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How to derive non relativistic Euler equations from the perfect fluid stress tensor

  1. Jan 12, 2010 #1
    Hi,

    I would like to start from the stress energy tensor for the perfect fluid:

    [tex]
    T^{\mu\nu}=\begin{pmatrix} \rho c^2 & 0 & 0 & 0\cr 0 & p & 0 & 0\cr 0 & 0 & p & 0\cr 0 & 0 & 0 & p\cr\end{pmatrix}
    [/tex]

    where [tex]\rho[/tex] is the mass density and [tex]p[/tex] is the pressure, and I would like to derive the nonrelativistic Euler equations: continuity equation, momentum equation and the energy equation:

    [tex]
    \partial_t\rho + \partial_i(\rho v^i) = 0
    [/tex]
    [tex]
    \partial_t (\rho v^i) + \partial_j (\rho v^i v^j + p\delta^{ij}) = 0
    [/tex]
    [tex]
    \partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0
    [/tex]
    in the above, [tex]v^i[/tex] is the nonrelativistic velocity, [tex]E={1\over2}\rho v^2 + \rho e[/tex] the kinetic plus internal energy density, [tex]\partial_t\equiv {\partial\over\partial t}[/tex] is the time derivative and [tex]\partial_i\equiv {\partial\over \partial x^i}[/tex] are the spatial derivatives and we sum over [tex]i[/tex].
    Those are the equations found for example on wikipedia:

    http://en.wikipedia.org/wiki/Euler_equations_(fluid_dynamics)

    Here are my attempts to derive it:

    http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids [Broken]
    http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#relativistic-derivation-of-the-energy-equation [Broken]

    essentially, I start from

    [tex]\partial_\nu T^{\mu\nu} = 0 [/tex]

    and for [tex]\mu=i[/tex] we get the momentum equation in the limit [tex]c\to\infty[/tex], and for
    [tex]\mu=0[/tex] we get:

    [tex]
    \partial_t \left(\left(\rho c^2 + p{v^2\over c^2}\right)\gamma^2\right)
    +
    \partial_i\left(\left(\rho c^2 + p\right)v^i \gamma^2\right)
    =0
    [/tex]

    where [tex]\gamma=(1-{v^2\over c^2})^{-1/2}[/tex]. If we let [tex]E=\rho c^2[/tex], and neglect the term [tex]p {v^2\over c^2}[/tex], we almost get the right energy equation, the only problem is that this energy [tex]\rho c^2[/tex] also contains the rest mass energy, and thus the pressure [tex]p[/tex] is negligible, and the nonrelativistic limit of this equation just gives the equation of the continuity.

    How do I derive the energy equation? Somehow, I guess I need to separate the rest mass, use the equation of continuity for the rest mass (that follows from the conservation of the baryon number), and then I should be just left with the kinetic+internal energy and the right equation for nonrelativistic energy. Does anyone know how to do that in all details?

    I tried to find this question already answered, but didn't find exactly what I want (I am new to the forums, so I might have missed that --- please point me to the right direction).

    Thanks,
    Ondrej Certik
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 12, 2010 #2

    George Jones

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  4. Jan 12, 2010 #3
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Thanks for the link and the hint.

    So I think the energy equation must follow from this:

    [tex]
    \dot{\varrho} + (\varrho+p)\nabla_au^a = 0
    [/tex]

    but that's where I got stuck, because if I do the nonrelativistic approximation, I have to introduce units, so I get:

    [tex]
    \dot{\varrho c^2} + (\varrho c^2+p)\nabla_au^a = 0
    [/tex]

    then I divide by [tex]c^2[/tex] and neglect [tex]{p\over c^2[/tex] compared to [tex]\rho[/tex] and I obtain the equation of continuity, but not the equation for energy, see my original post.

    I also believe it must be possible somehow to do this right, but so far I wasn't successful.

    Ondrej
     
  5. Jan 12, 2010 #4

    George Jones

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    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    This looks right. What happens when you work with

    [tex]
    (\varrho+p)\dot{u}^a = \nabla^ap - u^a\dot{p}?
    [/tex]
     
  6. Jan 12, 2010 #5
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Then I get the following equation:

    [tex]
    \partial_t (\rho u^i) + \partial_j (\rho u^i u^j + p\delta^{ij}) = 0
    [/tex]

    which is just the Euler momentum equation:

    [tex] {\partial (\rho{\bf u})\over\partial t} + \nabla \cdot
    (\rho {\bf u}{\bf u}^T) + \nabla p = 0
    [/tex]

    and that is fine too. However, I also need to obtain the energy equation somehow and that's what I don't know how to do it.
     
