Lie derivative of contraction and of differential form

In summary, the conversation discusses the process of proving the identity L_X\langle \omega , Y \rangle=\langle L_X \omega, Y \rangle + \langle \omega, L_X Y \rangle for given vector fields X, Y and one-form ω, where LX denotes the Lie derivative with respect to field X. The individual is working through the steps and has reached an expression involving local components, but is unsure how to proceed. They consider using the definition of the Lie derivative and the pull back, but encounter difficulties in expressing quantities at 0. Ultimately, the conversation concludes with a potential solution involving coordinates.
  • #1
Blazejr
23
2
Hello. I'm learning about Lie derivatives and one of the exercises in the book I use (Isham) is to prove that given vector fields X,Y and one-form ω identity [itex]L_X\langle \omega , Y \rangle=\langle L_X \omega, Y \rangle + \langle \omega, L_X Y \rangle[/itex] holds, where LX means Lie derivative with respect to field X.
My idea was to just expand everything in terms of local components. This is where I got (skipping some intermediate steps, because I'm pretty sure I got them right):
[tex]L_X \langle \omega , Y \rangle = \langle \omega , L_X Y \rangle + Y^{\mu}(\omega_{\nu}X^{\nu}_{\; \; ,\mu}+\omega_{\mu , \nu}X^{\nu})[/tex]
where Einstein convention is assumed.
Now if I were able to prove that expression in the brackets is actually equal to [itex](L_X \omega)_{\mu}[/itex] (which is listed as the next exercise in book, by the way) the task would be finished - I have no idea how to do that, though. Alternatively if I knew how to prove the original formula without expanding it in terms of components I could use my results so far in solving the next task. Any tips will be much appreciated.

Edit: I noticed that I wrote in wrong subforum (for which I'm sorry) after wrting this post and I'm not able to delete it now. If whoever is in charge could move it instead of deleting I would be grateful.
 
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  • #2
So, what are you allowed to use to "prove" that the expression in brackets is equal to the Lie derivative of a one form? The way I've seen it done, usually one derives that result ##(L_X \omega)_\mu=\omega_\nu\partial_\mu X^\nu + X^\nu\partial_\nu \omega_\mu## from requiring the identity you posted at the beginning...which is just requiring the Lie derivative to be Leibniz.
 
  • #3
By contracting a one-form with vector field we get a scalar function. Lie derivative of a scalar field [itex]f[/itex] is defined to be [itex]L_X f = Xf[/itex] in Isham. From that I deduced that
[tex]L_X \langle \omega , Y \rangle = X^{\mu}\partial_{\mu} (\omega_{\nu} Y^{\nu})=\langle \omega , L_X Y \rangle + Y^{\mu}(\omega_{\nu}X^{\nu}_{\; \; ,\mu}+\omega_{\mu , \nu}X^{\nu})[/tex]
I have no idea what else can I do with these expressions.
 
  • #4
So how has Isham defined ##L_X \omega##? I do not have this book so I don't know which way he is approaching this problem. However, if he asks you to prove the statement in your OP, surely he has defined ##L_X\omega## by this point?
 
  • #5
[itex]L_X \omega := \left. \frac{\mathrm{d}}{\mathrm{d}t}\phi_t^{*}\omega \right|_{t=0}[/itex]
Where [itex]\phi _t[/itex] is local flow associated with field X.
 
  • #6
As it is 3 in the morning, I can't guarantee my answer. I apologize if I got the derivation wrong, but let's examine the expression ##L_X\omega(Y)##.

We can use the definition to get:

$$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}\left.\frac{\phi^*_{t+\Delta t}\omega-\phi^*_t \omega}{\Delta t}\right|_{t=0}(Y)$$

We know that ##\phi^*_0## is the identity transformation, so we can simplify the above to:$$(L_X \omega)(Y) = \lim_{\Delta t\rightarrow 0}\left.\frac{(\phi^*_{\Delta t}\omega)(Y)-\omega(Y)}{\Delta t}\right|_{t=0}$$

Now we use the definition of the pull back to obtain:

$$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}\frac{\left.\omega([\phi^*_{\Delta t}]^{-1}Y)\right|_{t=\Delta t}-\left.\omega(Y)\right|_{t=0}}{\Delta t}$$

The difficult part now is to figure out what ##\left.\omega([\phi^*_{\Delta t}]^{-1}Y)\right|_{t=\Delta t}## is equal to. The problem is of course that all the quantities in this expression are based at a distance ##\Delta t## up the congruence of ##X##, whereas we would like to be able to work with quantities based at 0.

