What does this expression involving Partial Derivatives mean?

AI Thread Summary
The discussion centers on understanding the expression involving partial derivatives, specifically in the context of a symmetric function with three parameters. Participants clarify that the indices used (i, j, k) are interchangeable with (1, 2, 3) but do not specify which corresponds to which. It is emphasized that the function's symmetry allows for this substitution, and the partial derivatives can be computed accordingly. A suggestion is made to use a brute force method to derive the expression, while also noting the importance of maintaining cyclic order among the indices. The conversation highlights the complexity of the problem and the need for careful consideration of the indices involved.
physicss
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Homework Statement
Hello, what does this expression mean?
Relevant Equations
(Picture)
I already solved w x x/|x|
For (w1,w2,w3) and (x1,x2,x3)
2E486A9A-524A-4515-AC6C-71F2B9313E92.jpeg
 
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physicss said:
Homework Statement: Hello, what does this expression mean?
Relevant Equations: (Picture)

I already solved w x x/|x|
For (w1,w2,w3) and (x1,x2,x3) View attachment 327170
Then you just have to take the partial derivative wrt ##x_i## and again wrt ##x_j##.
 
haruspex said:
Then you just have to take the partial derivative wrt ##x_i## and again wrt ##x_j##.
Thanks for the answer. Would xi and xj be x1 and x2 in this case?
 
physicss said:
Thanks for the answer. Would xi and xj be x1 and x2 in this case?
No. Because the function is symmetric in the three parameters, you can replace them with ##x_i##, ##x_j##, ##x_k##, where it is understood that {i,j,k}={1,2,3}, but which is which is unspecified.
For example, suppose you had the function ##x_1x_2x_3## then its partial derivative wrt ##x_i## and ##x_j## would be ##x_k##.

Edit, you might also need to assume that i, j, k are in the same cyclic order as 1, 2, 3.

Edit 2: Just realised my posts may be off the mark. I need to solve it myself first.

Edit 3:
Rereading the question, I see it does not refer to indices 1, 2, 3. That is something you assumed. So my correct answer to your post #3 is:

Yes, they are using i, j, k as the indices, not 1, 2, 3.
 
haruspex said:
No. Because the function is symmetric in the three parameters, you can replace them with ##x_i##, ##x_j##, ##x_k##, where it is understood that {i,j,k}={1,2,3}, but which is which is unspecified.
For example, suppose you had the function ##x_1x_2x_3## then its partial derivative wrt ##x_i## and ##x_j## would be ##x_k##.

Edit, you might also need to assume that i, j, k are in the same cyclic order as 1, 2, 3.

Edit 2: Just realised my posts may be off the mark. I need to solve it myself first.

Edit 3:
Rereading the question, I see it does not refer to indices 1, 2, 3. That is something you assumed. So my correct answer to your post #3 is:

Yes, they are using i, j, k as the indices, not 1, 2, 3.
Thank you
 
Presumably ##\vec {\omega}## is constant and does not depend on the ##x_i##. I would try the brute force method which is always safe.
  1. Write ##\dfrac{\vec x}{|\vec x|}=\dfrac{x_1~\hat{x}_1+x_2~\hat{x}_2+x_3~\hat{x}_3}{\left[x_1^2+x_2^2+x_3^2 \right]^{1/2}}.##
  2. Find ##\dfrac{\partial^2}{\partial x_1\partial x_2}\left( \dfrac{x_1~\hat{x}_1+x_2~\hat{x}_2+x_3~\hat{x}_3}{\left[x_1^2+x_2^2+x_3^2 \right]^{1/2}} \right)##.
  3. Do a cyclic permutation of indices to find the other two terms.
  4. Take the cross product.
There might a simpler way to do this but I can't see what it is. I assume that in your original expression you have "off-diagonal" elements only, i.e. it is stipulated somewhere that ##i\neq j##.
 
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