lie derivative


by Terilien
Tags: derivative
Terilien
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#1
Apr16-07, 08:19 PM
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i was curious as to what exactly this is and more importantly, what motivates it. what are its applications?
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ObsessiveMathsFreak
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#2
Apr17-07, 03:33 AM
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Vectors to a surface/manifold lie in the tangent spaces to the surface. Each tangent space is a vector space with the same dimension as the manifold(most of the time), and you can add and subtract vectors etc without leaving this tangent space.

However, what happens if you take the derivative of a vector field on the surface in a direction along the surface. In other words if [tex]\mathbf{v}[/tex] is teh vector field and [tex]\mathbf{w}[/tex] is the direction, then the rate of change of [tex]\mathbf{v}[/tex] in the direction of [tex]\mathbf{w}[/tex] is:

[tex]\nabla_{\mathbf{w}} \mathbf{v}[/tex]

The result of this operation is a vector, but in general, in fact nearly always, this vector will not lie in any tangent space to the manifold. You might say, so what. After all if the manifold is embedded in a higher dimensional space we can still consider such vectors as normal. However, for one reason or another in differential geometry, people prefer not to think of the manifold as being embedded in a higher dimensional space, and instead having intrinsic properties. This point of view runs straight into a problem when it turns out that in general, second derivatives lie "outside" the surface and are no intrinsic.

The Lie derivative offers a way out of this dilemma. As it turns out if you compute
[tex]\nabla_{\mathbf{w}} \mathbf{v}-\nabla_{\mathbf{v}} \mathbf{w}[/tex]

Then by a great stroke of luck, the final result is always in the tangent space to the manifold. Those terms that point the vector out of the tangent space and into higher dimensional space cancel out and you are left with a vector that lies soley in the tangent space. Hence you have a sort of "intrinsic second derivative" and can once again pretend that the higher dimensional space does not matter/exist (sometimes it in fact really may not exist at all!)

To sum up, the Lie derivatibe is useful for the following two reasons:
1) It is a second derivative
2) It is always in the tangent space
explain
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#3
Apr17-07, 11:49 AM
P: 28
Sorry, but the Lie derivative is a first derivative rather than a second derivative.
Also, the covariant derivative is also always in the tangent space.

I think the real reason the Lie derivative is useful is because it represents the action of the flow of a vector field. A vector field on a manifold generates a flow (a diffeomorphism) which transports points and thus also transports every vector and tensor. The Lie derivative [tex]\mathcal{L}_\mathbf{v}\mathbf{u}[/tex] measures the change in a given vector field [tex]\mathbf{u}[/tex] with respect to what this vector field would have been were it transported by the flow of [tex]\mathbf{v}[/tex].

Chris Hillman
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#4
Apr17-07, 12:16 PM
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Arrow

lie derivative


Unfortunately, OMF got many things wrong in his reply. Anyone interested would do well to consult a good textbook, such as the ones suggested here:
http://math.ucr.edu/home/baez/RelWWW....html#mathback
See in particular Flanders, Differential Forms, with Applications to the Physical Sciences and John M. Lee, Introduction to Smooth Manifolds. The book by Nakahara, Geometry, Topology and Physics can also be useful for keeping all those definitions sorted. The Wikipedia can sometimes be useful for "advanced math" topics such as http://en.wikipedia.org/wiki/Lie_derivative but printed textbooks are far more reliable than anything you'll find on the web. Given the delicacy of mathematical reasoning, IMO a serious math student would be foolish to choose less reliable sources when better ones are as close as the local math library (or an on-line bookseller).
Hurkyl
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#5
Apr17-07, 04:58 PM
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Quote Quote by Terilien View Post
i was curious as to what exactly this is and more importantly, what motivates it. what are its applications?
Well, as its name suggests, it's a derivative. Directional derivatives were very useful in calculus, weren't they? So you can surely appreciate that a notion of directional derivative would be useful when doing calculus on a manifold.

