﻿ Is there a formal relationship between the flow related term in the Transport Theorem and the Lie derivative? | PhysicsOverflow
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+ 3 like - 0 dislike

As discribed in Anthony Zee's "Einstein Gravity in a Nutshell" Ch V Appendix 2, thinking about a vector field $V^{\mu}$ as the velocity field of a fluid in spacetime $V^{\mu}(X(\tau))$ where the worldline $X(\tau)$ takes the role of the trajectory of a fluid parcel, the Lie derivative of a vector (or more general tensor) field $W^{\mu}(x)$ along the direction $V^{\mu}$ can be derived from the mismatch between the "conventional" form applying the ordinary derivative expected (analog to the Eulerian difference in fluid dynamics) difference between W at two points $P(x)$ and $Q(\tilde{x})$ and the (called Lagrangian in fluid dynamics) change of $W^{\mu}(x)$ when following the flow from $x$ to $\tilde{x}$ along the worldline (or trajectory)

\(\mathcal{L}_V W^{\mu}(x) = V^{\nu}(x)D_{\nu}W^{\mu}(x) - W^{\nu}(x)D_{\nu}V^{\mu}(x)\)

where $D_{\nu}$ is the covariant derivative.

In fluid dynamics, a Lagrangian conserved quantity can be described by means of the transport theorem

\(\frac{d}{dt}\int\limits_{\mathcal{G}(t)} A(t)dV = \int\limits_{\mathcal{G}(t)} ( \frac{\partial A(\vec{r,t})}{\partial t} + \nabla\cdot\vec{v}(\vec{r},t)A(\vec{r},t))dV = 0\)

where $\frac{dA}{dt} = \frac{\partial A}{\partial t} + (\vec{v}\nabla))A$ is the material derivative, $\mathcal{G}(t)$ is the volume of a fluid parcel, A(t) is a flow variable, and $\vec{v}(\vec{r},t)$ is the flow field.

The Lie derivative and the second term on the r.h.s. of the transport theorem seem both to describe some kind of correction due to the flow which has to be added to obtain the full derivative seen by an observer moving with the flow. In addition, if the second part on the r.h.s. of the transport theorem vanishes, the quantity $A$ does not change with time which seems similar to the fact that a tensor  quantity W does not change with (proper) time if it is Lie transported.

So could the second term in the transport theorem formally be seen as some kind of a Lie derivative along the flow field $\vec{v}(\vec{r},t)$?

asked Oct 13, 2014 in Theoretical Physics by Dilaton (5,240 points) [ revision history ]
edited Oct 19, 2014 by Arnold Neumaier

1 Answer

+ 3 like - 0 dislike

Yes. Indeed, for a conserved vector field $W$ in place of $A$, your integral formula would need the Lie derivative; simply substituting $A$ by $W$ would not give a correct result.

In general, $L_VA:=V\cdot \nabla A$ is the Lie derivative of the scalar field $A$ along $V$, and $L_VW:=V\cdot \nabla W -W\cdot \nabla V$ is the Lie derivative of the vector field $W$ along $V$. Higher order tensors have similar formulas, with even more terms.

By the way, your notation $\nabla \cdot VA $ is misleading, as it suggests to multiply $A$ with the divergence of $V$, while one wants to multiply $V$ with the gradient of $A$.

answered Oct 19, 2014 by Arnold Neumaier (13,969 points) [ revision history ]

Thanks Arnold. What confuses me is that the Transport Theorem can also be stated as that the change of A in a control volume moving with the flow is given by the local change of A plus the A transported in or out of the volum through its boundary

\(\frac{d}{dt}\int \limits_{\mathcal{G}(t)} A(r,t)dV = \int \limits_{\mathcal{G}(t)} \frac{\partial X(r,t)}{\partial t} + \oint\limits_{\partial\mathcal{G}(t)} X(r,t)(v(r,t)\cdot df)\)

The secont term can then the by means of Gauss' theorem be converted into a volume integral too and the divergence appears...

and what is confusing in this line of thought?

The last sentence in your answer, that there should not be the divergence (of v) but the gradient of A instead ... :-/

On a general manifold, one cannot take a volume integral of a vector field, so something is stranger about the computation in your comment. Do you have a concrete example with a physical meaning?

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