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Lie Derivative

The Lie derivative evaluates the change in a given tensor as it moves along the flow of some other vector field. An example may be a flowing river where we consider two neighboring points connected by a vector. The Lie derivative tells us how this vector - the separation between the two points - changes as the water in the river flows.

Consider a curve CC with coordinates xμ(λ)x^{\mu} (\lambda) parametrized by λ\lambda. The tangent vector

uμ=∂xμ∂λu^{\mu} = \frac{\partial x^{\mu}}{\partial \lambda}

represents the flow.

Now, consider a point PP represented by the coordinates xμx^{\mu} and a point QQ represented by the coordinates xμ+dxμx^{\mu} + dx^{\mu} that are infinitesimally separated.

Another vector field with components vμv^{\mu} is defined. The goal is to find how vμv^{\mu} changes as it moves between the points PP and QQ. The setup looks as follows:

The setup

Let

x~μ=xμ+dxμ=xμ+uμdλ\tilde{x}^{\mu} = x^{\mu} + dx^{\mu} = x^{\mu} + u^{\mu} d \lambda

be the coordinates of point QQ. The tensor transformation law requires:

v~μ(x)=∂x~μ∂xνvν(x)=(∂xμ∂xν+∂uμ∂xνdλ)vν(x)=δνμvν(x)+∂uμ∂xνvν(x)dλ=vμ(x)+∂uμ∂xνvν(x)dλ. \begin{align*} \tilde{v}^{\mu} (\boldsymbol{x}) = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x}) &= \left(\frac{\partial x^{\mu}}{\partial x^{\nu}} + \frac{\partial u^{\mu}}{\partial x^{\nu}} d \lambda\right) v^{\nu} (\boldsymbol{x}) \\ &= \delta^{\mu}_{\nu} v^{\nu} (\boldsymbol{x}) + \frac{\partial u^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x}) d \lambda \\ &= v^{\mu} (\boldsymbol{x}) + \frac{\partial u^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x}) d \lambda. \end{align*}

The value of vμv^{\mu} at point QQ is:

vμ(x~)=vμ(x+dx)=vμ(x)+dvμ(x)=vμ(x)+∂vμ∂xν(x)dxν=vμ(x)+∂vμ∂xν(x)uνdλ. \begin{align*} v^{\mu} (\boldsymbol{\tilde{x}}) = v^{\mu} (\boldsymbol{x} + d\boldsymbol{x}) &= v^{\mu} (\boldsymbol{x}) + d v^{\mu} (\boldsymbol{x}) \\ &= v^{\mu} (\boldsymbol{x}) + \frac{\partial v^{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) dx^{\nu} \\ &= v^{\mu} (\boldsymbol{x}) + \frac{\partial v^{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} d\lambda. \end{align*}

The Lie derivative of vector v\boldsymbol{v} along the curve CC is defined as:

Luvμ(x)=lim⁡dλ→0vμ(x~)−v~μ(x)dλ=lim⁡dλ→0vμ(x)+∂vμdxν(x)uνdλ−vμ(x)−∂uμ∂xνvν(x)dλdλ=lim⁡dλ→0(∂vμdxν(x)uν−∂uμ∂xνvν(x))=∂vμ∂xν(x)uν−∂uμ∂xνvν(x). \begin{align*} \mathcal{L}_{\boldsymbol{u}} v^{\mu} (\boldsymbol{x}) &= \lim_{d\lambda \to 0} \frac{v^{\mu} (\boldsymbol{\tilde{x}}) - \tilde{v}^{\mu} (\boldsymbol{x})}{d\lambda} \\ &= \lim_{d\lambda \to 0} \frac{v^{\mu} (\boldsymbol{x}) + \frac{\partial v^{\mu}}{dx^{\nu}} (\boldsymbol{x}) u^{\nu} d\lambda - v^{\mu} (\boldsymbol{x}) - \frac{\partial u^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x}) d \lambda}{d\lambda} \\ &= \lim_{d\lambda \to 0} \left(\frac{\partial v^{\mu}}{dx^{\nu}} (\boldsymbol{x}) u^{\nu} - \frac{\partial u^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x})\right) \\ &= \frac{\partial v^{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} - \frac{\partial u^{\mu}}{\partial x^{\nu}} v^{\nu} (\boldsymbol{x}). \end{align*}

If the Lie derivative is zero, it is said that the object was Lie transported.