(X and Y will denote tangent vector fields)

I suppose the only mystery is why the Lie derivative of X w.r.t. Y should be [Y, X]; we already know what the derivative of a scalar is, and we can define the derivative of higher tensors with the product rule. (And with the metric, if we really want to differentiate cotangent vector fields, and tensors built from them)

I suppose you could work it out from the notion that it should be a "directional derivative" -- I'm lazy, and I will invoke the fact that derivatives are intimately tied to antisymmetry, and there is a natural antisymmetric product of vector fields: if I represent X and Y as differential operators, then [Y, X](f) = (YX - XY)(f). (Multiplication here is composition, as you might expect) And happily, the higher derivatives cancel, so that [Y, X] is a vector field.
ObsessiveMathsFreak
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#6
Apr17-07, 05:03 PM
P: 406
Apologies to all. I hadn't had my coffee before writing that post ! And I was working from (my poor) memory.

I've used the nabla symbol to mean the whole derivative in the embedding space, forgetting that in differential geometry, it is used to denote only the part derivative that is in the tangent space.

And I should have clarified that the Lie derivative is a second derivative of "functions" on the manifold, not of vectors. It is a first derivate of vector fields, which are themselves first derivatives of functions.

I'll fix the symbols in that post once I've confirmed what the symbol for the whole derivative is supposed to be.
Chris Hillman
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Apr17-07, 06:26 PM
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Quote Quote by ObsessiveMathsFreak View Post
I should have clarified that the Lie derivative is a second derivative of "functions" on the manifold, not of vectors. It is a first derivate of vector fields, which are themselves first derivatives of functions.
OMF, this is still misleading IMO. By definition, the Lie derivative takes a vector fields to a new vector field. A vector field takes a function to a new function. Levels of structure are well explained in textbooks like Boothby or Lee, and should not be confused.

Even worse, I think you are missing the point. To compute the Lie derivative in terms of a coordinate basis, traditionally one says something like this: set [itex]A = a \, \partial_x, \; B = b \, \partial_x[/itex] where [itex]a, \, b[/itex] are certain smooth functions, the "components" of the vector fields wrt [itex]\partial_x[/itex] (you can add more dimensions if desired) and compute the effect of [itex][A,B][/itex] on an arbitrary smooth function:
[tex] [A, \, B] \, f = A \, B \, f - B \, A \, f = a \, \partial_x \left( b \, f_x \right) - b \, \partial_x
\left(a \, f_x \right) [/tex]
[tex]
\hspace{2cm}
= a \, b_x \, f_x + a \, b \, f_{xx} - b \, a_x \, f_x - a \, b \, f_{xx}
= \left( a \, b_x - b \, a_x \right) f_x
[/tex]
But [tex]f[/tex] was arbitrary, so
[tex] L_A \, B = \left[ A, \, B \right] = \left( a \, b_x - b \, a_x \right) \, \partial_x [/tex]
(Some authors use the opposite sign in defining the Lie derivative.) That is, the component of [itex][A,B][/itex] wrt [itex]\partial_x[/itex] is
[tex] a \, b_x - b \, a_x [/tex]
In conventional tensor analysis, this is written
[tex]
[A, \, B] = \left( A^n \, {B^m}_{,n} - B^n \, {A^m}_{,n} \right) \; \partial_{x^m}
[/tex]
See for example Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry, 2nd ed., Academic Press, 1986. So like any other vector field, [itex]L_A \, B[/itex] is in fact a first order homogenous linear differential operator on functions.
garrett
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#8
Apr17-07, 08:35 PM
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http://deferentialgeometry.org/#%5B%...rivative%5D%5D
DeadWolfe
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#9
Apr17-07, 08:56 PM
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Why is Differential misspelled in the name of your website?
explain
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#10
Apr18-07, 02:42 AM
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nice site design - but extremely complicated notation...
ObsessiveMathsFreak
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#11
Apr18-07, 06:07 AM
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Quote Quote by Chris Hillman View Post
OMF, this is still misleading IMO. By definition, the Lie derivative takes a vector fields to a new vector field. A vector field takes a function to a new function. Levels of structure are well explained in textbooks like Boothby or Lee, and should not be confused.
If vectors are considered to be derivatives, then surely the derivative of a vector is in some sense a second derivative. It's construction is such that it lies in the tangent plane, and therefore is itself "only" a first derivative. But you did have to take a second derivative to get it, so it is in some sense a degenerate second derivative.

