How can a metric connection have torsion?

[Matterwave]

Hi, I'm reading this wikipedia article on the metric connection, and it says that the Levi-Civita connection is a metric connection without torsion. If the metric connection is defined so that the covariant derivative of the metric is 0, how can there be torsion? Doesn't this condition force the connection to be the Christoffel symbols and therefore there is no torsion?

 

[TrickyDicky]

See http://en.wikipedia.org/wiki/Einstein–-Cartan_theory

 

[Matterwave]

I just realized that Einstein-Cartan theory must use a non-symmetric metric connection (if it is to at least preserve the equivalence principle), so apparently these things can exist.

Do we take the torsion free connection in GR because these are the ones that give us metric geodesics which coincide with the affine geodesics?

In Einstein-Cartan theory, are the geodesics defined by the connection it uses different than the geodesics defined by the metric (in the sense of maximizing the interval)?

 

[Ben Niehoff]

The geodesics are the same for all metric connections, whether they have torsion or not. The only difference is that parallel transport along these geodesics has extra twisting terms.

In general, for a class of connections all differing only by their torsions, the geodesics will be the same. You can see this easily in the geodesic equation

\frac{d^2 x^\mu}{d \lambda^2} + \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d \lambda} \frac{d x^\rho}{d \lambda} = 0
Notice that the connection coefficients get symmetrized by contracting with the velocity vector. So the geodesic equation doesn't care about the antisymmetric part (i.e. the torsion).

 

[dextercioby]

The cancellation of the nonmetricity tensor fixes only the symmetric components of the connection, while the antisymmetric ones are linked to the torsion tensor by the computation of the commutator of covariant derivatives of a vector field.

See also the discussion in the thread referenced by my blog entry below including the books mentioned there.

https://www.physicsforums.com/blog.php?b=2565

 

[Matterwave]

But I thought that one could prove, from maximizing the interval that the geodesic equation that the Christoffel symbols are defined from the metric:

\Gamma^i_{jk}=\frac{1}{2} g^{l i}\left(\frac{\partial g_{lk}}{\partial x^{j}}+\frac{\partial g_{lj}}{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right)

This is obviously symmetric, no? Is it that hidden in here is an anti-symmetric term that the books don't usually mention?

EDIT: I have no idea why latex is not rendering the first partial sign...

 

[Matterwave]

Looking at the derivation here: http://people.uncw.edu/hermanr/gr/geodesic.pdf

I don't see him throwing out any anti-symmetric parts. Do we just arbitrarily add them since they won't affect our geodesic equations?

 

[DrGreg]

EDIT: I have no idea why latex is not rendering the first partial sign...

The vBulletin software that runs this forum doesn't understand \LaTeX (which is handled by an add-on), thinks you've typed a very long word that will mess up line-wrapping, so inserts an extra space which upsets the TEX.

The fix is for you to insert spaces into your TEX at reasonably frequent intervals in places where it doesn't matter, i.e. between symbols.

 

[Ben Niehoff]

But I thought that one could prove, from maximizing the interval that the geodesic equation that the Christoffel symbols are defined from the metric:

\Gamma^i_{jk}=\frac{1}{2}g^{li} \left( \frac{\partial g_{lk}}{\partial x^{j}} + \frac{\partial g_{lj}}{\partial x^{k}} - \frac{\partial g_{jk}}{\partial x^{l}} \right)

Not quite. Assuming a metric-compatible connection, then extremizing the length functional will give you the coefficients that appear in the geodesic equation. It will not, as you have pointed out, give you the antisymmetric part of \Gamma^\mu_{\nu\rho}.

Note that for non-metric connections, geodesics do not extremize the length functional. In fact, one can have the notion of a "connection" without even having a metric at all. Extremizing the length functional is not the most fundamental way to define what a geodesic is. The most fundamental way is to say that geodesics are autoparallels; that is, they parallel-transport their own tangent vector along themselves. This definition applies for any connection, metric-compatible or otherwise, including when you don't have a metric.

 

[Matterwave]

Ok, that makes sense. Thanks!

 

[jfy4]

My fellow forum users, I learned so much from this thread just by lurking... thank you, my god...

 

[UndeadCat]

Also:
http://en.wikipedia.org/wiki/Christ...f_the_second_kind_.28asymmetric_definition.29