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

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Discussion Overview

The discussion revolves around the vector identity involving the velocity vector ##\mathbf{V}##, its gradient, and the vorticity ##\boldsymbol{\omega}## in the context of aerodynamics. Participants explore the mathematical implications and identities related to this equivalence, focusing on the operations involved and their interpretations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the identity ##\mathbf{V} \cdot \nabla \mathbf{V} = \nabla\left(\frac{V^{2}}{2}\right)-\mathbf{V} \times \boldsymbol{\omega}## and questions how ##\nabla \times \boldsymbol{\omega}## fits into it.
  • Another participant emphasizes the importance of parentheses in the expression, clarifying that ##\mathbf{V} \cdot \nabla \mathbf{V}## should be interpreted as ##(\mathbf{V} \cdot \nabla)\mathbf{V}##.
  • There is a suggestion to start from the expression ##\mathbf{V} \times \nabla \times \mathbf{V}## to analyze the identity further.
  • One participant admits to limited knowledge of Levi-Civita identities and expresses familiarity with Calculus II, indicating some struggle with Einstein notation.
  • Another participant clarifies that ##\mathbf{V} \cdot \nabla## is a "vector-dot-del" operator and provides a detailed component-wise expansion of both sides of the identity to illustrate differences.
  • Participants share basic calculus identities they remember, such as ##\nabla \times \nabla f = 0## and ##\nabla \cdot \nabla \times f = 0##, suggesting these could be relevant to the discussion.
  • There is a recommendation to calculate all components of ##\mathbf{V} \times (\nabla \times \mathbf{V})## to compare with the other expressions, highlighting the complexity of the calculations involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the vector identity, and multiple viewpoints regarding the operations and identities involved remain present throughout the discussion.

Contextual Notes

Participants express uncertainty about specific mathematical identities and operations, indicating that their understanding may depend on further exploration of vector calculus techniques.

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
 
Last edited:
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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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