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|

1 Answer

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  • kb
    Lv 7
    7 years ago
    Best Answer

    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!

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