# How was absolute 0 calculated?

1. Oct 22, 2008

### janakiraman

Hi, according to 3rd law of thermodynamics, absolute 0 can never be attained. But i really dont understand how was the exact value of absolute 0 calculated almost 100 years back? I think it comes from the thermodynamic relationships, but i dont understand how. Can anyone explain me the derivation of the absolute 0 value?

2. Oct 22, 2008

### Staff: Mentor

Extrapolation?

3. Oct 23, 2008

### janakiraman

Even i understand that we can graphically extrapolate the ideal gas scenario and then find the 0K. but how do u justify that extrapolation? Ideal gas is not obeyed even if the gas reaches its boiling point, in that case i'm sure all the gases would freeze before we reach 0K and how can you justify the assumption of ideal gas to be correct and extrapolate it to find 0K?

Last edited: Oct 23, 2008
4. Oct 23, 2008

### Staff: Mentor

It is not just ideal gas. There are many formulas that show how different things change with the temperature - like Clausius-Clapeyron relation, Arrhenius equation - if you draw plots they suggest there must exist some specific temperature where "something changes". And, what is important, all these equations seems to predict this specific temperature to be in about the same place.

Or think about it this way - you have several equations relating some phenomenas to temperature. In all cases the relation became nicely linear if you assume that temperature is zero at -273.15 deg C. If zero is put in different place, all relations become much more complicated. For me that's enough to extrapolate

5. Oct 23, 2008

### HallsofIvy

Staff Emeritus
For an ideal gas, "PV= nRT" where T has to be measured in "degrees Kelvin"- T= 0 at absolute 0 and the size of a degree is the same as in Celcius. That means if we take T0 to be absolute 0 in degrees Celcius and measure T in degrees Celcius, we have PV= nR(T- T0). Keeping P and n constant, measure V for several different temperatures and determine the "least squares" line. You can get T0 from that.

6. Oct 23, 2008

### janakiraman

@ above

thank you very much for the explanation. But still i'm not convinced with the physical interpretation, i cannot do a series of curve fitting and just extrapolate to find absolute 0. i mean that doesn't sound anything physics to me

7. Oct 23, 2008

### Staff: Mentor

Physical interpretation comes later. First you find out that there exist some specific temperature with interesting (even if supported only by algebraic tricks) properties.

That's what scientists do all the time. They look through available data looking for interesting patterns and points. Sometimes these characteristic points don't have any particular meaning, sometimes it turns out they are fundamental.

8. Oct 23, 2008

### atyy

9. Oct 23, 2008

### lalbatros

Obviously, the perfect gas law is a very strong hint.
But the foundation of the absolute temperature definition is the second law of thermodynamics.
A first consequence is the existence of an absolute scale.
A second consequence is a principle to "measure it".

That's the all point of thermodynamics, I believe.
It is actually about collecting data and organizing them scientifically and efficiently.
It is obviously better to use the thermodynamic scale than any other empirical scale.
It is at least more convenient to apply the rules of thermodynamics.

Let's assume I have created an empirical scale of temperture, based on some property of a material.
Is there not general rule that could be applied to derive the thermodynamic scale, and locate the "absolute zero" on the empirical scale?

Let me know your findings.

10. Oct 23, 2008

### janakiraman

Yeah i do agree that the empirical scale based on property of material (for example the Celsius scale based upon water) and my ides is, i had thought that the absolute scale is based upon the ideal gas scenario. If as told by you if they are based upon rigorously only on the laws of thermodynamics then they must be true for all material since all of them obey the thermodynamic laws, but on other hand if they are based upon the extrapolation of the ideal gas or fermi or bose gas scenario, i don't think the resultant doesn't have to be necessarily the absolute 0 since we know that most of materials don't obey the ideal gar or bose gas laws under all temperature and pressure.

A better idea would be is to discuss the actual way how is it calculated here if possible so that we can debate over it.

11. Oct 23, 2008

### Andy Resnick

Assigning who determined absolute zero is tricky- most people think it was Kelvin, in 1854, but in addition to Kelvin, Joule and Rankine also had parts to play, as did Carnot. I think the modern terminolgy is that Joule and Thompson, as well as Rankine, performed experiments to measure Boltzman's constant 'k'. At that time, there was not Boltzman constant- it was called "Carnot's function $\mu$", with $\mu \equiv \frac{1}{\Lambda_{V}} \frac{\partial p}{\partial \theta}$, where $\Lambda_{V}$ is the latent heat at constant volume, and $$\theta$$ the ideal-gas temperature.

In any case, the goal was to determine an absolute scale of temperature- one that was material independent. This is no small thing; temperature is measured by thermometers which are made of definite materials; at that time it was not obvious how to compare one thermometer to another.

In any event, experiments were performed measuring the compressibility and coefficients of expansion of various gases by Regnault, and Joule and Kelvin used a thermodynamic equation of state similar to Rankine and collapsed Regnault's data (and added some of their own), which lead to an 'absolute zero' point at -273.72 degress below the freezing point of water.

At least that's my understanding of the historical record.

12. Oct 23, 2008

### f95toli

I am note sure you could even talk about an exact numerical value for absolute zero until the redefinition of the temperature scale (some 50 years ago), before that there was simply no agreed scale; in the early days of low temperature physics many labs actually maintained their own scales for low temperatures (e.g. Kammerling Onnes lab in Leiden).

Nowadays absolute zero is exactly -273.15 degrees C, but that is a definition (or to be precise: 0 degrees C is the same as 273.15K, since SI uses the Kelvin scale).

Moreover, even today temperature is not actually properly defined below 0.65K, this is the lower limit of the latest published temperature scale (ITS-90) so it is impossible to perform traceable calibrations of thermometers below that temperature (although this is not really a problem).

