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I Variation of perfect fluid and Lie derivative

  1. Jun 12, 2017 #1
    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. jcsd
  3. Jun 12, 2017 #2

    fresh_42

    Staff: Mentor

    I read it as follows:

    ## \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).
     
  4. Jun 12, 2017 #3
    Thank you for your reply.

    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.
     
  5. Jun 12, 2017 #4

    fresh_42

    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.
     
  6. Jun 13, 2017 #5
    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
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