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Consider the shm for a single particle. Then the particle's position is given by (assume zero initial phase):

$$x = a \times \sin(\omega t)$$

The infinitesimal probability of finding a particle between $x$ and $x + dx$ is:

$$dp(x) = \frac{dt}{T} = \frac{dt \times \omega}{2\pi} = \frac{dx}{2\pi \sqrt{a^{2}-x^{2}}}$$

hence the probability density function is given by (I hope that so far derivations are correct):

$$pdf(x) = \frac{1}{2\pi \sqrt{a^{2}-x^{2}}}$$

Let's now imagine that, instead of a single particle performing shm, we have an ensemble of particles performing shm with different amplitudes. Let us also assume that $pdf(x)$ is known for the whole ensemble and we call it $pdf_{E}(x)$. So, $pdf_{E}(x)$ is known. How can we calculate $pdf_{E}(a)$?

Here's my attempt: $$pdf_{E}(x) = pdf_{E}(a \geqslant x) \times pdf(x) = \frac{pdf_{E}(a \geqslant x)}{2\pi \sqrt{a^{2}-x^{2}}}$$ then, rearranging we get: $$pdf_{E}(a \geqslant x)=2\pi \sqrt{a^{2}-x^{2}} \times pdf_{E}(x)$$ This is where I get stuck. I'm strugling to figure out how to proceed in order to be able to extract and plot $pdf_{E}(a)$ from the above.

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  • $\begingroup$ Maybe this is a typo but when you say "Let us also assume that $pdf(x)$ is known for the whole ensemble and we call it $pdf_E(x)$.", does it mean that $pdf(x) = pdf_E(x)$? $\endgroup$ – caverac Mar 9 '18 at 10:25
  • $\begingroup$ When I've used $pdf(x)$ before, was in the context of having a single particle. So, in the context of having an ensemble of particles I called the distribution $pdf_{E}(x)$. They're not the same, no. $\endgroup$ – unkown Mar 9 '18 at 10:33
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Consider one period $T=2\pi/\omega$ and suppose that the random time $t$ at which you measure the displacement $x$ of oscillator is uniformly distributed over the period of oscillation: $$f(t)=\frac{1}{T},\quad t\in[0,T)$$

By symmetry the p.d.f. $g(x)$ for displacement $x$ is an odd function i.e. $g(x)=g(-x)$ for $x\in[-a,a]$, so it suffices to work out $g(x)$ for $x\in[0,a]$. A given value of $x$, except the endpoints $x=\pm a$, is obtained at two distinct time instants in $[0,T)$. For $x\in[0,a)$ let the two time instants $t_1,t_2,$ be such that $\omega t_1=\sin^{-1}(x/a)\in[0,\pi/2]$ and $\omega t_2=\sin^{-1}(x/a)\in[\pi/2,\pi]$. Then we have: \begin{align} g(x)dx&=f(t_1)dt+f(t_2)dt\\ g(x)&=\frac{2}{T}\frac{dt}{dx}=\frac{1}{\pi\sqrt{a^2-x^2}},\quad x\in[-a,a] \end{align} in which I have also included the endpoints $x=\pm a$ because the probability measure over a finite set of points is zero anyway. You may verify that $g(x)$ is correctly normalised.

Now let us consider an ensemble of oscillators. The oscillators have random amplitude $a$ whose p.d.f. is $q(a)$. Let $h(x)$ be the corresponding p.d.f. for displacement $x$; $h(x)$ is supposed given and we want to find $q(a)$. The p.d.f. $g(x)$ found previously I shall write as $g(x|a)$ which is to be read as the p.d.f. for displacement given that the oscillator has amplitude $a$. From Bayes theorem we have: $$h(x)=\int_0^\infty da~g(x|a)q(a)$$ The integral needs to be inverted somehow to obtain $q(a)$ and here is a trick which I think will work. Consider the transformation $\eta=a^2$. Then since $a\geq 0$ there is one-to-one correspondence between $\eta$ and $a$, which allows us to relate $q(a)$ to p.d.f. of $\eta$: $$q(a)=q^*(\eta)\frac{d\eta}{da}=2a~q^*(\eta)$$ Therefore $$h(x)=\int_0^\infty d\eta~\frac{1}{\pi\sqrt{\eta-x^2}}\times q^*(\eta)$$

We now similarly transform $x$. By symmetry we have $h(x)=h(-x)$, so let us limit ourselves to $x\in[0,\infty)$. Transforming to $X=x^2$ then gives: $$h(x)=h^*(X)\frac{dX}{dx}=2x~h^*(X)=2\sqrt{X}~h^*(X)$$ Therefore $$2\sqrt{X}~h^*(X)=\int_0^\infty d\eta~\frac{1}{i\pi\sqrt{X-\eta}}\times q^*(\eta)$$

We have a convolution integral above. Applying Laplace transform $\mathbf{L}\equiv\int_0^\infty ds~\exp(-sX)$ gives: $$ q^*(\eta)=\mathbf{L}^{-1}\left[ \frac{\mathbf{L}\left[ 2\sqrt{X}~h^*(X) \right]}{\mathbf{L}\left[ \frac{1}{i\pi\sqrt{X-\eta}} \right]} \right]$$ from which $q(a)$ may be obtained.

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  • $\begingroup$ I don't understand how you got to $q(a)=q^{*}(\eta)\frac{d\eta}{da}$. Is this some sort of chain rule? I totally agree with $h(x)=\int g(x|a)q(a)da$ and the Bayesian interpretation. Maybe the Abel transform is suitable in this case. I'll dig some more. $\endgroup$ – unkown Mar 12 '18 at 15:12
  • $\begingroup$ That is how random variables transform. Essentially it says that the two probabilities $q(a)da$ and $q^*(\eta)d\eta$ are equal. $\endgroup$ – Deep Mar 13 '18 at 3:14

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