Understanding Index Notation and Tensor Operations in Vector Calculus

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SUMMARY

The discussion centers on the validity of the equation involving vector calculus: \(\frac{1}{2}\nabla v^2 - v \cdot \nabla v = v \times \nabla \times v\). The user argues that the left-hand side can be simplified to \(v_j \partial_i v_j\), which they believe equals zero. However, the response clarifies that \(v_j \partial_i v_j\) is not equivalent to \((v \cdot \nabla) v_i\) due to the distinction between free and dummy indices in tensor notation. Additionally, the user inquires about the identity \(\nabla \cdot (\nabla u) = \nabla (\nabla \cdot u)\), seeking confirmation of its validity.

PREREQUISITES
  • Understanding of vector calculus and tensor notation
  • Familiarity with the concepts of partial differentiation
  • Knowledge of free and dummy indices in tensor operations
  • Basic principles of vector fields and operations like divergence and curl
NEXT STEPS
  • Study the properties of free and dummy indices in tensor calculus
  • Learn about the divergence and curl operations in vector fields
  • Explore the implications of the identity \(\nabla \cdot (\nabla u) = \nabla (\nabla \cdot u)\)
  • Review advanced vector calculus topics, including the use of index notation in physics
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with vector calculus and tensor operations, particularly those focusing on fluid dynamics or electromagnetism.

Niles
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Homework Statement


Hi

I have a vector v. According to my book, the following is valid:
<br /> \frac{1}{2}\nabla v^2-v\cdot \nabla v = v\times \nabla \times v <br />
I disagree with this, because the first term on the LHS I can write as (partial differentiation)
<br /> \frac{1}{2}\partial_i v_jv_j = v_j\partial_i v_j <br />
which is just v\cdot \nabla v. So IMO it should equal 0.

What is wrong here?
 
Last edited:
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Niles said:

Homework Statement


Hi

I have a vector v. According to my book, the following is valid:
<br /> \frac{1}{2}\nabla v^2-v\cdot \nabla v = v\times \nabla \times v <br />

I disagree with this, because the first term on the LHS I can write as (partial differentiation)
<br /> \frac{1}{2}\partial_i v_jv_j = v_j\partial_i v_j <br />
which is just v\cdot \nabla v. So IMO it should equal 0.

What is wrong here?

v_j \partial_i v_j \neq v_j \partial_j v_i = (v \cdot \nabla) v_i

Note that on the left the index on the derivative is i (a free index), but on the right the index on the derivative is j (a dummy index).
 
Thanks. I hope there is omething else you can help me with (you seem to have experience with this kind of notation). Is the following always true for some vector u?

<br /> \nabla \cdot (\nabla u) = \nabla (\nabla \cdot u)<br />
 

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