Re: Why is pressure tensor in hydrodynamics purely dissipative?
- From: "Igor Khavkine" <igor.kh@xxxxxxxxx>
- Date: Wed, 8 Feb 2006 10:58:41 +0000 (UTC)
Max wrote:
In hydrodynamics, the pressure tensor PI(i,j) is defined as
PI(i,j) = <(Vi-<Vi>)(Vj-<Vj>)>,
where the averageis taken for the velocity distribution, i,j are
indexes as x,y, z.
The diagonal components, e.g.PI(x,x), are internal energy density.
The interesting parts are the off-diagonal components, namely the
dissipative pressure tensor. According to the Navier-Stokes equation,
the off-diagonal components are purely dissipative, with transport
coefficients such as shear viscosity and bulk viscosity.
My question is: why are off-diagonal components of the pressure tensor
PURELY dissipative, as opposed to a mixture of non-dissipative and
dissipative terms?
Why do you say that these terms are purely dissipative? If you look in
Landau's book on Fluid Dynamics (section 16, vol. 6 of his Course on
Theoretical Physics), you'll find relevant calculation. In the
simplified case of an incompressible fluid, he calculate the total
kinetic energy dissipated due to internal friction (viscosity).
Landau finds that the rate of disspiation is proportional to the
velocity gradient and part of the generalized pressure tensor, which
you denote by PI. However, the same part of PI figures not only in this
dissipation formula, but also in the formula giving the rate of
transport of kinetic energy orthogonal to the velocity field, as is
expected when viscosity is present.
Here's how the relevant part of the pressure tensor defined. If PI(i,j)
is the tensor you defined above (although it is not necessary to give
its microscopic definition, it comes up naturally when writing down the
Navier-Stokes equations in full generality), then it can be split into
independent parts:
PI(i,j) = P delta(i,j) - sigma'(i,j),
where delta(i,j) is the Kronecker-delta, P is the scalar pressure,
while -sigma'(i,j) is the traceless part of PI(i,j). It is this
sigma'(i,j) tensor that figures in the energy dissipation formula.
As such, the components of sigma'(i,j) can be said to generate
dissipative terms in the Navier-Stokes equations. However, since the
same tensor figures in the formula for the energy transport transverse
to the fluid's motion, they can also be said to generate conservative
terms in the Navier-Stokes equations. Moreover, sigma'(i,j) may have
non-zero diagonal elements, it is only required to be traceless. It
does contain all the off-diagonal elements of PI(i,j), but it does not
appear so easy to separate them from the components on PI's diagonal.
So, unless you have more convincing evidence, it appears that your
conclusion about the character of the off-diagonal terms of PI(i,j) has
been premature.
Hope this helps.
Igor
.
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