Question about the definition of Q-criterion

TurbJet · February 3, 2019, 16:21

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,

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

sxpsxp007 · November 27, 2019, 01:11

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.

TurbJet · November 27, 2019, 01:31

Quote:

Originally Posted by sxpsxp007

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.

sxpsxp007 · November 27, 2019, 01:40

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.

sxpsxp007 · November 27, 2019, 02:05

Anyway,

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

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

The ordering in tensor calculus does make sense.

February 3, 2019, 16:21	Question about the definition of Q-criterion	#1
TurbJet Senior Member Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 9	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, 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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November 27, 2019, 01:11		#2
sxpsxp007 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.

November 27, 2019, 01:40		#4
sxpsxp007 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.

November 27, 2019, 02:05		#5
sxpsxp007 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.