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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    $\begingroup$ 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. $\endgroup$ Oct 14, 2012 at 4:36
  • $\begingroup$ Related question on Math.SE: math.stackexchange.com/q/1426755/11127 $\endgroup$
    – Qmechanic
    Sep 8, 2015 at 21:28

4 Answers 4


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.

Here are 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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    $\begingroup$ Great links and answer! Can we think of a quantum version of the dome?What would the energy Eigen state be, which would have exactly the energy required for a classical particle to rest on top (Assuming that angular momentum of the state is null)? $\endgroup$ Oct 31, 2017 at 19:26

In addition to Lubos Motl's correct answer, I would like to make a few 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 $$ z~=~-\frac{2K}{3g}s^{3/2}. \tag{1}$$ 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$^1$ $$ 0~=~\frac{E}{m}~=~\frac{\dot{s}^2}{2}+gz.\tag{2}$$
    In the first equality of (2), we used the initial conditions $$ s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0.\tag{3}$$ 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 $$ \dot{s}\ddot{s}~\stackrel{(2)}{=}~-g\dot{z}.\tag{4}$$ Division on both sides of eq. (4) with $\dot{s}$ yields$^2$ $$ \ddot{s}~\stackrel{(4)}{=}~-g\frac{\dot{z}}{\dot{s}}~=~-g\frac{dz}{ds}~\stackrel{(1)}{=}~K\sqrt{s}~.\tag{5}$$ Equation (5) is the sought-for equation of motion. Alternatively, combining eqs. (1) and (2) yield the following first-order ODE $$\dot{s} ~\stackrel{(1)+(2)}{=}~\sqrt{\frac{4K}{3}} s^{\frac{3}{4}}. \tag{6}$$

  2. Norton's initial value problem (IVP) is $$ \begin{align} \ddot{s}(t)~=~&K\sqrt{s(t)}, \cr s(t\!=\!0)~=~&0, \cr \dot{s}(t\!=\!0)~=~&0, \cr t~\geq~&0. \end{align}\tag{7} $$ The IVP (7) has two solution branches$^3$ $$s(t) ~=~\frac{K^2}{144}t^4\qquad\text{and}\qquad s(t) ~=~0~, \tag{8} $$ 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 IVP $$\begin{align} \dot{s}(t) ~=~&\sqrt{\frac{4K}{3}} s(t)^{\frac{3}{4}}, \cr s(t\!=\!0)~=~&0,\cr t~\geq~&0.\end{align}\tag{9}$$ Not surprisingly, the IVP (9) has the same two solution branches (8), and thus also demonstrates 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. 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 alters Norton's IVP (7).

$^3$ For each solution $s$, which is defined for non-negative times $t\geq 0$, let us for convenience extend in a trivially fashion $s(t<0):=0$ for negative times $t<0$. Then if we time-translated a solution $t\mapsto s(t)$ into the future, we get another solution $t\mapsto s(t-T)$ for some moduli parameter $T\geq 0$. Therefore strictly speaking, the first branch in eq. (8) generates a 1-parameter solution with a moduli parameter $T\geq 0$. So in fact, the IVP (7) has infinitely many solutions! Note that the second trivial solution branch (8) can be viewed as the $T\to \infty$ moduli limit of the first solution branch (8).

  • $\begingroup$ Excellent that you added some relevant math terminology, too. ;-) +1. $\endgroup$ Oct 13, 2012 at 6:06

The dome equation is expressed in terms of arc length and height, which hides a lot of bad behaviour. As Luboš has pointed out, beyond a certain point it is no longer physical so we need to constraint it before that happens.

You can actually work out the Cartesian equation for it, which is plottable: Norton's dome in Cartesian Coordinates

This isn't the only constraint that should be applied though. It's said that because of the non-analytical nature of the dome curve (it's not differentiable at the apex beyond the first derivative) it admits more than one solution. In fact, it's the square root that allows us to easily intuit the other solution which represents the trajectory of a particle reaching or leaving the apex with velocity $v$ at the limit $v = 0$.

Norton tries to pass off his other solution as Newtonian but in fact it's not (at the apex, anyway). This in itself isn't the source of his claim of non-determinism though. That comes from him stitching two solutions together that have different initial conditions at some arbitrary time T. This makes zero sense and has no physical justification.

I've written a detailed analysis of why Norton's dome does not prove Newtonian mechanics is non-deterministic here.


Can we think of it another way, which is that T is any non-negative real number, and we can think of it as taking a random number in a non-negative number, but we know that the probability that this number is greater than any certain number n is 1, that T can be 0,3,5, the square root of 29, etc., etc., but the probability is zero, so from a macro point of view, the ball never falls, Although he could fall at any time, the probability of falling at any time is zero.


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