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Old   February 3, 2019, 17:21
Question Question about the definition of Q-criterion
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Greetings,

I am trying to understand the definition of Q-criterion, which is the 2nd invariants of the velocity gradient tensor
Q = \frac{1}{2}\left[\text{tr}(D)^2 - \text{tr}(D^2)\right] = \frac{1}{2}(D_{ii}D_{jj} - D_{ij}D_{ji}) = \frac{1}{2}(u_{i,i}u_{j,j} - u_{i,j}u_{j,i})
where the D_{ij} = \frac{\partial u_i}{\partial x_j} is the velocity gradient tensor. Under the incompressible condition, will reduce to Q = -\frac{1}{2}u_{i,j}u_{j,i}

Well, this part I can totally get it. But go further will have Q = \frac{1}{2}\left(||\Omega||^2 - ||S||^2\right) (from Eq.(2) in [1]), where \Omega is the vorticity tensor, S is the rate-of-strain tensor, and the magnitude is defined as ||\Omega||^2 = tr(\Omega\Omega^T) = \Omega_{ij}\Omega_{ji}.

However, if start from this one and go back to the definition stated at the top, I just can't reach the same result, which I have
Q = \frac{1}{2}[(u_{i,j} - u_{j,i})(u_{j,i} - u_{i,j}) - (u_{i,j} + u_{j,i})(u_{j,i} + u_{i,j})] = -\frac{1}{2}(u_{i,j}u_{i,j} + u_{j,i}u_{j,i}) \neq -\frac{1}{2}u_{i,j}u_{j,i}
or
\neq \frac{1}{2}(u_{i,i}u_{j,j} - u_{i,j}u_{j,i})
which apparently different from the one after applying incompressible condition.

I am not sure where it goes wrong. So can anybody give me some hints?

Thanks a lot!

PS: most of the definitions come from here
[1]. J.Jeong, F.Hussain, On the identification of vortex, J. Fluid Mech. (1995), vol. 285, pp. 69-94
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Old   November 27, 2019, 02:11
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Hi,

Have you figured it out? I just drop by this post. I guess that the problem lies in the defintion of the magnitude of Omega, mag(Omega)^2=Omega(ij)Omega(ij). Then you can get the correct Q.
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Old   November 27, 2019, 02:31
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Quote:
Originally Posted by sxpsxp007 View Post
Hi,

Have you figured it out? I just drop by this post. I guess that the problem lies in the defintion of the magnitude of Omega, mag(Omega)^2=Omega(ij)Omega(ij). Then you can get the correct Q.
I just ignore the definition and directly use their conclusion.

The definition of mag(Omega) is given in the paper. But it seems to me the one you proposed is the same as theirs.
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Old   November 27, 2019, 02:40
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Hi,

Please correct me if I am wrong.

The inner product of two tensors: C(ij)=A(ik)B(kj)
A(ik)=Omega(ik) B(kj)=Transpose[Omega(kj)]=Omega(jk)

Then, C(ij)=Omega(ik)Omega(jk)

tr[C(ij)]=C(ii)=C(jj)=Omega(ik)Omega(ik)=Omega(jk)Omega(jk)

Their definition seems correct.
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Old   November 27, 2019, 03:05
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Anyway,

tr(A^2)=A(ij)A(ji)

tr(AA^T)=A(ij)A(ij)

The ordering in tensor calculus does make sense.
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