# Calculus derivation of a surprising result of Galileo’s

An interesting paper by Herman Erlichson discusses Galileo’s attempt to prove that the minimum time path for a particle falling from a point on the lower quadrant of a circle is via the circle itself. (I attach the paper here: Erlichson, H, 1998, Galileo’s Work on Swiftest Descent from a Circle And How He Almost Proved the Circle Itself Was the Minimum Time Path, The American Mathematical Monthly, Vol 105, No 4, 338-347). I was particularly intrigued by equation (1) in this paper which asserts that the time taken for a particle to fall along a straight line starting at the circle is independent of the starting point. Thus, for example, a particle following a longer chord starting higher up on the circle would take exactly the same time to reach the bottom as a particle following a shorter chord starting lower down on the circle. Galileo explained it in his book, The Two New Sciences, as follows:

(I attach the English translation of Galileo’s book here: galileo two new sciences).

Underneath equation (1) in Erlichson’s paper, Erlichson gives a quick heuristic derivation of this result but I couldn’t resist exploring it more deeply using calculus methods (which were not available in Galileo’s time – calculus was invented by Isaac Newton, who was born almost at the same time that Galileo died). I want to record these calculations in the present note, in particular showing first how the time taken along an arbitrary straight line definitely does depend on the starting point, and then showing how this dependence instantly `disappears’ when the starting point of the straight line is taken to lie on a circle. I find it remarkable how this happens algebraically – nothing fundamentally changes about the straight line other than assuming that it does or does not start on a circle!

For the purposes of this note I have amended Figure 1 in Erlichson’s paper by imposing a coordinate system as shown below:

The origin of the coordinate system is taken to be at C and we suppose that the point D has coordinates $(X, nX)$. The equation of the straight line DC is $y = nx$. Let us ignore the circular arc connecting D and C for the moment and work out the time it would take for a particle of mass $m$ to fall from D to C along the straight line. We can do this by considering the energy of the particle at point D. If it starts at rest (i.e., its initial velocity is zero), the energy at D consists only of the particle’s potential energy given by $mgy(X) = mgnX$. Potential energy is transferred to kinetic energy as the particle moves down the line and the total energy at each $x$-coordinate is given by

$\frac{1}{2}mv^2 + mgnx = mgnX$

$\iff$

$v = \sqrt{2gn(X - x)}$

Given that $y = nx$ we see that $y^{\prime} = n$ so the element of distance along the straight line is $\mathrm{d}x\sqrt{1 + (y^{\prime})^2} = \mathrm{dx}\sqrt{1 + n^2}$. Therefore the element of speed is

$\frac{\mathrm{d}x}{\mathrm{d}t}\sqrt{1+n^2}$

Setting this equal to $v$ and solving for $\frac{\mathrm{d}x}{\mathrm{d}t}$ we get

$\frac{\mathrm{d}x}{\mathrm{d}t} = \sqrt{\frac{2gn(X-x)}{1+n^2}}$

We now observe that the time of descent $T$ will be given by the integral

$T = \int_0^T \mathrm{d}t = \int_0^X \frac{\mathrm{d}t}{\mathrm{d}x} \mathrm{dx} = \int_0^X \sqrt{\frac{1+n^2}{2gn(X-x)}}dx$

$= \sqrt{\frac{1+n^2}{2gn}} \int_0^X (X - x)^{1/2}\mathrm{d}x$

$= 2\sqrt{X} \sqrt{\frac{1+n^2}{2gn}}$

$= \sqrt{\frac{2X}{gn}}\sqrt{1+n^2}$

Thus, in this case, the formula for $T$ definitely depends on the starting point $X$. But watch what happens when we now assume that the point D lies on a circle of radius $L$ as shown in the diagram. The circle has equation

$x^2 + (y - L)^2 = L^2$

so the equation of the circular arc joining D and C is

$y = L + \sqrt{L^2 - x^2}$

as indicated in the diagram. At the point D we therefore have

$nX = L + \sqrt{L^2 - X^2}$

$\iff$

$n = \frac{L}{X} + \sqrt{\frac{L^2}{X^2} - 1}$

so

$n^2 = \frac{L^2}{X^2} + \frac{2L}{X}\sqrt{\frac{L^2}{X^2} - 1} + \frac{L^2}{X^2} - 1$

and therefore

$1 + n^2 = \frac{2L^2}{X^2} + \frac{2L}{X}\sqrt{\frac{L^2}{X^2} - 1}$

Substituting these expressions for $n$ and $1 + n^2$ in the formula for $T$ we get

$T = \sqrt{\frac{2X}{gn}}\sqrt{1+n^2}$

$= \sqrt{\frac{2X\big(\frac{2L^2}{X^2} + \frac{2L}{X}\sqrt{\frac{L^2}{X^2} - 1}\big)}{g\big(\frac{L}{X} + \sqrt{\frac{L^2}{X^2} - 1}\big)}}$

$= \sqrt{\frac{4L\big(\frac{L}{X} + \sqrt{\frac{L^2}{X^2} - 1}\big)}{g\big(\frac{L}{X} + \sqrt{\frac{L^2}{X^2} - 1}\big)}}$

$= \sqrt{\frac{4L}{g}}$

$= 2\sqrt{\frac{L}{g}}$

which is equation (1) in Erlichson’s paper. The dependence on the starting point $X$ has now vanished, so all starting points on the lower quadrant of the circle will yield the same time of travel to the origin!