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Landau fluid mechanics- mathematical passage

  1. Nov 3, 2009 #1
    In the book about fluid-mechanics Landau in the first pages in the isentropic case from

    [tex] dw=\frac{dp}{\rho}[/tex]

    deduces

    [tex] \nabla w=\frac{\nabla p}{\rho}[/tex]

    but I can't understand... in its derivation dw and dp are material differential ([tex]dw=w(x+vdt,t+dt)-w(x,t)[/tex]) in my view) so writing it explicitely I was only able to deduce:

    [tex](\nabla w-\frac{\nabla p}{\rho})\vec{v}=\partial_t(w-\frac{p}{\rho})[/tex]

    but nothing more...

    what am I missing?
     
  2. jcsd
  3. Nov 3, 2009 #2
    I don't have the book so may need a little more context. I have not been taught about material differential, only the material derivative. What is w? What flow assumptions have been made (eg. is it steady?), what is the aim of the derivation?

    Thanks
     
  4. Nov 4, 2009 #3
    I admit that the notation 'material differential' is a notation of mine that I find it expressive... if you want I wrote its definition in the first message... if f=f(x,t)

    [tex]df=f(x+vdt,t+dt)-f(x,t)[/tex]

    where I used the notation commonly used by physicist that talk about 'infinitesimal quantities'...
    mathematically speaking it is the differential of f at a point (x,t) applied to a certain direction...

    anyway following Landau fluid-mechanics page 4. he wants to study a fluid in the isentropic case (entropy costant for every fluid element and equal for all elements, so equal in all space)... so he considers w, the heat function per unit volume and writes following a fluid element:

    [tex]dw=Tds+Vdp[/tex]

    (V specific volume) (S specific entropy) but ds=0, so he arrives at the first equation of my first message...

    Going further he says 'and so', arriving at the second equation of my first message... but actually I can't understand how he worked it out because dw and dp are not equal to [tex]w(x+k,t)-w(x,t)[/tex] (it that case I would understand) but are different...

    I hope the problem is clear now...
     
  5. Nov 4, 2009 #4

    DrDu

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    I don't have my Landau at hand, momentarily. But Landau often leaves away some "obvious" assumptions. E.g. the equation you want to prove would hold in the stationary case where the partial derivative with respect to t does vanish.
     
  6. Nov 4, 2009 #5
    I didn't notice that in the case of stationarity (which I think is more than 'steady flow') all was ok :).......

    but actually Landau doesn't consider that case in these pages (in his equations time derivatives are present)... but maybe the way of thinking you suggest (that is, putting to zero the time derivative in the second equation of my first message) is the rigth one to understand what Landau is really doing... that at the moment I still don't understand...
     
  7. Nov 5, 2009 #6

    DrDu

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    I had a closer look at Landau yesterday evening. The problem is that there is nothing like a "material differential". A differential is a differential, independently of the parameterization of the functions.
     
  8. Nov 5, 2009 #7
    what parametrization are you talking about?... I think that the first law of thermodynamics can't describe the hole differential of w(x,t) but can only describe its behaviour in a certain direction... I know that a differential is a differential but in this case (as often in physics and in Landau's books) the expression must be 'explained'... I describe it better...

    I thought landau wanted to apply the first law of thermodynamics to a fluid element.... in this case I think that the thermodynamic quantity must be referred to the fluid element flowing in space so the correct interpretation of dw I thought was the one I gave up, that is: w(x+vdt,t+dt)-w(x,t)... I don't see how you can deduce anything else on the function w(x,t)...

    with the last expression I mean the heat function of the fluid element in the position x+vdt at the time t+dt minus the heat function of the fluid element in the position x at time t... times and positions are written in order to make the fluid element considered composed by the same molecules!... I thought this was a regular procedure (found a similar one somewhere)... but I'm ready to change my mind if you explain me what's wrong with that!
     
  9. Nov 5, 2009 #8

    DrDu

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    Got your point now. But I think Landau really wants to compare different fluid elements with each other. How could this be done? I think the key is to assume that the different fluid elements are described by the same equation of state (that is, they have the same composition, e.g. pure water). w (the enthalpy) T, S, p, V are all state functions. If the last four are equal, then also w is equal for two different volume elements. If p and V are different in two different volumes 1 and 2, I get the difference as [tex]w(2)=w(1)+\int_1^2 V(x) dp/dx dx [/tex] or, if the state functions depend contiuously on the co-ordinates, dw=Vdp.
     
  10. Nov 5, 2009 #9
    perfect!... I see the point now... so actually his dw is w(x+dx,t)-w(x,t) and he compares different parts of fluids at the same time... I really wasn't able to see that he could do that just with the assumption that the equation of state was the same everywhere!... thank you very much! :smile:
     
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