I'm going through the textbook "Emmy Noether's Wonderful Theorem" by Dwight Neuenschwander. Therein, the author defines the coordinate transformation (infinitesmal rotation of orthogonal axes) given by $$x'=x\cos\varepsilon+y\sin\varepsilon,\quad y'=-x\sin\varepsilon+y\cos\varepsilon,\quad z'=z.$$ He goes on to say that the distance functional $$s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$ becomes, under the infinitesmal rotation of axes, $$s'=\int_{a'}^{b'}\sqrt{1+\left(\frac{dy'}{dx'}\right)^2}\,dx'.$$
Can someone please explain in detail how the distance functional takes the same form under this transformation? I feel like this should be easy, but I cannot see how. I tried to express $dy'/dx'$ in terms of $x$ and $y$:
$$\frac{dy'}{dx'}=\frac{dx}{dx'}\frac{dy'}{dx}=\frac{dy}{dx}-\tan\varepsilon,$$
and use this to compute the integrand, but it turns into a mess. I've tried other things in a similar vein. Any suggestions?
Simple answer: Because rotations are an isometry of the euclidean metric $g_{uv} = \delta_{uv}$. The integral:
$$L = \int_a^b \sqrt{g_{uv} \frac{dx^u}{dt} \frac{dx^v}{dt} } dt =\int_a^b \sqrt{(\frac{dx^1}{dt})^2 + (\frac{dx^2}{dt})^2 }$$
Will be the same even in rotated coordinates. More detailed: After rotation, our metric is still $g_{u'v'}= \delta_{u'v'}$ , our components are:
$$L= \int_{a'}^{b'} \sqrt{ g_{u'v'} \frac{dx^{u'} }{dt} \frac{dx^{v'} }{dt}} = \int_a^b \sqrt{ (\frac{dx^{1'} }{dt})^2 + (\frac{dx^{2'} }{dt})^2} dt$$
Where $x^{1'}$ and $x^{2'}$ are $x^1$ and $x^2$ after rotation.
Previous answer on the invariance of arc length integral
The issue is you misunderstood how we are taking the functions here. We make $x,y$ depend on $\epsilon$, I'll show it for the $x$ coordinate:
$$x_{rotated} = \cos \epsilon x + \sin(\epsilon)y $$
For small values of $\epsilon$, the above becomes:
$$ x_{rotated} \approx x +\epsilon y$$
Rearranging,
$$ \delta x = x_{rot}-x = \epsilon y +\mathbb{O}(\epsilon^2)$$
Essentially, $\delta x$ here is the difference between $x$ before and after rotation. Not the 'differential step' induced in the rotated coordinate by a step in the input plane.
Similarly, we have:
$$ \delta y= -x \epsilon +O(\epsilon^2)$$
We can write:
$$ \frac{dy'}{dx'} = \frac{d( y + \delta y)}{d(x+ \delta x)} = \frac{\frac{dy}{dx} + \frac{d (\delta y)}{dx}}{1+ \frac{d (\delta x)}{dx}}= \frac{\frac{dy}{dx} -\epsilon}{1 +\epsilon \frac{dy}{dx}}$$
The last equality, I got through plugging back previous expression for $\delta x $ and $\delta y$. Now the rest is done on page-66 of the book which involves using a lot of series expansion.