Deriving the Vector Identity: $\nabla(\vec{A} \cdot \vec{B})$

AI Thread Summary
The discussion centers on deriving the vector identity $\nabla(\vec{A} \cdot \vec{B})$, with a specific focus on the equation provided. A participant expresses frustration after reaching a dead end using analytical methods and seeks alternative approaches due to an impending exam. There is a suggestion to organize the work better and use LaTeX for clarity, but the participant feels pressed for time and unable to format their work. The conversation highlights the complexity of the derivation and the need for clear communication in mathematical discussions. Overall, the thread emphasizes the challenges of mastering vector calculus identities under time constraints.
Xsnac
Messages
32
Reaction score
1

Homework Statement


I'm trying to derive the vector identity:
$$\nabla(\vec{A} \cdot \vec{B})$$

Homework Equations


$$ \nabla(\vec{A} \cdot \vec{B})=(\vec{B} \cdot \nabla) \vec{A} + ( \vec{A} \cdot \nabla ) \vec{B} + \vec{B} \times (\nabla \times \vec{A})+ \vec{A} \times ( \nabla \times \vec{B})$$

The Attempt at a Solution


I tried to do it using analitical methods and I think I hit a dead end.
I tried everything, even the reverse start from the $$(\vec{B} \cdot \nabla) \vec{A} + ( \vec{A} \cdot \nabla ) \vec{B} + \vec{B} \times (\nabla \times \vec{A})+ \vec{A} \times ( \nabla \times \vec{B})$$ part but this is the best I could get

Scan.jpg
At this point I'm even willing to learn a totaly new method .. I have an exam tomorrow and this is the only one I can't get right.​
 
Physics news on Phys.org
Xsnac said:

Homework Statement


I'm trying to derive the vector identity:
$$\nabla(\vec{A} \cdot \vec{B})$$
This isn't an identity. Identities generally have equal signs in them.

Homework Equations


$$ \nabla(\vec{A} \cdot \vec{B})=(\vec{B} \cdot \nabla) \vec{A} + ( \vec{A} \cdot \nabla ) \vec{B} + \vec{B} \times (\nabla \times \vec{A})+ \vec{A} \times ( \nabla \times \vec{B})$$

The Attempt at a Solution


I tried to do it using analytical methods and I think I hit a dead end.
I tried everything, even the reverse start from the $$(\vec{B} \cdot \nabla) \vec{A} + ( \vec{A} \cdot \nabla ) \vec{B} + \vec{B} \times (\nabla \times \vec{A})+ \vec{A} \times ( \nabla \times \vec{B})$$ part but this is the best I could get

Scan.jpg
At this point I'm even willing to learn a totaly new method .. I have an exam tomorrow and this is the only one I can't get right.​
Speaking personally, I'm not really prone to putting in the effort to follow your chicken scratches. Why don't you try organizing your work in digestible chunks and posting it in LaTeX.
 
vela said:
This isn't an identity. Identities generally have equal signs in them.Speaking personally, I'm not really prone to putting in the effort to follow your chicken scratches. Why don't you try organizing your work in digestible chunks and posting it in LaTeX.

I don't have 3 hours to format a text... I'm practicing for tomorrow's exam... I put a lot of effort to write the small pieces of latex code in this post aswel. (forgot all the syntax and got to relearn it today..)
And I'm looking for some other way since mine I think is a dead-end.
 
Well, the approach you're taking is the one I would use. Omit the unit vector stuff. It just clutters up the derivation.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top