Poincare Algebra -- Quick Question

  • Thread starter binbagsss
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  • #1
binbagsss
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Homework Statement



Does ##x_p\partial_v\partial_u-x_v\partial_p\partial_u=0##

Homework Equations



I need this to be true to show a poincare algebra commutator.

We have just shown that ##[P_u, P_v] =0 ##, i.e. simply because partial derivatives commute.

Where ##P_u=\partial_u##

The Attempt at a Solution


[/B]
I'm guess it's using that partial derivatives commute and renaming the ##v## and ##p## for each other, however I'm a bit confused with this, because, I know this isn't really the context here, but these are free indicies not dummy indices, i..e not an index summing over, so I'm a bit unclear how you could simply rename.

Many thanks
 

Answers and Replies

  • #2
stevendaryl
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Homework Statement



Does ##x_p\partial_v\partial_u-x_v\partial_p\partial_u=0##

Homework Equations



I need this to be true to show a poincare algebra commutator.

We have just shown that ##[P_u, P_v] =0 ##, i.e. simply because partial derivatives commute.

Where ##P_u=\partial_u##

The Attempt at a Solution


[/B]
I'm guess it's using that partial derivatives commute and renaming the ##v## and ##p## for each other, however I'm a bit confused with this, because, I know this isn't really the context here, but these are free indicies not dummy indices, i..e not an index summing over, so I'm a bit unclear how you could simply rename.

Many thanks

Free indices means that it must be true for every possible choice of [itex]u,p,v[/itex], and for all possible values of [itex]x_p[/itex] and [itex]x_v[/itex]. So try the particular case where [itex]u,p,v[/itex] are all different indices, and [itex]x_p = 0[/itex], and [itex]x_v \neq 0[/itex].
 

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