• Support PF! Buy your school textbooks, materials and every day products Here!

Gradient of vector

  • Thread starter enricfemi
  • Start date
  • #1
191
0
[tex]\nabla[/tex][tex]\stackrel{\rightarrow}{A}[/tex]

when a gradient operater act on a vector,what is it stand for ?
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
[tex]\nabla[/tex][tex]\stackrel{\rightarrow}{A}[/tex]

when a gradient operater act on a vector,what is it stand for ?
Visually, what you wrote looks like

[tex]\nabla_{\vec A}[/tex]

The title of the thread and your LaTeX suggests you meant

[tex]\nabla \vec A[/tex]

These are two different things. The first is an operator, the gradient with respect to the components of [itex]\vec A[/itex], rather than the normal gradient which is take with respect to spatial components. The second form is the gradient of a vector. It is a second-order tensor. If [tex]\vec A = \sum_k a_k \hat x_k[/tex],

[tex](\nabla \vec A)_{i,j} = \frac{\partial a_i}{\partial x_j}[/tex]

BTW, it is best not to separate things the way you did in the original post. Here is your original equation as-is:

[tex]\nabla[/tex][tex]\stackrel{\rightarrow}{A}[/tex]

Now look at how this appears when written as a single LaTeX equation:

[tex]\nabla\stackrel{\rightarrow}{A}[/tex]
 
Last edited:
  • #3
9
0
The second form is the gradient of a vector. It is a second-order tensor. If [tex]\vec A = \sum_k a_k \hat x_k[/tex],

[tex](\nabla \vec A)_{i,j} = \frac{\partial a_i}{\partial x_j}[/tex]
Does this make a matrix using row i and column j for the entries?
 
  • #4
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
Yes.
 
  • #5
9
0
Thank you.
 

Related Threads for: Gradient of vector

Replies
0
Views
3K
  • Last Post
Replies
1
Views
647
  • Last Post
Replies
8
Views
61K
  • Last Post
Replies
5
Views
14K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
4
Views
5K
Replies
9
Views
1K
Top