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I was watching this video What is Temperature?. It states that when we measure temperature we are measuring $dU\over dS$ at equilibrium. But at equilibrium, how the entropy and the internal energy are changing? Both shouldn't be constant?

Please can someone clarify it for me?

joseph f. johnson
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Tonylb1
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    Second law of thermodynamics: heat flows only from hot to cold (unless something else happens... like the refrigerator is being supplied with electricity from the power plant). Two temperatures are equal when there is no heat flow. No need for entropy and all that. Just measure heat flow. In practice we use small thermodynamic machines called thermocouples. When they show zero voltage difference between two thermal baths, then the temperatures are equal. – CuriousOne Feb 09 '16 at 08:32
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    The proper definition would rather be dS/dU = 1/T. There's a slight difference in the meaning of those derivatives. Now to your question, why should those values be constant at equilibrium? I can add heat from outside the system, energy (and normally entropy) goes up and the system relaxes back. Thermodynamic equilibrium doesn't mean, that there are no fluctuations. On small scales and amounts entropy can even decrease. Those states decay pretty fast under normal conditions, but still should be acknowledged. – rtime Feb 09 '16 at 08:37
  • The actual metrological temperature scale, by the way, is defined in a rather complicated way by a bunch of (mostly) phase transitions: https://en.wikipedia.org/wiki/Scale_of_temperature#International_temperature_scale_of_1990. Ugly? You bet... but dU/dS just doesn't work in real life, no matter how great it sounds on paper. – CuriousOne Feb 09 '16 at 08:40
  • @t.rathjen So at equilibrium we still have a small variation, thus we can calculate the derivative right? – Tonylb1 Feb 09 '16 at 08:45
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    The first part of my answer to this other question might help. Temperature tells you how "badly" a system wants to get rid of energy. – DanielSank Feb 09 '16 at 08:55
  • You are puzzling over the idea, that if you don't add any energy from outside, why should entropy change with respect to the energy, so why is this derivative not equal zero? Consider the case explained in the video you posted. You can cut your system in two halves with different energies. Now those systems will relax to equilibrium via transfer of energy => dS/dU = 1/T is defined. Those fluctuations I was talking about come directly from the energy distribution (also mentioned in the video) and are of order 1/(N)^(1/2). They hardly count. I just wanted to emphasize they still exist – rtime Feb 09 '16 at 09:05
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    @DanielSank: I don't understand how you connect the two questions... for one thing one can define temperature without all the statistical mechanics hocus pocus (no physicist has ever been counting states when measuring temperature!). Secondly, even in statistical mechanics the system doesn't have a way of "getting rid of energy" without a second heat bath or some mechanical process, so the concept of "how badly it wants to do that" seems a bit strange, unless we are building a heat engine, after all. – CuriousOne Feb 09 '16 at 09:09
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    @CuriousOne I do wish you would use responsible language. Referring to statistical mechanics as "hocus pocus" may lead less experienced readers to actually believe the implication that statistical mechanics is not worth their time. Secondly, if you would please spend five minutes researching experimental realizations of negative temperature in ultracold gasses it may help you overcome the notion that temperature is always measured with an alcohol thermometer. – DanielSank Feb 09 '16 at 09:26
  • @t.rathjen Great. If we take a room for example, the temperature is well defined, and should be equal to dU/dS right? but how sorry I didn't understand. – Tonylb1 Feb 09 '16 at 09:31
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    @DanielSank: High school students may use alcohol thermometers, physicists use temperature normals and calibrated thermocouples, which, as you may know, are physical implementations of the proper definition of thermodynamic temperature. I am sure you will be able to explain to the OP how you have been counting states in gases... even though a few hundred molecules have enough phase space to dance around for the entire life of the universe without repeating themselves. Seriously, before anybody tries their hand on statistical mechanics, they should be told how nature actually does things. – CuriousOne Feb 09 '16 at 09:34

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Your question is how the differential quotient $\frac {dU}{dS}$ can mean anything in equilibrium when the quantities $U$ and $S$ are supposed to be constant, it is equilibrium after all... Indeed, $dU$ or $dS$ do not mean changes over time in a physical sense, ie., over time during some process, Instead they mean the differentials of the respective quantities between neighboring equilibrium states. Hence, the term a "quasi-static process" that goes through on equilibrium states and it is not a "process" in a conventional sense at all, because it does not happen in time.

hyportnex
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How do we define temperature?

