Frankel 2.1 Problems


If \(\mathbf{v}\) is a vector and \(\alpha\) is a covector, then

\sum a_i^V v_V^i &= \sum_i \left[ \sum_j \frac{\partial x_U^j}{\partial x_V^i}a_j^U \right] \left[ \sum_k \frac{\partial x_V^i}{\partial x_U^k}v_U^k \right] \\
&= \sum_{i,j,k} \frac{\partial x_U^j}{\partial x_V^i} \frac{\partial x_V^i}{\partial x_U^k}a_j^U v_U^k.


$$\frac{\partial x_U^j}{\partial x_V^i} \frac{\partial x_V^i}{\partial x_U^k} = \delta_k^j,$$

we have

$$\sum a_i^V v_V^i = \sum a_i^U v_U^i.$$

If instead we were to look at the sum \(\sum v^i w^i\) where \(\mathbf{w}\) is another vector, then we would get a coordinate dependent result.



Let \(x^1\), \(x^2\), and \(x^3\) be the usual cartesian coordinates in \(\mathbb{R}^3\) and let \(u^1 = r\), \(u^2 = \theta\) (colatitude), and \(u^3 = \phi\) be spherical coordinates.

(i)    The metric tensor for spherical coordinates is defined

$$g_{ij} = \langle \partial_{i}, \partial_{j} \rangle.$$

Since we know \(g_{ij} = \delta_{ij}\) in cartesian coordinates, we can easily get the metric in spherical coordinates by the transformation

$$g_{ij} = \sum_{rs} \frac{\partial x^r}{\partial u^i}\frac{\partial x^s}{\partial u^j}\delta_{rs}.$$

Therefore the metric tensor components are

g_{11} &= 1, \\
g_{22} &= r^2, \\
g_{33} &= 2r^2\cos^2\theta, \\
g_{12} &= g_{21} = 0, \\
g_{13} &= g_{31} = 0, \\
g_{23} &= g_{32} = 0.

(ii)   We can use this to compute the components of the gradient vector \(\nabla f\),

$$(\nabla f)^i = \sum_j g^{ij}\frac{\partial f}{\partial u^j}.$$

So then we have

(\nabla f)^1 &= 1, \\
(\nabla f)^2 &= 1/r^2, \\
(\nabla f)^3 &= 1/2r^2\cos^2\theta.\\

(iii)   The orthogonality of the spherical basis is evident from the diagonality of \(g\), and the non-normality is evident from the diagonal itself. If we rescale the coordinate system we can make it orthonormal. To do this we divide every coordinate vector by its magnitude. This makes the gradient 

$$\nabla f = \frac{\partial f}{\partial r} + \frac{1}{r}\frac{\partial f}{\partial \theta} + \frac{1}{2r\cos\theta}\frac{\partial f}{\partial \phi}.$$