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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?

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I wonder if this problem goes away if you allow only $C^\infty$ (infinitely differentiable) solutions to the equations of motion, or if a dome can be contrived so that a Bump Function of time is a solution. –  James Turton Oct 14 '12 at 4:36

2 Answers 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


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:


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.

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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 Norton's equation of motion (7). I prefer to call the (non-negative) arc length $r$ for $s$, and the vertical 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$^1$ $$\tag{2} 0~=~\frac{E}{m}~=~\frac{\dot{s}^2}{2}+gz.$$
    In the first equality of (2), we used the initial conditions $$\tag{3} \qquad s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0.$$ We assume that $t\mapsto s(t)$ is twice differentiable wrt. time $t\geq 0$. (In detail, at the initial time $t=0$ we assume that the function is one-sided twice differentiable from right.) Differentiation of eq. (2) wrt. time $t$ leads to $$\tag{4} \dot{s}\ddot{s}~\stackrel{(2)}{=}~-g\dot{z}.$$ Division on both sides of eq. (4) with $\dot{s}$ yields$^2$ $$\tag{5} \ddot{s}~\stackrel{(4)}{=}~-g\frac{\dot{z}}{\dot{s}}~=~-g\frac{dz}{ds}~\stackrel{(1)}{=}~K\sqrt{s}~.$$ Equation (5) is the sought-for equation of motion. Alternatively, combining eqs. (1) and (2) yield the following first-order ODE $$ \tag{6}\dot{s} ~\stackrel{(1)+(2)}{=}~\sqrt{\frac{4K}{3}} s^{\frac{3}{4}}.$$

  2. Norton's initial value problem is $$ \tag{7} \ddot{s}(t)~=~K\sqrt{s(t)}, \qquad s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0, \qquad t~\geq~0. $$ The initial value problem (7) has two solution branches $$\tag{8} s(t) ~=~\frac{K^2}{144}t^4\qquad\text{and}\qquad s(t) ~=~0~, $$ as can be easily checked. The failure to have local uniqueness of the ODE (7), which leads to indeterminism of the classical system, can from a mathematical perspective be traced to that the square root $\sqrt{s}$ in eq. (7) fails to be Lipschitz continuous at $s=0$.

  3. Alternatively, from mechanical energy conservation (6), one can consider the initial value problem $$ \tag{9} \dot{s}(t) ~=~\sqrt{\frac{4K}{3}} s(t)^{\frac{3}{4}}, \qquad s(t\!=\!0)~=~0,\qquad t~\geq~0.$$ Not surprisingly, the initial value problem (9) has the same two solution branches (8), and thus also demonstrate failure to have local uniqueness.


$^1$ I imagine that the point particle is sliding with no friction. (The rolling ball in Norton's figure is slightly misleading and presumably only for illustrative purposes.). A more complete derivation would check that the point particle doesn't loose contact with the doom. If one would like to avoid such an analysis, one may for simplicity assume that the dome is a two-sided constraint.

$^2$ Division with $\dot{s}$ is only valid if $\dot{s}\neq 0$. Now recall that the mechanical energy $E=0$ is zero. Therefore if $\dot{s}=0$ then $z=0$ and hence $s=0$ must be zero, cf. eqs. (1) and (2). Hence the division-by-zero issue is limited to the tip of the dome. Ultimately, it turns out that the $\dot{s}=0$ branch does not lead to new solutions not already included in eq. (8), nor alter Norton's initial value problem (7).

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Excellent that you added some relevant math terminology, too. ;-) +1. –  Luboš Motl Oct 13 '12 at 6:06
If I'm not mistaken, shouldn't that factor of 1/2 be 7/10 if the marble is a solid sphere? There's an additional kinetic term from its rotation as well as centre of mass motion. –  Chay Paterson Apr 10 '13 at 19:39
In the answer I have just assumed a point particle sliding with no friction. I think Norton originally does the same for simplicity. The rolling ball in Norton's figure is slightly misleading and presumably only for illustrative purposes. –  Qmechanic Apr 10 '13 at 19:53
I have a doubt about this answer: is it licit to derive the equations of motion as you did in point (1) if the initial velocity is zero? And if yes, why? –  pppqqq Jan 2 '14 at 15:34
I updated the answer with a third comment. –  Qmechanic Sep 20 '14 at 18:46

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