Are Lie Brackets and Their Derivatives Equivalent in Vector Field Calculations?

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pleasehelpmeno
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Hi i have two questions:
1) When asked to prove [itex]\mathcal{L}_{u}\mathcal{L}_{v}W - \mathcal{L}_{v}\mathcal{L}_{u}W = \mathcal{L}_{[u,v]}[/itex].

I achieved [itex][u,v]w = \mathcal{L}_{[u,v]}[/itex]. This was found by appliying a scalar field <b> to the LHS and simplifying and expanding using + and scalar linearitys to get [itex][u,v]w[/itex] but I am not sure if these are equivalent.

2) When asked to calculate the Lie bracket [X,Y] where [itex]X=5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x}[/itex] and [itex]Y= y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y}[/itex] is this equivalent to:
[itex]\left((5x^{2}\frac{\partial}{\partial t} -4t\frac{\partial}{\partial x})\frac{\partial (y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})}{\partial x^{a}}-(y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})\frac{\partial (5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x})}{\partial x}\right)\frac{\partial}{\partial x^{b}}[/itex]

and if so can it be expanded any further I am not so sure but i don't fully understand [itex]\frac{\partial}{\partial x^{a}}[/itex] derivatives.
 
on Phys.org
never mind figured it out
 
I'm sure you're missing out the vector field W in the RHS of 1).
 
yeah i figured them out and yeah missed it off, thx
 

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