Does the presence of torsion require a nonsymmetric metric?
If torsion = antisymmetric part of the connection coefficients, and
[tex]\Gamma_{\alpha\beta\gamma}=\frac{1}{2}(g_{\alpha\beta,\gamma}+g_{\alpha \gamma,\beta}g_{\beta\gamma,\alpha})[/tex] then doesn't the metric have to have an antisymmetric component? The first two terms on RHS are together necessarily symmetric in beta and gamma. The only asymmetry in the beta and gamma indeces can come from the third term. 
Re: Does the presence of torsion require a nonsymmetric metric?
You can probably find old references that refer to antisymmetric components of the metric (Einstein was one that used that language). Today, we'd distinguish a connection with torsion from the LeviCivita connection (which is the unique torsionfree Riemannian connection) above by writing
[tex] \Gamma_{\alpha\beta\gamma}=\frac{1}{2}(g_{\alpha\b eta,\gamma}+g_{\alpha\gamma,\beta}g_{\beta\gamma,\alpha}) + \gamma_{\alpha\beta\gamma}, [/tex] where [tex]\gamma_{\alpha\beta\gamma}[/tex] contains the torsion. 
Re: Does the presence of torsion require a nonsymmetric metric?
Thank you, fzero. Can you justify that some from a GR point of view. Mathematically it's cool because we can do anything we want, but in GR everything is contained in the metric. Do you know what the physical interpretation of an extra term added to the connection coefficient would be?

Re: Does the presence of torsion require a nonsymmetric metric?
Torsion implies a new kind of geometry different from riemann's. And it causes vectors to rotate in specific circumstances which does not happen in corresponding nontorsional geometry(riemannian) in same circumstance. Take connection coefficient to be [tex]\Gamma_{ijk}= a_{ijk}+b_{ijk} [/tex]. I have changed some notations...a is riemann connection coefficient made out of symmetric metric, and b is the one with torsion. Now define a transport of a vector [tex]n_{i}[/tex]along a vector [tex]m_{i}[/tex] to be defined in terms of [tex]\Gamma[/tex] as
[tex]n_{i}(t+dt)= n_{i}(t) + m^{j}\nabla_{j}n_{i} dt[/tex] where [tex]\delta_{i}[/tex] is covariant derivative with respect to torsional connection. Here 't' parametrizes the curve along which [tex]m_{i}[/tex] tangent. It is easy to see the effect of torsion now. If [tex]a_{ijk}[/tex] were zero, that is metric were constant, above transport could be written as [tex]n_{i}(t+dt)= n_{i}(t) + b_{ijk}m^{j}n^{k} dt[/tex] Thus we are now in 'minkowski' spacetime with a torsion. Its just to demonstrate the effect of torsion, so things be better simple. Now choose [tex]m^{j}= (1,0,0,0)[/tex] or vector pointing in 'time'(no absolute notion of time intended), and let [tex]n^{i}(t)[/tex] be [tex](0,x(t),y(t),0)[/tex]. Note that I dont care about covariant and contravariant components since metric is constant, and can be assumed to be [tex](1,1,1,1)[/tex]. So write everything as [tex]x(t+dt) = x + b_{101}x dt + b_{102}y dt[/tex] [tex]y(t+dt) = y + b_{201}x dt + b_{202}y dt[/tex] [tex]z(t+dt) = b_{301}x dt + b_{302}y dt[/tex] z is the last component of vector n. Since b is antisymmetric in indices i,k, being the antisymmetric component of connection, choose it in following way(as an example)  [tex]b_{101}=b_{202}=A, b_{102}=b_{201}=B[/tex], Rest other terms zero. Thus equations are [tex]x(t+dt) = x(t)*[1+Adt] + y(t)Bdt[/tex] [tex]y(t+dt)= x(t)*Bdt + y(t)[1 + Adt][/tex] [tex]z(t+dt)=0[/tex] Thus rotation is clear here, along with possible change in length of vector. Thus torsion causes rotation even in minkowski spacetime, when transported along a vector, which does not happen in torsionfree minkowski spacetime. Generalization to general spacetime is easy. Choose locally flat metric. In physical sense, I am not sure, but I guess people tried to incorporate spin using torsion. It is easy to see why. We have [tex]G_{uv}=T_{uv}[/tex] . If 'intrinsic spin' is present, then we need stress tensor to have an antisymmetric part. This means [tex]G_{uv}[/tex] must have antisymmetric part, which is provided by torsion. 
Re: Does the presence of torsion require a nonsymmetric metric?
Quote:

Re: Does the presence of torsion require a nonsymmetric metric?
fzero...
It seems really interesting. Could you give more details. Particularly gravity to antisymmetric 2form coupling by Einstein and all. 
Re: Does the presence of torsion require a nonsymmetric metric?
