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Calculating the coordinate basis

Physics Asked on May 19, 2021

In Robert M. Wald’s General Relativity the definition of the "coordinate basis" (of the tangent space) of a manifold is given by:

Let $psi: O to U subset mathbb{R}^n$ be a chart with $p in O.$ If $f in mathcal{F},$ then by definition $f circ psi^{-1}: U to mathbb{R}$ is $C^{infty}.$ For $mu = 1, … , n$ define $X_mu: mathcal{F} to mathbb{R}$ by

$$X_{mu} (f) = frac{partial}{partial x^ mu} (f circ psi^{-1})Big|_{psi(p)}.$$


The basis ${ X_mu }$ of $V_p$ is called a coordinate basis. Had we chosen a different chart, $psi^{‘},$ we would have obtained a different coordinate basis ${ X’_nu }.$ We can, of course, express $X_mu$ in terms of the new basis ${ X’_nu }.$ Using the chain rule of advanced calculus, we have

$$X_{mu} = sum_{nu =1}^{n} frac{partial x’^nu}{partial x^ mu} Big|_{psi(p)} {X’}_nu.$$

However, this definition does not seem to be "right".

Let $O$ be $mathbb{R}^2$ and let $Psi$ be the identity chart. In this case, we get that
$$X_1 (f) = frac{partial f}{partial x}$$
$$X_2 (f) = frac{partial f}{partial y}$$
which is fine. If we now choose a different chart though, $psi^{‘}$, which is the "polar coordinate system" given by $psi(r, theta) = (r cos theta, r sin theta);$ $psi^{-1}(x,y) = (sqrt{x^2 + y^2}, arctan {y/x}).$

From the chain rule it seems like what we want is
$$X_1 = frac{partial f}{partial x} = frac{partial f}{partial r}frac{partial r}{partial x} + frac{partial f}{partial theta}frac{partial theta}{partial x} = frac{partial r}{partial x} X’_r + frac{partial theta}{partial x}X’_theta $$

Or, in other words, $X’_1 = X’_r = frac{partial f}{partial r}, X’_2 = X’_theta = frac{partial f}{partial theta}.$ But if we calculate $X’_1$ and $X’_2$ from the definition we get:

$$X’_1 = frac{partial}{partial x} (tilde{f} circ psi^{‘-1})Big|_{psi^{‘}(p)}.$$
(where $f(x,y) = tilde{f} (rcos theta, r sin theta) = tilde{f}(r, theta).$)
$$X’_1 = frac{partial tilde{f}}{partial r}frac{partial r}{partial x} + frac{partial tilde{f}}{partial theta}frac{partial theta}{partial x}.$$

How exactly are we supposed to apply the definition given for the polar coordinate chart?

2 Answers

The diagram on page 15 is helpful. The axes of your chart are not x and y. Your axes are labeled $theta$ and $r$. It is very important to understand that the chart is literally just a labeling in $R^n$.

$f$ is defined on the manifold. Not the chart. But you have a mapping from the manifold to the chart.

Therefore,

$f|_p$ = $f(r,theta)_ |_{psi^{-1} (r,theta)}$.

${X}_theta (f)= frac{partial}{partial x^theta} (f circ psi^{-1} (r,theta))$

or

$frac{partial f}{partial theta}= frac{partial}{partial theta} (f circ psi^{-1} (r,theta))$

If $g:(r,theta) rightarrow (x(r,theta),y(r,theta))$ then

$frac{partial f}{partial theta}= frac{partial}{partial theta} (f circ g circ psi^{-1} (r,theta))$

Then apply the chain rule to get $frac{partial f}{partial theta}= frac{partial f}{partial x} frac{partial x}{ partial theta}+ frac{partial f}{partial y} frac{partial y}{ partial theta}$.

Correct answer by user288901 on May 19, 2021

While @ExpertNonexpert is right that $tilde{f}$ is an unnecessary misdirection, the mistake that actually matters is in this line:

$$X'_1 = frac{partial}{partial x} (tilde{f} circ psi^{'-1})Big|_{psi^{'}(p)}.$$

Specifically, the derivative is taken with respect to the wrong parameter. It should be

$$X'_1 = frac{partial}{partial {x'}^1} (f circ psi^{'-1})Big|_{psi^{'}(p)}.$$

And since ${x'}^1$ is just $theta$, everything works out as you'd expect.

Answered by Mike on May 19, 2021

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