Why is arc length independent of parametrization?

Question

For example, say a curve $C$ can be parametrized as $x_1(t)$, $y_1(t)$ over an interval $A$ and $x_2(t), y_2(t)$ over an interval $B$. Why is the arc length computed using the first parameterization the same as the length computed using the second parameterization? Is this always the case, or are there any exceptions?

Answer 1

Assuming $A = [c,d]$ and $B=[e,f]$, there is a change of parameter diffeomorphism $h : [c,d] \to [e,f]$ such that $x_1(t)=x_2(h(t))$ and $y_2(t)=y_2(h(t))$, $t \in [c,d]$. Letting $s = h(t) \in [e,f]$ we get \begin{align*} \int_{s=e}^{s=f} \sqrt{(dx_2/ds)^2 + (dy_2/ds)^2} \, ds &= \int_{t=c}^{t=d} \sqrt{\left(\frac{dx_1/dt}{h'(t)}\right)^2 + \left(\frac{dy_1/dt}{h'(t)}\right)^2} \, h'(t) \, dt \\ &= \int_{t=c}^{t=d} \sqrt{(dx_1/dt)^2 + (dy_1/dt)^2} \, dt \end{align*}

Answer 2

Let $\phi = (x_1, y_1)$, $\psi = (x_2, y_2)$. Note the definition of the surface integral claims that, $$\int_{C}1\,dS := \int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt$$ is well defined, independent of the parameterization $(A, \phi)$. That is, it claims that $$\int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt = \int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds.$$ Set $F = \psi^{-1} \circ \phi \colon A \to B$. Note that $F$ is a diffeomorphism. By the change of variables formula for integration substituting $s = F(t)$, we have \begin{align} \int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds = \int_{A}\sqrt{\det(D\psi(F(t))^TD\psi(F(t)))}|\det(DF(t))|\,dt. \end{align} We have $\phi(t) = \psi(F(t))$, so $D\phi(t) = D\psi(F(t))DF(t)$. Thus \begin{align} \sqrt{\det(D\phi(t)^TD\phi(t))} &= \sqrt{\det(DF(t)^TD\psi(F(t))^TD\psi(F(t))DF(t))} \\ &= \sqrt{\det(D\psi(F(t))^TD\psi(F(t)))}|\det(DF(t))|. \end{align} Thus $\int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt = \int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds$, so the surface integral is well defined, independent of the parameterization. The above proof works for surfaces of arbitrary dimension, and with small modification, works for integrals of any function $f \colon C \to \mathbb{R}$, not just $f = 1$. Technically the sets $A$ and $B$ should be open though.