The gradient in spherical coordinates is given by $$\left(\partial_r f, \frac{1}{r} \partial_\theta f, \frac{1}{r \sin \phi}\partial_\phi f\right)$$

However, I get a wrong answer if I try to compute it a different way, by lowering the index of the differential using the metric in spherical coordinates. The metric in spherical coordinates is $$g = \begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2 \phi \end{pmatrix}$$ So if I take $g^{-1} (df) = g^{-1} (\partial_r f \; dr + \cdots)$, then I get $$\left(\partial_r f, \frac{1}{r^2} \partial_\theta f, \frac{1}{r^2 \sin^2 \phi}\partial_\phi f\right)$$

What's going wrong here?


1 Answer 1


Differential geometry seldom users orthonormal bases the way vector calculus does. Your expression for the gradient to start with is in terms of an orthonormal basis, but the metric you used is incompatible with that; it uses the actual coordinate basis. Try writing the gradient in terms of the same basis that you use for the metric and try again.

  • $\begingroup$ I'm sorry, I don't understand what you mean. The basis is $\partial_r, \partial_\theta, \partial_\phi$ for the tangent space, and $dr, d\theta, d\phi$ for the cotangent space? To get from cotangent to tangent I just multiply by the inverse of the metric right? $\endgroup$
    – user182973
    Commented Jun 28, 2014 at 6:16
  • $\begingroup$ @user182973 The expression you gave for "The gradient in spherical coordinates is given by..." isn't written in terms of either of those bases; that's why you're getting the wrong answer. There metric concerts from tangent to cotangent fine, but you're comparing against a vector written in neither basis, so you appear to get the wrong answer. $\endgroup$
    – Muphrid
    Commented Jun 28, 2014 at 6:27
  • $\begingroup$ Aha, ok. So one basis is normalized and the others aren't. To normalize, we multiply by $1/r, 1/r \sin \theta$, and then we see that these vectors have length 1 under this metric. For future reference, there is another thread that explains something similar here math.stackexchange.com/questions/261830/… $\endgroup$
    – user182973
    Commented Jun 28, 2014 at 6:43

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