Bending Energy of regular planar curves is parameterization invariant

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I need to show that the Bending Energy of a Planar Curve

$$\int_I {\kappa^{2}||\gamma^{'}||}$$

is invariant under a reparameterization of $\gamma$

I'm not really sure how exactly I go about doing this. If I can say choose a parameterization for $\gamma(t)$ like $<\cos t, \sin t>$ then calculate the bending energy then just reparameterize like, say, $<\cos(-t), \sin(-t)>$ or like $<\cos(t^{2}), \sin(t^{2})>$ and then recalculate and show these two bending energies are the same?

Or would I need to prove it in a more general sense (as opposed to specific examples) and if so, how exactly would I go about doing that?

Thanks in advance to anyone who can help me out on this. Looking forward to a lively and interesting discussion!

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EDIT1:

I thought the total context should be comprehensively clear. So I shall attempt to reply more about how it comes about so directly in classical mechanical energy considerations. (I was never a teacher).

Stretching and Bending are the two fundamental phenomena of elastic motions.They are subjects of first two Mechanics chapters in a Strength of Materials course for linear elements. In classical mathematical surface theory we have as surface extensions manifolds dealing with first and second fundamental forms.

In Stretching for a "line manifold", the linear spring we have energy conservation

$$ P_s + K_s = \frac12 k x \cdot x + \frac12 m {\dot x}^2 $$

which when differentiated with respect to time $t$ yields harmonic motion along $x$ as

$$ { m \ddot x} + k \,x=0 $$

Next in Bending energy of elstic curves like the Elastica, the bending potential energy and bending kinetic energy sums

$$ P_b + K_b = \frac12 k_b\theta \, \theta + \frac12 I {\dot \theta }^2 $$

which by differentiation with time

$$ k_b \theta + I \, \ddot \theta =0 $$

depicting harmonic rotary motion. Now your question is about potential energy. We have Moment or Torque $ M= k_b \, \theta $

$$ dP_b = \frac12\ M d \theta = \frac12\, M \frac{d \theta}{ds} ds =\frac12\ M \,\kappa \,ds \tag1 $$

by virtue of Euler Bernoulli theorem Moment Curvature relation with flexural rigidity $EI$ determines bending geometry

$$ \frac{M}{EI}=\kappa \tag2$$

Plug in

$$ dP_b = \frac{M^2}{2EI} ds \tag3$$

So the total physical strain energy of bending is

$$ P_b = \int \frac{M^2}{2EI} ds = \frac{EI}{2} \int \kappa^2 ds\tag4$$

And this maximum potential bending energy is conserved in any arc based parametrization of the curve under dynamic motion when kinetic energy is zero in extreme positions.. i.e., when $ \omega= \dot \theta =0$.

It can be noted that it is the total energy $ P_b + K_b$ that is conserved but never $ P_b $ alone at all instants of time due to energy conservation principle.

Whatever be the parametrization you bend it and the curve stores static flexural energy.

EDIT2:

Isometric invariance relations of curves built on first fundamental form differential coefficients

Freedom from parametrization for quantities depending on quantities dependent on first fundamental form. They can be as a class described as Natural Equations E.g.,

$ \frac{d\phi}{ds}=\kappa = f(s)$ is intrinsic equation for any curve in the plane for all f, the Cesàro equation.

$ \frac{d\phi}{ds}=\kappa =1 $ is intrinsic equation for a Circle.

$ \phi = $ constant is intrinsic equation for a straight line.

$ \frac{d\phi}{ds}=\kappa = a\cdot s $ is intrinsic equation for Cornu's spiral.

$ \frac{d\phi}{ds}=\kappa = 1/R, \, R^2 = 2 a s ,\quad, R^2+ s^2 =1. $ for well known curves.

The parametrization can be done upto Euclidean motions translation and rotation that is coordinate free.