But, I'm not entirely solid with all the concepts in differential geometry so I won't labour this point. It made sense to me at the time to justify its construction in this way, because there didn't seem to be much use for it in and of itself, and it's use as a poisson bracket seems to only be of much use in 2n phase spaces. It seems to me to be more than just a simple first derivative, or a first derivative with some additional properties.

As my appeal to authority for the day, In Mathematical Methods of Classical Physics, Arnold originally constructs the Lie Derivative as the second order ,mixed, partial derivative of a function on the manifold. He does later point out that it is a first order operator though, and goes on from this to point out that because of this Lie differentiation as an operator turns vector fields into a group. But what use all this is, I'm not sure.

Quote Quote by explain View Post
nice site design - but extremely complicated notation...
The maintainer of that site is undergoing an attempt to fix differential geometry notation, which is admittedly broken. Wish him well.
Chris Hillman
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Apr18-07, 10:50 AM
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Quote Quote by ObsessiveMathsFreak View Post
As my appeal to authority for the day, In Mathematical Methods of Classical Physics, Arnold originally constructs the Lie Derivative as the second order ,mixed, partial derivative of a function on the manifold. He does later point out that it is a first order operator though, and goes on from this to point out that because of this Lie differentiation as an operator turns vector fields into a group. But what use all this is, I'm not sure.
That's because you entirely missed the point of what Arnold was trying to tell you, which is the same thing I am trying to tell you: yes, prima facie you'd expect [itex][A, \, B][/itex] to give a second order operator, which would prevent it from being a vector field, since vector fields are first order linear homogeneous operators. So it is a surprise that the bracket gives a vector field. IOW, [itex] B \mapsto L_A \, B[/itex] takes a vector field to a vector field, rather than to something more mysterious.

And while I don't have that book in front of me, I've read it and I'll bet that if you read whatever you glanced at more carefully, you'll find that he says that the real vector fields on M form a Lie algebra (not a Lie group!) under the bracket.
ObsessiveMathsFreak
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#13
Apr18-07, 05:52 PM
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Quote Quote by Chris Hillman View Post
That's because you entirely missed the point of what Arnold was trying to tell you, which is the same thing I am trying to tell you: yes, prima facie you'd expect [itex][A, \, B][/itex] to give a second order operator, which would prevent it from being a vector field, since vector fields are first order linear homogeneous operators.
The fact that the second derivative terms canceled out in the Lie derivative did not escape me. Nevertheless, the fact remains that second derivatives were taken in order to compute it. The higher derivatives canceled to leave only first order terms, which was the whole point of the operation, but nonetheless the derivative can be regarded as one of second order, albiet a degenerate case. If your functions are once but not twice differentiable, you will not be able to obtain the Lie derivative, despite being able to obtain the vector derivative.

If the Lie derivative is simply just another first order vector field, there seems to be little point to it beyond the usual opaqueness. I looked at it as a the "best worst choice" for an intrinsic second derivative, because it seemed to serve little other function besides notational convienience. This may be a matter of complete pedantics, but it makes sense to me to regard it as such.

Edit: For some bizzare reason, I can't edit that second post anymore. In any case, it may not be such a dredful loss as the notation for the derivative of one vector in the direction of asnother turned out to be:
[tex]\mathbf{w}(\mathbf{v})[/tex]
Hurkyl
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Apr18-07, 09:06 PM
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Quote Quote by ObsessiveMathsFreak View Post
If your functions are once but not twice differentiable, you will not be able to obtain the Lie derivative, despite being able to obtain the vector derivative.
Huh? The Lie derivative of a scalar field exists for any once-differentiable function. The Lie derivative of a tangent vector field doesn't involve scalar fields at all. So just what do you mean?

In case it's relevant, I would like to point out that if X and Y are once-differentiable tangent vector fields, and f is a once-differentiable scalar field, then
[X, Y](f)
exists, even if f is not twice-differentiable.


