- #1

quantum123

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Is it really independent of physical substance? Why?

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- Thread starter quantum123
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- #1

quantum123

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Is it really independent of physical substance? Why?

- #2

Gerenuk

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[tex]pV\propto T[/tex]

So basically you keep on variable constant and thus find the relative change of the other variables.

Experimentally

https://www.physicsforums.com/showpost.php?p=2270246&postcount=10

The ideal gas law is valid whenever the energy and the density of states are power laws of the momentum.

There might be better experts who can answer this...

- #3

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Is it really independent of physical substance? Why?

I'm not sure what temperature 'is', but thermodynamics uses 'temperature' as a real, positive, number that (somehow) relates to the energy of a system. The specifics of that relationship have not yet been worked out for the general case, only for equilibrium conditions.

Kelvin and Rankine scales are 'absolute' in the sense that the measured temperature is independent of the thermometer- something that was not the case in the early days of thermometry.

In practice, the temperature of an object tells you how hot it is.

- #4

netheril96

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[tex]pV\propto T[/tex]

So basically you keep on variable constant and thus find the relative change of the other variables.

Experimentally

https://www.physicsforums.com/showpost.php?p=2270246&postcount=10

The ideal gas law is valid whenever the energy and the density of states are power laws of the momentum.

There might be better experts who can answer this...

What you say is the ideal gas temperature scale.It is equivalent to the thermodynamical scale in practice,but NOT in the fundamentals.The ideal gas scale depends on the universal behavior of gas,but the thermodynamical scale is independent of all material.

In theory,the thermodynamical scale is based on the efficiency of Carnot engine because according to the Second Law of Thermodynamics,all reversible cycles between two reservoirs have the same efficiency.Then we choose a standard,in practice the triple point of water as 273.16K.

In practice,it is realized using different ways in different range of temperature.See http://www.bipm.org/utils/common/pdf/its-90/SUPChapter1.pdf

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- #5

netheril96

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I'm not sure what temperature 'is', but thermodynamics uses 'temperature' as a real, positive, number that (somehow) relates to the energy of a system. The specifics of that relationship have not yet been worked out for the general case, only for equilibrium conditions.

Kelvin and Rankine scales are 'absolute' in the sense that the measured temperature is independent of the thermometer- something that was not the case in the early days of thermometry.

In practice, the temperature of an object tells you how hot it is.

Even thermodynamic temperature can be

See http://en.wikipedia.org/wiki/Negative_temperature

- #6

Gerenuk

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Could you please describe the way with which they get a temperature reading, other than using the ideal gas law? But don't forget that it should be a fundamental principle not relying on data parameters which is collected by ideal gas law thermometers.In practice,it is realized using different ways in different range of temperature.See http://www.bipm.org/utils/common/pdf/its-90/SUPChapter1.pdf

I believe it's impossible to practically apply the definition of temperature you used.

Also theoretically the reversible cycle definition seems impractical. I actually went through that definition in detail and found that you need to be able to identify

- reversible processes and also

- isothermal processes

Both of which you cannot do in a strictly rigorous way.

- #7

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Even thermodynamic temperature can benegative

See http://en.wikipedia.org/wiki/Negative_temperature

I assume you are referring to the statistical-mechanical description of a two-level system. While that's an interesting result, it cannot honestly represent a thermodynamic temperature- after all, then there would be a large flow of *heat*, which is not observed.

- #8

quantum123

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- #9

Mr.Miyagi

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I believe the usual approach to defining temperature is as follows: Considering two compartments which are divided by a diathermal wall (and isolated as a whole), one can define a temperature if the compartments are in thermal equilibrium. This occurs when

[tex]\frac{\partial S_1}{\partial E_1}=\frac{\partial S_2}{\partial E_2}[/tex],

where the volume of each compartment and the number of particles in each compartment is kept constant. S is the entropy.

Since the compartments are in thermal equilibrium, they have the same temperature. So

[tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex].

That is, the partial derivatives define a function of the temperature. When we choose f(T)=1/T, this definition becomes identical with the perfect gas scale.

That said, I'm just a recent student of the subject. There might be a better definition. But this one is pretty general, since it makes no assumptions about specific properties of the system. It basically only uses the second law of thermodynamics.

