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

[pleasehelpmeno]

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

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

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

and if so can it be expanded any further I am not so sure but i don't fully understand $\frac{\partial}{\partial x^{a}}$ derivatives.

 

[pleasehelpmeno]

never mind figured it out

 

[dextercioby]

I'm sure you're missing out the vector field W in the RHS of 1).

 

[pleasehelpmeno]

yeah i figured them out and yeah missed it off, thx

 

[paranormal]

Hello,

Thank you for your questions. Let's start with the first one. The Lie bracket [u,v] represents the commutator of two vector fields u and v, and it is defined as [u,v] = u(v) - v(u), where u(v) denotes the action of u on v. In your proof, you have correctly shown that [u,v]w = \mathcal{L}_{[u,v]}w, where \mathcal{L}_{[u,v]} denotes the Lie derivative of w along the vector field [u,v]. This means that the two expressions are equivalent.

For your second question, the Lie bracket [X,Y] is not equivalent to the expression you have written. The Lie bracket is a vector field, while the expression you have written is a differential operator. To calculate [X,Y], you need to use the definition of the Lie bracket mentioned above. First, calculate X(Y) and Y(X), and then subtract Y(X) from X(Y) to get [X,Y]. I am not sure what you mean by expanding it further, but you can apply the same process to calculate higher order Lie brackets.

I hope this helps clarify your doubts. If you have any further questions, please feel free to ask.