13. Oct 23, 2008

### atyy

Phillips's Nobel lecture mentions a 40 micro-kelvin temperature - is it possible that another reasonable definition of absolute temperature will be found where that value doesn't hold, or is his method likely to become a standard?

14. Oct 23, 2008

### f95toli

People use temperatures well below 0.65K all the time (I do measurement down to about 15 mK), but the whole concept of temperature becomes as bit iffy at very low temperatures. You have to be VERY precise about what you mean and WHICH temperature you are refering to (i.e. what did you actually measure?). Phillips is refering to the mean kinetic energy of the particles in a gas, which is obviously perfectly legit (it is just basic kinetic theory of gases) as long as there are enough ions around (which isn't obvious in his experiments).
However, the problem is that there is no obvious way to transfer this to e.g. the temperature of the phonons in a solid. One reason being that the electronic temperature (the temperature of the "electron gas") can differ quite significantly from the temperature of the lattice and at low temperatures most people use resistance thermometry (or in some cases noise thermometry); there are also other problems.

There have been some recent attempts to extend the scale down to about 20mK using other methods (e.g. the melting curve of He-3 and nuclear orientation thermometry, I actually use the latter method sometimes) but those projects were closed down a few years ago.
Note that you can buy resistance thermometers that have been "calibrated" using nuclear orientation thermometry* down to 20 mK (from Lakeshore), but the calibration isn't "offical"; AFAIK you can't trace it back to e.g. NIST, NPL or PTB and there is therefore no guarantee that your thermometer will agree with one calibrated by another company (and chances are pretty good that the error could be significant, say a few percent).

*NOT is a method which measures the distribution of "up" and "down" spins in a Co-60 crystal held in a strong magnetic field, i.e. it essentially measures the lattice temperature

15. Oct 23, 2008

### atyy

So did Phillips have enough ions in his gas?

Are we able to compare the temperature in the Phillips experiments with say the Bose-Einstein condensation experiments?

16. Oct 23, 2008

### Andy Resnick

That's exactly correct- because the equations of state (constitutive relations) used to compare various thermometers break down at those low temperatures. High temperatures pose measurement problems too, I bet.

But the OP was asking a very specific historical question- how was absolute zero (whatever the value) originally defined and arrived at?

17. Oct 24, 2008

### janakiraman

Andy

Precisely yes, i want to know how the result was arrived at. As per the information you provided, the comparison was done between different kinds of thermometers and the value was extrapolated to the point where the pressure becomes 0. but most of the thermometers that were used at high temperatures where they behaved like ideal gases and there is no real reason why they need to behave the same way under the low temperatures. So how was the extrapolation justified?

18. Oct 24, 2008

### Renge Ishyo

Hi there Mr. Resnick, good to see a familiar face while I drop by.

Here's the derivation Kelvin gave for the definition of the absolute temperature scale (paraphrazing this from the treatment in "Modern Thermodynamics: from heat engines to dissipative structures," Wiley and Sons). It is based on the following two facts from the Carnot reversible heat engine:

1. All reversible heat engines will behave the same way regardless of the material they are made out of.

2. The maximum efficiency of a heat engine (which is the efficiency of a reversible heat engine) cannot exceed 1.

Start with Carnots theorem:

efficiency = 1 - Q2/Q1 = 1 - f(t2,t1)

were Q2 = heat of the cold reservoir and Q1 = heat of the hot reservior and where
f(t2,t1) is a function of the two temperatures cold and hot respectively. It can be shown by analyzing two consecutive Carnot cycles (the values from the second will be denoted with a ') that f(t2,t1) can be written as a ratio:

efficiency = 1 - (Q2/Q')/(Q'/Q1) = 1- f(t2,t')/f(t',t1)

efficiency = 1- Q2/Q1 = 1 - f(t2)/f(t1)

Now define a new temperature scale T = f(t) so that:

efficiency = 1- Q2/Q1 = 1 - T2/T1

It can be seen that on this new temperature scale when T2 (the temp of the cold reservoir) = zero the efficiency is at a maximum (meaning that this is the lowest attainable temperature possible, or that this T is an "absolute zero"). Furthermore, because Carnot's analysis showed that reversible heat engine behavior is independent of the material properties of the engine, this "absolute zero" is independent of any material properties of the thermometer. Once this definition is known, you can do other things to relate this new temperature scale to the ideal gas law and such both analytically and experimentally.

Hope this helped.

Last edited: Oct 24, 2008
19. Oct 24, 2008

### janakiraman

@ Renge

Thank you very much. Indeed it did clear a lot of doubts of mine related to absolute 0. But still i'm curious to know about the extrapolation techniques used. Just in case if anyone can throw more light on it.

20. Oct 27, 2008

### Andy Resnick

I haven't read the original papers, only a survey of them. But the measurements were not conducted as the pressure was taken to very low values; Regnault measured the specific heats of gases in 1853 (translated in Phil. Mag (4) 5, 473-483.) but also pointed out that the equation of state for gases is not rigorously true.

Joule and Kelvin, in 1854 (Phil. Trans of the Roy. Soc. 144, 321-364) used a form of Rankine's equation of state: pV = AT + B + (C+D/T + G/T^2) F/V, where F is a 'standard volume', taken to be 12.387 cubic feet, corresponding to 1 pound of air under atmospheric pressure p_0 at the ice point.

So, Regnault and Joule/Kelvin measured the isothermal compressibility and isopycnic coefficients of expansion for various gases at two temperatures: 4.75 C and 17 C, and fit the data to the equation of state above, which can be simplified by assuming the terms containing C, D, and G are small compared to pV: F/V = p(AT_0 + B)/p_0(AT+B) (p_0 is atmospheric pressure, T_0 the absolute temperature at the ice point). Fitting the data to that equation gives T_0 = 273.72.

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