About negative temperatures: Defining temperature this way, negative temperatures arise quite naturally. If the partial derivative is negative, so is the temperature. This can happen, for example, in a paramagnetic solid in a magnetic field. Consider the interaction between the dipoles and the magnetic field. The system has a minimal energy if all the dipoles in the solid are parallel with the field and maximal energy if they are anti-parallel with the field. Both the minimum and maximum energy correspond to an entropy of 0 (that is, the arrangement of the dipoles is completely known).

If we'd plot the entropy against the energy, we would see a curve that goes up from zero at the minimal energy and goes down from some point to reach zero again at maximum energy. When the curve goes down, we have a negative slope and therefore a negative temperature.

The key to negative temperature is the maximum energy, which forces the slope to become negative at some point. When we consider the complete system (not just the interaction between the dipoles and the magnetic field) there can be no maximum energy, since the kinetic energy of particles is unbounded. But we can still talk about these separate aspects of the system if we assume that the different aspects interact very weakly with each other, which is the case in this situation.

[tex]\frac{\partial S_1}{\partial E_1}=\frac{\partial S_2}{\partial E_2}[/tex],

where the volume of each compartment and the number of particles in each compartment is kept constant. S is the entropy.

Since the compartments are in thermal equilibrium, they have the same temperature. So

[tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex].

That is, the partial derivatives define a function of the temperature. When we choose f(T)=1/T, this definition becomes identical with the perfect gas scale.

That said, I'm just a recent student of the subject. There might be a better definition. But this one is pretty general, since it makes no assumptions about specific properties of the system. It basically only uses the second law of thermodynamics.

About negative temperatures: Defining temperature this way, negative temperatures arise quite naturally. If the partial derivative is negative, so is the temperature. This can happen, for example, in a paramagnetic solid in a magnetic field. Consider the interaction between the dipoles and the magnetic field. The system has a minimal energy if all the dipoles in the solid are parallel with the field and maximal energy if they are anti-parallel with the field. Both the minimum and maximum energy correspond to an entropy of 0 (that is, the arrangement of the dipoles is completely known).

If we'd plot the entropy against the energy, we would see a curve that goes up from zero at the minimal energy and goes down from some point to reach zero again at maximum energy. When the curve goes down, we have a negative slope and therefore a negative temperature.

The key to negative temperature is the maximum energy, which forces the slope to become negative at some point. When we consider the complete system (not just the interaction between the dipoles and the magnetic field) there can be no maximum energy, since the kinetic energy of particles is unbounded. But we can still talk about these separate aspects of the system if we assume that the different aspects interact very weakly with each other, which is the case in this situation.

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- #10

f95toli

Science Advisor

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The truth is that there IS no good

While this might not be very satisfactory from a philosophical point of view it does have the advantange that it works...

Also, some primary thermometers (e.g. nuclear orientation thermometers and noise thermomenters) are based on statistical principles that are quite difficult to connect to more classical thermodynamics.

- #11

Mr.Miyagi

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The truth is that there IS no goodgeneraldefinition of temperature, which is why the international temperature scale (ITS-90) uses many different "definitions" (depending onf which fixed point is used for a particular range); although there is e.g. no fixed point at very low temperatures (ITS-90 starts at 0.65K).

While this might not be very satisfactory from a philosophical point of view it does have the advantange that it works...

Also, some primary thermometers (e.g. nuclear orientation thermometers and noise thermomenters) are based on statistical principles that are quite difficult to connect to more classical thermodynamics.

Sure, the scale is arbitrary to some degree. But temperature can be related to entropy and energy. By doing so, we see that f(T)=1/T (in the derivative in my previous post) is a reasonable choice for the scale: Defined this way, heat goes from hot bodies to cold bodies, which is consistent with our idea of hot and cold. f(T)=T would imply heat goes from cold bodies to hot bodies. So, in my opinion, not all scales are created equally. Perhaps like reference frames in mechanics, some are more useful than others.

That, and we may at least say that temperature IS independent of the physical substance, which was one of the OP's questions. It just depends on the second law of entropy (and conservation of energy).

