4
$\begingroup$

or a phenomenon where we can only measure the standard deviation ($\sigma_w$) of a variable $w$ and not the mean $\overline{w}$

$\endgroup$
  • $\begingroup$ I'm surprised nobody has mentioned heat capacity (variance of the energy in a thermodynamic system, relevant for phase transitions). It'd be great if some experimentalist could say a few words on it. $\endgroup$ – Arturo don Juan Sep 24 '17 at 1:53
5
$\begingroup$

One example which comes to mind is Brownian motion. Here you observe a small particle in a fluid (or more generally a small dynamical system coupled to a heat bath). The particle will jiggle back and forth in all directions, which is due to the kicks delivered by the thermal motion of the fluid around it. One then takes the average over many particles (or over time for one single particle if one is careful with correlations).

The important quantity here is the distance a particle has moved after a time delay $\tau$, which Einstein showed has a standard deviation that grows as $\sqrt\tau$. The mean of that distance is not very relevant and strictly speaking, in a vector sense, it is zero.

$\endgroup$
  • $\begingroup$ Very good example, but I was thinking "is it true that in order to measure the std, the mean must be zero?" $\endgroup$ – llllllllllllllllllllllllllllll Jul 6 '13 at 14:12
  • $\begingroup$ Another thing is that when measuring $\tau$, you'll be measuring the variance and you deduce the std by a square root; can't we measure "directly" the std? $\endgroup$ – llllllllllllllllllllllllllllll Jul 6 '13 at 14:13
3
$\begingroup$

One does not measure the standard deviation nor the mean. One makes many measurements of a value and from that they compute the mean of the measurements and the standard deviation of the measurements from the mean. So in a word, no.

The mean is the average value of all measurements (appropriately weighted). If you make more than one measurement of a variable, there is always a mean. The standard deviation (stdev) is a numerical representation of how tightly packed the measurements tend to be around the mean. A low stdev indicates most of the measurements are very close to the mean and a high stdev indicates the measurements are very spread out around the mean. Again anytime there is more than one measurement, there is a stdev associated. Therefore, it does not make sense to say you are "measuring the standard deviation" since the standard deviation is a descriptor of your measurements.


Often people will refer to values as an average over something else. Such as temperature of a gas being an average of the kinetic energy of the particles that make it up. It's true that we are referring to the same type of mean, however, it would still not be possible to make one measurement of the standard deviation. The stdev in this example would be the closeness of the kinetic energy of each particle to the average kinetic energy. However, this is not measurable because it does not represent something physical.

The mean/average value is something physical; it is the best approximation of a specified value of a large system that is composed of smaller systems (thus measuring the large system once should result in the mean value of the smaller systems). The stdev, on the other hand, is purely statistical; it is the best approximation of the precision of the data and is not measurable.

Having said that, as pointed out, there are occurrences where the stdev or distribution of the data is more interesting (not directly measurable still, but can provide more insight). As said, the Cosmic Microwave Background is one of the instances where the deviation of the data from the mean is more interesting than the mean of the data. In this case, it is because theories predict that the early universe should have been homogeneous; non-homogeneous distributions of variations in the CMB (our earliest view of the universe) would therefore give us better insight into the different processes and events that occurred during that early time.

$\endgroup$
  • $\begingroup$ You missed my point: I'm not taking several measurements and taking their mean and std; I'm doing one measurement of a macroscopic quantity that is the mean of something like the pressure or the temperature. $\endgroup$ – llllllllllllllllllllllllllllll Jul 5 '13 at 19:15
  • 2
    $\begingroup$ Perhaps I am still missing your point, but how would one measure the mean of something with just one measurement? $\endgroup$ – Jim Jul 5 '13 at 19:34
  • $\begingroup$ Isn't the temperature of an ideal monoatomic gas a measure of the average kinetic energy of its atoms? $\endgroup$ – llllllllllllllllllllllllllllll Jul 5 '13 at 19:59
  • $\begingroup$ Ah, now I see what you meant. The answer is still the same, but I'll add a some specifics to clarify the difference. $\endgroup$ – Jim Jul 5 '13 at 20:01
  • 1
    $\begingroup$ There do exist quantities with distributions that are more interesting than their means. E.g. cosmic microwave background. Maybe someone with more expertise can elaborate. $\endgroup$ – innisfree Jul 5 '13 at 20:06
3
$\begingroup$

