Arc Length and Differential Forms

Question

Suppose $\gamma$ is circle in $\mathbb{R}^3$ defined by coordinates $\begin{pmatrix}r\cos\theta\\r\sin\theta\\0\end{pmatrix}$, and function $F: \gamma \rightarrow \mathbb{R}^3$ is defined by $F(\gamma(\theta)) = \begin{pmatrix}-\sin\theta \\ \cos\theta\\0\end{pmatrix}$, and let 1-form in $\mathbb{R}^3$$\lambda_F = F_1dx+F_2dy+F_3dz$, where $F_i$ is the component of $F$.

Why $\int_\gamma{\lambda_F}$ is the length of the circle $\gamma$?

I know that one can do some calculation like $\int_\gamma{\lambda_F} = \int_0^{2\pi}\vert\gamma'(\theta)\vert d\theta = 2\pi r$, what I'm asking is an intuitive explanation, i.e. why when you put that specific $\lambda_F$ there, it gives you the arclength, intuitively.

Thanks.

Answer 1

Let $\gamma:[a,b]\longrightarrow\Bbb R^3$ be a parametrized curve. Near a point $(t_0, \gamma(t_0))$, $$\gamma(t)\approx \gamma'(t_0)(t-t_0)+\gamma(t_0)$$ i.e. $$\|\gamma(t)-\gamma(t_0)\|\approx\|\gamma'(t_0)\||t-t_0|$$ or, $$\Delta(\text{length})\approx\|\gamma'(t)\|\Delta t.$$ What happens in the limit?

Answer 2

Use Pythagoras.

The arc length between the points $A$ and $B$ is approximated by the length of the straight segment $AB$. If $A=(x_1,y_1,z_1)$ and $B=(x_2,y_2,z_2)$ then $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$.

When $A$ and $B$ are close together then $x_2-x_1\sim (F\circ\gamma_1)'d\theta=F_1d\theta$. Similarly for $y_2-y_1$ and $z_2-z_1$.

Answer 3

set $\phi_x=\frac{\partial \phi}{\partial x}, \phi_y=\frac{\partial \phi}{\partial y}, dx=\phi_xdt, dy=\phi_ydt$

as $\frac{\phi_x}{\sqrt{\phi_x^2+\phi_y^2}}dx+\frac{\phi_y}{\sqrt{\phi_x^2+\phi_y^2}}dy=\sqrt{\phi_x^2+\phi_y^2}dt$

and this is the definition of the arc-length. From this, you can know the answer of your question since $cos(\theta),sin(\theta)$are in fact the $\phi$ here.