# Is there an uncertainty associated with the value 0 K for absolute zero?

When I say absolute zero, I’m not talking about the hypothetical temperature 0 K; I’m talking about the temperature at which a thermodynamic system has the lowest energy.

Everywhere I look, sites maintain that this should be –273.15 ºC, but I can’t find any uncertainty quoted with this prediction. Could someone shed some light on this?

I know that the Celsius scale is defined to be an easy conversion from the kelvin scale. What I mean to ask is the following: The Boltzmann constant $$k_{\rm B}$$ has a fixed value; the reference temperature $$273.15\;\rm K$$ is also a fixed value used in the relation $$t/\mathrm{°C}=T/\mathrm{K}-273.15$$. Therefore, how can we know that a thermodynamic system has the lowest energy at exactly $$0\;\rm K$$? If the triple point of water $$T_{\rm TPW}=273.16\;\rm K$$ has a standard relative uncertainty of $$3.7\times10^{-7}$$, then why shouldn’t $$T_{\rm A0}$$, especially since it has never been achieved? Something has to give and not be by definition, so to speak.

As another answer explains, absolute zero is defined in the Kelvin temperature scale as precisely $$\pu{0 K}$$. But this is not the entire story, as all measurements have an associated uncertainty (all thermometers are imprecise), see e.g. the wikipedia article describing the International Temperature Scale of 1990 and the Provisional Low Temperature Scale of 2000. For instance the second wikipedia article links to a International Committee for Weights and Measures' document describing the uncertainties associated with the calibration of the temperature scale based on measurements of the melting pressure of $$\ce{^3He}$$.

Recently the SI value of the Kelvin was changed. This is explained in the wikipedia:

The definition of the kelvin underwent a fundamental change. Rather than using the triple point of water to fix the temperature scale, the new definition uses the energy equivalent as given by Boltzmann's equation.

Previous definition: The kelvin, unit of thermodynamic temperature, is 1/273.16 of the thermodynamic temperature of the triple point of water.

2019 definition: The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $$1.380649×10^{−23}$$ when expressed in the unit $$\pu{J⋅K−1}$$, which is equal to $$\pu{kg⋅m2⋅s−2⋅K−1}$$, where the kilogram, metre and second are defined in terms of h, c and $$\Delta ν_{Cs}$$.

The kelvin may be expressed directly in terms of the defining constants as:

$$\pu{1 K} = \frac{1.380 649 \times 10^{23}}{(6.626 070 15 \times 10^{-34})(9 192 631 770)} \frac{h \Delta \nu _{Cs}}{k}$$

So what does this mean for the definition of $$\pu{0 K}$$ ? That doesn't change, that is an absolute, thermodynamic definition. However, evidently the triple point of water is now no longer exactly $$\pu{273.16 K}$$. You can think of this as shifting the uncertainty from the size of a step to the number of steps that take you from absolute zero to the triple point. But both before and after the redefinition, $$\pu{0 K}$$ has meant $$\pu{0\pm 0 K}$$, that hasn't changed.

$$^\ast$$you can't actually reach absolute zero, so you can only extrapolate to 0 K and provide an uncertainty for this extrapolation, i.e. you could estimate how far your measured temperature is from absolute zero, and provide uncertainties associated with your estimate.

• Our temperature scale is fixed around the reference temperature $T_0=273.15\;\rm K$ and a change of one kelvin of temperature results in exactly a change of $1.380\,648\times10^{-23}\;\rm J$ of thermal energy according to the International Bureau of Weights and Measures. – gen-z ready to perish Oct 23 '19 at 5:00
• It's all well explained in International Committee for Weights and Measures documentation such as: bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-EN.pdf – Buck Thorn Oct 23 '19 at 14:13
• @gen-zreadytoperish That's not true, as Buck Thorn quoted. Kelvin is a unit of absolute temperature - zero means zero, just like zero meters means no distance and zero seconds means no time. There's no uncertainty, it's not a result of measurement, it's zero by definition (regardless if it's possible to reach or measure e.g. zero distance). The "reference temperature" you're talking about is the uncertain part: "At the time of adopting the present definition TTPW was equal to 273.16 K with a relative standard uncertainty of 3.7E-7 based on measurements of k made prior to the redefinition". – Luaan Oct 24 '19 at 7:27
• @gen-zreadytoperish So, the zero value is fixed. The Boltzmann costant is fixed. The thing that changes with improved measurements and what-not is the triple point of water. If we suddenly realized we were all stupid and the triple point of water is actually 283.16 K (oops!), that's the only thing that changes - not the Kelvin scale, not the zero-point, not the Boltzmann constant. – Luaan Oct 24 '19 at 7:30
• This might also be helpful: en.wikipedia.org/wiki/Boltzmann_constant; also, look up the ideal gas thermometer – Buck Thorn Oct 24 '19 at 7:59

