What is the Jacobi amplitude?

Question

This has been bothering me for a while now, and it keeps giving me trouble - how does one explain the Jacobi amplitude, and how does one calculate it?

I've tried just about everything there is to wrap my head around it, but it still seems to confuse me.

I've seen am(u|k), and have tried to calculate u, but I end up running in circles, as calculating u requires that I find phi (aka the Jacobi Amplitude) first.

I don't have MATLAB available to me, but I do have Wolfram|Alpha and Desmos at the ready.

score 3 · Accepted Answer · edited Mar 31 '18 at 18:37

3

Some relations:

\begin{align*} u &= F(\phi| m) \\ &= \int_{0}^{\phi} \frac{d\theta}{\sqrt{1-m\sin^2 \theta}} \\ &= \int_{0}^{\sin \phi} \frac{dx}{\sqrt{1-mx^2}} \\ &= \int_{0}^{\operatorname{sn} u} \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}} \\ x &= \sin \phi \\ &= \operatorname{sn} (u|m) \\ &= \sin [F^{-1}(u|m)] \\ \operatorname{sn}^{-1} (x|m) &= F(\sin^{-1} x|m) \\ \operatorname{am} (u|m) &= \phi \\ &= F^{-1}(u|m) \\ \operatorname{sn} (u|m) &= \sin \phi \\ \end{align*}

Note that Mathematica or WolframAlpha uses $m=k^2$.

edited Mar 31 '18 at 18:37

J. M. ain't a mathematician

75,051

answered Aug 03 '16 at 00:33

Ng Chung Tak

18,990

So, given that x= sin(phi), then arcsin(x)= phi, which is also equal to the Jacobi amplitude function am(u|k)? – AstroFox Aug 03 '16 at 00:48
Yes. For details, please look up here. – Ng Chung Tak Aug 03 '16 at 00:50

What is the Jacobi amplitude?

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