Torsion tensor definition doubt

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  • Thread starter Shirish
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Let's say we have any two covariant derivative operators ##\nabla## and ##\nabla'##. Then there exists a tensor ##C^{\alpha}_{\mu\nu}## such that for all covariant vectors ##\omega_{\nu}##,$$\nabla_{\mu}\omega_{\nu}=\nabla'_{\mu}\omega_{\nu}-C^{\alpha}_{\mu\nu}\omega_{\alpha}$$
Now I'm quoting the relevant section on torsion tensor definition:
What if the no-torsion requirement is dropped? Set ##\omega_{\nu}=\nabla_{\nu}\phi=\nabla'_{\nu}\phi##: (which gives) ##\nabla_{\mu}\nabla_{\nu}\phi=\nabla'_{\mu}\nabla'_{\nu}\phi-C^{\alpha}_{\mu\nu}\nabla_{\alpha}\phi##. Antisymmetrize over ##\mu## and ##\nu##, and assume ##\nabla'## is torsion free, but ##\nabla## is not. In that case ##\nabla_{[\mu}\nabla_{\nu]}\phi=-C^{\alpha}_{[\mu\nu]}\nabla_{\alpha}\phi##. The torsion tensor is defined as ##T^{\alpha}_{\mu\nu}\equiv 2C^{\alpha}_{[\mu\nu]}##, implying that $$(\nabla_{\mu}\nabla_{\nu}-\nabla_{\nu}\nabla_{\mu})\phi=-T^{\alpha}_{\mu\nu}\nabla_{\alpha}\phi$$
I don't understand why this is so. I mean the LHS is can also be notationally represented as ##\nabla_{[\mu}\nabla_{\nu]}\phi##, so either there should be a factor of ##1/2## on the RHS, or the torsion tensor should be defined as ##T^{\alpha}_{\mu\nu}\equiv C^{\alpha}_{[\mu\nu]}##, or am I missing something?
 

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Ibix
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Looks OK to me.$$\begin{eqnarray*}
\nabla_\mu\nabla_\nu\phi-\nabla_\nu\nabla_\mu\phi&=&-C^\alpha{}_{\mu\nu}\nabla_\alpha\phi+C^\alpha{}_{\nu\mu}\nabla_\alpha\phi\\
2\nabla_{[\mu}\nabla_{\nu]}\phi&=&-2C^\alpha_{[\mu\nu]}\nabla_\alpha\phi\\
&=&-T^\alpha{}_{\mu\nu}\nabla_\alpha\phi
\end{eqnarray*}$$You appear to be defining anti-symmetrisation as$$\nabla_{[\mu}\nabla_{\nu]}\phi=\nabla_\mu\nabla_\nu\phi-\nabla_\nu\nabla_\mu\phi$$Carroll, at least, defines it as$$\nabla_{[\mu}\nabla_{\nu]}\phi=\frac 12\left(\nabla_\mu\nabla_\nu\phi-\nabla_\nu\nabla_\mu\phi\right)$$Carroll is consistent with your OP (note: in general the prefactor is ##1/n!## when ##n## indices are being summed over). I seem to recall reading that not all sources put the ##1/n!## prefactor in - are you possibly taking a definition of anti-symmetrisation from a source that doesn't include it and a definition of torsion from a source that does?
 
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Ah you're right. I didn't know that the antisymmetrization operator is defined along with a factor of ##1/2##
 
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Ibix
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Ah you're right. I didn't know that the antisymmetrization operator is defined along with a factor of ##1/2##
It certainly appears to have been defined that way here. The argument for the prefactor comes from the notion that if you have an antisymmetric tensor ##A^{\mu\nu}## then ##A^{[\mu\nu]}=\frac 12\left(A^{\mu\nu}-A^{\nu\mu}\right)=A^{\mu\nu}## and antisymmetrising an already antisymmetrised tensor does nothing. A similar argument can be made for wanting a symmetrised symmetric tensor to be unchanged, and hence placing the same prefactor there.

However, it's purely a convention. Having had a quick glance at Carroll's notes confirms my recollection - he warns that not everyone puts in the prefactor. So you do have to keep an eye on what convention is in use!
 
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