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February 3, 2019, 17:21 |
Question about the definition of Q-criterion
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#1 |
Senior Member
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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 where the is the velocity gradient tensor. Under the incompressible condition, will reduce to Well, this part I can totally get it. But go further will have (from Eq.(2) in [1]), where is the vorticity tensor, is the rate-of-strain tensor, and the magnitude is defined as . 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 or 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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November 27, 2019, 02:11 |
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#2 |
New Member
Shang
Join Date: Jan 2018
Posts: 5
Rep Power: 8 |
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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November 27, 2019, 02:31 |
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#3 | |
Senior Member
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Quote:
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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November 27, 2019, 02:40 |
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#4 |
New Member
Shang
Join Date: Jan 2018
Posts: 5
Rep Power: 8 |
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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November 27, 2019, 03:05 |
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#5 |
New Member
Shang
Join Date: Jan 2018
Posts: 5
Rep Power: 8 |
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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q criterion |
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