Let $S$ be a surface parameterized by variables $u,v$ and $\alpha(t)=(u(t),v(t))$ be a curve on the surface.

I am of the understanding that we can find the arc length of $\alpha$ by integrating it's speed:

$L(\alpha) = \int_{t_0}^{t_1} |\alpha'(t)|dt= \int_{t_0}^{t_1} \sqrt{u'(t)^2+v'(t)^2}dt$

however, at the bottom of page 37 on Fomins book on calculus of variations, it gives the arc length of $\alpha$ as:

$\int_{t_0}^{t_1} \sqrt{Eu'(t)^2+2Fu'(t)v'(t)+Gv'(t)^2}dt$

Where $E,F$ and $G$ and the coefficients of the first fundamental form of the surface.

What gives? Why are there two different formulas? Which one should I use here?

Is the first formula only valid on flat surfaces?

I hope this isn't a stupid question...

  • 1
    $\begingroup$ To supplement AlkaKadri's (+1) answer, the second formula reduces to the first when $E = G = 1$ and $F = 0$; particularly, the formulas are identical in the Euclidean metric. $\endgroup$ Commented Jun 23 at 1:42

1 Answer 1


The first formula is incorrect. $u(t)$ and $v(t)$ are the coordinates of the curve, not the actual position, so you've calculated the speed incorrectly.

Let $\sigma = \sigma(u,v)$ be the surface parametrization. Then the speed is $$\begin{align} \left|\frac{d}{dt} \sigma(u(t),v(t))\right| &= \left|\sigma_u u'(t) + \sigma_v v'(t)\right|\\ &= \sqrt{\left(\sigma_u u'(t) + \sigma_v v'(t)\right) \cdot \left(\sigma_u u'(t) + \sigma_v v'(t)\right)}\\ &= \sqrt{(\sigma_u\cdot\sigma_u) u'(t)^2 + 2(\sigma_u\cdot\sigma_v)u'(t)v'(t) + (\sigma_v\cdot\sigma_v)v'(t)^2}\\ &= \sqrt{E u'(t)^2 + 2F u'(t)v'(t) + G v'(t)^2} \end{align}$$ which recovers the second formula.

Related to your question about the validity of the first formula, if $\alpha(t) = (u(t),v(t)) \in \mathbb{R}^2$ is a curve in two-dimensional space, then the first formula is valid. That means that indeed, in some sense, the first formula is only valid on flat surfaces (and when the coordinate system is Cartesian).

  • $\begingroup$ Ugh duh... $\sigma$ is a function in itself!!! Thx for this $\endgroup$ Commented Jun 24 at 14:13
  • $\begingroup$ $\int |\alpha| dt$ is the length of the curve in the uv-plane (parameterizing space, preimage of the chart), right? $\endgroup$ Commented Jun 24 at 14:16
  • 1
    $\begingroup$ @PhysicsIsHard It certainly is! That first formula basically leaves out the "Jacobian" factor related to stretching/twisting this curve into whatever shape it takes in three-dimensional space under the image of $\sigma$. $\endgroup$
    – AlkaKadri
    Commented Jun 24 at 16:13
  • $\begingroup$ AH wow!! So the first fundamental form is the jacobian of the transformation? That actually is very obvious now that I see it... $\endgroup$ Commented Jun 24 at 19:30

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