The Lie derivative of a scalar function is just the ordinary derivative:

Luf=dfdλ.\mathcal{L}_{\boldsymbol{u}} f = \frac{df}{d \lambda}.

To find the Lie derivative of a covector, we first need to consider how it transforms. From the previous tranformation, we may find:

x~μ=xμ+dxμ=xμ+uμdλ,xμ=x~μ−uμdλ. \begin{align*} \tilde{x}^{\mu} &= x^{\mu} + dx^{\mu} = x^{\mu} + u^{\mu} d \lambda, \\ x^{\mu} &= \tilde{x}^{\mu} - u^{\mu} d \lambda. \end{align*}

The inverse Jacobian is as follows:

∂xν∂x~μ=∂∂x~μ(x~ν−uνdλ)=∂x~ν∂x~μ−∂uν∂x~μdλ=δμν−∂uν∂xσ∂xσ∂x~μdλ=δμν−∂uν∂xσdλ(δμσ−∂uσ∂x~μdλ)=δμν−∂uν∂xμdλ+O(dλ2)=δμν−∂uν∂xμdλ. \begin{align*} \frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} &= \frac{\partial}{\partial \tilde{x}^{\mu}} (\tilde{x}^{\nu} - u^{\nu} d\lambda) \\ &= \frac{\partial \tilde{x}^{\nu}}{\partial \tilde{x}^{\mu}} - \frac{\partial u^{\nu}}{\partial \tilde{x}^{\mu}} d\lambda \\ &= \delta^{\nu}_{\mu} - \frac{\partial u^{\nu}}{\partial x^{\sigma}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\mu}} d\lambda \\ &= \delta^{\nu}_{\mu} - \frac{\partial u^{\nu}}{\partial x^{\sigma}} d\lambda \left(\delta^{\sigma}_{\mu} - \frac{\partial u^{\sigma}}{\partial \tilde{x}^{\mu}} d\lambda\right) \\ &= \delta^{\nu}_{\mu} - \frac{\partial u^{\nu}}{\partial x^{\mu}} d\lambda + O(d\lambda^2) \\ &= \delta^{\nu}_{\mu} - \frac{\partial u^{\nu}}{\partial x^{\mu}} d\lambda. \end{align*}

Note: the higher order terms of dλd\lambda are negligible.

The covector is transformed as follows:

a~μ(x)=∂xν∂x~μaν(x)=(δμν−∂uν∂xμdλ)aν(x)=aμ(x)−∂uν∂xμaν(x)dλ. \begin{align*} \tilde{a}_{\mu} (\boldsymbol{x}) = \frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} a_{\nu} (\boldsymbol{x}) &= \left(\delta^{\nu}_{\mu} - \frac{\partial u^{\nu}}{\partial x^{\mu}} d\lambda\right) a_{\nu} (\boldsymbol{x}) \\ &= a_{\mu} (\boldsymbol{x}) - \frac{\partial u^{\nu}}{\partial x^{\mu}} a_{\nu} (\boldsymbol{x}) d\lambda. \end{align*}

