# Why is it impossible to measure the quantisation of the energy loss of a damped simple small angle pendulum?

A simple pendulum is damped by friction.

Initial maximum amplitude is $\frac{\pi}{30} rad$
(6 degrees)

Planck's quantized form for the average energy of a mode of frequency $\nu$ $$\langle\epsilon\rangle = \frac{h \nu}{e^{\frac{h \nu}{k T}}}$$

energy loss in a damped simple pendulum is an exponential decaying Cosine amplitude.

It has a constant frequency.

## Question

Show with a suitable calculation that it would be impossible to measure the quantisation of the energy loss of the pendulum?

Any sort of tips hints welcome? I am just starting quantum mechanics and I don't really understand what the question is asking?

I guess one could say that Planck quantised his modes of oscillation of cavity radiation by making their average energy dependent on frequency. But the classical pendulum is independent of frequency. But this doesn't prove watertight why you cannot measure it.

Tell me if this is rubbish

• Where exactly is it, when it's k.e. is at at a maximum ? – user108787 Oct 8 '16 at 11:34
• It is vertical at 0 degrees – Conor Cosnett Oct 8 '16 at 11:35
• Your argument doesn't explain why the quantisation cannot be measured. It seems to me that the statement you've been asked to prove might actually be false. I don't know. I'm curious to see the responses. – garyp Oct 8 '16 at 11:51

This is an estimation type problem. You are being asked to understand the relative scale of things.

The level spacing of a harmonic oscillator (and it is a reasonable first approximation to simple treat the system as harmonic for the purposes of a BotE calculation) is $$\Delta E = h f = \hbar \omega \;,$$ where $f$ is cyclic the frequency of the system and $\omega$ is the angular frequency.

Try computing the level spacing for this oscillator. Ask yourself what instruments you would use to even begin to measure the energy of such an object to such astounding precision.

I think that the answer to the measurement problem of the quantized energy loss is easy! The quantized energy loss $\hbar\omega$ of the damped pendulum is orders of magnitude smaller than the mean thermal energy of the pendulum. Assuming that the pendulum with small angle amplitude has small damping and is essentially a harmonic oscillator with angular frequency $\omega$, you can calculate the quantized energy change of the pendulum to be $$\Delta E=\hbar\omega=\hbar\sqrt{\frac{g}{L}}=7.35\cdot10^{-34}\ \mathrm{J}.$$ For the mean thermal energy of the pendulum with one degree of freedom you get classically $$\langle E\rangle=kT=4.14\cdot10^{-21}\ \mathrm{J}.$$ This thermal energy is 13 orders magnitude larger than the quantized energy change $\Delta E$ of the pendulum. Therefore, there is no chance that one can ever measure it. The quantized energy loss of the pendulum is completely lost in the thermal noise of the pendulum.

• I have deleted my earlier answer which erroneously referred to the quantization problem of a pendulum with dissipation. – freecharly Oct 8 '16 at 20:31
• @Bosoneando - Thank you for editing the equations. I have to tackle the equation editor. – freecharly Oct 8 '16 at 20:33

I'm note entirely sure what type of argument you are looking for, but can for instance note that the occilation amplitude of the pendulum is exponentially decaying with time.

This means that the energy quanta that are being dissipated are also becoming exponentially small as a function of time. Because of that, there is no experiment that will be able to resolve the energy quanta after some characteristic time.

• I'm not sure about your argument. First, accepting your premise, when the oscillations are large, the quanta will be large. If you can measure the quantization at all, you should be able to when the quanta are large. But I don't see why the quanta have to decrease in size at all. The quanta could stay the same size (the system does resemble an harmonic oscillator for short intervals of time) but the rate of creation decrease. – garyp Oct 8 '16 at 11:46
• @garyp I accept you objection. I have a naggi feeling that the correct agument here has something to with that the system is open, and so you will never be able to meassure the transition between two energy eigenstates, – Mikael Fremling Oct 8 '16 at 11:48
• It seems to me that one needs to make some assumptions about the loss mechanism. I await responses to this one! – garyp Oct 8 '16 at 11:51
• @garyp just "friction" – Conor Cosnett Oct 8 '16 at 11:52
• @ConorCosnett Ther eis no such thing in the quantum world as 'just friction' ;) – Mikael Fremling Oct 8 '16 at 11:55

## protected by Qmechanic♦Oct 8 '16 at 18:34

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).