Probability density function in classical mechanics

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The discussion explores the concept of a probability density function in classical mechanics, drawing parallels to quantum mechanics. It highlights that while classical mechanics provides deterministic trajectories through Hamiltonian equations, a probability function can be constructed to indicate the likelihood of finding a particle in a specific spatial interval. For a simple harmonic oscillator (SHO), the probability of locating the particle is higher near its endpoints, with a derived function showing this distribution. The mathematical formulation involves the oscillator's motion and velocity, ultimately leading to a probability distribution that accounts for the particle's behavior during oscillation. This approach allows for the calculation of average energy, momentum, and position similar to quantum mechanics.
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Probability density function plays fundamental role in qunatum mechanics. I wanted to ask if there is any analogous density function in classical mechanics. Obviously if we solve Hamilton equations we get fully deterministic trajectory. But it should be possible to find function which shows probability of finding classical particle in given interval of space (or generalized coordinate). For example it is known that if we have got harmonic oscillator it is most probable to find it nearby farthest place from "0". Do you know how to construct such function? If yes, can we calculate avarage energy, momentum, possition etc. as we do it in qunatum mechanics (by integral)?
 
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In the case of a simple harmonic oscillator (SHO); your intuition is right, it is more likely to encounter it near its endpoints of oscillation.

If we choose our coordinate frame such that the origin is at the equilibrium point for the SHO and starts the oscillation at the displacement ##A##, then the equation of motion is \begin{align}x(t) = A\cos(\omega t)\end{align} where \begin{align}\omega = \sqrt{\frac{k}{m}}\end{align} and ##k## is the spring constant. The velocity is \begin{align}\frac{dx}{dt} = -A\omega\sin(\omega t) \end{align} and the period of oscillation is ##T = 2\pi/\omega##. Then the probability for encountering it during the time interval ##dt## is ##\vert dt/T\vert##, thus the probability for encountering in the interval ##dx## is \begin{align}\bigg\vert\frac{dt}{T}\bigg\vert &= \frac{dx}{2\pi A}\csc(\omega t) \\ &= \frac{dx}{2\pi}\frac{1}{\sqrt{A^2 - x^2}}.\end{align} You can therefore think of \begin{align}\frac{1}{2\pi}\big(A^2-x^2\big)^{-\frac{1}{2}}\end{align} as the probability distrubution for this SHO.EDIT: The probability for encountering it is of course twice that given in (6) as it passes through the interval ##dx## twice (one time in each direction) during a full period of oscillation. This can be fixed in the derivation by only considering half a period of oscillation instead of a full period.
 
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