# Probability density for velocity in mechanical energy

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

Consider a 2D 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-indefinitely.

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.

• Your particle isn't rolling, it's sliding. It needs a $\frac{1}{2}I\omega^2$ term to complicated things. – JEB Nov 2 '17 at 1:32
• Nitpick: the sign of the potential energy U is off: U increases with increasing h, while v decreases comeasurately. – Cosmas Zachos Nov 2 '17 at 14:23

I assume you are only interested in 2d motion in the (x,h) plane, so you are taking vanishing angular momentum, which would only complicate the problem.

Your essential error is that the "intuitive" form you have is wrong--it would make sense as a probability density in arclength, but not in v. That is, if you can review for the much simpler classical oscillator, the system spends more time at some xs or vs than in others, in general disproportionately so: consider t(x) or t(v).

For a half-period τ/2, the infinitesimal probability of the marble with v in the interval dv is
$$\frac {2dt}{\tau}=\frac{2}{\tau} \frac{dv}{\dot{v}}=P(v)dv ~,$$ that is, you need $\dot{v}$ as a function of v and E which is definitely a function of the shape of the bowl. The process flows linearly in time, but nonlinearly w.r.t. functions of the time, such as x(t), v(t), etc, and must be adjusted by the corresponding Jacobian determinants in the measure.

I will set up the problem for a parabolic bowl, $h=kx^2$, based on this answer, but won't pursue the general bowl shape case -- maybe there is a slick solution. $$T= \frac{m}{2}(\dot{x}^2 + \dot{h}^2) = \frac{m}{2} v^2= \frac{m}{2}(\dot{x}^2 + (2kx\dot{x})^2)=\frac{m}{2}\dot{x}^2(1 + 4k^2x^2) , ~~~~~~U= mgh =mgkx^2.$$ h has been eliminated. The Euler-Lagrange equation of motion reduces to
$$(1+4k^2x^2)\ddot{x}+4k^2\dot{x}^2x+2gkx=0 .$$

Since $E=\frac{m}{2} v^2+ mgkx^2$, and $$\frac{d}{dt} v^2= 2v \dot{v}=2 \dot{x} \ddot{x}(1 + 4k^2x^2) + 8k^2 x \dot{x}^3\\ = -4gkx\dot{x}=-4gkv\frac{x}{\sqrt{1+4k^2x^2}},$$ all you have to do is eliminate x in favor of v and E, to find $\dot{v}$ as a function of v and E, $$\dot{v}=2gk~/\sqrt{4k^2+mgk/(E-mv^2/2)}~.$$ As t goes from 0 to τ/4, v goes from 0 to $\sqrt{2mE}$.

I doubt if what you're trying to achieve is possible without defining the shape of the well.

One well behaved system is the parabolic potential well, for which:

$$h\propto x^2$$

And thus:

$$U(x)=kx^2$$

That can be Classically treated or (for very small systems) we can use the Schrödinger equation (here time-independent and in the $x$-dimension only):

$$\Big[-\frac{\hbar}{2m}\nabla^2+U(x)\Big]\Psi(x)=E\Psi(x)$$

The wave functions $\Psi_n(x)$ and energy levels $E_n$ are well known and on display in the link, as it is the case of the Quantum Harmonic Oscillator. (A full derivation can be found here).

Find the velocity by dividing the momentum operator by mass $m$:

$$\hat{v}=-\frac{i}{m}\hbar \nabla$$

Without defining $U(x)$, no precise quantum information can be obtained because the Schrödinger equation cannot be solved.