Norton's dome and its equation

Question

Norton's dome is the curve $$h(r) = -\frac{2}{3g} r ^{3/2}.$$ Where $h$ is the height and $r$ is radial arc distance along the dome. The top of the dome is at $h = 0$.

Via Norton's web.

If we put a point mass on top of the dome and let it slide down from the force of gravity (assume no friction, mass won't slide off dome), then we will get the equation of motion $$\frac{d^2r}{dt^2} ~=~ r^{1/2}$$ (Not just me, lots of sources give this answer).

But this equation of motion doesn't make sense. Because as $r$ becomes large, the tangential force is also becoming large. The tangential force should always be less than or equal to the drive force from gravity. What am I seeing wrong?

Answer 1

In addition to Lubos Motl's correct answer, I would like to make two comments related to Norton's dome:

1. First a brief derivation of the equation of motion. I prefer to call the (non-negative) arc length $r$ for $s$, and the height $h$ for $z$. Like Lubos Motl, I will introduce a proportionality factor $K$ for dimensional reasons, so that the equation for Norton's dome reads $$\tag{1} z~=~-\frac{2K}{3g}s^{3/2}. $$ Here the constant $(g/K)^2$ has dimension of length. Equation (1) is only supposed to be valid for sufficiently small (but finite) arc lengths $s\geq 0$. Since there is no friction, we have mechanical energy conservation $$\tag{2} \frac{E}{m}~=~\frac{\dot{s}^2}{2}+gz.$$
   Differentiation wrt. time $t$ leads to $$\tag{3} \dot{s}\ddot{s}~=~-g\dot{z},$$ which after dividing both sides with $\dot{s}$ yields $$\tag{4} \ddot{s}~\stackrel{(3)}{=}~-g\frac{\dot{z}}{\dot{s}}~=~-g\frac{dz}{ds}~\stackrel{(1)}{=}~K\sqrt{s}~.$$ Equation (4) is the sought-for equation of motion.

2. The initial value problem $$ \tag{5} \ddot{s}(t)~=~K\sqrt{s(t)}, \qquad s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0, $$ has two solution branches $$\tag{6} s(t) ~=~\frac{K^2}{144}t^4\qquad\text{and}\qquad s(t) ~=~0~, $$ as can be easily checked. The failure to have local uniqueness of the ODE (5), which leads to indeterminism of the classical system, can from a mathematical perspective be traced to that the square root $\sqrt{s}$ in eq. (5) fails to be Lipschitz continuous at $s=0$.

Answer 2

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing.

However, let me answer your question:

The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and you won't be able to find points that are "deeper" below the summit than the total length from the summit along the dome. Well, the dome is actually truncated earlier than that.

See e.g. this presentation

http://philsci-archive.pitt.edu/3195/1/NortonDome.pdf

of the problem. Note that Norton's goal was study the behavior near $h=0$ and $r=0$ which he called an "example of indeterminism in Newtonian classical physics" because the particle may sit at the top for any amount of time, and suddenly freely decide and get rolling. That's why the truncation of the dome isn't important.

My more general comments about Norton's dome and its harmlessness in quantum physics:

http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html?m=1

In that article, I also calculated that the dome has to end up at the point where $dh/dr=1$ because it's the sine of an angle which implies $r_{\rm max}=(9/4)g^2=h_{\rm max}$; I also use an additional coefficient $K$ to make the formulae dimensionally correct.