First we examine the push forward of ##Y## term. What this term is doing is it's pushing forward ##Y## along a congruence of ##X##. At the parameter ##\Delta t##, there is already a vector ##Y(\Delta t)## there. This vector is different from the vector ##[\phi^*_{\Delta t}]^{-1}Y(0)## by exactly the Lie derivative ##L_X Y## as ##\Delta t\rightarrow 0##. We can therefore say:

$$[\phi^*_{\Delta t}]^{-1}Y(0)\approx Y(\Delta t)-L_X Y(0)\Delta t$$

We can then expand ##Y## because it is presumably an analytic vector field: ##Y(\Delta t)\approx Y(0)+ \Delta t \frac{d}{dt}Y(0)## we keep only first order terms of course. Additionally we can expand the one form field ##\omega## to read ##\omega(\Delta t)\approx \omega(0)+\Delta t\frac{d}{dt}\omega(0)##.

Now we have expressed everything in terms of objects at 0. We get finally then that:$$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}
\left. \frac{(\omega+\Delta t \frac{d}{dt}\omega)(Y+\Delta t \frac{d}{dt}Y-L_X Y\Delta t)-\omega(Y)}{\Delta t}\right|_{t=0}$$

Taking this limit (so that we can drop all terms of order ##(\Delta t)^2##) we get:

$$(L_X \omega) (Y) =\left(\frac{d}{dt}\omega\right)(Y)+\omega(\frac{d}{dt} Y)-\omega([X,Y])$$

We know of course that if we define ##Y=\frac{d}{ds}## then:

$$\omega([X,Y])=\omega\left(\frac{d}{dt} Y - \frac{d}{ds} X\right)$$

We can cancel one of the terms to finally get:

$$(L_X \omega) (Y) =\left(\frac{d}{dt}\omega\right)(Y)+\omega\left(\frac{d}{ds} X\right)$$

And in coordinate notation, this is indeed as desired:

$$(L_X \omega) (Y)=Y^\nu X^\mu \partial_\mu \omega_\nu+\omega_\mu Y^\nu\partial_\nu X^\mu$$

My derivation might be completely wrong however...hopefully someone better at this can help you out.
 
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1. What is a Lie derivative of contraction?

The Lie derivative of contraction is a mathematical operation that describes the rate of change of a tensor field along a vector field. It is used to measure the change of a tensor field in the direction of a given vector field.

2. How is the Lie derivative of contraction calculated?

The Lie derivative of contraction is calculated by taking the contraction of the tensor field with the Lie derivative of the vector field. This is represented mathematically as L_X(T) = i_X(dT) + d(i_X(T)), where L_X(T) is the Lie derivative of contraction of the tensor field T along the vector field X, i_X is the interior product, and d is the exterior derivative.

3. What does the Lie derivative of contraction measure?

The Lie derivative of contraction measures the change of a tensor field along a given vector field. It can also be thought of as a directional derivative that describes the change of the tensor field in the direction of the vector field.

4. What is the relationship between Lie derivative of contraction and differential forms?

The Lie derivative of contraction and differential forms are closely related because they both describe the change of a tensor field. The Lie derivative of contraction is used to measure the change along a vector field, while differential forms are used to measure the change along a tangent space.

5. What are the applications of Lie derivative of contraction and differential forms?

The Lie derivative of contraction and differential forms have many applications in mathematics and physics. They are used in the study of differential geometry, relativity, and fluid mechanics, among others. They are also used in fields such as computer vision and robotics for understanding and modeling the movement of objects in space.

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