# Magnitude of a vector in general relativity

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

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