Quote:
And is the connection coefficient above still consistent with [tex]\nabla_\mu g=0[/tex]? 
Re: Does the presence of torsion require a nonsymmetric metric?
I think its consistent becos
[tex]\nabla_{i}g_{jk}= \partial_{i}g_{jk} + \Gamma_{ijx}g^{x}_{k} + \Gamma_{ikx}g^{x}_{j} + \gamma_{ijx}g^{x}_{k} + \gamma_{ikx}g^{x}_{j}[/tex] Contract with metric, and last two terms vanish out of antisymmety. SO its the same condition on metric as in non torsional case. I think physical interpretation of torsion relates to rotating vector when transported along another vector (my previous post). I would be happy to know if there is a deeper interpretation of it. 
Re: Does the presence of torsion require a nonsymmetric metric?
This thread might be a tad old, but I'm just adding my two cents for anyone that is still interested.
Generically speaking, there are 2 independent notions that can be defined on a (RiemannCartan) manifold: 1) The metric, for determining infinitesimal lengths 2) The connection, for determining parallel transport. (Note that parallel transport is determined in its entirety. This includes the possible effects of there being rotation of a vector when parallel transported) While we're at it, we can also determine the covariant derivative of the metric, [itex] Q_{\mu\nu\rho} = \Delta_\mu g_{\nu\rho} [/itex] This is the nonmetricity tensor. One can work with manifolds where this tensor is nonzero. If, however, [itex] Q =0 [/itex] then that means that under parallel transport, infinitesimal lengths and angles are preserved under parallel displacement. Also, this is the condition for local Minkowskian/ Euclidean structure to be valid. If the nonmetricity tensor were not zero, one could not presume that the manifold were locally Minkowskian/ Euclidean. The LeviCivita connection, on the other hand, is the special case of the (locally Minkowskian) manifold which has 0 torsion. This need not be so at all in more general manifolds. In fact, in a torsionfree locally Minkowskian space, one can go to the noncoordinate basis (or the tetrad) locally, not by a coordinate transformation, but by postulating that we have such a basis and that we can express it in terms of the coordinates at the point being considered (as every vector in a linear space can). Hence, the tetrad field is nothing but the coefficients of these linear combinations. Notice how, when the metric is reduced (locally, i.e. at a point, not in an open set about the point) to the Minkowskian form, we can also make it such that its derivatives = 0. So, if we were to do this in "standard" GR, where we work only with locally Minkowskian, torsionfree manifolds, we'd get the connection to be 0 (because the connection, which is the LeviCivita connection in this case, is determined entirely by the metric and its derivatives). That such a transformation is indeed possible (making metric trivial and derivatives =0) is a nontrivial proposition, and a good (sketchy) proof is given in the book by Carroll. (on pgs 7374) While it is possible to get rid of the connection in this manner in standard GR, we do not get rid of the curvature because we do not have enough "degrees of freedom" to remove the second derivatives of the metric as well. Now we return to more general spaces. While in standard GR, the connection = 0 locally, in spaces with torsion, it is not possible to get rid of the connection. This is because the connection also depends on the torsion tensor (the dependence is denoted by [itex] \gamma_{\alpha\beta\gamma} [\itex] in one of the posts above). For nonzero torsion, the local connection does not vanish, even though the metric is locally Minkowskian. This is the origin of the spin connection. It is possible to connect spinors to GR (and potentially describe rotating black holes with spin and mass) which is not possible to do easily with the usual coordinate basis. For those interested, there is an excellent article in Rev. Mod. Phys. by Hehl et al. The link is here: http://rmp.aps.org/abstract/RMP/v48/i3/p393_1 A very good exposition in my opinion and definitely worth reading even if you are mildly interested. We now return to the tetrads and can attempt a sort of conceptual understanding of things. In spaces with torsion, we clearly have more degrees of freedom. If we parallel transport a vector on the manifold, and it rotates, then this degree of rotation yields these additional degrees of freedom. The socalled integrability conditions are then a measure of "noncommutativity" of the tetrad basis  what happens when we transport one way and then the other? So, in answer to one of the earlier questions  there are now additional degrees of freedom  in addition to the specification of the metric, we'd also need to specify the torsion (or the spin connection) It is of historical interest to note that what the Cosserat brothers (and henceforth adapted to GR by Cartan) did was to assume that any elastic medium had not just the displacement field, but also a rotation field. Assume we had a perfectly (linearly) elastic, isotropic body, obeying Hooke's law. If we stretched it in one direction, the usual laws of elasticity would determine the displacements in the transverse directions as being equal (depending on the Poisson ratio). If, in addition, the medium were such that at each point a local moment were present (as opposed to just a nonzero force density) then we'd clearly need to describe the response of the body to an external moment. So we'd need further constitutive relations in addition to Hooke's law for determining the complete state, which now contains both the displacement field and the rotation field (rotation at each point) 
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