  7. Jan 12, 2010 #6
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    I have a little problem here with the derivation of energy tensor from Lagrange equations. How'd one be possibly able to convert the equation
    http://certik.github.com/theoretical-physics/book/images/math/cba9e5e539e0fd54c7fa70def2082965767e1bbe.png [Broken]
    into
    http://certik.github.com/theoretical-physics/book/images/math/56db3e95c06721e962f220763e3c8240cf581d9b.png [Broken]?[/URL]
    This is obtained here: http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#perfect-fluids [Broken]

    Thanks
    AB
     
    Last edited by a moderator: May 4, 2017
  8. Jan 12, 2010 #7
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Just look at the first equation, take the d L / dx^\mu from the left hand side, put it to the right hand side, and then insert it into the brackets, where you differentiate with respect to \nu, so you also have to insert the \delta_\mu^\nu. then you get the second equation. It's a standard derivation of the Noether conserved current.

    This is unrelated to this discussion though, I just put it into my notes, because I thought I could derive the tensor somehow from the Lagrange equations, but it turns out it is not easy.
     
    Last edited by a moderator: May 4, 2017
  9. Jan 12, 2010 #8
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    My point was ahead to motivate you to start from a Lagrangian to get that equation. But as you have noticed, it does not sound so easy. I don't know what else can be considered an alternative here to derive energy equation other than Lagrange. Maybe George knows...
     
  10. Jan 12, 2010 #9
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    No, I appreciate the suggestion. :) I run out of all ideas myself, so any fresh insight from you and others is appreciated.

    The way I understand it is that there are 3 nonrelativistic equations (continuity, momentum and energy) and they follow from the conservation law of the stress energy tensor [tex]\partial_\nu T^{\mu\nu}[/tex] and the conservation of the baryon number [tex]\partial_\mu (n U^\mu)[/tex]. The spatial compents of the stress energy tensor give the momentum equation. So we are left with the temporal component of the stress energy tensor and the baryon number law. Somehow we need to get the continuity + energy equations from them, I guess by somehow combining them.

    I don't think anything else is needed. Of course, you can also derive the energy equation just purely from nonrelativistic considerations, that is shown here:

    http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#energy-equation [Broken]

    and I am sure there are other ways to get it too. But I just want to see how it follows from the relativistic conservation laws.
     
    Last edited by a moderator: May 4, 2017
  11. Jan 12, 2010 #10
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Look at the pages 49 and 50 of Weinberg's "Gravitaion and Cosmology Principles" book...
    There you can find a relativistic proof of energy equation.

    AB
     
    Last edited by a moderator: May 4, 2017
  12. Jan 13, 2010 #11
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    I obtained the book. Weinberg essentially derives the same equation as Schutz:

    [tex]u^\alpha \partial_\alpha S = 0[/tex]

    which says, that the flow conserves specific entropy. This equation is obtained from a thermodynamic relation, e.g. eq. 2.10.18 in Weinberg:

    [tex]k T d \sigma = pd \left(1\over n\right) + d \left(\rho\over n\right)[/tex]

    where the right hand side follows from the 0th component of

    [tex]\partial_\nu T^{0\nu}[/tex]

    However, do you know how to obtain

    [tex]

    \partial_t E + \partial_j\left(v^j\left(E + p \right)\right) = 0

    [/tex]

    from that? One might need the equation of state for that, so let's use the ideal gas.
     
  13. Jan 13, 2010 #12
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    The below picture shows the page the page 98 of Warner's "Into to Differential Geometry and GR".

    http://img508.imageshack.us/img508/3976/scanjo.jpg [Broken]

    Without having the internal energy involved, this 'energy equation' has been obtained through a geometric definition of flow of energy in three directions plus the direction of time, and this can be combined with the components of stress-energy tensor to give us the 'energy equation' as in the required form. But I am really out of any idea about how internal energy can be considered here...

    I've spent much of today on this thing, and checked more than 40 books regarding relativity, but none of them obtain a good and vivid derivation and most of them address the reader to have a look at another book for a proof (e.g. Weinberg's proof) which makes it even worse because you learn at the end that it is just your time getting wasted...
     
    Last edited by a moderator: May 4, 2017
  14. Jan 13, 2010 #13
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Take a look, Certik:

    http://img509.imageshack.us/img509/5189/97652996.jpg [Broken]

    AB
     
    Last edited by a moderator: May 4, 2017
  15. Jan 13, 2010 #14

    George Jones

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    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Great!

    A few details: [itex]0 = \partial_\beta T^{\alpha \beta}[/itex] is true in all coordinate systems, and, in particular, is true in the global inertial coordinate system of an inertial observer who watches the fluid stream by. The result follows when this equation is contracted with the 4-velocity of the observer instead of the 4-velocity of the fluid. In the observer's inertial coordinate system with [itex]x_0 = ct[/itex], [itex]u^0 = c \gamma[/itex], and [itex]u^i = \gamma v^i[/itex]. This gives

    [tex]T^{00} = \rho' \left( u^0 \right)^2 = \rho' \gamma^2 c^2 = \rho c^2[/tex]

    and

    [tex]T^{0i} = \left( \rho' + p'/c^2 \right) u^0 u^i = \left( \rho' + p'/c^2 \right) \gamma^2 c v^i = \left( \rho + p/c^2 \right) c v^i[/tex]

    I have taken the unprimed frame to be the observer's frame and the primed frame to be a frame (momentarily) comoving with the fluid.
     
