V * grad(V) = grad(V^2/2) - rotor(omega)

  • #1
Rikyuri
3
2
Hi, while studying for my aerodynamics class, I encountered this equivalence that my professor gave us as a vector identity:
$$
\mathbf{V} \cdot \nabla \mathbf{V} = \nabla\left(\frac{V^{2}}{2}\right)-\mathbf{V} \times \boldsymbol{\omega}
$$
where ## \boldsymbol{\omega} = \nabla \times \mathbf{V} ##I tryed to expand the operator and found that ## \mathbf{V} \cdot \nabla \mathbf{V} = \nabla(\frac{V^{2}}{2}) ## but that can't be true.
I really don't understend how ## \nabla \times \boldsymbol{\omega} ## fits into the equivalence.
If someone can explain how this works, it would be great.

PS: I hope that the LaTeX insertions work; if not, how do you insert LaTeX code in a post? (solved)

Edit: Latex insertion correction
 
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  • #2
Rikyuri said:
how do you insert LaTeX code in a post?
You put it between ## ... ## for inline or between $$ ... $$ for outline.
 
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  • #3
Be careful as ## \sum_i v_i \partial_i v_k \neq \sum_i v_i \partial_k v_i ##. You should use parentheses, here ##\mathbf V \cdot \nabla \mathbf V## means ##(\mathbf V \cdot \nabla)\mathbf V##. I would start from ##\mathbf V\times \nabla \times \mathbf V##. What techniques do you know ? Do you know Levi-Civita identities?
 
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  • #4
pines-demon said:
Be careful as ## \sum_i v_i \partial_i v_k \neq \sum_i v_i \partial_k v_i ##. You should use parentheses, here ##\mathbf V \cdot \nabla \mathbf V## means ##(\mathbf V \cdot \nabla)\mathbf V##. I would start from ##\mathbf V\times \nabla \times \mathbf V##. What techniques do you know ? Do you know Levi-Civita identities?
Oh, so it is the divergence, not the gradient. I don't really know Levi-Civita identities; I've only heard about them.
Regarding the techniques, I have knowledge of Calculus II.
If can be usefull I started to use Einstein notation for this class, sometimes I struggle a bit with it, but I can uderstand it. Thanks a lot for the help.
 
  • #5
Rikyuri said:
Oh, so it is the divergence, not the gradient. I don't really know Levi-Civita identities; I've only heard about them.
Regarding the techniques, I have knowledge of Calculus II.
If can be usefull I started to use Einstein notation for this class, sometimes I struggle a bit with it, but I can uderstand it. Thanks a lot for the help.
It is not even the divergence or gradient, it is the "vector-dot-del" (##\mathbf V \cdot \nabla ##) operator. If ##\mathbf V = (v_x,v_y,v_z)## then
$$(\mathbf V \cdot\nabla)\mathbf V = (v_x \partial_x+v_y \partial_y+v_z \partial_z)\mathbf V =\begin{pmatrix}[v_x \partial_x+v_y \partial_y+v_z \partial_z]v_x\\
[v_x \partial_x+v_y \partial_y+v_z \partial_z]v_y\\
[v_x \partial_x+v_y \partial_y+v_z \partial_z]v_z\end{pmatrix} $$
Compare with
$$\nabla \left(\frac12 V^2\right)=\begin{pmatrix}
v_x \partial_x v_x+v_y \partial_x v_y+v_z \partial_x v_z\\
v_x \partial_y v_x+v_y \partial_x v_y+v_z \partial_y v_z\\
v_x \partial_z v_x+v_y \partial_z v_y+v_z \partial_z v_z\\
\end{pmatrix}$$
which is totally different.Which calculus identities do you know? You can also just write it in components as I did and see if the relation holds.
 
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  • #6
pines-demon said:
It is not even the divergence or gradient, it is the "vector-dot-del" (##\mathbf V \cdot \nabla ##) operator.
Which calculus identities do you know?
Not much:
$$\nabla \times \nabla f = 0; \nabla \cdot \nabla \times f = 0$$
Those are the only one I remember using a part for this new one.
 
  • #7
Rikyuri said:
Not much:
$$\nabla \times \nabla f = 0; \nabla \cdot \nabla \times f = 0$$
Those are the only one I remember using a part for this new one.
Then I suggest that you just calculate all 3 components of $$\mathbf V \times (\nabla \times \mathbf V)$$ and compare with the other too. There is no easy calculation without other calculus identities.

[Note to mentors: can the title of this thread be changed to (V dot del)V=grad(V^2/2)-(V cross omega)?]
 
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