Direct Proof of Div, Grad, and Curl Operator Identities

  • Thread starter Thread starter unscientific
  • Start date Start date
  • Tags Tags
    Curl Grad
Click For Summary
SUMMARY

The discussion focuses on proving the identities of the divergence, gradient, and curl operators in vector calculus, specifically addressing the identity involving the triple cross product and the gradient operator. The participant successfully demonstrated that the triple cross product equals 2a using Einstein notation and established that 2∇(a·r) equals 2a, confirming the left-hand side of the equation. The conversation highlights a desire for a more elegant proof method, indicating that while the current approach is valid, participants seek a more streamlined solution.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically divergence, gradient, and curl operators.
  • Familiarity with Einstein notation and its application in vector identities.
  • Knowledge of vector operations, including the triple cross product.
  • Basic proficiency in mathematical proofs and identities in calculus.
NEXT STEPS
  • Research the properties and applications of the divergence operator in vector fields.
  • Study the gradient operator and its role in scalar field analysis.
  • Explore the curl operator and its significance in fluid dynamics and electromagnetism.
  • Learn advanced techniques for proving vector identities, including alternative proof methods and shortcuts.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are looking to deepen their understanding of vector calculus and improve their skills in proving operator identities.

unscientific
Messages
1,728
Reaction score
13

Homework Statement



16ixf0z.png



Homework Equations





The Attempt at a Solution



For part (c), I showed that the tripple cross product = 2a using einstein notation. Then, I showed that 2∇(a.r) = 2a which is the same as LHS. I don't think this is as elegant as it can get..

How do I prove it directly? I've looked up some operator identities:
d5b8119de472d7a6d75c1672e93ccd34.png
 
Physics news on Phys.org
Your method is fine. I would have done the same thing. I know what you mean though. It feels like there should be some kind of shortcut method.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
10K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K