Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Recovering forces from Christoffel symbols

  1. Apr 24, 2013 #1


    User Avatar
    Staff Emeritus
    Science Advisor

    I don't think I've ever seen this discussed in a textbook, this is an attempt to throw some light on the connection between Christoffel symbols and forces.

    In particular I want to derive the later as an approximation of the former, with some limitations on choices of coordinate systems (inertial frames).

    I'll start off with assuming a cartesian coordinate system (t,x,y,z).
    Next, we will add some "flatness" assumptions. Basically, to reduce the rank 3 Christoffel symbols to a rank 1 vector, we need the majority of them to be zero, or at least "small".

    [itex]\Gamma^x{}_{tt}, \Gamma^y{}_{tt}, and \Gamma^z{}_{tt}[/tex] can be nonzero, the rest must be "small".

    We will assume this to be the case, and check to see if it's sufficient for starters.

    Now that we have the right rank, we still need to worry about the fact that Christoffel symbols aren't tensors. So we take a close look at the transformation law:

    Wiki http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=551829920

    writes this as:

    [tex]\overline{\Gamma^k{}_{ij}} =
    \frac{\partial x^p}{\partial y^i}\,
    \frac{\partial x^q}{\partial y^j}\,
    \frac{\partial y^k}{\partial x^r}
    \frac{\partial y^k}{\partial x^m}\,
    \frac{\partial^2 x^m}{\partial y^i \partial y^j} [/tex]

    We can substitute in [itex]\partial y^i = \partial y^j = \partial t[/itex], since i and j must be equal to the index of t (usually zero) by our assumption.

    [tex]\overline{\Gamma^k{}_{ij}} =
    \frac{\partial x^p}{\partial t}\,
    \frac{\partial x^q}{\partial t}\,
    \frac{\partial y^k}{\partial x^r}
    \frac{\partial y^k}{\partial x^m}\,
    \frac{\partial^2 x^m}{\partial t^2} [/tex]

    We see that we need [itex] \frac{\partial^2 x^m}{\partial t^2} [/itex] to be zero - this is true if we transform between inertial frames.

    This then leaves us with basically the basic tensor transformation law, except that forces transform quadratically with time.

    This is a bit troublesome, apparently we can't do arbitrary time transformations t->t' if we wish to use forces. This makes sense, though, intiutively - it means as part of choosing an "inertial frame" we're stucking with choosing a regularly ticking clock if we wish to use forces. Then this factor becomes a simple scale factor.

    I think that's pretty much the basics, though due to the limitations on time transformation, some extra attention should be paid to Lorentz transforms to demonstrate why four-fources work relativistically.

    I feel there are some things lacking in this approach - it doesn't quite demonstrate that Lorentz transforms work, nor does it demonstrate, really, why you can use forces in polar coordinates. But I think it's a start.

    If anyone has seen anything similar to this in the literature, I'd be interested in hearing about it.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted