Yes, geodesics are minimizers of the distance functional. The latter is defined generally on a manifold with a metric. Associated to every diffusion is a metric which is the inverse of the variance-covariance matrix of the process. The article calculates the geodesics for the 2D example where the diffusion is the Heston process.

I give an attempt of intuitive explanation.

I neglect any form of friction.

We can prove that, using kinetic energy theorem, that it will remain the same for two points. Therefore the speed difference should come from the fact that we have balls and not points.

Thus, the rotational effects must be considered. From a kinetic viewpoint, the ball advances to the right because all the points on the top right of the ball drive the ball to the right in addition to its gravity center's inertia. Besides, in addition to the latter, the rotational inertia adds.

An intuitive idea of a ball's rotational inertia is the following: each dot of the ball, except its center, pushes its immediate neighbor to the direction orthogonal to the rotation axis according to the three fingers' rule. Each point is thus pushed, killing more equilibrium and enhancing the adequate movement. It turns out that the effect comes from the fact that each point has a mass.

When the ball goes down, rotation inertia is increased because of the gravity exercised to each point of the ball, increasing rotational inertia of the ball, which gains speed. When the ball goes up, rotational inertia decreases but the cycloid is done such that the ball has an overall increased rotational inertia.

Specifically, in case of no friction, the ball has a rotational inertia higher when it goes up than if the trajectory was a line.

Note that, in presence of friction, this explanation is rejected but I give it to you as an exercise.

JR

EDIT : I leave the comment originally as it was to point out a mistake I said. The kinetic energy theorem is not needed and the statement I wrote about it is wrong. So ignore it.

I neglect any form of friction.

We can prove that, using kinetic energy theorem, that it will remain the same for two points. Therefore the speed difference should come from the fact that we have balls and not points.

Thus, the rotational effects must be considered. From a kinetic viewpoint, the ball advances to the right because all the points on the top right of the ball drive the ball to the right in addition to its gravity center's inertia. Besides, in addition to the latter, the rotational inertia adds.

An intuitive idea of a ball's rotational inertia is the following: each dot of the ball, except its center, pushes its immediate neighbor to the direction orthogonal to the rotation axis according to the three fingers' rule. Each point is thus pushed, killing more equilibrium and enhancing the adequate movement. It turns out that the effect comes from the fact that each point has a mass.

When the ball goes down, rotation inertia is increased because of the gravity exercised to each point of the ball, increasing rotational inertia of the ball, which gains speed. When the ball goes up, rotational inertia decreases but the cycloid is done such that the ball has an overall increased rotational inertia.

Specifically, in case of no friction, the ball has a rotational inertia higher when it goes up than if the trajectory was a line.

Note that, in presence of friction, this explanation is rejected but I give it to you as an exercise.

JR

EDIT : I leave the comment originally as it was to point out a mistake I said. The kinetic energy theorem is not needed and the statement I wrote about it is wrong. So ignore it.

I want to give another explanation based on the point only, still with no friction.

When the point goes down, it gains speed therefore the duration to hit downhill (ie middle of trajectory) is lower than the one to hit middle point if the trajectory was a line.

In addition, the only force that (positively) works is the weight.

When the point goes up, the duration to hit uphill is the same. Indeed, the trajectory is supposed to be symmetric. In addition, the only force that works (negatively) is the weight, and is exactly the opposite of the weight work when the point went downhill.

Bearing all above in mind, the durations are the same.

It turns out that the duration for the point to get downhill and to return uphill is lower than the one if the trajectory was a line.

This explanation would be sufficient in a finance brainteaser, I believe, but is physically meaningless to me since they are balls and not points.

When the point goes down, it gains speed therefore the duration to hit downhill (ie middle of trajectory) is lower than the one to hit middle point if the trajectory was a line.

In addition, the only force that (positively) works is the weight.

When the point goes up, the duration to hit uphill is the same. Indeed, the trajectory is supposed to be symmetric. In addition, the only force that works (negatively) is the weight, and is exactly the opposite of the weight work when the point went downhill.

Bearing all above in mind, the durations are the same.

It turns out that the duration for the point to get downhill and to return uphill is lower than the one if the trajectory was a line.

This explanation would be sufficient in a finance brainteaser, I believe, but is physically meaningless to me since they are balls and not points.

