I'm having issues using Einstein notation in polar coordinates in flat space, I must be missing something basic.

Consider the following example. Take the following metric on a 2+1 spacetime; $ds^2 = -dt^2+dr^2+r^2d\theta^2$. Define the four-vector field $A^\mu = (0,0,f(r ))$ and scalar field $\phi=g(\theta)$.

The gradient in polar coordinates is $\nabla \equiv \vec{r}\partial_r+\vec{\phi}\frac{1}{r}\partial_{\phi} $.

There are two ways of calculating $A^\mu\partial_\mu\phi$ that give me different results:

$$A^\mu \partial_\mu \phi = A^i \partial_i \phi = \vec{A}\cdot \nabla \phi = fg'(\theta)/r$$ or $$A^\mu \partial_\mu \phi = A^i \partial_i \phi = A^\theta\partial_\theta \phi=fg'(\theta)$$

And I get different $r$-dependence... (which you get from the gradient). I know the first answer is correct, so what am I doing wrong in component notation?

Edit The first one is wrong. This only works in an orthonormal basis.

  • $\begingroup$ How do you know the first answer is correct? $\endgroup$
    – knzhou
    May 9 '19 at 9:58
  • $\begingroup$ @knzhou heh because if I use the latter, all my equations become extremely ugly. Proof by aesthetics :-). $\endgroup$ May 9 '19 at 10:14
  • $\begingroup$ What I assume is going on is that you got the definition of $A$ wrong, it's actually $A_\mu$ you gave the components of. So you want to compute $A_\mu \partial^\mu \phi$ and pick up the $1/r$ by raising the index on the partial derivative. $\endgroup$
    – knzhou
    May 9 '19 at 10:18
  • $\begingroup$ @knzhou I thought about that, but If you're given a vector $\vec{A}$, you put it into your 4-vector $A^\mu$ because you want vectors to transform like coordinates, right? $\endgroup$ May 9 '19 at 10:22
  • $\begingroup$ I have no idea what that means. If you're given a 3-vector, it only means that it transforms like a 3-vector under rotations. It could be the spatial part of a 4-vector, but it could also be the spatial part of a one-form (the classic example being $\vec{A}$ itself), and it could also be part of an antisymmetric two-form (one example being $\vec{B}$), and so on. $\endgroup$
    – knzhou
    May 9 '19 at 10:23

The metric


is singular, and therefore not appropriate. So let us instead consider that in Minkowski space

$$ds^2=-dt^2+dr^2+r^2d\theta^2 ,$$

which does not affect the problem you are encountering.

For the above metric, one has $g_{\mu\nu}=(-1,1,r^2)$, which implies its inverse $g^{\mu\nu}=(-1,1,1/r^2)$.

Now, for the contraction, we note that

$$A^\mu B_\mu=g^{\mu\nu}A_\mu B_\nu=g_{\mu\nu}A^\mu B^\nu .$$

Consider the case,

$$A^\mu =(0,0,f(r)) ,$$ $$B_\nu=(\partial_t\phi,\partial_r\phi,\partial_\theta \phi) ,$$

where $$\partial_\mu \phi\equiv \frac{\partial\phi}{\partial x^\mu}$$ with $x^\mu=(t,r,\theta)$.

By putting all the pieces together one finds

$$A^\mu B_\mu=g^{\mu\nu}A_\mu B_\nu=g_{\mu\nu}A^\mu B^\nu=f(r)\partial_\theta\phi .$$

The key point is that the quantity $\nabla\phi$ you wrote above is something else. It is not a convariant neither a contravariant tensor, to be more precise, there is nothing wrong about $\nabla\phi$ being a vector, which is coordinate independent as well as basis independent, but its components in its form given above, namely, $$\nabla \phi=\hat{r}\frac{\partial\phi}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial\phi}{\partial\theta}\to \nabla\phi\stackrel{?}{=}(\partial_r\phi,\frac{1}{r}\partial_\theta\phi),$$ are not those of a covariant or contravariant tensor. As a result, the "contraction" carried out subsequently, in terms of those components of the tensor, does not make sense.

This is because its bases are unit vectors according to the convention where it is introduced (in calculus). Those basis vectors ($\hat{r},\hat{\theta}$) are unitary as well as orthonormal, but are noncoordinate! For all the tensor manipulation to be meaningful, one must work with coordinate bases, and usually the latter are not unit vectors. See for example, the discussions of section 5.5 noncoordinate bases of the textbook A first course in general relativity by Schutz.

This is indeed a confusing point if one is not aware of the difference. If the above explanation is not clear to you, take a look at the textbook mentioned above, and maybe from the beginning of chapter 5, where covariant derivative is introduced by a generic fashion. To me, the textbook is quite well written regarding those basic concepts with a minimal amount of mathematics.

  • 1
    $\begingroup$ The metric is singular in 2+1 spacetime, but ok for pure 2 dimensional space case. But the question was somehow posed in Minkowski spacetime, while $\nabla \phi$ was understood only as spatial gradient. Thus further adds to the ambiguity. $\endgroup$
    – gamebm
    May 9 '19 at 11:53

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