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I am reading statistical mechanics from Concepts in Thermal Physics, the author states the following after deriving the equipartition theorem.

A mass on a spring has energy $E$ which is given as the sum of two quadratic terms:

$$E = \frac{1}{2}mu^2 + \frac{1}{2}kx^2$$

and by Equipartition theorem we have:

$$ E = 2\cdot\frac{1}{2}k_\mathrm{B} T$$

What troubles me is the following:

How big is this energy? At room temperature, $k_\mathrm{B} \approx 0.025$ eV, which is a tiny energy. This energy isn’t going to set a $10$ kg mass on a stiff spring vibrating very much!

Isn't the energy of the system,

$$E = \sum_{i=1}^n E_i =\sum_{i=1}^{n} (K_i + P_i)$$

where $n$ is the number of particles composing the system? Are the two expressions for energy equal?

Edit

What I can't understand is what are the degrees of freedom of the system. When we have a box filled with gas particles we don't examine the macroscopic motion of the box (or at least this is not subject of statistical mechanics). Why the author considers the macroscopic velocity of the mass on the spring? Shouldn't we take only into account the (solid) particles that constitute the mass on the spring?

Also if there is vibration then some force must have initiate this motion. For example, we must first stretch the string in order to start vibrating.

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  • $\begingroup$ Does the author say "macroscopic"? I don't see why a "10 kg mass" couldn't be a point mass (which has no physical extent whatsoever). Furthermore, this is clearly a toy example, so not sure what you real concern is... $\endgroup$
    – hft
    Aug 5, 2022 at 19:02
  • $\begingroup$ The author doesn't use the word "macroscopic". I used that word to make the question more clear. Isn't the motion of a mass on spring or a box considered macroscopic? $\endgroup$ Aug 5, 2022 at 20:39

2 Answers 2

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The two expressions

$$E = \frac{1}{2}mu^2 + \frac{1}{2}kx^2;$$ $$E = \sum_{i=1}^{n} (K_i + P_i)$$

are aren't quite equivalent; they have different reference values. (Energy has no absolute value; we always need to specify a reference, and we often use different reference levels for convenience in different analysis frameworks.)

The first equation ignores thermal energy, or alternatively sets the reference value of energy as a motionless mass and an undeformed spring at room temperature. Here, we've also implicitly assumed that the mass is rigid (i.e., that any energy change of interest is associated only with its bulk kinetic energy, or the kinetic energy of its center of mass) and that the spring is massless (i.e., that any energy change of interest consists of only strain energy).

The second equation dispenses with the lumped-component assumptions of a mass and spring, which makes it more general but also less amenable to problems involving such idealized macroscale objects.

A practical example is spontaneous vibration of atomic-force-microscopy cantilevers. These extremely thin beams—deliberately designed to be compliant enough to deform from atomic-scale asperities and attractive forces on surfaces—are also compliant enough to detectably vibrate at room temperature from thermal energy alone:

Sevim, S., Shamsudhin, N., Ozer, S. et al. An atomic force microscope with dual actuation capability for biomolecular experiments. Sci Rep 6, 27567 (2016).

Here, the feature of interest is the 18 kHz peak in air, indicating the excitation of natural resonance from thermal motion. (The peak is dampened and shifted to 4 kHz in water.)

From the equipartition theorem, we can estimate the amplitude of the resulting motion or the speed of the cantilever tip at any particular temperature. For the former, we set $\frac{1}{2}k_BT=\frac{1}{2}k\widehat{x^2}$, where $\widehat{x^2}$ (sometimes $\langle{x^2}\rangle$) is the mean squared thermal deflection, as discussed by Butt & Jaschke in "Calculation of thermal noise in atomic force microscopy."

For more discussion and references in this context, see Ma et al., "Thermal noise in contact atomic force microscopy".

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  • $\begingroup$ My question was about how is it possible thermal energy to set in motion a macroscopic object. Can you elaborate on your last paragraph? From what I understand a sum of both expression would be more appropriate? $\endgroup$ Aug 5, 2022 at 8:41
  • $\begingroup$ Please see the revised answer. I'm not sure what "sum of both expression" means. $\endgroup$ Aug 5, 2022 at 13:42
  • $\begingroup$ @AntoniosSarikas The author did not say he is only concerned with macroscopic objects. Yes, the author goes on to mention a 10kg mass, but he did not say the mass is macroscopic. (Can't an idealized point mass weight 10 kg?) Anyways, it is a toy example for sure, so not really worth getting worked up about... $\endgroup$
    – hft
    Aug 5, 2022 at 18:59
  • $\begingroup$ @Chemomechanics Would it have meaning to calculate the heat capacity based on the first expression? $\endgroup$ Aug 5, 2022 at 20:57
  • $\begingroup$ I don't really see how that would be possible. What method did you have in mind? $\endgroup$ Aug 5, 2022 at 22:18
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Short answer is: Mass will oscillate with an amplitude given by the following expression:

enter image description here

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    $\begingroup$ Please use Mathjax syntax to typeset equations (very close to LaTeX syntax). Screenshots, especially difficult to read like this one, are far less usable. $\endgroup$
    – Miyase
    Aug 5, 2022 at 14:31
  • $\begingroup$ I am not asking why the amplitude is small. I am asking why there is motion at all (even if small). $\endgroup$ Aug 5, 2022 at 18:31
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    $\begingroup$ Aakash, your answer is fine in principle, but it gets a downvote from me because it is just a link to an image. That link is subject to link rot. Please writeup this answer in Latex/Mathjax so I can give it an upvote (or maybe some nice person will edit it for you, but I wouldn't count on it). $\endgroup$
    – hft
    Aug 5, 2022 at 18:53
  • $\begingroup$ @AntoniosSarikas there is motion because it is in contact with a heat bath. At normal room temperatures the motion is too small to see. In reality, at higher temperature the spring will probably just melt so you won't see anything with the naked eye in that case either. $\endgroup$
    – hft
    Aug 5, 2022 at 18:53
  • $\begingroup$ @hft Why the thermal energy of the constituent particles are neglected though? If I was asked to specify the degrees of freedom for a box full of gas particles I would say that it has $3N$ (three for each $u_x, u_y, u_z$ of the particles). Why we must include the velocity of the center of mass as an extra degree of freedom? $\endgroup$ Aug 5, 2022 at 19:02

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