Q: Consider a function f = r^2 given in the spherical coordinate system. Using the expression for the scale factors, as well as the formula for the Laplacian compute the Laplacian ∇^2 f . The result is a number times r^(n-2) . State this number.
Not sure where I've gone wrong with this, I think I'm close to the answer but I can't seem to get r^(n-2), or get rise of cosec(theta). I'll show my workings:
f = r^2 hr = 1 htheta = r hphi = rsintheta
grad f = 1/hr * ∂f/∂r * er + 1/htheta * ∂f/∂theta * etheta + 1/hphi * ∂f/∂phi * ephi
= 2r er
∇ . A = 1/h1h2h3 [∂(h2h3A1)∂u1 + ∂(h1h3A2)/∂u2 + ∂(h1h2A3)/∂u3]
∇ = 2rer = A1e1 where A1 = Ar = 2r
∇ . A = 1/hrhthetahphi (∂(r * rsintheta * A1)/∂r + ∂(1 * rsintheta * A2)/∂theta + ∂(1 * r * A3)/∂phi)
Now A2 = A3 = 0, A1 = 2r
∇ . A = 1/r^2*sin^2theta * ∂/∂r (r^2 * sin(theta) *2r)
= 1/r^2*sin^2theta * ∂/∂r (2r^3 sin(theta))
= 1/r^2*sin^2 (theta) * 6r^2 sin(theta)
= 6/sin(theta)
Now I know that the correct answer is 6, so am I correct? Is my working correct?
I must admit I find our OP Sam Connell's question a little unclear; nevertheless I feel I can make enough sense out of it to give it a try, so here goes:
We first find $\nabla^2 r^2$ in three dimensions:
By definition we have
$\nabla^2 r^2 = \nabla \cdot (\nabla r^2); \tag 1$
it is both well-known and easy to see (e.g. by working in cartesian coordinates) that
$\nabla r^2 = \dfrac{d(r^2)}{dr} \nabla r = 2r \nabla r; \tag 2$
$\nabla r = \dfrac{\partial r}{\partial r} \hat e_r = \hat e_r, \tag 3$
where $\hat e_r$ denotes the unit vector field tangent to the radial lines and pointing in the outward direction; combining (2) and (3) yields
$\nabla r^2 = 2r \hat e_r; \tag 4$
thus,
$\nabla^2 r^2 = \nabla \cdot (2r \hat e_r) = \nabla (2r) \cdot \hat e_r + 2r \nabla \cdot \hat e_r$ $= 2\nabla r \cdot \hat e_r + 2r \nabla \cdot \hat e_r = 2 \hat e_r \cdot \hat e_r + 2r \nabla \cdot \hat e_r = 2 + 2r \nabla \cdot \hat e_r, \tag 5$
where we have exploited the well-known identity $\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X$, holding for any differentiable function $f$ and vector field $X$; here $f = 2r$ and $X = \hat e_r$; now,
$\hat e_r = r^{-1} \vec r, \tag 6$
and it is easy to compute that
$\nabla \cdot \vec r = 3, \tag 7$
since $\vec r = (x, y, z) \in \Bbb R^3$; therefore,
$\nabla \cdot \hat e_r = \nabla \cdot (r^{-1} \vec r) = \nabla (r^{-1}) \cdot \vec r + r^{-1} \nabla \cdot r$ $= -r^{-2} \hat e_r \cdot \vec r + r^{-1}\nabla \cdot \vec r$ $= -r^{-2} r + 3r^{-1} = 2r^{-1}; \tag 8$
$\nabla^2 r^2 = 2 + 2r\nabla \cdot \hat e_r = 2 + 2r(2r^{-1}) = 6. \tag 9$