Velocity gradient decomposition of a fluid flow

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  • #1
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If the velocity gradient decomposition is done by symmetric and antisymmetric parts then ##\frac{\partial v^i}{\partial x^j}=\sigma_{ij}+\omega_{ij}## where ##\sigma _{ij}=\frac{1}{2}(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i})## and ##\omega_{ij}=\frac{1}{2}(\frac{\partial v^i}{\partial x^j}-\frac{\partial v^j}{\partial x^i})## where ##\omega_{ij}## is the component of vorticity e.g ##\omega=\frac{1}{2}(\nabla × v),\omega_{ij}=-\epsilon_{ijk}\omega_k## and ##\sigma_{ij}## is the shear part as it doesn't include rotation (e.g if ##v=\Omega × x=\epsilon_{pqr}\Omega_q x_r## putting this in ##\sigma_{ij}## gives 0 and ##\omega_{ij}## gives ##\Omega_k## where ##\Omega ## is the angular velocity. here we have used that ##\epsilon_{ijk}=1## for ##i,j,k## placed in antisymmetric order.
But in the paper quoted in eq 2(a) the velocity gradient contains an extra term ##\frac{1}{3}\delta_{ij}\theta## where ##\theta## is the trace of ##\sigma_{ij}## can anyone please explain why this extra term is added??
 

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  • #3
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They are defining ##\sigma## in a different way than you are defining it.
 
  • #4
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They are defining ##\sigma## in a different way than you are defining it.
Will you please suggest what it would be...
If that is the case then for ##\frac{\partial v^i}{\partial x^i}## will contain terms ##\sum_k\frac{1}{3}\frac{\partial v^k}{\partial x^k}## and for ##i\ne j \frac{\partial v^i}{\partial x^j}=\sigma_{ij}+\omega_{ij}## where ##\sigma_{ij}## has to be ##\frac{1}{2}(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i})##....isn't it??
 
  • #5
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They define ##\sigma## as the so-called deviatoric rate of deformation tensor , given by $$\sigma_{ij}=\frac{1}{2}\left(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i}\right)-\delta_{i,j}\sum_k\frac{1}{3}\frac{\partial v^k}{\partial x^k}$$
 
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  • #6
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They define ##\sigma## as the so-called deviatoric rate of deformation tensor , given by $$\sigma_{ij}=\left(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i}\right)-\delta_{i,j}\sum_k\frac{1}{3}\frac{\partial v^k}{\partial x^k}$$
There should be a ##\frac{1}{2}## in front of the 1st term isn't it??
Okk but in the paper in eq 2(a) it is written that ##\sigma_{[ij]}=0## e.g trace of ##\sigma=0## then how it is coming??
 
  • #7
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There should be a ##\frac{1}{2}## in front of the 1st term isn't it??
Yes. I've edited it in.
Okk but in the paper in eq 2(a) it is written that ##\sigma_{[ij]}=0## e.g trace of ##\sigma=0## then how it is coming??
Check it out. The trace of the deviatoric rate of deformation tensor is exactly equal to zero (mathematically).
 
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  • #8
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@Chestermiller
Sir one thing that if the fluid element is undergoing rotation with angular velocity ##\Omega## therefore the velocity at any point will be ##v=\Omega×x## or ##v_k=\Omega_i x_j-\Omega_j x_i##,putting in the expression of vorticity ##\omega=\frac{1}{2}\nabla×v## comes as ##\omega_k=\Omega_k## e.g ##k^{th}## component of vorticity is equal to ##k^{th}## component of angular velocity.
Therefore is it that vorticity arises only out of rotation of a fluid element??
Vorticity is the angular velocity of the fluid...(i,j,k are anticlockwise following each other)
 
  • #9
wrobel
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that if the fluid element is undergoing rotation with angular velocity Ω\Omega therefore the velocity at any point will be v=Ω×xv=\Omega×x or vk=Ωixj−Ωjxiv_k=\Omega_i x_j-\Omega_j x_i,p
fluid is not the same as rigid body
er in eq 2(a) it is written that σ[ij]=0\sigma_{[ij]}=0 e.g trace of σ=0\sigma=0
##\sigma_{[ij]}=0## does not imply ##tr\,\sigma=0##
 
  • #10
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fluid is not the same as rigid body

##\sigma_{[ij]}=0## does not imply ##tr\,\sigma=0##
Although the trace of ##\sigma## is coming 0 ,but then what the term ##\sigma_{[ij]}=0## implies??
 
