I Minkowski to Euclidean line element from coordinate changes?

[Pencilvester]

TL;DR

I haven’t studied coordinate change law— what am I doing here that’s illegal?

I would guess there’s some subtlety in the relationship between basis vectors and coordinates that I’m ignoring, but I really have no idea.

$$ ds^2 = -dt^2 + d\tilde{x}^2 $$


$$ d\tilde{t} = dt / \sqrt{\tilde{x}} $$
$$ \downarrow $$
$$ ds^2 = -\tilde{x} ~ d\tilde{t}^2 + d\tilde{x}^2 $$


$$ dx = -d\tilde{x} ~ \Rightarrow ~ \frac{d\tilde{x}}{dx} = -1 ~ \Rightarrow ~ \tilde{x} = -x ~ (+ C) $$
$$ \downarrow $$
$$ ds^2 = x ~ d\tilde{t}^2 + dx^2 $$


$$ dy = \sqrt{x} ~ d\tilde{t} $$
$$ \downarrow $$
$$ ds^2 = dx^2 + dy^2 $$

 

[Ibix]

Given your definitions of $d\tilde t$, $dy$ and $dx$ you have that $dy=\sqrt{x/\tilde x}\ dt=i\ dt$. So you have effectively picked up the old $ict$ convention in $c=1$ units.

 

[Orodruin]

$$ d\tilde{t} = dt / \sqrt{\tilde{x}} $$

Apart from what was already said, what kind of coordinate transformation do you imagine that would satisfy this? Coordinate differentials by their nature as the differential of coordinate functions must be exact. The RHS here is not exact and therefore there can be no function $\tilde t(t,\tilde x)$ that satisfies this relation. (That it is not exact is clear from the fact that it is not closed and all exact forms are closed.)

 

[PeterDonis]

The RHS here is not exact and therefore there can be function $\tilde t(t,\tilde x)$ that satisfies this relation.

I assume you mean cannot here.

To rephrase what you are saying in a way that might be more accessible to the OP, the function $\tilde{t} (t, \tilde{x} )$ must be a function of both $t$ and $\tilde{x}$ since $1 / \sqrt{\tilde{x}}$ appears in the $dt$ term; but that means $d \tilde{t}$ must have both a $dt$ and a $d \tilde{x}$ term in it. So the formula given in the OP for $d \tilde{t}$ can't be right as it stands.

 

[Orodruin]

I assume you mean cannot here.

I think I intended to write out ”can be no function”, but yes, the same idea. Fixed it.

Edit: Indeed, more hands-on even, if the factor in front of $dt$ is $1/\sqrt{\tilde x}$ then $\partial \tilde t/\partial t = 1/\sqrt{\tilde x}$ implying that $\tilde t = t/\sqrt{\tilde x} + f(\tilde x)$. But this in turn would imply that
$$
d\tilde t = dt/\sqrt{\tilde x} +\left[f’(\tilde x) - \frac{t}{2\sqrt{\tilde x}^3}\right] d\tilde x
$$
where there is no way to get rid of the $t$-dependent term in front of $d\tilde x$.

 

[Pencilvester]

Apart from what was already said, what kind of coordinate transformation do you imagine that would satisfy this? Coordinate differentials by their nature as the differential of coordinate functions must be exact. The RHS here is not exact and therefore there can be no function $\tilde t(t,\tilde x)$ that satisfies this relation. (That it is not exact is clear from the fact that it is not closed and all exact forms are closed.)

I think this is the crucial point of which I was ignorant. Coordinate differentials were new to me, and I was having trouble contextualizing them with coordinate transformations. The fact that they need to be exact seems obvious now that you’ve told me, but it’s definitely what I was missing, so thanks!