2
$\begingroup$

To be sure my basic physics isn't rusty...

Consider a 2-D bowled shaped classical potential well within which a classical particle of mass m is rolling. In this system the conservation of energy holds so the particle of mass m would roll from one end to another.

Because conservation of energy holds, we expect the mechanical energy to be

$E=T + U = \frac{1}{2}mv^{2}-mgh$

where v is the velocity of the particle

g is the gravitational acceleration

and

h is the heigh relative to the ground.

In classical physics, the maximum velocity of the particle occurs when the particle is at $r=\left ( x,h=0 \right )$ and the minimum velocity occurs when the particle is at some position $r=\left ( x,h \right )$ \exists h on both end of the well such that its kinetic energy T is 0 and potential energy U is at maximum.

Again, this follows from the conservation of energy:

$\Delta T= - \Delta U$

Now, I would like to construct a mathematical equation describing the probability of finding this particle of mass m as a function of its velocity. Intuitively, the greater the velocity of the particle at some point the lower the probability to find the particle and the smaller its velocity is the higher the probability to find the particle.

Solving $ E=T + U = \frac{1}{2}mv^{2}-mgh$ for v:

$v=\sqrt{\frac{2 \left ( E+mgh \right ) }{m}}$

If we want to explictly determine the probability of finding the particle as a function of its velocity, we should expect the probability density as a function of velocity to be of the form

$P=P\left ( v \right ) \propto \frac{1}{v}$

which comports to our common sense intuition.

How can I go about constructing a more explicit and informative equation that would enable to me determine the probability of finding the particle as a function of its velocity?

Any help is appreciated.

$\endgroup$
  • $\begingroup$ as Erwin's answer below shows, you can't do this abstractly. But if you know the particular potential, then it amounts to finding how much time is spent at a given velocity in each cycle (the motion is periodic) - that is, if the probability is based on randomly selecting a moment in time, which seems most natural to me. But maybe this is not what you meant, since your intuitive conclusion seems to be based on a different assumption, so first of all, you should define clearly what is meant by "finding" or what rules is the "finding" process obeying. $\endgroup$ – Nick Pavlov Nov 1 '17 at 15:06
2
$\begingroup$

I think with the information given, you cannot construct such a probability density function, because it very much depend on the shape of the 2-D bowl in which the ball is moving.

Compare the 'bowls' A and B with shape $y=x^2$ and $y=\min((x+1)^2, 0, (x-1)^2)$, respectively. Bowl B is similar to bowl A, except that its 'bottom' is on the interval $[-1, 1]$ rather than a single point. In both cases, the maximum speed of the balls will be the same (given that the balls started from the same height). But the speed profile will be different, i.e. the ball in bowl B will maintain its maximum speed for a while. As a result the probability density function $P(v)$ will be different.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.