  • #11
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Although the trace of ##\sigma## is coming 0 ,but then what the term ##\sigma_{[ij]}=0## implies??
It means that, as defined by these authors, ##\sigma_{ij}## represents the anisotropic part (shear part, non-volumetric part) of the rate of deformation tensor.
 
  • #12
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@Chestermiller
Sir one thing that if the fluid element is undergoing rotation with angular velocity ##\Omega## therefore the velocity at any point will be ##v=\Omega×x## or ##v_k=\Omega_i x_j-\Omega_j x_i##,putting in the expression of vorticity ##\omega=\frac{1}{2}\nabla×v## comes as ##\omega_k=\Omega_k## e.g ##k^{th}## component of vorticity is equal to ##k^{th}## component of angular velocity.
Therefore is it that vorticity arises only out of rotation of a fluid element??
Vorticity is the angular velocity of the fluid...(i,j,k are anticlockwise following each other)
Yes. Vorticity represents the rotational part of the local fluid movement. This is factored out of the fluid viscous response behavior because, as reckoned from the frame of reference of an observer rotating with the vorticity angular velocity of the fluid, it could not contribute to the stresses developed in the fluid. That is, the behavior of the fluid must be independent of the translation and rotation of the observer.
 
  • #13
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##\sigma_{[ij]}=0## does not imply ##tr\,\sigma=0##
These authors are using a different definition of the parameter ##\sigma## than the one you are accustomed to.
 
  • #14
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It means that, as defined by these authors, ##\sigma_{ij}## represents the anisotropic part (shear part, non-volumetric part) of the rate of deformation tensor.
Is it therefore that the total shear is decomposed into two parts one is the deviatoric shear ##\sigma_{ij}## and the other is the volumetric shear ##\frac{1}{3}(\nabla • v)\delta_{ij}## where ##\nabla • v## is the trace of the total shear say ##s_{ij}## and this total shear is not contributed out of any rotation.
Isn't it??
 
  • #15
wrobel
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These authors are using a different definition of the parameter ##\sigma## than the one you are accustomed to.
I do not understand your remark ##\sigma_{[ij]}:=\frac{1}{2}(\sigma_{ij}-\sigma_{ji})##
 
  • #16
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Is it therefore that the total shear is decomposed into two parts one is the deviatoric shear ##\sigma_{ij}## and the other is the volumetric shear ##\frac{1}{3}(\nabla • v)\delta_{ij}## where ##\nabla • v## is the trace of the total shear say ##s_{ij}## and this total shear is not contributed out of any rotation.
Isn't it??
Yes.
 
  • #17
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I do not understand your remark ##\sigma_{[ij]}:=\frac{1}{2}(\sigma_{ij}-\sigma_{ji})##
Ya that means this deviatoric shear is symmetric ,as the total shear is symmetric ##\frac{1}{2}(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i})##
 
  • #18
vanhees71
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What you write down is not the shear. As @Chestermiller already explained, you have to decompose the tensor in its irreducible components, which are the antisymmetric part (as you correctly say that's vorticity) (transforming according to the fundamental representation of the rotation group, ##\ell=1##), the traceless part of the symmetric part (transforming according to the representation ##\ell=2##), and the trace part (transforming according to the trivial representation.

Geometrically it's also clear: As you said correctly yourself the antisymmetric part refers to rotations of the fluid element as a whole. However of the symmetric part only the traceless part refers to shear, i.e., deformation of the fluid element without changing its volume, while the trace part refers to the volume-changing deformations. That's why in general there are two viscosities in the first-order-gradient expansion (aka Navier-Stokes approximation) of the full kinetic equation: Shear and bulk viscosity.
 
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  • #19
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Okk sir in this paper in eq.2(a) where the decomposition is ##\sigma_{ij}+\frac{1}{3}(\delta_{ij}\theta)+\omega_{ij}## ,
Now from the stress tensor components e.g
##T_{ii}=\mu(\partial_i v_i-p)##
And ##T_{ij (i\ne j)}=\frac{1}{2}\mu(\partial_j v_i+\partial_ i v_j)##
Togetherly ##T_{ij}=\mu[\frac{1}{2}(\partial_j v_i+\partial_ i v^j)-p\delta_{ij}]##
If ##p=\frac{1}{3}\nabla • v=\frac{1}{3}\theta##
Then the diagonal components of ##T_{ij}## includes the normal stress and the off diagonal terms contains the shear part .
Therefore here ##\sigma_{ij}## is the contribution of total stress and the term containing ##\sigma_{ij}+\frac{1}{3}\theta## is the contribution of the shear part isn't it??
 
