Norton's dome is a non-problem in quantum physics
A user at the Physics Stack Exchange asked a question about a fun system in classical physics:
Norton's dome and its equationThe question was why it was apparently possible for the acceleration of a marble to exceed any bound, including \(g\) itself. I answered: the dome is actually finite, truncated at some distance from the bottom.
Click the dome above for Norton's web page giving the original presentation of the problem or try this David Malament's exposition.
What's the problem? Well, the problem is that classical physics admits seemingly innocent potentials for which a marble may sit at the summit for an arbitrary amount of time and then ? unexpectedly, surprisingly, unpredictably, freely ? it may decide it's time to roll.
Let me say in advance that if you simply ban similar "pathological enough" shapes of the potential and domes in Newton's physics, and/or ban some extraordinarily fine-tuned initial conditions for the marble, the problem is gone. But are you really allowed to ban them?
To be sure what the problem is, the dome above has the shape given by an equation relating the "depth" of the marble beneath the summit, \(h(r)=y(r)\lt 0\) and a radial coordinate \(r\) which is nothing else than the length of the straight trajectory of the marble measured from the summit along the dome. The relationship is simple:\[
h = -\frac{2K}{3g} r^{3/2}
\] With this power law, the shape near the summit is smoother than a sharp cone \(h=-|r|\) but less smooth than \(h=-r^2\). Only the latter "naturally" appears near maxima in various potentials we know in physics. But we may actually produce cones and Norton's domes out of steel so you can't really say they're impossible. ;-)
You may also notice that for too high values of \(r\), you will get too high values of \(|h(r)|\), perhaps even \(|h(r)|\gt r\) which is geometrically impossible because we clearly have \(|h(r)|\lt r\) from the definition of the quantities. We could actually make the inequality stronger than that.
I inserted an extra constant \(K\) because the original equation didn't even pass the dimensional analysis, not even if one sets \(m=1\) for the mass. But I will generally not focus on the numerical prefactors although I have already fixed them in an updated version of this blog entry.
(My opinion is that the constant \(K\) with units \(\sqrt{\rm meter}/{\rm second}^2\) is arbitrary ? it determines the size of the dome ? but it has to be inserted for dimensional reasons. If I did my Mathematica calculation correctly, the maximum value of \(\rho\) is then \(\rho\approx 1.62466 g^2/K^2\) giving the maximum depth \(h_{\rm max}=r_{\rm max}\) as \(h_{\rm max}=2.25 g^2/K^2\).)
Now, how will the marble move? You may parameterize it by a single function, e.g. \(r(t)\). Assume \(h(r)\ll r\) because that's what we really care about. In this approximation, we see that the downward angle at the location \(r(t)\) is \({\dd h(r)}/{\dd r}\) which is \((K/g)r^{1/2}\) with the same constant \(K\): I just differentiated the \(3/2\)-th power (and the factor \(2/3\) canceled). This angle is tiny due to the approximation we made.
The downward gravitational acceleration \(g\) may be decomposed to a component that is normal to the motion ? that will be cancelled by the pressure of the dome ? and one that is tangential ? and that's what we care about. This tangential acceleration will accelerate the marble along the dome. Therefore, we have the equation\[
\frac{\dd^2 r(t)}{\dd t^2} = g\cdot \sin\alpha \approx K r^{1/2}.
\] This \(\approx\) relation is actually exact even for finite angles because we need the sine for a decomposition of the force but \(Kr^{1/2}\) was actually a sine as well because \(r\) was measured along the dome (it would be \(\tan\alpha\) if we were using \(\rho\), the straight distance from the axis). The second time derivative of the function \(t\) is the square root of this function. Cute: \(r(t)=0\) is clearly a solution. However, after the moment we call \(t=0\), and it can be any moment, the solution may suddenly continue as \[
r(t) = \frac{K^2}{144} t^4.
\] I could have incorporated the factor \(1/144\) into a new constant \(K'\) but I wanted to keep Norton's conventions ? he puts \(K=1\) although I wasn't quite able to understand what convention for the masses and times this is supposed to be.
At any rate, it's easy to see why this fourth power obeys the equation: if you differentiate it twice with respect to time \(t\), you reduce the power from \(t^4\) to \(t^2\) but also pick the factor \(4\times 3=12\) from the derivatives of the fourth and third powers. This will reduce \(1/144\) to \(1/12\), the same change of the coefficient as if you take the square root of \(r(t)\). The square root of \(t^4\) is \(t^2\), too.
Note that Newton's equations are obeyed at the "transitional moment" \(t=0\), too. You may also write down the time reversed solution: the marble shoots from the bottom towards the top with the "marginal velocity" that allows it to return immediately after reaching the summit; immediately continue on the other side of the dome; or do either of these two things after an arbitrary amount of time, perhaps even infinite.
It's weird but sort of "infinitely unlikely" so I wouldn't go far enough to claim that the setup proves that classical physics is "indeterministic".
Quantum mechanics is healthier than classical physics
However, I want to add some new comment. In classical physics, the marble on Norton's dome seems to have some extra degree of unpredictability ? one that is far more unknowable than the unpredictability in which you may actually calculate the probabilities. The marble may sit there for an arbitrary period of time and then starts to roll, sparking a war, for example. In quantum mechanics, this extra unpredictability is impossible.
If you have an initial state \(\ket\psi\) in quantum mechanics, it always predicts the odds that something will happen. In particular, if you tried to place a marble near the summit of a dome, the uncertainty principle guarantees that you can't have \(r=0\) and \(v=0\) at the same moment. At most, you may choose a compromise in which both of these quantities are small enough, scaling like \(\sqrt{\hbar}\) (times some functions of macroscopic parameters), and it will take a certain finite time interval for the particle to get to the bottom of the dome.
The conclusion is much more general and I may give you an explanation why. In the pathological behavior of \(r(t)\) above, we relied on a strange differential equation that had a square root of \(r(t)\) on the right hand side. However, quantum mechanics allows us to rewrite all the evolution in Schr?dinger's picture whose dynamical equation is Schr?dinger's equation. And the key point is that it is a linear equation in \(\ket\psi\). Linear equations don't allow the Norton-dome-type pathological behavior. A wave function can't "be constant" and suddenly start to roll. Schr?dinger's equation always tells it how much it must change. This "order" is strengthened by the fact that the equations governing evolution in quantum mechanics are first-order, not second-order (in any picture).
Many people think that quantum physics adds lots of pathologies to physics and physics has gotten confusing and ambigiuous. But once you actually understand how both classical physics and quantum mechanics work, you will see that classical physics has lots of potential systemic failures while quantum mechanics offers us a protection against them.
In general relativity, we could mention many extra examples. In a new paper, Leonard Susskind tells us that black hole firewalls ? something only allowed because of quantum mechanics ? are needed to protect us against de facto time machines. Alice could otherwise send the information about the early Hawking radiation to herself who is already in the black hole.
Well, I don't think that the firewalls exist or are needed but there are other examples in which quantum physics protects the overall consistency of physics in the presence of relativistic gravity. If you really need an example, the black holes' huge entropy ? which depends on \(\hbar\) and therefore on the quantum phenomena ? is needed to protect the second law of thermodynamics in the presence of black holes.
String theory offers us a much vaster array of "cures" for problems that would spoil "less quantum" theories than string theory. In some sense, string theory is the "most quantum theory" you may have. It's why it's also the most consistent one and the only one capable of providing us with the final picture.
And that's the memo.
Source: http://motls.blogspot.com/2012/10/classical-physics-is-sometimes-more.html
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