In the opening chapters of typical QFT books, the covariant coordinates $x_\mu = g _{\mu\nu}x^\nu$ $x^\mu = (t,x,y,z)$ and the differential operator $\partial^\mu = \frac{\partial}{\partial x_{\mu}}=(\frac{\partial}{\partial t}, -\nabla)$ are introduced to facilite the evaluation of spacetime interval $x^2$ and the magnitude square of a four gradient of some scalar field $\partial ^\mu \phi \partial _\mu \phi$. On the other hand, we do not introduce the $x_\mu$ in general relativity where the spacetime is curved and our coordinate is curvilinear, in general.

My questions are the followings:

  1. What is the reason for the absence of $x_{\mu}$ in general relativity where curvilinear coordinates are necessary for describing a curved spacetime?

2.Does $x_\mu = g_{\mu\nu}x^\nu$ make sense only in the flat spacetime where the coordinate $x^\mu$ themselves are contravariant component of some vector, and the $x^\mu$ transform like $x^{\prime\mu} = \frac{\partial x^{\prime\mu}}{\partial x^\nu}x^\nu$ in this special circumstance?

I would like to share some of my opinions for further discussion. (Please evaluate and point out the mistakes of my arguments.) In fact, $\partial^\mu $ is fully equivalent to $g^{\mu\nu}\partial_\nu $ in flat spacetime characterized by a constant metric. ($x_\mu = g_{\mu\nu}x^\nu$, $x^\mu = g^{\mu\nu}x_\nu$ so $\partial^{\mu} = \frac{\partial}{\partial x_{\mu}}=\frac{\partial x^{\nu}}{\partial x_{\mu}}\frac{\partial}{\partial x^{\nu}}=g^{\mu\nu}\partial_{\nu}$ We can always work only with $x^\mu$ and $\partial_\mu $ at the expense of always carrying the inverse metric $g^{\mu\nu}$ in the kinetic term of the Lagrangian.

However, if $x^\mu$ is a set of curvilinear coordinates describing a curved spacetime, then the coordinates $x^\mu$ cease to carry any contravariant vectorial significance and they are merely a label of a point in spacetime without any vectorial meaning. (Euclidean analogy: $(x,y,z)$ are the components of a position vector $\vec{r}$ but a general curvilinear coordinate describing the same Euclidean space are not components of a vector, in general.) Consequently, it no longer makes sense to define $x_{\mu} = g_{\mu\nu}x^{\nu}$ for a general coordinate $x^\mu$ describing a curved spacetime. (We could do this in a Euclidean Space with $x,y,z$ but this is redundant as the metric is identity. On the other hand, $x_\mu$ is not equal to $x^\mu$ in a flat-spacetime due to the pseudo-Euclidean nature of Minkowskian spacetime. This is the reason why we need $x_{\mu}$ and $x^{\mu}$ in relativistic QFT.)

On the other hand, a vector $\vec{V}$ always have two kinds of components $V^\mu$ and $ V_\mu$, regardless of the spacetime geometry and the coordinate system describing it. Now the covariant components of the gradient of a scalar field $\phi$ is naturally obtained by $\frac{\partial \phi}{\partial x^\mu}$. If there is no such thing as $x_\mu$ for a set of general coordinate $x^ \mu$ describing a curved spacetime, then the only way to obtain the contravariant component of the gradient is to employ the inverse metric $g^{\mu\nu}\frac{\partial}{\partial x^ \nu}$. (And we have to raise the lower index that comes from taking derivative with respect to coordinate $x^\mu$ in this manner using the inverse metric when manipulating other field variables.) This might be the reason why there is no $x_\mu$ in general relativity. In summary, the introduction and usage of $x_\mu$ depend on the contravariant vectorial nature of $x^{\mu}$ which depends on the flatness of spacetime and linearity of the coordinate system describing it. (Am I correct?)

  • $\begingroup$ You are essentially entirely correct, which makes an answer somewhat redundant. $\endgroup$ – Erik Jörgenfelt Oct 21 '17 at 10:23

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