  16. Jan 13, 2010 #15
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Excellent, I really appreciate your help. You derived the equation that I have here, eq. (3):

    http://certik.github.com/theoretical-physics/book/src/fluid-dynamics/general.html#relativistic-derivation-of-the-energy-equation [Broken]

    But as I write there (and also above in this thread), the problem with it is that [tex]\rho c^2[/tex] is the total energy, including the rest mass, and thus the "p" next to it is completely negligible. So what you got is just the continuity equation for rho.

    The E in the energy equation in the euler equations on the other hand is equal to the kinetic energy [tex]{1\over2}\rho v^2[/tex] and the internal energy.

    We would get what we want, if we could substract the continuity equation for rho, from the equation that you just derived. If you see what I mean. But my question is then --- how do we derive the continuity equation for rho? Surely we cannot use the same 0th component of the stress energy tensor to derive *both* the continuity equation and the energy equation, can we?

    Somehow we need to use the conservation of particles (baryon) law, but in there there is the number of particles [tex]n[/tex] and so we need to somehow connect [tex]n[/tex] with [tex]\rho[/tex], so we need to use the equation of state somehow. It's really complex. :(


    Let me know what you think. I think we should start from the equations in Weinberg or Schutz, and write down the thermodynamic relations for the ideal gas, connecting the number of particles "n", density "rho", entropy "s", etc. And somehow, we might get it.
     
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  17. Jan 13, 2010 #16

    George Jones

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    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Here is something sloppy and simple-minded, probably too sloppy and simple-minded.

    Represent the energy density of the internal and kinetic energy by (changing the meaning of [itex]E[/itex]) by [itex]E = \rho c^2 - \rho' c^2[/itex] and use [itex]\rho c^2 = E + \rho' c^2[/itex] in what we have. This gives

    [tex]0 = \frac{\partial E}{\partial t} + \nabla \cdot \left[ \left( E + p \right) \vec{v} \right] + \frac{\partial \rho' c^2}{\partial t} + \nabla \cdot \left[ \left( \rho' c^2 \right) \vec{v} \right][/tex]

    and by the a non-relativistic approx. to the continuity equation???

    I haven't had time to think if this is legit., as I'm scrambling to prepare to put food on my table (i.e., do real work.)

    Back in a few hours
     
  18. Jan 13, 2010 #17
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens


    Yes, that's my idea too --- but now we need to get rid of the two terms on the right --- the continuity equation for rho --- who do we derive that those are 0?
     
  19. Jan 13, 2010 #18
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Nice attempt. Your derivation is even true in GR but locally not globally. As you know, any metric tensor at a given point would be cast into the form of a Minkowski metric (not by a coordinates transformation but rather by manipulation of the Jaccobi matrices or, to better put it into words, by making an involutive matrix out of the metric tensor as all eigenvalues of this new matrix consist of-1 and +1).
     
  20. Jan 13, 2010 #19
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    And what about a transformation?!! Gotta think of it...

    AB
     
  21. Jan 13, 2010 #20
    Re: How to derive non relativistic Euler equations from the perfect fluid stress tens

    Also a related question -- the nonrelativistic energy [tex]E[/tex] is composed of the kinetic and an internal one.

    The kinetic energy can be obtained by:

    [tex]\rho c^2 \gamma={\rho c^2\over \sqrt{1-{v^2\over c^2}}} = \rho c^2 \left(1 + {v^2\over 2 c^2}+\cdots\right) = \rho c^2 + {1\over 2} \rho v^2 + \cdots[/tex]

    now I am not sure what exactly the internal energy is, but I think it is composed of the random motion of the molecules in the liquid (e.g. a temperature), i.e. everything besides the rest mass and the kinetic energy, which is obtained by boosting into some other frame by [tex]\gamma[/tex]. How do we obtain the rest mass energy? Well, I would guess it's:

    [tex] \rho' = n m \gamma ^2[/tex]

    where [tex]n[/tex] is the particle number, [tex]m[/tex] is the mass of one particle and the two [tex]\gamma[/tex] factors are necessary (see e.g. Schutz). We then substract [tex]\rho'[/tex] as George did. What remains is to prove, the continuity equation for [tex]\rho'[/tex]. From the conservation of the Baryon number, it follows the continuity equation for [tex]n[/tex]:

    [tex]
    0 = \frac{\partial (n\gamma)}{\partial t} + \nabla \cdot \left[ n\gamma \vec{v} \right]
    [/tex]

    so by multiplying by [tex]m[/tex], you get the continuity equation for [tex]\rho_0[/tex], well, up to one [tex]\gamma[/tex] factor, but I think it might be ok, since we want a nonrelativistic limit at the end anyway.

    So we are almost there. Do you think the kinetic+internal energy is equal to [tex]\rho c^2 - nm\gamma^2 c^2 = \rho c^2 - \rho' c^2[/tex]?
     
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