Probability density function in classical mechanics

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SUMMARY

The discussion centers on the concept of probability density functions in classical mechanics, particularly in relation to simple harmonic oscillators (SHO). It establishes that while classical mechanics is deterministic, a probability distribution can be derived for the position of a particle in a SHO. The probability density function is given by the equation 1/(2π) * (A² - x²)^{-1/2}, which indicates a higher likelihood of finding the particle near its maximum displacement. The discussion also touches on calculating average energy and momentum using integrals similar to quantum mechanics.

PREREQUISITES
  • Understanding of Hamiltonian mechanics
  • Familiarity with simple harmonic oscillators (SHO)
  • Knowledge of probability density functions
  • Basic calculus for integration and differential equations
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  • Explore the derivation of probability density functions in classical systems
  • Learn about energy and momentum calculations in classical mechanics
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Physicists, mechanical engineers, and students of classical mechanics seeking to understand the probabilistic interpretations of particle motion in deterministic systems.

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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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