I Explore Coordinate Dependent Statements in Orodruin's Insight

[George Keeling]

TL;DR Summary

Peculiar Coordinate conversion in Orodruin's Insight "Explore Coordinate Dependent Statements in an Expanding Universe"

I am studying @Orodruin's Insight "Explore Coordinate Dependent Statements in an Expanding Universe". It looks pretty interesting. About three pages in it reads "expanding $x^a$ to second order in $\xi^\mu$ generally leads to$$
x^a=e_\mu^a\xi^\mu+c_{\mu\nu}^a\xi^\mu\xi^\nu+\mathcal{O}_3
$$where we have introduced the notation $\mathcal{O}_n$ for terms that are of order three [$n$?] or higher in the coordinates." I don't know why one would expand $x^a$ like that.

$x^a$ are general curved coordinates, $\xi^\mu$ are local Minkowski coordinates at some point $p$. $e_\mu$ are orthonormal vectors at $p$ in the $\xi^\mu$ system. They might be basis vectors, I am not sure. $e_\mu^a$ are the coordinates of $e_\mu$ in the $x^a$ system. $c_{\mu\nu}^a$ is a mysterious thing to be discovered. It turns out to be not a tensor but more like a Christoffel symbol. I think. We are also told earlier that $$
e_\mu^a=\frac{\partial x^a}{\partial\xi^\mu}
$$so the first part of the first equation is $$
x^a=\frac{\partial x^a}{\partial\xi^\mu}\xi^\mu
$$which is just the tensor transformation law.

Where does the first equation come from and why don't we use the ordinary tensor transformation equation?

 

[vanhees71]

The $\underline{u}_{\mu}$ denote a tetrad, i.e., four orthonormal vectors (in the sense of the Lorentzian metric) with the $\xi_{\mu}$ Gaussian coordinates at a spacetime point. The $e_{\mu}^a$ are the components of the $\mu$-th tetrad vector wrt. the holonomous basis defined by the coordinates $x^a$.

 

[DrGreg]

Where does the first equation come from and why don't we use the ordinary tensor transformation equation?

Because $x^a$ are general curved coordinates, not tensor components. In general coordinates, differentials of coordinates are tensors but coordinates are not.

The first equation is, in effect, the first few terms of a Taylor series.

 

[George Keeling]

Because $x^a$ are general curved coordinates, not tensor components.

I realized that last night . Luckily I didn't get up. It is the second time in as many weeks that I have tripped up on the T T equation!

The first equation is, in effect, the first few terms of a Taylor series.

I had some old notes on the Taylor series for a function of two variables and, when extended to four, it works for an expansion at the origin and gives Orodruin's equation perfectly . Thank you!

 

[vanhees71]

To remember the Taylor expansion in multiple coordinates note that you can write it in terms of the operator exponential function as
$$f(\vec{x}+\vec{a})=\exp(a^j \partial_j) f(\vec{x}).$$
Now just Taylor expand the exp and note that all $\partial_j$ commute with each other (in the appropriate domain of these differential operators of course ;-)):
$$\exp(a^j \partial_j) = \sum_{k=0}^{\infty} \frac{1}{k!} (a^j \partial_j)^k = \sum_{k=0}^{\infty} \frac{1}{k!} a^{j_1} \partial_{j_1} \cdots a^{j_k}\partial_k.$$
Then the Taylor series around $\vec{x}$ reads [Edit in view of #6]
$$f(\vec{x}+\vec{a})=\sum_{k=0}^{\infty} \frac{1}{k!} a^{j_1} \cdots a^{j_k} \partial_{j_1} \cdots \partial_{j_k} f(\vec{x}).$$

 

[George Keeling]

I guess the last equation under

Then the Taylor series around $\vec{x}$ reads

should have less brackets on the left and more arrows on the right:$$
f\left(\vec{x}+\vec{a}\right)=\sum_{k=0}^{\infty}{\frac{1}{k!}a^{j_1}\cdots a^{j_k}\partial_{j_1}\cdots\partial_{j_k}f\left(\vec{x}\right)}
$$I expanded the $k=2$ term in that for two dimensions and it came out just like my old notes on the Taylor series for a function of two variables. My notes improve. Thank you!
PS I originally found the expansion on math.libretexts.org. It's at equation (6) there.