lobo
Lv 6
lobo asked in Science & MathematicsMathematics · 7 years ago

# how do you show that the divergence of a unit vector r(hat) = 2/|r|?

note that r(hat) = r(vector)/|r|

Relevance
• kb
Lv 7
7 years ago

Let r = (x(1), x(2), ..., x(n)).

So, |r| = √[(x(1))^2 + (x(2))^2 + ... + (x(n))^2].

Hence, r(hat)

= (x(1)/√[(x(1))^2 + (x(2))^2 + ... + (x(n))^2], ..., x(n)/√[(x(1))^2 + (x(2))^2 + ... + (x(n))^2]).

For the divergence, we need to compute for each k = 1, 2, ..., n:

(∂/∂x(k)) x(k)/√[(x(1))^2 + (x(2))^2 + ... + (x(n))^2]

= (∂/∂x(k)) [x(k) * ((x(1))^2 + (x(2))^2 + ... + (x(n))^2)^(-1/2)]

= 1 * ((x(1))^2 + ... + (x(n))^2)^(-1/2) + x(k) * -x(k) ((x(1))^2 + ... + (x(n))^2)^(-3/2)

= [((x(1))^2 + ... + (x(n))^2) - (x(k))^2] ((x(1))^2 + ... + (x(n))^2)^(-3/2).

Therefore, div(r(hat))

= Σ(k = 1 to n) (∂/∂x(k)) x(k)/√[(x(1))^2 + (x(2))^2 + ... + (x(n))^2]

= Σ(k = 1 to n) [((x(1))^2 + ... + (x(n))^2) - (x(k))^2] ((x(1))^2 + ... + (x(n))^2)^(-3/2)

= ((x(1))^2 + ... + (x(n))^2)^(-3/2) * Σ(k = 1 to n) [((x(1))^2 + ... + (x(n))^2) - (x(k))^2]

= ((x(1))^2 + ... + (x(n))^2)^(-3/2) * [n ((x(1))^2 + ... + (x(n))^2) - (x(1))^2 + ... + (x(n))^2)]

= ((x(1))^2 + ... + (x(n))^2)^(-3/2) * (n-1) ((x(1))^2 + ... + (x(n))^2)

= (n-1) * ((x(1))^2 + ... + (x(n))^2)^(-1/2)

= (n-1)/|r|.

In the case where n = 3, this reduces to 2/|r|.

I hope this helps!