If the Lie derivative is simply just another first order vector field, there seems to be little point to it beyond the usual opaqueness.
It's a directional derivative. You think directional derivatives are pointless? And what is this "opaqueness" of which you speak? And why is it a bad thing, as you seem to imply?
ObsessiveMathsFreak
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#15
Apr19-07, 04:14 AM
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Quote Quote by Hurkyl View Post
Huh? The Lie derivative of a scalar field exists for any once-differentiable function. The Lie derivative of a tangent vector field doesn't involve scalar fields at all. So just what do you mean?
That's not my understanding from the derivation. As derived above
Quote Quote by Chris Hillman
[tex] [A, \, B] \, f = A \, B \, f - B \, A \, f = a \, \partial_x \left( b \, f_x \right) - b \, \partial_x
\left(a \, f_x \right) [/tex]
[tex]
\hspace{2cm}
= a \, b_x \, f_x + a \, b \, f_{xx} - b \, a_x \, f_x - a \, b \, f_{xx}
= \left( a \, b_x - b \, a_x \right) f_x
[/tex]
The second derivatives ordinarily cancel. But if the second derivative does not exist, then there is no defined way to cancel the terms. For example let

[tex]f(x) = x^{\frac{3}{2}}[/tex]]
[tex]f(x) =\frac{3}{2} x^{\frac{1}{2}}[/tex]
[tex]f(x) = \frac{3}{4}x^{-\frac{1}{2}}[/tex]

Using the final derived formula [tex]\left( a \, b_x - b \, a_x \right) f_x[/tex] gives
the Lie derivative to be zero at x=0. However, the derivation from the line above requires that we are able to evaluate and then find the difference of the two second derivative terms. We cannot do this at x=0 and so you can see that the final derived formula for the Lie derivative is not always correct. This is similar to the problem of not being able to evaluate the value of
[tex]f(x)=\frac{x-1}{x-1}[/tex]
at x=1, despite the fact that it seems to have the value 1 because of the cancellation of terms.

You can get around this for many functions by introducing generalised functions, however, the very fact that you need to do this, when no problem exists for first derivatives, indicates that there is more to the Lie derivative than simply being another first derivative.

I suppose you could also get around it by defining the Lie derivative to be the final formula, but if you do this I think the meaning will be irrevocably lost.

Quote Quote by Hurkyl View Post
It's a directional derivative. You think directional derivatives are pointless? And what is this "opaqueness" of which you speak? And why is it a bad thing, as you seem to imply?
Introducing news ones would be if they are serving no purpose. To be honest I've never come across a differential geometry text that actually uses the Lie derivative as anything other than notational compression, so I'm not sure exactly what use it is in practice. As to the opaqueness, I'm referring to the usual listing of properties, Jacobi identity,etc , which while interesting to know, offer no insight into what this derivative actually does, or what it is used for, which is probably why the original question was asked in the first place.
Doodle Bob
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Apr19-07, 05:02 AM
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Quote Quote by ObsessiveMathsFreak View Post
...To be honest I've never come across a differential geometry text that actually uses the Lie derivative as anything other than notational compression, so I'm not sure exactly what use it is in practice...
I completely understand this impression re: Lie derivatives, OMF. The role of Lie derivatives is often understated in the literature. My experience has been that its importance (and it is *very* important) gets passed on more verbally than any other way.

First of all, the usual method definition that is causing this once-or-twice-differentiability problem, is actually the "wrong" definition. The "correct" definition was mentioned (somewhat wrongly -- flows are a *path* of local diffeomorphisms) by explain: it's the instantaneous rate of change of the second vector field along the flow of the first vector field. Using this definition, one sees that L_X Y is well-defined for C^1 vector fields.

One also sees that this is indeed a dynamics object. Intuitively, it measures how the flows of the two vector fields interact at any given point. It'll be zero at a point, if the two flows commute "instantaneously" at that point.

The importance of [,] comes from Frobenius' Theorem, which tells us when a given distribution of vector fields in a manifold can be "integrated" to form the tangent bundle of a submanifold. This is a really, really, really important theorem. But, depending on the text you're reading, that importance is not always evident.
explain
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#17
Apr19-07, 07:07 AM
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Quote Quote by ObsessiveMathsFreak View Post
If the Lie derivative is simply just another first order vector field, there seems to be little point to it beyond the usual opaqueness. I looked at it as a the "best worst choice" for an intrinsic second derivative, because it seemed to serve little other function besides notational convienience. This may be a matter of complete pedantics, but it makes sense to me to regard it as such.
Your question seems to be that math books explain how to define the Lie derivative but not why it is defined in this way, and also they do not explain why this construction is useful at all. (It seems that you meant this by the words "usual opaqueness".) It was also my impression that in many books on general relativity people introduce the Lie derivative, differential forms, etc., but then never use them in actual calculations, except maybe using the Lie derivative to formulate the condition for the Killing vector field (but never actually calculating anything with the Lie derivative itself).

