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Problem with fluid mechanics part of GR derivation

  1. Aug 14, 2010 #1

    andrewkirk

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    Hello all

    I am trying to teach myself general relativity and am working through the text 'a first course in general relativity' by Bernard F Schutz. So far I have made slow but consistent progress but I am perplexed by a couple of things in the fluid mechanics part where it derives the consequences of the laws of conservation of energy and momentum for a 'perfect fluid' (fluid with no heat conduction or viscosity).

    In Schutz the conservation laws are expressed as

    [itex]

    T^{\alpha \beta},_{\beta} = [(\rho+p)U^{\alpha}U^\beta+p\eta^{\alpha\beta}],_\beta = 0

    [/itex] (4.39)
    where T is the stress-energy tensor of the fluid, [itex]\rho[/itex] is mass/energy density, p is pressure, U is four-velocity of the infinitesimal fluid element under consideration and [itex]\alpha[/itex] and [itex]\beta[/itex] are indices in Minkowski space hence can have values 0,1,2,3. [itex]\eta[/itex] denotes components of the metric tensor. A subscript preceded by a comma indicates partial differentiation with respect to the coordinate represented by that subscript.

    Schutz says 'first let us assume that [itex](nU^\beta),_beta = 0[/itex] '
    (where n is the particle density of the fluid, ie number of particles per unit volume)
    and then uses this assumption in the derivations that follow, but does not give any explanation of why this assumption is made or what its physical significance is. Nor does he appear to subsequently relax this assumption, as one might expect he would given his use of the words 'FIRST let us assume'.

    Can anyone explain what is the justification and the meaning of this assumption?

    Secondly, Schutz derives the following formula for a perfect fluid, valid for [itex]\alpha[/itex] = 0,1,2,3:

    [itex]nU^\beta(U^\alpha(\rho+p)/n) ,_\beta+p,_\beta\eta^{\alpha\beta} = 0[/itex] (4.45).

    He then observes that, in the MCRF (momentarily co-moving reference frame of the infinitesimal fluid element), for [itex]\alpha[/itex] = 1,2,3 (ie the three spatial dimensions), [itex]U^\alpha=0[/itex] and hence the above formula can be written:

    [itex](\rho+p)U^i,_{\beta} U^\beta+p,_{\beta} \eta^{i\beta}=0[/itex] (4.52)

    This is true for i =1,2,3 but not for i=0 as [itex]U^0=1[/itex] in the MCRF. (Schutz uses Roman letters instead of Greek when only considering the three spatial dimensions).

    Now comes the step I don't understand. Schutz says
    'Lowering the index i makes this easier to read and changes nothing. Since [itex]\eta_i^\beta=\delta_i^\beta[/itex] we get:
    [itex](\rho+p)U_i,_\beta U^\beta +p,_i=0[/itex] (4.53)'

    The trouble is that, to lower the index i, you need to multiply equation 4.52 by [itex]\eta_{ji}[/itex] and sum over i = 0,1,2,3. But the equation is only true for i = 1,2,3, not 0.

    So how is it possible to derive the last equation (which is required for what follows later) without cheating?

    Thank you very much to anyone who can help with this.

    Andrew
     
    Last edited: Aug 15, 2010
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  3. Aug 14, 2010 #2

    bcrowell

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    Welcome to Physics Forums!

    It will be easier for us all to communicate if you use LaTeX to mark up your math. For example, what you wrote as (nUbeta),beta = 0 can be done like this: [itex](nU^\beta)_{,\beta}[/itex]. To see how I accomplished that, hit the QUOTE button on my post. Basically you surround everything with itex tags, put backslashes before Greek letters, and use ^ and _ for superscripts and subcripts.

    I think this is just a statement that particles can't be created or destroyed. For example, suppose we describe cars on a freeway in a 1+1-dimensional space. If cars up ahead are going faster than those behind, then the density of cars has to decrease. If we had a constant n and a variable U, this would violate conservation of cars, and it would also violate the given equation.
     
  4. Aug 14, 2010 #3

    jtbell

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    Easier: simply click on the equation. You should get a popup window that contains the LaTeX code.
     
  5. Aug 15, 2010 #4
    For your second question, does Schutz' argument go through if, instead of trying to lower directly on index on 4.52, you first lower an index on 4.45 and then focus on the spatial components?
     
  6. Aug 15, 2010 #5

    andrewkirk

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    Thank you very much bcrowell and jtbell for the tip on Latex. I have updated the original question as per your suggestions and it looks so much better now!

    Thanks also bcrowell for the explanation re Schutz's assumption being about conservation of particles. I can see this now. If the traffic speed is higher further down the road then the density of vehicles reduces (I can imagine Tour de France riders going over the top of a mountain and they spread out as they go down the other side at high speed).

    Thank you yossell for your suggestion re how to solve the second problem. It does indeed work if you lower the index before limiting the indices to the spatial dimensions. It looks like Schutz got his arguments in the wrong order here. I have written out the correct derivation below, mostly for my own benefit.

    Start with:
    [itex]
    nU^\beta(U^\alpha(\rho+p)/n) ,_\beta+p,_\beta\eta^{\alpha\beta} = 0
    [/itex] (4.45)
    Then lower the index [itex]\alpha[/itex] by multiplying the equation by [itex]\eta_{\gamma\alpha}[/itex] and summing over [itex]\alpha[/itex] as follows:

    [itex]
    nU^\beta(\eta_{\gamma\alpha} U^\alpha(\rho+p)/n) ,_\beta+p,_\beta \eta_{\gamma\alpha} \eta^{\alpha\beta} = 0
    [/itex]
    Summing over [itex]\alpha[/itex] gives:
    [itex]
    nU^\beta(U_\gamma(\rho+p)/n) ,_\beta+p,_\beta \delta_{\gamma}^{\beta} = 0
    [/itex]
    Whence, relabelling [itex]\gamma[/itex] back to [itex]\alpha[/itex]. we get:
    [itex]
    nU^\beta(U_\alpha(\rho+p)/n) ,_\beta+p,_\alpha = 0
    [/itex]
    NOW we can observe that, if we restrict consideration to [itex]\alpha[/itex]=1,2,3, relabelling it accordingly as i, we can use the fact that [itex]U^i=0[/itex] to conclude that:
    [itex]
    U^\beta U_i ,_\beta (\rho+p) ,_\beta+p,_i = 0
    [/itex] (4.53)
    And then, using the fact that [itex]U_i,_\beta U^\beta[/itex] is the definition of the i-spatial component of the four-acceleration [itex]a_i[/itex] we transform 4.53 to:
    [itex]
    (\rho+p) a_i+p,_i = 0
    [/itex] (4.54)
    which is a generalisation to of Newton's law F=ma. Beautiful!

    Thank you very much to all.
     
    Last edited: Aug 15, 2010
  7. Aug 15, 2010 #6
    As bcrowell says, this is a statement of conservation of particle number. There are two assumptions here. The first is that the particle current is given by [itex]j^\mu = nU^\mu[/itex], or in other words, there is no net particle flow in the local MCRF (where U = [1,0,0,0], which was implicitly defined in terms of the stress-energy tensor). The second is that particle number is conserved--no particles are created or destroyed, which results in the local conservation equation [itex](j^\mu),_\mu = 0[/itex]
     
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