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covariant derivative of metric tensor


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#21 ajb

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Posted 15 December 2007 - 08:16 PM

The exterior derivative is a derivation on differential forms. Locally you can always represent differential forms as \omega = \omega_{0}(x) + \omega_{\mu} dx^{\mu} + \frac{1}{2!}\omega_{\mu \nu} dx^{\mu} dx^{\nu}  \cdots

where you think of the dx^{\mu} as formally being Grassmann odd. (they are functions on a particular supermanifold). Thus we have the antisymmetry properties of differential forms;
dx^{\mu}dx^{\nu} = - dx^{\nu} dx^{\mu} etc.

The exterior derivative (aka de Rham differential) is the homological vector field d which in local coordinates looks like

d = dx^{\nu} \frac{\partial}{\partial x}

As it is homological we have

d^{2} = \frac{1}{2}[d,d] = 0

So you define the action of the exterior derivative on a differential form as

d[\omega] = d\omega
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#22 Norman Albers

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Posted 18 December 2007 - 03:16 AM

Thank you, my friend.

You could have an antisymmetric part in general, although the definition of a metric is that it is symmetric.

If there were an antisymmetric part could it interpret the non-commutation of quantum mechanics?
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#23 ajb

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Posted 18 December 2007 - 10:50 AM

Thank you, my friend.

If there were an antisymmetric part could it interpret the non-commutation of quantum mechanics?


The antisymmetric part of g^{ab} (note upper indexes) is a bi-vector and as such could be used to define a bracket on functions on the manifold. If this bi-vector satisfies the condition [g,g]=0 where  [,] is the Schouten-Nienhuis bracket (a generalisation of the Lie bracket), then we have a Poisson structure, i.e. a Poisson bracket on functions. This bracket would now also satisfy the the Jacobi identity.

This is still classical. To make it quantum you would want to deform the Poisson algebra or use Dirac's canonical quantisation.

What is true is that the commutator of quantum mechanics can be formulated as a classical Poisson bracket if you take the manifold to the the Hilbert space of states. (Not sure if the infinite dimensions makes this difficult in practice). Everything would look like classical mechanics.

Now, if you were to further insist that the antisymmetric part of the metric were non-degenerate, i.e. t^{a}g_{ab}t^{b}= 0 \implies t^{a} , t^{b} = 0 for all vectors t, then you would have a symplectic structure. (We assume non-degeneracy for the symmetric part).

So simply, I don't think you can view it as the quantum commutator, but it is the "classical origin" of it.
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#24 Norman Albers

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Posted 18 December 2007 - 03:53 PM

I very much appreciate your well-considered answer. I am slowly but steadily feeling my way into further understanding. As with many of my exchanges with solidspin, this will percolate in and maybe a few weeks later I'll find something intelligent to say! I have read a little on the exterior differential form having to do with considering a normal vector to a surface; the order determines the sense:  dx^udx^v = -dx^vdx^u, as you offered.
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#25 Norman Albers

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Posted 21 December 2007 - 04:51 PM

ajb, here's a question for you. I am working with the degenerate metric form which is used to describe the Schwarzschild metric:   g_{ab}=\eta_{ab} -2mk_ak_b , where \eta_{ab} is the flat-space Lorentzian form, and k_a is a null four-vector.. Is this form still useful when we look at interior solutions? I have the example of a ball of constant density, which is solved for interior form and matched at the boundary condition to the exterior solution. I am working to combine a Kerr exterior to the angular momentum source expressed in my electron near-field model.
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#26 Norman Albers

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Posted 26 December 2007 - 02:41 AM

I work with a near-field which has no residues at the origin.
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#27 Norman Albers

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Posted 2 January 2008 - 09:46 PM

ajb, the answer about the degenerate metric just fell on my head like a brick. There is generated in my scheme a set of terms on the RHS with no m-dependence, namely the original Lorentzian flat-space statement of the E&M source tensor. There are no such terms to balance them on the LHS since in Cartesian coordinates the Christoffel symbols of the first kind are linear in m. Therefore I cannot use this presumed form of the metric tensor in the interior. This is an important change of direction for me.
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#28 ajb

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Posted 7 January 2008 - 09:49 PM

Sorry Albers, been away for christmas.

I am no expert in building solutions. I assume that you can indeed have very different metrics in different regions, but you will have to match them in a smooth way. I beleive that this matching can be difficult.
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#29 Norman Albers

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Posted 7 January 2008 - 10:41 PM

Thanks, ajb. There is plenty I can do learning to walk, yet. My text humbles me by assigning the low-field Lense-Thirring problem as an exercise, so I'm working on this for one thing.
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