As far as I understand, the Lie derivative is useful mainly because it represents the infinitesimal action of the flow of a vector field. A vector field on a manifold generates a flow (a diffeomorphism) which transports points along the flow lines. Since vectors are "arrows between nearby points" and tensors are defined through vectors, the flow also transports every vector and tensor. This is a very natural construction, defined independently of any metric on the manifold.

Now suppose we are given a vector field [tex]\mathbf{v}[/tex] and we want to compute its derivative in a direction given by a vector [tex]\mathbf{u}[/tex] at some point [tex]p[/tex]. The vector [tex]\mathbf{u}[/tex] is an arrow from the point [tex]p[/tex] towards a neighbor point [tex]p'[/tex]. But we can't simply compute the derivative as [tex]\mathbf{v}(p')-\mathbf{v}(p)[/tex], because the vectors [tex]\mathbf{v}(p)[/tex] and [tex]\mathbf{v}(p')[/tex] are in tangent spaces at different points. Somehow we need to transport [tex]\mathbf{v}(p')[/tex] into the tangent space at [tex]p[/tex]. Let us transport it using the flow of [tex]\mathbf{u}[/tex]. Note that we need to use an entire vector field [tex]\mathbf{u}[/tex], not just its value at the point [tex]p[/tex]. The result is the Lie derivative [tex]\mathcal{L}_\mathbf{u} \mathbf{v}[/tex]. So the Lie derivative is the "flow derivative" in this sense.

The Lie derivative is useful in many computations, if one tries to avoid using index notation and tries to concentrate on the geometric meaning of every object in a computation. For instance, I can list the following good uses of the Lie derivative in general relativity:

1) A formula for the Levi-Civita connection can be written through the Lie derivative, the exterior differential, and the metric. In this way, one has an explicit representation of the covariant derivative through the metric combined with the metric-free operations (exterior differential and the Lie derivative).

[tex]g(\nabla _x v, y)=\frac{1}{2} (d \hat{g} v) (x,y) + \frac{1}{2} (\mathcal{L}_v g)(x,y).[/tex]

Here [tex]\hat{g}v[/tex] is the 1-form corresponding to the vector [tex]v[/tex] through the metric [tex]g[/tex].

This formula is easier to derive than the standard Koszul formula.
Then one can easily analyze various cases (integrable or geodesic Killing vectors, etc.)

2) The equation of motion for a particle in curved spacetime can be derived quickly using the Lie derivative. (No Christoffel symbols needed!)

3) The Raychaudhury equation in general relativity is about the rate of change of volume under a diffeomorphism. The Lie derivative is a natural starting point for the derivation.

4) Killing vectors can be computed in tetrad formalism easily if one uses the Cartan homotopy formula,

[tex]\mathcal {L}_v = \iota_v d + d \iota_v [/tex]

5) An infinitesimal gauge transformation in GR (i.e. infinitesimal change of coordinates) can be easily written through the Lie derivative. Calculations with gauge transformations are then also simplified.

These are just the examples that I can find now.

Unfortunately, the style of many books in mathematics is that "here are some definitions and constructions - first learn them all and then we'll talk". It is assumed that the reader will figure out what these definitions mean and why these constructions are useful. "Lie derivative is the operator L satisfying the following properties... The reader will prove in an exercise that the operator L is unique." This is similar to telling students at grade school that "Numerical fractions, such as 8/3, are equivalence classes on pairs of integers, with the following axioms... Now let us prove that every fraction has a unique canonical form." Understandably, this style leads to frustration for many students.
Doodle Bob
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#18
Apr19-07, 12:58 PM
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Quote Quote by explain View Post
...A vector field on a manifold generates a flow (a diffeomorphism) ...
Careful: a flow is a path of local diffeomorphisms.


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