And similarly the value of aa at point x~\tilde{x}:

aμ(x~)=aμ(x+dx)=aμ(x)+daμ(x)=aμ(x)+∂aμ∂xν(x)dxν=aμ(x)+∂aμ∂xν(x)uνdλ. \begin{align*} a_{\mu} (\boldsymbol{\tilde{x}}) = a_{\mu} (\boldsymbol{x} + d\boldsymbol{x}) &= a_{\mu} (\boldsymbol{x}) + da_{\mu} (\boldsymbol{x}) \\ &= a_{\mu} (\boldsymbol{x}) + \frac{\partial a_{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) dx^{\nu} \\ &= a_{\mu} (\boldsymbol{x}) + \frac{\partial a_{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} d\lambda. \end{align*}

Finally, the Lie derivative may be found:

Luaμ=lim⁡dλ→0aμ(x~)−a~μ(x)dλ=lim⁡dλ→0aμ(x)+∂aμ∂xν(x)uνdλ−aμ(x)+∂uν∂xμaν(x)dλdλ=lim⁡dλ→0(∂aμ∂xν(x)uν+∂uν∂xμaν(x))=∂aμ∂xν(x)uν+∂uν∂xμaν(x). \begin{align*} \mathcal{L}_{\boldsymbol{u}} a_{\mu} &= \lim_{d\lambda \to 0} \frac{a_{\mu} (\boldsymbol{\tilde{x}}) - \tilde{a}_{\mu} (\boldsymbol{x})}{d\lambda} \\ &= \lim_{d\lambda \to 0} \frac{a_{\mu} (\boldsymbol{x}) + \frac{\partial a_{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} d\lambda - a_{\mu} (\boldsymbol{x}) + \frac{\partial u^{\nu}}{\partial x^{\mu}} a_{\nu} (\boldsymbol{x}) d\lambda}{d\lambda} \\ &= \lim_{d\lambda \to 0} \left(\frac{\partial a_{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} + \frac{\partial u^{\nu}}{\partial x^{\mu}} a_{\nu} (\boldsymbol{x})\right) \\ &= \frac{\partial a_{\mu}}{\partial x^{\nu}} (\boldsymbol{x}) u^{\nu} + \frac{\partial u^{\nu}}{\partial x^{\mu}} a_{\nu} (\boldsymbol{x}). \end{align*}

Lastly, consider the Lie derivative of the metric tensor. When doing the above transformation, the metric tensor is transformed as follows:

g~μν(x)=∂xα∂x~μ∂xβ∂x~νgαβ(a)=(δμα−∂uα∂xμdλ)(δνβ−∂uβ∂xνdλ)gαβ(a)=(δμαδνβ−δμα∂uβ∂xνdλ−δνβ∂uα∂xμdλ+∂uα∂xμ∂uβ∂xνdλ2)gαβ(x)=(δμαδνβ−δμα∂uβ∂xνdλ−δνβ∂uα∂xμdλ)gαβ(x)=δμαδνβgαβ(x)−δμα∂uβ∂xνdλgαβ(x)−δνβ∂uα∂xμdλgαβ(x)=gμν(x)−∂uβ∂xνgμβ(x)dλ−∂uα∂xμgαν(x)dλ. \begin{align*} \tilde{g}_{\mu \nu} (\boldsymbol{x}) &= \frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial x^{\beta}}{\partial \tilde{x}^{\nu}} g_{\alpha \beta} (\boldsymbol{a}) \\ &= \left(\delta^{\alpha}_{\mu} - \frac{\partial u^{\alpha}}{\partial x^{\mu}} d\lambda\right) \left(\delta^{\beta}_{\nu} - \frac{\partial u^{\beta}}{\partial x^{\nu}} d\lambda\right) g_{\alpha \beta} (\boldsymbol{a}) \\ &= \left(\delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\mu} \frac{\partial u^{\beta}}{\partial x^{\nu}} d\lambda - \delta^{\beta}_{\nu} \frac{\partial u^{\alpha}}{\partial x^{\mu}} d\lambda + \frac{\partial u^{\alpha}}{\partial x^{\mu}} \frac{\partial u^{\beta}}{\partial x^{\nu}} d\lambda^2\right) g_{\alpha \beta} (\boldsymbol{x}) \\ &= \left(\delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\mu} \frac{\partial u^{\beta}}{\partial x^{\nu}} d\lambda - \delta^{\beta}_{\nu} \frac{\partial u^{\alpha}}{\partial x^{\mu}} d\lambda\right) g_{\alpha \beta} (\boldsymbol{x}) \\ &= \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} g_{\alpha \beta}(\boldsymbol{x}) - \delta^{\alpha}_{\mu} \frac{\partial u^{\beta}}{\partial x^{\nu}} d\lambda g_{\alpha \beta}(\boldsymbol{x}) - \delta^{\beta}_{\nu} \frac{\partial u^{\alpha}}{\partial x^{\mu}} d\lambda g_{\alpha \beta} (\boldsymbol{x}) \\ &= g_{\mu \nu}(\boldsymbol{x}) - \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta}(\boldsymbol{x}) d\lambda - \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} (\boldsymbol{x}) d\lambda. \end{align*}

