Editing Thermodynamic temperature (section)

==== High speeds of translational motion ====
Although very specialized laboratory equipment is required to directly detect translational motions, the resultant collisions by atoms or molecules with small particles suspended in a [[fluid]] produces [[Brownian motion]] that can be seen with an ordinary microscope. The translational motions of elementary particles are ''very'' fast<ref>Even room–temperature air has an average molecular translational ''speed'' (not vector-isolated velocity) of 1822&nbsp;km/hour. This is relatively fast for something the size of a molecule considering there are roughly {{val|2.42|e=16}} of them crowded into a single cubic millimeter. Assumptions: Average molecular weight of wet air = 28.838 g/mol and {{mvar|T}} = 296.15&nbsp;K. Assumption's primary variables: An altitude of 194 meters above mean sea level (the world–wide median altitude of human habitation), an indoor temperature of 23&nbsp;°C, a dew point of 9&nbsp;°C (40.85% relative humidity), and {{cvt|lk=in|760|mmHg|kPa}} sea level–corrected barometric pressure.</ref> and temperatures close to [[absolute zero]] are required to directly observe them. For instance, when scientists at the [[National Institute of Standards and Technology|NIST]] achieved a record-setting cold temperature of 700&nbsp;nK (billionths of a kelvin) in 1994, they used [[optical lattice]] laser equipment to [[Adiabatic process|adiabatically]] cool [[caesium|cesium]] atoms. They then turned off the entrapment lasers and directly measured atom velocities of 7&nbsp;mm per second to in order to calculate their temperature.<ref>{{cite journal |title=Adiabatic Cooling of Cesium to 700&nbsp;nK in an Optical Lattice |first=A. |last=Kastberg |display-authors=etal|journal=Physical Review Letters |volume=74 |issue=9 |date=27 February 1995 |pages=1542–1545 |doi=10.1103/PhysRevLett.74.1542 |pmid=10059055 |bibcode=1995PhRvL..74.1542K }} A record cold temperature of 450&nbsp;[[Kelvin#SI prefixes|pK]] in a Bose–Einstein condensate of sodium atoms (achieved by A. E. Leanhardt ''et al.''. of [[Massachusetts Institute of Technology|MIT]]){{cn|{{subst:DATE}} equates to an average vector-isolated atom velocity of 0.4&nbsp;mm/s and an average atom speed of 0.7&nbsp;mm/s.</ref> Formulas for calculating the velocity and speed of translational motion are given in the following footnote.<ref name="Boltzmann">The rate of translational motion of atoms and molecules is calculated based on thermodynamic temperature as follows:
<math display="block">\tilde{v} = \sqrt{\frac {\frac{k_\text{B}}{2} \cdot T}{\frac{m}{2}}}</math>
where
* <math display="inline">\tilde{v}</math> is the vector-isolated mean velocity of translational particle motion in m/s
* {{math|''k''<sub>B</sub>}} ([[Boltzmann constant]]) = {{val|1.380649|e=-23|u=J/K}}
* {{mvar|T}} is the thermodynamic temperature in kelvins
* {{mvar|m}} is the molecular mass of substance in kg/particle

In the above formula, molecular mass, {{mvar|m}}, in kg/particle is the quotient of a substance's [[molar mass]] (also known as ''atomic weight'', ''[[atomic mass]]'', ''relative atomic mass'', and ''[[Atomic mass unit|unified atomic mass units]]'') in [[Gram|g]]/[[Mole (unit)|mol]] or [[Atomic mass unit|daltons]] divided by {{val|6.02214076|e=26}} (which is the [[Avogadro constant]] times one thousand). For [[diatomic]] molecules such as [[hydrogen|H<sub>2</sub>]], [[nitrogen|N<sub>2</sub>]], and [[oxygen|O<sub>2</sub>]], multiply atomic weight by two before plugging it into the above formula.

The mean ''speed'' (not vector-isolated velocity) of an atom or molecule along any arbitrary path is calculated as follows:
<math display="block">\tilde{s} = \tilde{v} \cdot \sqrt{3}</math>
where <math display="inline">\tilde{s}</math> is the mean speed of translational particle motion in m/s.

The mean energy of the translational motions of a substance's constituent particles correlates to their mean ''speed'', not velocity. Thus, substituting <math display="inline">\tilde{s}</math> for {{mvar|v}} in the classic formula for kinetic energy, {{math|1=''E''<sub>k</sub> = {{sfrac|2}}''mv''{{i sup|2}}}} produces precisely the same value as does {{math|1=''E''<sub>mean</sub> = 3/2''k''<sub>B</sub>''T''}} (as shown in {{section link|#Nature of kinetic energy, translational motion, and temperature}}). The Boltzmann constant and its related formulas establish that absolute zero is the point of both zero kinetic energy of particle motion and zero kinetic velocity (see also ''[[#Notes|Note 1]]'' above).</ref>

[[File:Argon atom at 1E-12 K.gif|thumb|left|upright=1.4|'''Figure 2.5''' This simulation illustrates an argon atom as it would appear through a 400-power optical microscope featuring a reticle graduated with 50-micron (0.05&nbsp;mm) tick marks. This atom is moving with a velocity of 14.43 microns per second, which gives the atom a kinetic temperature of one-trillionth of a kelvin. The atom requires 13.9 seconds to travel 200 microns (0.2&nbsp;mm). Though the atom is being invisibly jostled due to zero-point energy, its translational motion seen here comprises all its kinetic energy.]]It is neither difficult to imagine atomic motions due to kinetic temperature, nor distinguish between such motions and those due to zero-point energy. Consider the following hypothetical thought experiment, as illustrated in ''Fig.&nbsp;2.5'' at left, with an atom that is exceedingly close to absolute zero. Imagine peering through a common optical microscope set to 400 power, which is about the maximum practical magnification for optical microscopes. Such microscopes generally provide fields of view a bit over 0.4&nbsp;mm in diameter. At the center of the field of view is a single levitated argon atom (argon comprises about 0.93% of air) that is illuminated and glowing against a dark backdrop. If this argon atom was at a beyond-record-setting ''one-trillionth'' of a kelvin above absolute zero,<ref>One-trillionth of a kelvin is to one kelvin as the thickness of two sheets of kitchen aluminum foil (0.04&nbsp;mm) is to the distance around Earth at the equator.</ref> and was moving perpendicular to the field of view towards the right, it would require 13.9 seconds to move from the center of the image to the 200-micron tick mark; this travel distance is about the same as the width of the period at the end of this sentence on modern computer monitors. As the argon atom slowly moved, the positional jitter due to zero-point energy would be much less than the 200-nanometer (0.0002&nbsp;mm) resolution of an optical microscope. Importantly, the atom's translational velocity of 14.43 microns per second constitutes all its retained kinetic energy due to not being precisely at absolute zero. Were the atom ''precisely'' at absolute zero, imperceptible jostling due to zero-point energy would cause it to very slightly wander, but the atom would perpetually be located, on average, at the same spot within the field of view. This is analogous to a boat that has had its motor turned off and is now bobbing slightly in relatively calm and windless ocean waters; even though the boat randomly drifts to and fro, it stays in the same spot in the long term and makes no headway through the water. Accordingly, an atom that was precisely at absolute zero would not be "motionless", and yet, a statistically significant collection of such atoms would have zero net kinetic energy available to transfer to any other collection of atoms. This is because regardless of the kinetic temperature of the second collection of atoms, they too experience the effects of zero-point energy. Such are the consequences of [[statistical mechanics]] and the nature of thermodynamics.