Component of Lie Derivative expression vector field

[binbagsss]

1. Homework Statement


Hi,

I have done part a) by using the expression given for the lie derivative of a vector field and noting that if $w$ is a vector field then so is $wf$ and that was fine.
In order to do part b) I need to use the expression given in the question but looking at a component of the lie derivative of $u$ w.r.t $v$ instead..And I'm unsure how to get this expression .

Homework Equations

The Attempt at a Solution

I've looked it up and see it is:

$$(\mathcal L_vu)^a=[v,u]^a=v^c\partial_cu^a-u^c\partial_c v^a = v^cu^a_{,c}-u^cv^a_{,c}$$

for the $a$th component of $\mathcal L_v w$

this is what I need to use right?

I am unsure how you get this expression for the ath component from the vector field expression we are given , working with the $(0,0)$ rank tensors: $V=V^u \partial_u$. Should it be obvious via a index/dimension analysis or is it even more obvious than that?

I see that the expression is consistent with being covariant , if $\partial_u \to \nabla_u$ then the expression holds and is covariant, so could almost trial my way to the correct expression however, but would like a better understanding.

(For example say if we wanted the lie derivative of $a$component of $w$ wrt to $v$ my guess would be to do something like :
$V^uW^v\partial_v - W^v\partial_v V^u$

loosing the derivative associated with the vector field V- which I know isn't what we seek anyway. But I'm confused because it looks like the dummy indices have been interchanged in the second term too, before loosing the derivative in that term)

many thanks

 

[Orodruin]

Why do you need to look it up? You have been given the definition of the Lie derivative in the problem formulation. A priori, you should be able to find an expression for the components of the Lie derivative starting from that expression.

The derivatives of the vector components are normal partial derivatives. The point is that the asymmetry anyway leads to the Lie derivative being a vector field.

 

[binbagsss]

Why do you need to look it up? You have been given the definition of the Lie derivative in the problem formulation. A priori, you should be able to find an expression for the components of the Lie derivative starting from that expression.

yeh i recognised that, that's why I'm asking, and pretty much what i said in my op...

i don't understand how. and gave my thoughts. I'm asking for help..

 

[Orodruin]

For (a) you just need the definition, not any coordinate expression for the Lie derivative. For (b) you will need the coordinate expression for the Lie derivative of a vector field. I do not understand where the problem in obtaining this coordinate expression comes from. To compute $w(f)$, just apply the coordinate expression and the same for $v(w(f))$, where you should note that $w(f)$ is just a scalar field so the vector $v$ acting on that scalar field is given by the usual expression.

 

[binbagsss]

For (a) you just need the definition, not any coordinate expression for the Lie derivative. For (b) you will need the coordinate expression for the Lie derivative of a vector field. I do not understand where the problem in obtaining this coordinate expression comes from.

As I've said in the other thread, I don't really understand the definitions of vector field here or what a coordinate expression means. Other than the coordinate expression will be invariant I guess, whereas the vector field expression I'm guessing stands covariant?

As I've said in the other thread my GR notes pretty much say that a scalar is a rank $(0,0)$ tensor and $(1,0)$ rank is a vector, and as far as I can see $V^u \partial_u$ is a rank $(0,0)$ , so I'm unsure why it's termed a vector field... which is my confusion no one has seemed to reply to but instead just stated their knowledge on the subject and not help me un-confuse any concepts or tell me what I am doing wrong.

Anyhow, I have just received a reply in the other thread saying that $V_p = V^u \partial_u |_p$, well a lower index to me is a covector so perhaps if I could have this as $V^p = V^u \partial_u |_p$ then I can take the definition given in the question and get the b^th component by evaluating the last vector on each term at $b$ :

$(L_v u)^b = v^c \partial_c u^a \partial_a |_b - u^c \partial_c v^a \partial_a |_b = v^c \partial_c u^b - u^c \partial_c v^b$

 

[Orodruin]

As I've said in the other thread, I don't really understand the definitions of vector field here or what a coordinate expression means. Other than the coordinate expression will be invariant I guess, whereas the vector field expression I'm guessing stands covariant?

Then the correct way of going about things is to understand those definitions first, before you try delving deeper into the theory. Unless you first make a solid groundwork to stand on, everything will be shaky.

As I've said in the other thread

Which other thread?

my GR notes pretty much say that a scalar is a rank $(0,0)$ tensor and $(1,0)$ rank is a vector, and as far as I can see $V^u \partial_u$ is a rank $(0,0)$

This is wrong. It is a (1,0) tensor, i.e., a tangent vector. Why do you think it is a scalar? The tangent space is the vector space of directional derivatives at the given point and what you have is obviously a directional derivative.