And the value of the metric tensor at point x~\boldsymbol{\tilde{x}}:

gμν(x~)=gμν(x+dx)=gμν(x)+dgμν(x)=gμν(x)+∂gμν∂xσ(x)dxσ=gμν(x)+∂gμν∂xσ(x)uσdλ. \begin{align*} g_{\mu \nu} (\boldsymbol{\tilde{x}}) = g_{\mu \nu} (\boldsymbol{x} + d\boldsymbol{x}) &= g_{\mu \nu}(\boldsymbol{x}) + d g_{\mu \nu} (\boldsymbol{x}) \\ &= g_{\mu \nu}(\boldsymbol{x}) + \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} (\boldsymbol{x}) dx^{\sigma} \\ &= g_{\mu \nu}(\boldsymbol{x}) + \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} (\boldsymbol{x}) u^{\sigma} d\lambda. \end{align*}

Finally, the Lie derivative may be found:

Lugμν=lim⁡dλ→0gμν(x~)−g~μν(x)dλ=lim⁡dλ→0gμν(x)+∂gμν∂xσ(x)uσdλ−gμν(x)+∂uβ∂xνgμβ(x)dλ+∂uα∂xμgαν(x)dλdλ=lim⁡dλ→0(∂gμν∂xσ(x)uσ+∂uβ∂xνgμβ(x)+∂uα∂xμgαν(x))=∂gμν∂xσ(x)uσ+∂uβ∂xνgμβ(x)+∂uα∂xμgαν(x). \begin{align*} \mathcal{L}_{\boldsymbol{u}} g_{\mu \nu} &= \lim_{d\lambda \to 0} \frac{g_{\mu \nu} (\boldsymbol{\tilde{x}}) - \tilde{g}_{\mu \nu} (\boldsymbol{x})}{d\lambda} \\ &= \lim_{d\lambda \to 0} \frac{g_{\mu \nu}(\boldsymbol{x}) + \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} (\boldsymbol{x}) u^{\sigma} d\lambda - g_{\mu \nu}(\boldsymbol{x}) + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta}(\boldsymbol{x}) d\lambda + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} (\boldsymbol{x}) d\lambda}{d\lambda} \\ &= \lim_{d\lambda \to 0} \left(\frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} (\boldsymbol{x}) u^{\sigma} + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta}(\boldsymbol{x}) + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} (\boldsymbol{x})\right) \\ &= \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} (\boldsymbol{x}) u^{\sigma} + \frac{\partial u^{\beta}}{\partial x^{\nu}} g_{\mu \beta}(\boldsymbol{x}) + \frac{\partial u^{\alpha}}{\partial x^{\mu}} g_{\alpha \nu} (\boldsymbol{x}). \end{align*}

Infact, there is a pattern, for an (m,n)(m, n)-tensor, in addition to the first partial derivative, there will be mm negative terms and nn positive terms. This pattern is opposite to the covariant derivative.