- #1
- 10,397
- 1,577
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".
\Gamma^x{}_{tt}, \Gamma^y{}_{tt}, and \Gamma^z{}_{tt}[/tex] can be nonzero, the rest must be "small".<br /> <br /> We will assume this to be the case, and check to see if it's sufficient for starters.<br /> <br /> 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:<br /> <br /> Wiki <a href="http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=551829920" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=551829920</a><br /> <br /> writes this as:<br /> <br /> \overline{\Gamma^k{}_{ij}} =<br /> \frac{\partial x^p}{\partial y^i}\,<br /> \frac{\partial x^q}{\partial y^j}\,<br /> \Gamma^r{}_{pq}\,<br /> \frac{\partial y^k}{\partial x^r}<br /> + <br /> \frac{\partial y^k}{\partial x^m}\, <br /> \frac{\partial^2 x^m}{\partial y^i \partial y^j}<br /> <br /> We can substitute in \partial y^i = \partial y^j = \partial t, since i and j must be equal to the index of t (usually zero) by our assumption.<br /> <br /> \overline{\Gamma^k{}_{ij}} =<br /> \frac{\partial x^p}{\partial t}\,<br /> \frac{\partial x^q}{\partial t}\,<br /> \Gamma^r{}_{pq}\,<br /> \frac{\partial y^k}{\partial x^r}<br /> + <br /> \frac{\partial y^k}{\partial x^m}\, <br /> \frac{\partial^2 x^m}{\partial t^2}<br /> <br /> We see that we need \frac{\partial^2 x^m}{\partial t^2} to be zero - this is true if we transform between inertial frames.<br /> <br /> This then leaves us with basically the basic tensor transformation law, except that forces transform quadratically with time.<br /> <br /> 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.<br /> <br /> 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.<br /> <br /> 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.<br /> <br /> If anyone has seen anything similar to this in the literature, I'd be interested in hearing about it.
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".
\Gamma^x{}_{tt}, \Gamma^y{}_{tt}, and \Gamma^z{}_{tt}[/tex] can be nonzero, the rest must be "small".<br /> <br /> We will assume this to be the case, and check to see if it's sufficient for starters.<br /> <br /> 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:<br /> <br /> Wiki <a href="http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=551829920" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=551829920</a><br /> <br /> writes this as:<br /> <br /> \overline{\Gamma^k{}_{ij}} =<br /> \frac{\partial x^p}{\partial y^i}\,<br /> \frac{\partial x^q}{\partial y^j}\,<br /> \Gamma^r{}_{pq}\,<br /> \frac{\partial y^k}{\partial x^r}<br /> + <br /> \frac{\partial y^k}{\partial x^m}\, <br /> \frac{\partial^2 x^m}{\partial y^i \partial y^j}<br /> <br /> We can substitute in \partial y^i = \partial y^j = \partial t, since i and j must be equal to the index of t (usually zero) by our assumption.<br /> <br /> \overline{\Gamma^k{}_{ij}} =<br /> \frac{\partial x^p}{\partial t}\,<br /> \frac{\partial x^q}{\partial t}\,<br /> \Gamma^r{}_{pq}\,<br /> \frac{\partial y^k}{\partial x^r}<br /> + <br /> \frac{\partial y^k}{\partial x^m}\, <br /> \frac{\partial^2 x^m}{\partial t^2}<br /> <br /> We see that we need \frac{\partial^2 x^m}{\partial t^2} to be zero - this is true if we transform between inertial frames.<br /> <br /> This then leaves us with basically the basic tensor transformation law, except that forces transform quadratically with time.<br /> <br /> 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.<br /> <br /> 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.<br /> <br /> 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.<br /> <br /> If anyone has seen anything similar to this in the literature, I'd be interested in hearing about it.