You have to start with the zeroth law of thermodynamics which is all about bodies in thermal equilibrium.
The strange name is because after the first and second laws of thermodynamics were formulated suddenly somebody realised there was another law of thermodynamics which in some ways was more fundamental than one and two, so rather than change the name of one to two and two to three the law was given the name zero.

In simple terms it can be interpreted as follows.
If two bodies a joined together in thermal contact and no heat flows between them those two bodies are said to be in thermal equilibrium.
The zeroth law states that if body $A$ is in thermal equilibrium with body $C$ and body $B$ is in thermal equilibrium with body $C$ then if bodies $A$ and $B$ where joined together in thermal contact no heat would flow - they are in thermal equilibrium.
A statement of the obvious? Yes, but only after you have heard about the law?

Moving on one needs to find a parameter which will help you decide whether or not two bodies would be in thermal equilibrium is the were joined together in thermal contact.
The chosen parameter is temperature and the device use dto measure the temperature is called a thermometer.

So going back to the zeroth law the thermometer might be body $C$.
Joining the thermometer to body $A$ you might get a reading on the thermometer of 52 (no units yet). If you then join the thermometer to body $B$ and get the same reading of 52 then you know that if joined together $A$ and $B$ would be in thermal equilibrium.

To devise a thermometer you use a thermometric property of a substance; that is a property which changes with temperature and then decide on a scale of temperature so that when you take a temperature reading that reading is meaningful to others and you in the future.

In the video when the statement is made that "when we measure temperature we are measuring dU/dS at equilibrium" a thermometric property is being described which at present happens to be thought to be the best for a particular range of temperatures.
The problem is that although the kelvin scale of temperature is the favoured one at present its realisation in practice over many orders of magnitude of temperature is very difficult. No one thermometric property can be used to measure all temperatures and the problems multiply for very low and for very high temperatures relative to the temperature of the room you are sitting in.

Farcher
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  • There are no thermometers in thermodynamics and temperature is no more defined by thermometers than force is defined by force gauges in Newtonian mechanics. If you want to explain this to the OP, please use proper concepts. – CuriousOne Feb 09 '16 at 10:07
  • @CuriousOne I never said that temperature was defined by thermometers. I said that temperature was measured using thermometers which is very different. You can define temperature by whatever means you like but then you need to measure it? So the kelvin scale of temperature is defined but you then need a practical realisation of that scale? – Farcher Feb 09 '16 at 10:15
  • I think that we are arguing at cross purposes. To me a thermocouple is as much a thermometer as a mercury in glass thermometer. The thermometric property is the emf a thermocouple generates and that emf changes with temperature. The practical realisation of the kelvin scale using a thermocouple is to convert the emf reading on the thermocouple into a temperature on the Kelvin scale. The reason for not using the mercury in glass thermometer is that the temperature to be measured may be too high or too low or it is simply not accurate and reproducible enough. – Farcher Feb 09 '16 at 10:32
  • With your thermocouple you get a reading of voltage (power) how is that then converted into a temperature in kelvin? – Farcher Feb 09 '16 at 10:44
  • @CuriousOne The following has come statement has popped up "Please avoid extended discussions in comments. Would you like to automatically move this discussion to chat?" I leave it to you as to whether or not you want to continue the discussion as I really do not understand why you think that I have not answered, in part or whole, the original OP. From what you have written you can only decide accurately whether two temperatures are the same. It does not seem to tell me what the temperature is? – Farcher Feb 09 '16 at 11:21
  • Let us continue this discussion in chat. – CuriousOne Feb 09 '16 at 11:23
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Lets not complicate it with the use of thermodynamic terms.Temperature is a measure of the internal energy of a substance and internal energy is related with the change is the heat which is turns relate with entropy change.

Rohit Joshi
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