Why Does the Divergence of r-hat Over r-squared Equal Zero?

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Homework Help Overview

The discussion centers around the divergence of the unit vector \(\hat{r}\) divided by the square of its magnitude \(|r|^2\), specifically questioning why this expression equals zero. The subject area involves vector calculus and divergence operations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to understand their mistake in the calculation involving the divergence of \(\hat{r}\) and \(|r|^2\). Some participants question the differentiation process and suggest reviewing the definitions of the vectors involved.

Discussion Status

Participants are actively engaging with the original poster's attempt, with some providing feedback on the need for clearer calculations and others suggesting a reevaluation of the vector definitions. There is no explicit consensus yet, as various interpretations and corrections are being explored.

Contextual Notes

There is an indication that the original poster may have misinterpreted the relationship between the unit vector and the position vector, which is under discussion. The nature of the homework problem implies constraints on the level of assistance that can be provided.

manimaran1605
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Homework Statement



How divergence of r (cap)/ |r|^2 is equal to zero?



Homework Equations



r(cap)= x(cap)+y(cap)+z(cap)

|r|^2 as x^2+y^2+z^2


The Attempt at a Solution



I tried the problem and end up with with a different solution I took r(cap)= x(cap)+y(cap)+z(cap)
|r|^2 as x^2+y^2+z^2. Tell me where i went wrong?

where r(cap) is unit vector of r vector
|r|^2 is the square of modulus of r
 
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Could you please provide the calculation that lead you to the wrong answer? It is hard to pinpoint where you went wrong otherwise.
 
You didn't differentiate correctly. Show your work.
 
Here's my attempt to the problem
 

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Well ##\mathbf{\hat r} = \mathbf{\hat x} + \mathbf{\hat y} + \mathbf{\hat z}## seems false. You have ##\mathbf{r} = (x,y,z)##. Work from there.
 
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