3 down valleys but only 2 hills, the average altitude is lower

Yes but, refer to the video, the ball is faster at the first 'go down-go up' so it doesn't matter.

That's because the average height of the 'go down-go up' is lower. As it turns out the it's only valleys, I've marked them blue.

Ah alright I get what you said.

But watch out from the following point: I think you should consider as many hills as down valleys otherwise the two balls wouldn't be at the same final level, which would be a significant bias, and then no possible comparisons, shouldn't you?

But watch out from the following point: I think you should consider as many hills as down valleys otherwise the two balls wouldn't be at the same final level, which would be a significant bias, and then no possible comparisons, shouldn't you?

Yes, and we need to agree on "what's a hill and what's a valley". I would call the places where the right track is below the left track a valley, and if it's above a hill. Using this definition there are no hills!Ah alright I get what you said.

But watch out from the following point: I think you should consider as many hills as down valleys otherwise the two balls wouldn't be at the same final level, which would be a significant bias, and then no possible comparisons, shouldn't you?

IMO there is no simple answer. The extra distance travelled and the rotational kinetic energy affect the outcome. Because of that it's very good educational setup. You can teach many physical phenomena with it.

- Traden4Alpha
**Posts:**23951**Joined:**

Another way to look at this is to think of these two as pendulums swinging under gravity.

Both pendulums swing over the same horizontal distance but the one with the valleys swings a deeper arc. That deeper arc for a given horizontal distance is equivalent to a shorter pendulum which implies it will have a higher natural frequency (which is almost invariant with the total horizontal swing) than the valley-free one.

Both pendulums swing over the same horizontal distance but the one with the valleys swings a deeper arc. That deeper arc for a given horizontal distance is equivalent to a shorter pendulum which implies it will have a higher natural frequency (which is almost invariant with the total horizontal swing) than the valley-free one.

- katastrofa
**Posts:**9192**Joined:****Location:**Alpha Centauri

Ponderomotive force?

If you remove the wavy section and assuming the ball can bounce three times on the floor below instead, then they'll be both as fast again! (if you ignored rotational kinetic energy).

- Traden4Alpha
**Posts:**23951**Joined:**

Indeed!If you remove the wavy section and assuming the ball can bounce three times on the floor below instead, then they'll be both as fast again! (if you ignored rotational kinetic energy).

Yet the bouncing-ball scenario is a very different physical system than the rolling-ball one in that the drop-bounce-rebound does nothing to couple vertical motion with horizontal motion. An arc-tracked rolling ball is like a pendulum's swinging ball in that vertical potential energy is exchanged for horizontal kinetic energy and vice versa.

Moreover, if we replace the solid ball with a microscopic blackhole inside a styrofoam ball, then rotational energy would be nearly zero but I'd wager the results would be quite similar under the pendulum system analogy.

- Traden4Alpha
**Posts:**23951**Joined:**

Cognition at a distance?Ponderomotive force?

Cool, let's call that "T4A's law": the probability of a discussion tunneling toward a "blackhole inside a styrofoam ball gedankenexperiment" due to quantum random topic changes is non zero.

But it depends on the size of the black hole. If it's really small then will Hawking evaporate and end in either nothing or a remnant depending on your quantum gravity theory flavour.

But it depends on the size of the black hole. If it's really small then will Hawking evaporate and end in either nothing or a remnant depending on your quantum gravity theory flavour.

- Traden4Alpha
**Posts:**23951**Joined:**

According to http://xaonon.dyndns.org/hawking/ (the internet is freaking awesome!), a 100 gram blackhole would evaporate in 10^-20 seconds while emitting 10^34 watts!!!!!!!!!

The blackhole needs to be at least 1,000 metric tons to last more than the minute it takes to set up and film this demonstration. But it's still emitting 10^20 watts so the styrofoam (and track, and camera, and experimenter) are also evaporating!

A 10^13 metric ton black hole would emit a negligible 3.6 watts but the track is going to need to be a bit larger.

The blackhole needs to be at least 1,000 metric tons to last more than the minute it takes to set up and film this demonstration. But it's still emitting 10^20 watts so the styrofoam (and track, and camera, and experimenter) are also evaporating!

A 10^13 metric ton black hole would emit a negligible 3.6 watts but the track is going to need to be a bit larger.

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