In stellar dynamics, one of often tries to determine the mass distribution of gravitating systems (globular clusters, galaxies, clusters of galaxies, dark matter haloes,...) from the positions and velocities of their constituents. The simplest case are spherical systems with no net rotation. In this case the average velocity at every point is zero, while the velocity dispersions are linked to the gravitational potential through the spherical Jeans equation: $$ \frac{\text{d}\rho\sigma_r^2}{\text{d}r} + 2\frac{\rho\sigma_r^2}{r} - \frac{\rho\sigma_T^2}{r}= -\rho\frac{\text{d}\psi}{\text{d}r}, $$ where $\rho(r)$ is the density, $\sigma_r(r)$ and $\sigma_T(r)$ the radial and tangential velocity dispersion, and $\psi(r)$ the gravitational potential.

$\endgroup$
  • $\begingroup$ So you have to do a lot of measurement to find the stds? $\endgroup$ – llllllllllllllllllllllllllllll Jul 6 '13 at 14:22
  • $\begingroup$ @metacompactness A few dozen per radius. One way to do this is to select $n$ objects within a radius $[r,r+\Delta r]$. The dispersion is then $\sigma^2(r)\approx \sum_{i=1}^{n}\frac{(v_i - \langle v\rangle)^2}{n-1}$. The error bars on these dispersions are of the order $\sim \sigma/\sqrt{n}$. $\endgroup$ – Pulsar Jul 7 '13 at 13:35
  • $\begingroup$ I'm not taking several measurements and taking their mean and std; I'm doing one measurement of a macroscopic quantity that is the mean of something like the pressure or the temperature. (the same comment on Jim's answer) $\endgroup$ – llllllllllllllllllllllllllllll Jul 7 '13 at 13:39
2
$\begingroup$

In nuclear and particle physics, the primary data of interest are scattering cross sections. In these data, the structure of the particle(s) of interest shows up through various "resonances" in your data (i.e. big spikes in the data set). These resonances correspond to states of the system of particle(s) (e.g. nucleus, proton, etc.) of interest. The width of each one of those resonances gives, through the Heisenberg uncertainty principle for decay-energy and decay-time, a the decay time for that state. This width can be approximated by the FWHM of the state, from which typically a more accurate fitting of a relativistic Breit-Wigner function (Lorentzian function in the non-relativistic case) is carried out. At the surface, one can imagine that this width is similar to the standard deviation of a Gaussian distribution, though in interpretation they are different. This information is absolutely critical for understanding nuclear structure, and it also has tremendous importance in astronomy, nuclear engineering, etc.

enter image description here

$\endgroup$
  • 2
    $\begingroup$ It bears mention that as a first approximation these resonances are often Lorentzian (a.k.a. Breit-Wigner), so formally they don't have a well-defined standard deviation. I'm not completely sure what they use as a width measure in practice, though. $\endgroup$ – Emilio Pisanty Feb 11 '16 at 15:20
  • $\begingroup$ @EmilioPisanty Yeah you're right - I wasn't thinking too much about it. Formally one measures the width through an approximation of the FWHM. However, when I connect this quantity to the decay time of the state, my brain immediately thinks about Gaussians, which obviously led to a bit of an incorrect interpretation. $\endgroup$ – Arturo don Juan Feb 11 '16 at 15:42
  • $\begingroup$ @EmilioPisanty However, at the surface the $\sigma$ obtained from a roughly fitted Gaussian is related to the FWHM, which is related to the $\Gamma$. $\endgroup$ – Arturo don Juan Feb 11 '16 at 15:45
  • $\begingroup$ @EmilioPisanty I've edited my response. $\endgroup$ – Arturo don Juan Feb 11 '16 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.