# I Variation of perfect fluid and Lie derivative

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1. Jun 12, 2017

### TAKEDA Hiroki

In Hawking-Ellis Book(1973) "The large scale structure of space-time" p69-p70, they derive the energy-momentum tensor for perfect fluid by lagrangian formulation. They imply if $D$ is a sufficiently small compact region, one can represent a congruence by a diffeomorphism $\gamma: [a,b]\times N\rightarrow D$ where $[a,b]$ is some closed interval of $R^1$ and $N$ is some 3-dimensional manifold with boundary. The tangent vector of $\gamma$ is $W=(\partial/\partial t)_{\gamma}$. The Lagrangian is taken to be $$L=-\rho(1+\epsilon)$$ and the action $I$ is required to be stationary when the flow lines are varied and $\rho$ is adjusted to keep $j^a$ conserved where $\rho$ is a function and $\epsilon$ is the elastic potential as a function of $\rho$. A variation of the flow lines is a differentiable map $\alpha: (-\delta, \delta)\times[a, b]\times N\rightarrow D$ such that $$\alpha(0, [a,b],N)=\gamma([a,b],N).$$ They say "Then it follows that $$\Delta W=L_{K}W$$ where the vector $K$ is $K=(\partial/\partial u)_{\alpha}$."
I'm curious this equation is correct. I guess $\Delta W$ means its components is $(\partial W^i/\partial u)|_{u=0}$ in their book. However r.h.s components are calculated as follows.$$(L_{K}W)^i=\frac{\partial W^i}{\partial x^j}K^j-\frac{\partial K^i}{\partial x^j}W^j=\frac{\partial W^i}{\partial u}-\frac{\partial K^i}{\partial t}$$ So I wonder $$(\Delta W)^i=\frac{\partial W^i}{\partial u}\neq (L_{K}W)^i=\frac{\partial W^i}{\partial u}-\frac{\partial K^i}{\partial t}.$$ $(\partial K^i/\partial t)=0?$ Will you tell me where I am wrong?
This pdf file is Eur. Phys. J. H paper by S. Hawking. See page 19. But I'm sorry my notation is little different.
https://epjti.epj.org/images/stories/news/2014/10.1140--epjh--e2014-50013-6.pdf

2. Jun 12, 2017

### Staff: Mentor

$\Delta W \stackrel{p.17}{=} \pi (\left. \frac{\partial}{\partial u} \alpha \right|_{u=0}) \stackrel{p.19}{=} \pi(\left. K_\alpha \right|_{u=0})$ is the variation vector of the vector field $W$ in direction of $K_\alpha$, the variation of flow lines in direction $K_\alpha= \left( \frac{\partial}{\partial u} \right)_\alpha$ at the point $u=0$. Isn't this exactly the definition of the Lie derivative of $W$ along $K$ at this point? So $"$It then follows that $\Delta W = L_KW\,"$ is more a summary of the specific set-up of the example rather than a conclusion form previous statements. The conclusions come next (p.20).

3. Jun 12, 2017

### TAKEDA Hiroki

I see.. This equation is a definition rather than a derivation.
But I have a question. This paper is written more precisely by using bundle than Hawking-Ellis Book(1973). In the book, they denote $\partial\Psi_{(i)}(u,r)/\partial u)|_{u=0}$ by $\Delta\Psi_{(i)}$ where $\Psi_{(i)}(u,r)$ is a one-parameter family of fields, $u$ is a variation parameter and $r$ is a point of spacetime. The concept of bundle is not used. In this case, can I also understand $\Delta W=L_{K}W$ is exactly the definition of the Lie derivative of $W$ along $K$ at the point? And can both (r.h.s.) and (l.h.s.) components be $(\partial W^i/\partial u)$? Sorry I'm confused.

4. Jun 12, 2017

### Staff: Mentor

I'm not sure I understand you correctly. As soon as you have a (tangent) vector field all over the manifold, you also have vector bundles or even tensor bundles. It is a matter of viewpoint and language, not of a discrepancy regarding the manifold. To me the equation $\Delta W = L_KW$ is what variation calculus is all about, only expressed in terms of certain vector fields, $W$ and $K$.

The "missing" direction $K$ in $\Delta W$ is hidden in the definition of the variation vector (field) $\Delta$ of the vector field $W$ (p.17) which uses the direction $\left. K\right|_{u=0} = \left. \left( \left( \frac{\partial}{\partial u}\right) \circ \alpha \right) \right|_{u=0}$ and the variation $\alpha \, : \, \alpha(0,t,q)=\gamma(t,q)$ for $t \in [a,b]\; , \;q \in N\,$ by defining $\Delta W = \pi \left( \left. \frac{\partial}{\partial u}\right|_{u=0} \gamma(t,q) \right)$.

Perhaps
https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/
can help you to clarify the picture. It is only an overview and not especially about variations but is has some examples.

5. Jun 13, 2017

### TAKEDA Hiroki

Thanks a lot.
So Why is the (l.h.s) component $(\partial W^i/\partial u)$, though (r.h.s.) component is $(\partial W^i/\partial u)-(\partial K^i/\partial t)$ ?? I want you to explain without using the projection $\pi$ because I'm not familiar with the bundle. In the following calculation, this relation $(\partial W^i/\partial u)=(L_{K}W)^i=W^i{}_{;j}K^j-K^i{}_{;j}W^j$ is used. But $(L_{K}W)^i$ is also expressed as $(L_{K}W)^i=(\partial W^i/\partial u)-(\partial K^i/\partial t)$. Is this a contradiction?

Last edited: Jun 13, 2017