Christoffel symbols vanishing on a curve

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Hello!

Here and there I find that it is possible to make the Christoffel symbols vanish on a curve (e.g. lecture http://www.phys.uu.nl/~thooft/lectures/genrel_2010.pdf" [Broken] by 't Hooft).

The transformation law of the Christoffel symbols is relevant in this case:

[tex]\tilde{\Gamma}_{\kappa\lambda}^{\nu}(u(x))=u_{\:,\mu}^{\nu}\left[x_{\:,\kappa}^{\alpha}x_{\:,\lambda}^{\beta}\Gamma_{\alpha\beta}^{\mu}(x)+x_{\:,\kappa,\lambda}^{\mu}\right].[/tex]

We can fine-tune the second term in the square parentheses to cancel the first term and then [tex]\tilde{\Gamma}[/tex] vanishes. But how does one see it is only possible to do that only on a curve and not the whole space? What about more (or less) than four dimensions?

Thanks!
 
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  • #2
bcrowell
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There is no notion of intrinsic curvature in one dimension. Curvature measures the failure of parallelism, and there is no notion of parallelism in one dimension.
 
  • #3
nicksauce
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But how does one see it is only possible to do that only on a curve and not the whole space? What about more (or less) than four dimensions?
One way to think about it: Since the Riemann tensor is composed of purely the Christoffel symbols and its derivatives, if the Christoffel symbols vanished over the whole manifold, then so would the Riemann tensor. But we know that is only true in flat space.
 
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haushofer
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There is no notion of intrinsic curvature in one dimension. Curvature measures the failure of parallelism, and there is no notion of parallelism in one dimension.
But that's confusing; the holonomy is measured via closed curves.

To be honoust, I would have to think quite hard to understand why one can always choose coordinates in such a way that the connection disappears along a given curve.

But as I see it why exactly one is able to choose these kind of coordinates (in which the connection disappears, but not its derivative) only for points and curves, is because in dimension 2 and higher one can define curvature, whereas one-dimensional manifolds are always "flat" (vanishing Riemann tensor). Is that right?

To elaborate on that: that one can do it for a point is clear from the equivalence principe, but how can one understand that apparently one can always eliminate gravity locally also along a trajectory in spacetime?
 
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To elaborate on that: that one can do it for a point is clear from the equivalence principe, but how can one understand that apparently one can always eliminate gravity locally also along a trajectory in spacetime?
Hmm..beats me. But as long as you put it in terms of eliminating gravity--and I'm just guessing--the family of curves might consist of geodesics

[tex] \frac{d^2 x^\mu}{du^2} + \Gamma ^\mu _{\pi\rho} \frac{dx ^\pi}{du} \frac{dx^\rho}{du} =0[/tex]

where the second derivatives of x vanish.
 
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There's a reasonable chance that an inuitive answer can be derived from Leonard Susskind's youtube lecture # 5 here:

http://www.youtube.com/watch?v=WtPtxz3ef8U&p=6C8BDEEBA6BDC78D


where he derives the Christoffel symbol and provides some geometric insights while doing so.

Depending on your math skills, starting about minute 30 or later might get the job done.


I"m still just listening to his series of lectures for concepts before deciding if I really want to go back and take notes so I can understand and remember details. As a wild guess, I'd think the answer to the poster's question is related to the fact the Christoffel symbols are NOT tensors..so they do NOT transform as tensors.
 
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bcrowell
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bcrowell said:
There is no notion of intrinsic curvature in one dimension. Curvature measures the failure of parallelism, and there is no notion of parallelism in one dimension.
But that's confusing; the holonomy is measured via closed curves.
Sorry, I misunderstood your question.

I think what you want is this: http://en.wikipedia.org/wiki/Fermi-Walker_transport

Fermi-Walker transport is a generalization of Fermi transport. In Fermi transport, you basically just let a set of gyroscopes free-fall, and they define a set of coordinates such that the Christoffel symbols vanish along the geodesic trajectory of the gyroscopes. Fermi-Walker transport is a generalization to non-geodesic curves.
 
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haushoffer said:
... that one can do it for a point is clear from the equivalence principe, but how can one understand that apparently one can always eliminate gravity locally also along a trajectory in spacetime?
Is it because the coordinates constructed around the point ( a vector space ?) are transported isometrically by parallel transport ?

One thing that's not clear to me - along a curve the connections are the Ricci rotation coefficients, which can be transformed to the holonomic coords to give Christoffel symbols. Does this make a one-to-one correspondence between the RRCs and the CS ?
 
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bcrowell
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... that one can do it for a point is clear from the equivalence principe, but how can one understand that apparently one can always eliminate gravity locally also along a trajectory in spacetime?
Is it because the coordinates constructed around the point ( a vector space ?) are transported isometrically by parallel transport ?
I doubt that the explanation is this simple. Fermi published his paper on Fermi transport in 1922, and it wasn't until 1932 that Walker generalized it to non-geodesic curves. So it seems to have taken smart physicists at least a decade to figure it out.
 

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