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I have to prove that if $\alpha(t)$ is a regular curve in space and $\beta(s)$ is its reparametrization of unit speed then $\frac{d^2 t}{d s^2} = - \frac{\alpha ' (t) \cdot \alpha '' (t)}{|| \alpha ' (t) ||^4}$.

Attempt:

I know that the RHS is:

\begin{gather*} \alpha'(t)=\left(\frac{ds}{dt}\right)T(t) \\ \alpha''(t)= \left(\frac{ds}{dt}\right)T'(t)+\left(\frac{d^2s}{dt^2}\right)T(t) \end{gather*} where $T(t)$ is the unit tangent vector. When doing $\alpha'(t)\cdot \alpha''(t)$ we get \begin{equation*} \alpha'(t) \cdot \alpha ''(t) = \left(\frac{ds}{dt}\right) \left(\frac{d^2s}{dt^2}\right) T(t) \cdot T(t) = ||\alpha'(t)||\left(\frac{d^2s}{dt^2}\right) \end{equation*} And so \begin{equation*} -\frac{\alpha'(t)\cdot \alpha''(t)}{||\alpha'(t)||^4} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^3} \end{equation*} On the other hand, the LHS is: \begin{gather*} \frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{||\alpha'(t)||} = (||\alpha'(t)||)^{-1} \\ \Rightarrow \frac{d^2 t}{ds^2} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^2} \end{gather*} So there is a factor of $||\alpha'(t)||$ I am missing. Is the statement wrong or where am I wrong? Thank you in advance.

davidaap
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2 Answers2

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On the other hand, the LHS is: \begin{gather*} \frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{||\alpha'(t)||} = (||\alpha'(t)||)^{-1} \\ \Rightarrow \frac{d^2 t}{ds^2} = -\frac{\left(\frac{d^2s}{dt^2}\right)}{||\alpha'(t)||^2} \end{gather*} So there is a factor of $||\alpha'(t)||$ I am missing. Is the statement wrong or where am I wrong? Thank you in advance.

The ($\Rightarrow$) step is incorrect: $$ \frac{d}{ds}(\frac{ds}{dt})\neq\frac{d^2s}{dt^2}. $$

Note that $$ LHS=\frac{d}{ds}(\frac{dt}{ds})=\frac{d}{ds}(\frac{1}{ds/dt}) =\frac{-1}{(ds/dt)^2}\cdot\frac{d}{ds}(\frac{ds}{dt}) =\frac{-1}{\|\alpha'(t)\|^2}\cdot \frac{d}{ds}(\frac{ds}{dt})\tag{1} $$ Now, $$ \frac{d}{ds}(\frac{ds}{dt})=\frac{d}{dt}(\frac{ds}{dt})\cdot{\frac{dt}{ds}} =\frac{d^2s}{dt^2}\frac{1}{\|\alpha'(t)\|}\tag{2} $$ Combining (1) and (2) you get the desired LHS, which is equal to what you calculated for the RHS.

  • I noticed that, but can you explain more why $\frac{d}{ds}(\frac{ds}{dt}) = \frac{d}{dt}(\frac{ds}{dt})\cdot \frac{dt}{ds}$ ? Is it just the chain rule? – davidaap Sep 16 '18 at 18:10
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    @math4everyone: yes, it is just the chain rule. It comes in all the time in these differential geometric calculations, especially when dealing with both parameters $t$ and $s$. – Robert Lewis Sep 16 '18 at 18:35
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I think the introduction of $T(t)$ etc. confuses things somewhat.

It can be done as follows:

$\dfrac{ds}{dt} = \Vert \alpha'(t) \Vert = \langle \alpha'(t), \alpha'(t) \rangle^{1/2}; \tag 1$

$\dfrac{dt}{ds} = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}; \tag 2$

$\dfrac{d^2 t}{ds^2} = \dfrac{d}{ds} \dfrac{dt}{ds} = \dfrac{d}{ds} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2} = \dfrac{dt}{ds} \dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}; \tag 3$

$\dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle^{-1/2} = -\dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-3/2} \dfrac{d}{dt} \langle \alpha'(t), \alpha'(t) \rangle$ $= -\dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-3/2} \left ( 2 \langle \alpha'(t), \alpha''(t) \rangle \right ) = -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}}; \tag 4$

finally,

$\dfrac{d^2 t}{ds^2} = \dfrac{dt}{ds} \left ( -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}} \right ) = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\left ( -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^{3/2}} \right )$ $= -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\langle \alpha'(t), \alpha'(t) \rangle^2} = -\dfrac{\langle \alpha'(t), \alpha''(t) \rangle}{\Vert \alpha'(t) \Vert^4}. \tag 5$

Robert Lewis
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