If we consider a coordinate transformation defined by the rule $$x^{\mu}\rightarrow x'^{\mu}=x'^{\mu}(x)$$ If we consider the old coordinates as functions of the new ones, then the coordinate differentials transform according to the following rule (using the chain rule): $$dx^{\mu}=\frac{\partial x^{\mu}(x')}{\partial x'^{\nu}}dx'^{\nu}~~~~~~~~~~~~~~~(1).$$ The $dx^{\mu}$'s are type (1,0) contravariant tensors, and hence I can interpret it as a column vector (consider, for example, $\mu=1,2,3$): $$dx^{\mu}\equiv d\vec{x}=(dx^1,dx^2,dx^3)^T$$

I want to know how the covariant $dx_{\mu}$'s transform, and more importantly how to interpret them. Do they transform like (2) or like (3)?

$$ dx_{\mu}=\frac{\partial x^{\nu}(x')}{\partial x'^{\mu}}dx'_{\nu}~~~~~~~~~~~~~~~(2)\\ dx_{\mu}=\frac{\partial x_{\mu}(x')}{\partial x'^{\nu}}dx'^{\nu}~~~~~~~~~~~~~~~(3) $$ I obtain (2) by using the standard type (p,q) tensor transformation law and applying it to a type (0,1) tensor.

(3) is what I get If I write; $$dx_{\mu}=g_{\mu\gamma}dx^{\gamma}=g_{\mu\gamma}\frac{\partial x^{\gamma}(x')}{\partial x'^{\nu}}dx'^{\nu}=\frac{\partial x_{\mu}(x')}{\partial x'^{\nu}}dx'^{\nu}$$ Where $g_{\mu\gamma}$ is the metric tensor used to raise and lower indices. In this formula the $x_{\mu}$'s scare me a little as I don't know how to interpret them.

In (1) and (2) the $\frac{\partial x^{\mu}(x')}{\partial x'^{\nu}}$'s are easy to interpret since, for example, in a transformation to spherical coordinates I know what the $x^{\mu}(x')$'s look like (they are just $x=r\cos{\theta}\sin{\phi}$ etc). But what do the $x_{\mu}$'s look like in the case of spherical coordinates?

My understanding is this;

The $x^{\mu}$'s are type (1,0) tensors which are elements of one copy of the vector space, i.e. a (column) vector. Going back to the spherical coordinates example; $$x^{\mu}=(x^1,x^2,x^3)^T=(x(r,\theta,\phi),y(r,\theta,\phi),y(r,\theta,\phi))^T=(r\cos{\theta}\sin{\phi},r\sin{\theta}\sin{\phi},r\cos{\phi})^T$$ And so each component of the vector $x^{\mu}$ is a smooth function (on a manifold, with charts and all) which spits out a number in $\mathbb{R}$.

On the other hand, the covariant type (0,1) tensors $x_{\mu}$ are elements of the dual of the corresponding vector space and are hence linear functionals. These guys eat vectors and spit out real numbers. They too can be considered as (row) vectors. But what does such an $x_{\mu}$ explicitly look like in the case of spherical coordinates?

How can a metric tensor turn something familiar like $x^{\mu}$ into something unfamiliar which I don't even know how to express in the case of spherical coordinates. Further more, how would I carry out the partial differentiation in $\frac{\partial x_{\mu}(x')}{\partial x'^{\nu}}$ above?

Finally, I would like to point out that, to me, there doesn't seem to be any difference between the components of $x_{\mu}$ and a linear functional. In the spherical coordinates example, the first component of $x^{\mu}$ is $x^1=x(r,\theta,\phi)=r\cos{\theta}\sin{\phi}$. This is how I think of a functional; it eats a vector $x(\vec{r})=x((r,\theta,\phi))$ and spits out a real number (although I don't think it is a linear functional, perhaps infintesimally it is though). Is there a deeper meaning behind this?

Sorry for the lengthy question, but I wanted to make my confusion clear.

  • 1
    $\begingroup$ The coordinates $x^\mu$ are not a vector and you should not take the raised index on them too literally. As such, I don't think that $dx_\mu$ has any meaning. (Note that if you are using Cartesian coordinates in flat Minkowski space, then that is the sole exception where the coordinates can be treated as a vector. But then you wouldn't be doing coordinate transformations, except for linear ones.) $\endgroup$ – Ben Niehoff Apr 24 '17 at 11:29
  • $\begingroup$ In the derivation of (3) you cannot get g under the derivative symbol. $\endgroup$ – lalala Apr 24 '17 at 11:34
  • $\begingroup$ Echoing above comments to the post (v1): $dx_{\mu}$ in eqs. (2) & (3) is non-standard. Usually one only considers $dx^{\mu}$. $\endgroup$ – Qmechanic Apr 24 '17 at 12:47
  • $\begingroup$ Thank you for your comments. What is the best way to interpret the following: $$g_{\mu\nu}x^{\nu}=x_{\mu}$$ I am looking for a deeper explanation than the typical; "If $x^{\mu}$ are the components of a column vector then $g_{\mu\nu}x^{\nu}=x_{\mu}$ is just the components of the corresponding transpose" (in flat space).$$~$$ Is the best way to think of the difference between the two simply by the way in which they transform? $\endgroup$ – NormalsNotFar Apr 26 '17 at 11:10

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