[eesmith]

> The time-reversal of this motion ... at some arbitrary time.

The "ball rolling to the top of the sphere" requires infinite time. "Some arbitrary time" is an expression of a finite time.

You cannot simple mix ideas of finite and infinite and have the result make sense, as anyone who has stayed at the Hilbert Hotel knows. https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand...

[crazygringo]

> The "ball rolling to the top of the sphere" requires infinite time.

No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time. There's zero involvement of infinity, or mixing infinity, here.

[kergonath]

> No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time.

Can you explain why?

[crazygringo]

I don't know what kind of answer you're looking for. The equation was explicitly chosen/derived to have this property. I assume the mathematical proof of that isn't something that fits in a few sentences in an HN comment.

[eesmith]

Back when I was a little smithling who knew more math than physics, I complained about an assignment whose solution didn't make mathematical sense. My teacher commented that I needed to think like a physicist, that is, understand that certain mathematical issues didn't exist in the real world, so could be ignored.

The skit at https://www.youtube.com/watch?v=xPzR_D9qKeo gives some examples. The one at https://youtu.be/xPzR_D9qKeo?t=165 is pretty close to this example "if it's in physics, it's invertable."

That doesn't mean that if it's invertable it's in physics.

Are the inverse dynamics of this system still in Newtonian physics? For example, is is the inverse path actually on the described surface or does it detach? How does a moving mass have an instantaneous jerk with no change in velocity?

[xyzzyz]

The article explicitly discusses the fact that this is possible specifically because of the shape of the dome, and does not work on a hemisphere, precisely for reason you bring up.

[eesmith]

You are right - I misread it.

The next step would be to verify that the paths always stay on the surface. The mathematics shown says the point always follows the surface, but I don't see a demonstration that that's true.

I no longer have the skills to easily do this calculation.

EDIT: Oh man, I used to be a lot better at this. I remember the mgh = 1/2 m v^2 and the slope calculation, but can't figure out how tell when the falling point mass detaches from the slope. If it detaches at h=0 then there's no physically viable reversed path on the surface.

[kergonath]

The article explicitly discusses this without demonstrating anything. On it’s face this argument has the weight of these demonstrations.