I know that the magnitude of a vector $\mathbf{a}$ is given by,

$$ |\mathbf{a}|^2 = a_i a^i = g_{ij} a^i a^j$$

In 3 dimensions and Cartesian coordinates, this makes sense since then $g_{ij} = (1,1,1)$ and so,

$$ |\mathbf{a}|^2 = 1 \cdot x^2 + 1 \cdot y^2 + 1 \cdot z^2$$

which is the classic Pythagorean interval.

Now, suppose we are now in spherical polar coordinates $(r,\theta, \phi)$. We know apriori that the magnitude of the vector is just $r$. The flat 3d metric is now $g_{ij} = (1, r^2, r^2 \sin^2\theta)$. Then by the same logic as the Cartesian case,

$$ |\mathbf{a}|^2 = 1 \cdot r^2 + r^2 \cdot \theta^2 + r^2 \sin^2 \theta \cdot \phi^2$$

which clearly $\ne$ r.

Obviously I am misunderstanding something and would appreciate any guidance on what that is. Thanks

  • 2
    $\begingroup$ The position vector in polar coordinates does not have components $(r,\theta,\phi)$ $\endgroup$ – jacob1729 May 24 '18 at 18:49
  • $\begingroup$ That metric is for a small, local displacement about that point, not for the (huge) vector you're using. $\endgroup$ – Emilio Pisanty May 24 '18 at 18:49
  • $\begingroup$ Position is not a vector when you're working on a manifold. Only infinitesimal changes in position act like vectors. $\endgroup$ – Ben Crowell May 24 '18 at 19:03
  • $\begingroup$ I think you have your answer in the comments $\endgroup$ – ggcg May 24 '18 at 21:25
  • $\begingroup$ Can you explain what you mean by 'The position vector in polar coordinates does not have components (r,θ,ϕ)' $\endgroup$ – user1887919 May 29 '18 at 10:05

This is a common confusion, resulting from our Cartesian intuition about vectors. In a vector space, both vectors and points are given by a set of numbers, and one can identify the set of vectors originating from some point with the set of vectors originating from the origin, which is identified with the vector space itself. In a picture:

enter image description here

When one does differential geometry, the situation is different. The metric tensor $g$ is giving the inner product on the tangent space of each point in space, but those tangent spaces are different, as illustrated by this picture (red points to be explained soon):

enter image description here

The problem with the calculation you are doing, is that you are trying to compute a norm for $(r,\theta,\phi)$, which is a point rather than a tangent vector. Looking at this picture, what you are trying to do is to assign a norm to one of the red points - which does not make sense. The objects the metric do assign a norm to are the tangent vectors.

2nd Picture source: http://www.gregegan.net/ORTHOGONAL/03/WavesExtra.html


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