Generalized commutation relations

[jfy4]

I would like to work out the following commutation relations (assuming I have the operators right...)

(1) \left[\hat{p}^{\alpha},\hat{p}_{\beta}\right]

(2) \left[\hat{p}_{\alpha},\hat{L}^{\beta\gamma}\right]

(3) \left[\hat{L}^{\alpha\beta},\hat{L}_{\gamma\delta}\right]

where

\hat{p}^{\alpha}=i\nabla^{\alpha}

\hat{L}^{\alpha\beta}=i(x^{\alpha}\nabla^{\beta}-x^{\beta}\nabla^{\alpha})

where \nabla is the covariant derivative. I have managed to work out (1) I think, I got zero. The others I have started but I can't seem to reduce them down to a pretty form. Do I have the operators right (for general Einstein metric)?

EDIT:

I'll post what I got for (2)

=\delta_{\alpha}^{\beta}\partial^{\gamma}\psi-\delta_{\alpha}^{\gamma}\partial^{\beta}\psi+\Gamma^{\beta}_{\alpha\delta}x^{\delta}\partial^{\gamma}\psi-\Gamma^{\gamma}_{\alpha\delta}x^{\delta}\partial^{\beta}\psi

 

[Demystifier]

I have managed to work out (1) I think, I got zero.

Have you assumed that the covariant derivatives act on a scalar function?

See Eq. (121) in
http://www.mth.uct.ac.za/omei/gr/chap6/node10.html

 

[Demystifier]

Note also that, under GENERAL coordinate transformations, x^{\alpha} does NOT transform as a vector. Therefore, some of the expressions above you consider are not really general-covariant.

 

[jfy4]

Have you assumed that the covariant derivatives act on a scalar function?

See Eq. (121) in
http://www.mth.uct.ac.za/omei/gr/chap6/node10.html

Yes.

 

[jfy4]

Note also that, under GENERAL coordinate transformations, x^{\alpha} does NOT transform as a vector. Therefore, some of the expressions above you consider are not really general-covariant.

Geez! what a bummer... Why does it not transform that way?

Also, thanks for responding to these threads, you have been a great help.

 

[Demystifier]

Geez! what a bummer... Why does it not transform that way?

Because the differential dx^{\alpha} transforms as a vector and it is easy to show that dx^{\alpha} and x^{\alpha} do not transform equally, unless the coordinate transformation does not depend on x^{\alpha}. I leave it to you to show this explicitly, as a simple exercise.

 

[Demystifier]

An error correction: I meant ... unless the DERIVATIVES of coordinate transformations do not depend on x^{\alpha}.

 

[Matterwave]

A simple way to see that the "position" 4-vector is no longer a 4-vector under general coordinate transformation is to consider the case of transforming coordinates between Cartesian coordinates and spherical polar coordinates. One can see immediately that the components of a particle's position transforms non-linearly and therefore no longer obeys the linear transformation law. Differentials still DO obey the linear transformation laws though.

 

[jfy4]

Note also that, under GENERAL coordinate transformations, x^{\alpha} does NOT transform as a vector. Therefore, some of the expressions above you consider are not really general-covariant.

It appears that the coordinate 4-vector only transforms as a vector in local cases.

Given that, would it be appropriate to define the metric locally flat- however the second derivatives as non-zero. Then I could carry out the commutation relations and x^{\alpha} would transform appropriately but yet the curvature would be taken into account?

 

[jfy4]

This might sound silly, sorry for the naivete, but would replacing x^\alpha with \chi^\alpha in the above relations, where \chi is the displacement vector of separation between two particles on geodesic paths, make them generally co-variant?

 

[Demystifier]

No, because in differential geometry (on which general relativity is based) vectors are defined on POINTS. The path you mention is defined by more than one point, so there is no single natural point on which such a vector could be defined.