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

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SUMMARY

The forum discussion centers on the vector identity involving the velocity vector \(\mathbf{V}\) and its relationship with the gradient and curl operators. The identity presented is \(\mathbf{V} \cdot \nabla \mathbf{V} = \nabla\left(\frac{V^{2}}{2}\right) - \mathbf{V} \times \boldsymbol{\omega}\), where \(\boldsymbol{\omega} = \nabla \times \mathbf{V}\). Participants clarify that \(\mathbf{V} \cdot \nabla\) represents the vector-dot-del operator, not the gradient or divergence. They emphasize the importance of using parentheses in vector calculus expressions and suggest calculating components to verify the identity.

PREREQUISITES
  • Understanding of vector calculus, specifically vector identities.
  • Familiarity with the gradient (\(\nabla\)) and curl (\(\nabla \times\)) operators.
  • Knowledge of Einstein notation for tensor calculus.
  • Basic proficiency in Calculus II concepts.
NEXT STEPS
  • Study the Levi-Civita identities and their applications in vector calculus.
  • Learn how to compute the curl and divergence of vector fields.
  • Practice using the vector-dot-del operator in various vector calculus problems.
  • Explore the implications of vector identities in fluid dynamics and aerodynamics.
USEFUL FOR

Aerodynamics students, physics majors, and anyone interested in advanced vector calculus applications in fluid dynamics will benefit from this discussion.

Rikyuri
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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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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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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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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.
 
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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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.
 
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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