There is no uncertainty in this value, because the Celsius scale is defined in terms of absolute zero, rather than visa-versa. I.e, the Celsius scale is defined by specifying that -273.15 C (exactly) corresponds to absolute zero, and +0.1 C (exactly) corresponds to the triple point of Vienna Standard Mean Ocean Water (VSMOW). This locks in both the size (i.e., the spacing) of Celsius degrees (each Celsius degree is 1/273.16 of the way between absolute zero and the t.p. of VSMOW), and the offset between the Celsius and Kelvin temperature scales.

N.B.: This definition has been changed as of 2019 (see other answer).

• Well, the water triple point is at +0.01 Deg C, resp. at 273.16 K. – Poutnik Oct 23 '19 at 4:10
• I know that the Celsius scale is defined to be an easy conversion from the kelvin scale. What I mean to ask is the following: The Boltzmann constant $k_{\rm B}$ has a fixed value; the reference temperature $273.15\;\rm K$ is also a fixed value used in the relation $t/\mathrm{°C}=T/\mathrm{K}-273.15$. Therefore, how can we know that a thermodynamic system has the lowest energy at exactly $0\;\rm K$? If the triple point of water $T_{\rm TPW}=273.16\;\rm K$ has a standard relative uncertainty of $3.7\times10^{-7}$, then why shouldn’t $T_{\rm A0}$, especially since it has never been achieved? – gen-z ready to perish Oct 23 '19 at 4:51
• It is the other way. The Kelvin scale has been defined to be an easy conversion from the Celsius scale. The same for the Rankin scale from the Fahrenheit scale. – Poutnik Oct 23 '19 at 8:27
• @Poutnik Historically, it is the other way. But today, Kelvin is the fundamental scale. The scale is defined in terms of the Boltzmann constant, which is a fundamental constant in the SI system of units. But even then, from the very beginning, Kelvin was specifically designed to be an absolute temperature scale, while both Celsius and Fahrenheit are relative. Zero Kelvin was always zero by definition, "infinite cold". – Luaan Oct 24 '19 at 7:36
• @gen-zreadytoperish It's not the result of any measurement. The system at lowest thermodynamic energy has zero Kelvin by definition. That's the whole point of having an absolute temperature scale. The thing that changes is how that translates to Celsius or relative to the triple point of water or whatever. This was originally determined theoretically based on Kelvin's calculations; but regardless of what measurements happen, zero will always be zero, and one degree Kelvin will always be change of thermal energy of 1.380 649E−23 J, exactly, without uncertainties, by definition. – Luaan Oct 24 '19 at 7:45

If you want to define your own temperature scale you need at least to fixed (no uncertainty) points. Celcius assigned 0 degrees to a mixture of ice and water at atmospheric pressure and 100 degrees to the boiling point of water. If I understand your question correctly you are asking how can one put a "thermometer" at absolute zero if we can never reach that temperature? The answer is we can't. As mentioned by Buck Thorn the definition of the Kelvin changed recently (on May 20th of 2019 to be precise) and is now based on the Boltzmann constant. Before this definition, the Kelvin temperature scale was based on several reference temperatures which covered the range from 0.9 mK to 1357.77 K. By comparing with these reference temperatures, the absolute temperature could be determined within a certain measurements uncertainty. The scale was defined such that it provides a 0 temperature that is as close as possible to the absolute zero. The triple point of water was defined in such a way that an increase of 1 Kelvin corresponds to an increase of 1 Celcius. This consistency is an important part of redefining base units.

Since the new SI base units came into effect earlier this year the Boltzmann constant was fixed. To this end the committee charged with defining the new SI base units asked the Committee on Data for Science and Technology (CODATA) to collect very precise measurements of the Boltzmann constant that rely on several measurement techniques. The value that was obtained was very precise, even more precise than the SI committee needed. Why? Because the committee wanted to make sure that in the new system the triple point of water was still 273.16 K. So they rounded off the experimental Boltzmann factor to a relative uncertainty of about 1e-6. As a consequence, the new scale is compatible with the old scale, but extends over a much wider range as it is now linked to a measurement of the thermal energy $$E=k_\text{B}T$$. Of course when you now measure the thermal energy of the triple point of water you get a measurement uncertainty. Note that in principle we could get rid of temperature and use a measurement of the thermal energy instead. In that sense the Boltzmann constant is similar to the Avogadro constant: it's a conversion factor that has mostly historical (and practical) reasons and is not a "true fundamental constant of nature".