[Cryptography] Edwards roller coaster (a physical model)

Bill Cox waywardgeek at gmail.com
Fri Jun 10 01:28:27 EDT 2016

Grr... I knew I'd make at least one mistake: the speed is always r*w0, not
w0/r.  Some emails should be peer reviewed :>


On Thu, Jun 9, 2016 at 10:04 PM, Bill Cox <waywardgeek at gmail.com> wrote:

> This is a fun physical model for Edwards curves I came up with, which can
> represent most of the elliptic curves we use for crypto that are related to
> the Jacobi addition laws
> <https://en.wikipedia.org/wiki/Jacobi_elliptic_functions#Addition_theorems>.
>   I do not know of any relevance this has for crypto, but I like to have a
> physical model in my head when thinking about problems...
> When d < 0, we get a 4-pointed star inscribed in the unit circle.  This
> picture from Wikipedia is accurate, except that it flipped the sign of d,
> so negate the signs of d and you will be less confused, because then it
> will match the article:
> [image: Inline image 1]
> These squished circles where d < 0 are entirely within the unit circle.
> Place a unit hemisphere on top of it, and project the squished circle
> directly up onto the hemisphere.  Imagine that the resulting path is the
> track of a roller coaster.  At time t = 0, the roller coaster is moving at
> angular velocity w0, crossing the point (x, y, z) = (1, 0, 0).  w0 Is
> really just the speed at that point, but we need it to have units of
> 1/seconds for the units to work out.
> Now assume that the track is frictionless, and that there is a bizarre
> force acting on the roller coaster.  For d == -1, the force pushes away
> from the Z-axis with strength equal to e^(w0*r), where r is the distance
> from the Z-axis, in other words, sqrt(x^2 + y^2).  Got it?  What happens
> when you set off the roller coaster at t = 0, with a speed of w0?
> The roller coaster speeds around the track, moving faster near the corners
> of the star, and slower as it gets close to the Z-axis.  The speed is
> always w0/r, and the potential plus kinetic energy is constant.  In this
> model, the thing being added by the Edwards addition law is time.  Given
> the position at t1 and t2, we can use the addition law to compute the
> position at t1 + t2.
> I kind of want to build an animated gif :>
> Bill
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