# Lie derivatives

• Oct 23rd 2007, 08:20 AM
ppyvabw
Lie derivatives
Say you have a scalar field $\Phi(x)$ defined on a manifold M. Then imagine evaluating the field at a point an infitesimal distance away from x at $x'=x+\epsilon X$ where X is a illing vector of the manifold. Then

$\Phi(x')=\Phi(x)+\epsilon X^{\mu}\partial_{\mu}\Phi(x)$ to first order. Now the second term on the rhs is the Lie derivative of a scalar, so

$\Phi(x')=\Phi(x)+\epsilon L_{X}\Phi(x)$

Say now that $\Phi$ is not a scalar, say maybe a vector of spinor or tensor. Does the same thing follow, that

$\Phi^{\nu}(x')=\Phi^{\nu}(x)+\epsilon L_{X}\Phi^{\nu}(x)$ Where now $L_{X}$ is not the appropriate Lie derivative of whatever object $\Phi$ is.
• Oct 23rd 2007, 08:34 AM
topsquark
Quote:

Originally Posted by ppyvabw
Say you have a scalar field $\Phi(x)$ defined on a manifold M. Then imagine evaluating the field at a point an infitesimal distance away from x at $x'=x+\epsilon X$ where X is a illing vector of the manifold. Then

$\Phi(x')=\Phi(x)+\epsilon X^{\mu}\partial_{\mu}\Phi(x)$ to first order. Now the second term on the rhs is the Lie derivative of a scalar, so

$\Phi(x')=\Phi(x)+\epsilon L_{X}\Phi(x)$

Say now that $\Phi$ is not a scalar, say maybe a vector of spinor or tensor. Does the same thing follow, that

$\Phi^{\nu}(x')=\Phi^{\nu}(x)+\epsilon L_{X}\Phi^{\nu}(x)$ Where now $L_{X}$ is not the appropriate Lie derivative of whatever object $\Phi$ is.

Yes, though you might wind up with something uglier like
$A_{\nu}(x')=A_{\nu}(x)+\epsilon g_{\nu \lambda}X^{\mu}\partial_{\mu}A^{\lambda}(x)$
for your Lie derivative.

(That would be from General Relativity where $[g_{\mu \nu} ]$ is a metric. I'm not sure of the general structure if the embedding in the manifold isn't a space-time.)

-Dan
• Oct 23rd 2007, 09:37 AM
ppyvabw
Thanks very much for your quick reply.

The manifold is a spacetime, I forgot to mention.

Quote:

Yes, though you might wind up with something uglier like
http://www.mathhelpforum.com/math-he...bbe73204-1.gif
for your Lie derivative.
Maybe I didn't explain very well. That expression is exactly the same as the expression I wrote, because that derivative should be a covariant derivative when operating on a vector and so the metric commutes through it and lowers the index on $A_{\lambda}$ Is that right? That expression won't work for say a vector because we are comparing vectors in two different tangent spaces.

I meant would it still be the same expression, for say a vector

$A_{\nu}(x')=A_{\nu}+\epsilon L_{X} A_{\nu}(x)$

where now the expression for the Lie derivatie on a covariant vector is

$L_{X}A_{\nu}=X^{\mu}\nabla_{\mu}A_{\nu}+A_{\mu}\na bla_{\nu}X^{\mu}$ if I got the indices right.

I mean roughly can the Lie derivative be thought of as a generalisation of the directional derivative in vector calculus to more general settings.
• Oct 24th 2007, 04:15 AM
topsquark
Quote:

Originally Posted by ppyvabw
That expression won't work for say a vector because we are comparing vectors in two different tangent spaces.

Well, it's been (regrettably) about 10 months since I even looked at my QFT books, and I came up with that expression on the fly, so I'm not particularly surprised that it wasn't quite right for your needs. :o

Quote:

Originally Posted by ppyvabw
I meant would it still be the same expression, for say a vector

$A_{\nu}(x')=A_{\nu}+\epsilon L_{X} A_{\nu}(x)$

where now the expression for the Lie derivatie on a covariant vector is

$L_{X}A_{\nu}=X^{\mu}\nabla_{\mu}A_{\nu}+A_{\mu}\na bla_{\nu}X^{\mu}$ if I got the indices right.

I mean roughly can the Lie derivative be thought of as a generalisation of the directional derivative in vector calculus to more general settings.

Yes, as far as I know.

-Dan
• Oct 24th 2007, 04:35 AM
ppyvabw
Quote:

Well, it's been (regrettably) about 10 months since I even looked at my QFT books, and I came up with that expression on the fly, so I'm not particularly surprised that it wasn't quite right for your needs. :o
lol, I know. In the office last week we were discussing the CMB background and whether it violates relativity by defining a prefered reference frame. We decided it didn't, but later we were discussing the area of a paralelogram and the determinant of a two by two matrix, and also how to factorise cubics. We were totally lost.

So on the same day we saved general relativity, but failed to remember some A-level algebra lol.

Thanks for your help. I'll just go with it.