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Pressure tensor reduces to scalar pressure for isotropic dis

  1. Feb 10, 2016 #1
    1. Does anyone know why for an isotropic distribution function, pressure tensor reduces to a scalar pressure?

    For instance, for a Maxwellian distribution
    P=A ∫ vx vy exp-(vx2 + vy2 + vz2) dvx dvy dvz


    is not zero.

    I think everybody should realize how bogus some of the authors are. Google this problem, as I did, and you will see hundreds of books, repeating the same line " for isotropic, tensor is diagonal ....". You would think there would be at least one book that might show some derivation.. nope. It's almost like many of them write the books without knowing the equations and just repeating the other authors.
     
  2. jcsd
  3. Feb 10, 2016 #2
    Would it be correct to say that this equation is supposed to represent the shear stress on a plane of constant x in the y direction, or the shear stress on a plane of constant y in the x direction?
     
  4. Feb 10, 2016 #3
    The equation represent the Pxy component of the pressure tensor. Neither of x or y has to be constant.
     
  5. Feb 10, 2016 #4
    Pressure acts on the surface of a control volume, unlike a volumetric force such as gravity. It is also a scalar by nature as "pressure" cannot have direction, but multiplied by the area and dotted with the normal vector of the area yields a force. Try drawing a small control volume of a fluid and summing the forces in all directions. You'll see that the pressure can be subtracted out of your stress tensor.
     
  6. Feb 10, 2016 #5
    Physically, Pxy represents the pressure stress on a plane of constant x in the y direction, and also the pressure stress on a plane of constant y in the x direction. You can show this by dotting the pressure tensor with a unit normal on a plane of constant x, or with a unit normal on a plane of constant y. Suppose we are considering a plane of constant x. We are trying to determine the momentum flux component in the y direction impinging on a plane of constant x. What do you think the limits of integration should be in your equation for vx and vy if we are trying to find Pxy?
     
  7. Feb 10, 2016 #6
    He knows this. He is just trying to understand why the lack of a shear component is consistent with the Maxwell velocity distribtution.
     
  8. Feb 10, 2016 #7
    Oh, I didn't get that out of the question. Wouldnt it make more sense then to write the Maxwell distribution exponent as one of the sum of the energies of an ensemble of a fluid (eg, kinetic, potential, electromagnetic), and then show that P = rho*R*T/V. I think that might be straightforward in statistical mechanics. I have to take a look at my book.
     
  9. Feb 10, 2016 #8
    No. He wants to understand the forces, not the energies.
     
  10. Feb 10, 2016 #9
    The distribution function is in the velocity space and pressure tensor is the second moment of the distribution function. and (V V) in the integral gives you dyadic. I am just wondering if we are on the same page.
     
  11. Feb 10, 2016 #10
    Yes. I think we are on the same page. I've had lots of experience with stress tensors and dyadics. The point I'm trying to get at is that, on x the integral should be from zero to infinity, while, on y, the integral should be from minus infinity to plus infinity. The y integration will give you zero.
     
  12. Feb 10, 2016 #11
    Holy cow!... that works. thanks... I am not %100 clear though on why x limit should be from 0,+∞ but the fact that one of the inegration limits should be set to -∞,+∞ and that will give zero, makes sense.

    Thanks
     
  13. Feb 10, 2016 #12
    Going from 0 to infinity on vx means that you are only counting the molecules moving toward the surface. Going from - infinity to + infinity on vy means you are recognizing that there are molecules hitting the plane in equal numbers from both the +y direction and the -y direction. So their momenta cancel.
     
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