  • #20
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I kind of recognize what they are doing here, but am unfamiliar with any approach in which they multiply the pressure by viscosity, since this certainly doesn't have units of stress. And the equation equating the pressure to the divergence of the velocity vector makes no sense to me whatsoever.
 
  • #21
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@Chestermiller
If ##\sigma_{ij}=\frac{\mu}{2}(\frac{\partial v^i}{\partial x^j}+\frac{\partial v^j}{\partial x^i})## and ##p=\frac{\sigma_{ii}}{3}##the stress tensor becomes ##T_{ij}=\sigma_{ij}-p\delta_{ij}=\sigma_{ij}-\frac{\sigma_{kk}}{3}\delta_{ij}=\sigma_{ij}-\frac{\mu}{3}(\nabla • v)\delta_{ij}=\sigma_{ij}-\frac{\mu}{3}\theta \delta_{ij}## where ##T_{ii}## consist of normal stress and ##T_{ij(i\ne j)}## represents shear.
In this paper ##\sigma_{ij}## is the contribution of the velocity gradient of the total stress and ##\sigma_{ij}+\frac{1}{3}\theta \delta_{ij}## is the contribution of the velocity gradient to the shear part only
Isn't it??
 
  • #22
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What you write down is not the shear. As @Chestermiller already explained, you have to decompose the tensor in its irreducible components, which are the antisymmetric part (as you correctly say that's vorticity) (transforming according to the fundamental representation of the rotation group, ##\ell=1##), the traceless part of the symmetric part (transforming according to the representation ##\ell=2##), and the trace part (transforming according to the trivial representation.

Geometrically it's also clear: As you said correctly yourself the antisymmetric part refers to rotations of the fluid element as a whole. However of the symmetric part only the traceless part refers to shear, i.e., deformation of the fluid element without changing its volume, while the trace part refers to the volume-changing deformations. That's why in general there are two viscosities in the first-order-gradient expansion (aka Navier-Stokes approximation) of the full kinetic equation: Shear and bulk viscosity.
Of the total stress the matrix elements ##\sigma_{ij}(1-\delta_{ij})## (traceless) represents shear and another diagonal matrix having elements ##(\sigma_{ij}-p)\delta_{ij}## represents normal stress,isn't it??where ##p=\mu\frac{1}{3}\theta## and ##\sigma_{ij}=\frac{\mu}{2}(\partial_j v_i+\partial_i v_j)##
 
  • #23
vanhees71
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As usual, Landau Lifshitz vol. VI provides the most lucid explanation for the Navier-Stokes equation. It's a linear approximation for the shear stress, starting from the assumption that the shear stress is a linear function of the first-order gradients of the velocity fields, that are related to the relative motion of fluid cells, leading to friction between these fluid elements.

That means the shear stress must be a function of the symmetric piece of this gradients, i.e., ##(\partial_j v_k + \partial_k v_j)/2##. Now a fluid can be assumed to be isotropic, which implies that the proportionality constants (constants with respect to the fluid flow but still functions of the thermodynamic variables, e.g., density and temperature) are scalars. From a symmetric tensor you can form two irreducible pieces, the trace-free part (transforming under the ##\ell=2## irrep. of the rotation group) and the trace-part (transforming under the ##\ell=0## irrep. of the rotation group). Thus there are two scalar viscosities ##\eta## and ##\zeta##:
$$\sigma_{jk}'=\frac{\eta}{2} \left (\partial_j v_k + \partial_k v_j -\frac{2}{3} \delta_{jk} \vec{\nabla} \cdot \vec{v} \right) + \zeta \vec{\nabla} \cdot \vec{v} \delta_{jk}.$$
Since the first term refers to deformations of the fluid elements that don't change the volume, ##\eta## is the shear viscosity, while the 2nd term refers to the volume-changing deformations, ##\zeta## is the bulk viscosity.

For incompressible fluids (which is always a further approximation, of course) you have only shear stress and thus only the shear viscosity as constitutive parameter.
 

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