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Matterwave

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I have a question about Bernard Shutz's derivation of the Lie derivative of a vector field in his book Geometrical Methods for Mathematical Physics.

I will try to reproduce part of his argument here.

Essentially, we have 2 vector fields V and U which are represented by [itex]\frac{d}{d\lambda}[/itex] and [itex]\frac{d}{d\mu}[/itex] respectively. The Lie derivative [itex]L_{\vec{V}}\vec{U}[/itex] is defined so that we Lie drag the vector field U from a point Q with parameter [itex]\lambda_0+\Delta\lambda[/itex] back to a point P with parameter [itex]\lambda_0[/itex] on V, creating a new vector field U* and then "compare" it with U which is at the point [itex]\lambda_0[/itex] (by "compare" I of course mean the usual thing you do in calculus which is to take the delta lambda to 0).

To do this, Shutz Taylor expanded U*(Q) and then rearranged terms to get U*(P) in terms of U*(Q) since U*(Q)=U(Q) which makes sense. The part of his derivation that doesn't make sense to me is that in the Taylor expansion of U*(Q), you obviously get a term like

[tex]\Delta\lambda\frac{d}{d\lambda}\frac{d}{d\mu^*}[/tex] (evaluated at P)

Shutz uses the Lie dragging condition to say that the Lie bracket between U* and V is 0, so he just reverses the order of that derivative. In the end you get that the Lie derivative of U with respect to V is the Lie bracket [V,U]. What I don't get is, even though Shutz COULD reverse the order of the derivatives in the above expression (perfectly valid due to the Lie Bracket being 0), there's no obvious a priori reason to me that he SHOULD.

If he DIDN'T reverse the order, and you carry out the calculations, then at the end you don't get the Lie Bracket, you get 0 (it seems to me). This seems very weird to me. Can somebody shed some insight? Why did he HAVE TO switch the order?

I didn't want to type out the full derivation, so I may have bungled something somewhere...Hopefully this much information will be enough for you guys to help me.

I will try to reproduce part of his argument here.

Essentially, we have 2 vector fields V and U which are represented by [itex]\frac{d}{d\lambda}[/itex] and [itex]\frac{d}{d\mu}[/itex] respectively. The Lie derivative [itex]L_{\vec{V}}\vec{U}[/itex] is defined so that we Lie drag the vector field U from a point Q with parameter [itex]\lambda_0+\Delta\lambda[/itex] back to a point P with parameter [itex]\lambda_0[/itex] on V, creating a new vector field U* and then "compare" it with U which is at the point [itex]\lambda_0[/itex] (by "compare" I of course mean the usual thing you do in calculus which is to take the delta lambda to 0).

To do this, Shutz Taylor expanded U*(Q) and then rearranged terms to get U*(P) in terms of U*(Q) since U*(Q)=U(Q) which makes sense. The part of his derivation that doesn't make sense to me is that in the Taylor expansion of U*(Q), you obviously get a term like

[tex]\Delta\lambda\frac{d}{d\lambda}\frac{d}{d\mu^*}[/tex] (evaluated at P)

Shutz uses the Lie dragging condition to say that the Lie bracket between U* and V is 0, so he just reverses the order of that derivative. In the end you get that the Lie derivative of U with respect to V is the Lie bracket [V,U]. What I don't get is, even though Shutz COULD reverse the order of the derivatives in the above expression (perfectly valid due to the Lie Bracket being 0), there's no obvious a priori reason to me that he SHOULD.

If he DIDN'T reverse the order, and you carry out the calculations, then at the end you don't get the Lie Bracket, you get 0 (it seems to me). This seems very weird to me. Can somebody shed some insight? Why did he HAVE TO switch the order?

I didn't want to type out the full derivation, so I may have bungled something somewhere...Hopefully this much information will be enough for you guys to help me.

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