\mu\:,
    \qquad
    \dot{x}^\nu = J_{\nu'}^{\nu} \dot{x}^{\nu'}.
    [/tex]

    Differentiating the left equation of with respect to [itex]\lambda[/itex] then gives
    [tex]
    \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.
    [/tex]
    But since [itex]\ddot{x}^\mu = 0[/itex], this simplifies to
    [tex]
    \ddot{x}^{\mu'} = \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)\,\dot{x}^\mu.
    [/tex]
    We can compute the derivative of the Jacobian by swapping the order of derivatives as
    [tex]
    \frac{d}{d\lambda}\big(J_{\mu}^{\mu'}\big)
    =
    \frac{d}{d\lambda}\Big(\partial_\mu x^{\mu'}\Big)
    = \partial_{\mu}\dot{x}^{\mu'}.
    [/tex]
    Thus we have
    [tex]
    \ddot{x}^{\mu'} = \big(\partial_\mu \dot{x}^{\mu'}\big)\dot{x}^\mu.
    [/tex]
    Since we can write [itex]\dot{x}^{\mu'} = J_{\nu}^{\mu'}\dot{x}^\nu[/itex] and [itex]\dot{x}^\mu = J_{\nu'}^{\mu}\dot{x}^{\nu'}[/itex], we can write the equation above as
    [tex]
    \ddot{x}^{\mu'} = \partial_\mu\big(J^{\mu'}_{\nu}\dot{x}^\nu\big)J_{\nu'}^{\mu}\dot{x}^{\nu'}.
    [/tex]
    Writing [itex]\dot{x}^\nu = J^{\nu}_{\lambda'}\dot{x}^{\lambda'}[/itex], this equation becomes
    [tex]
    \ddot{x}^{\mu'} = \partial_\mu\big(J^{\mu'}_{\nu}J^{\nu}_{\lambda'}\dot{x}^{\lambda'}\big)J_{\nu'}^{\mu}\dot{x}^{\nu'}.
    [/tex]
    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.
     
    Last edited: Aug 26, 2014
  2. jcsd
  3. Aug 25, 2014 #2
    The rest of the problem is simple enough, but I'm stuck on part (c).
     
  4. Aug 25, 2014 #3
    Forgot to mention [itex]\eta_{\mu\nu}[/itex] is the metric of flat spacetime in cartesian coordinates, with signature - + + +.
     
  5. Aug 26, 2014 #4

    Orodruin

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    Staff Emeritus
    Science Advisor
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    Hint: The chain rule
    $$
    \frac d{d\lambda} = \frac{d x^\mu}{d\lambda} \partial_\mu
    $$
     
  6. Aug 26, 2014 #5
    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 [itex]\ddot{x}^\mu = 0[/itex], 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}
     
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