- #12

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That's different from claiming that the conceptual foundations of temperature are known and completely elucidated. I would disagree with that statement.

- #13

Mr.Miyagi

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- #14

Gerenuk

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Actually that is the most general definition that I support. But the question back to you: So you are given a piece of wood. How would you measure the temperature in practice, now?[tex]\frac{\partial S_1}{\partial E_1}=\frac{\partial S_2}{\partial E_2}[/tex],

where the volume of each compartment and the number of particles in each compartment is kept constant. S is the entropy.

Your definition only shifts the problem to a new (unknown) quantity, the entropy.

Could you mention the method they use to find a temperature (which is not based on ideal gas laws)? How can you guarantee that any other method doesn't cause inconsistencies? For example with a flawed method I could measure the temperature of one body, bring it on contact with another body, measure the other body and suddenly get a different temperature reading.The truth is that there IS no goodgeneraldefinition of temperature, which is why the international temperature scale (ITS-90) uses many different "definitions" (depending onf which fixed point is used for a particular range)

What do you mean by related? Of course if you change one quantity of a system, all others will most likely also change. But note that energy is a function of temperature alone (no p or V dependence) only in the special case that [itex]p\propto T[/itex] at constant V. That special case doesn't necessarily have to hold.Nobody here (IIRC) will disagree that temperature is related to energy.

- #15

Mr.Miyagi

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Actually that is the most general definition that I support. But the question back to you: So you are given a piece of wood. How would you measure the temperature in practice, now?

Your definition only shifts the problem to a new (unknown) quantity, the entropy.

Well, by sticking a thermometer in it. When the thermometer is in equilibrium with the wood, it gives a certain reading. Then calibrate the thermometer in a controlled experiment, where the value of the partial derivative is known. A gas that can be described as a perfect classical gas would do.

But now I'm really curious as to what the objections to this are, since I assume you've considered this as well. I can't find any hidden assumption in this process.

- #16

Gerenuk

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The hidden assumption is, that you need someone who has actually solved the problem of measuring temperature before you. And he must have used another method than you, because with your method alone you wouldn't be able to measure temperature.But now I'm really curious as to what the objections to this are, since I assume you've considered this as well. I can't find any hidden assumption in this process.

OK, so in the end you are using the ideal gas law definition all the time? You have a definition of temperature with entropy, but you never use it. Instead you bring all you objects in contact with an ideal gas and read off the temperature?Well, by sticking a thermometer in it. When the thermometer is in equilibrium with the wood, it gives a certain reading. Then calibrate the thermometer in a controlled experiment, where the value of the partial derivative is known. A gas that can be described as a perfect classical gas would do.

Now say you measure temperature of body A. Then you bring body A in contact with body B. Now you measure temperature of body B. Why are you sure that temperature B will give the same reading?

- #17

Mr.Miyagi

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I'm not really sure that I understand the objection.

I claim that

[tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex]

for compartiments in thermal equilibrium and I choose

[tex]f(T_i)=\frac{1}{T_i}[/tex].

When considering an ideal classical gas, this statement happens to be equivalent to pV=NkT. This is based on the second law, not on the ideal gas law. I don't see how that undermines the generality of the first equation.

The particular choice for the function is made, so that the ideal gas law is a special case in which the microscopic variables may be replaced by macroscopic ones. Again, the ideal gas law is be derived from this definition, not the other way around.

About the last part, I don't see why the second measurement would give the same reading. Could you elaborate on this situation? I don't get the issue you're trying put forward.

correction: I shouldn't say the ideal gas law is derived from the second law. It just corresponds to the definition of temperature that I gave.

I claim that

[tex]\frac{\partial S_i}{\partial E_i}=f(T_i)[/tex]

for compartiments in thermal equilibrium and I choose

[tex]f(T_i)=\frac{1}{T_i}[/tex].

When considering an ideal classical gas, this statement happens to be equivalent to pV=NkT. This is based on the second law, not on the ideal gas law. I don't see how that undermines the generality of the first equation.

The particular choice for the function is made, so that the ideal gas law is a special case in which the microscopic variables may be replaced by macroscopic ones. Again, the ideal gas law is be derived from this definition, not the other way around.

About the last part, I don't see why the second measurement would give the same reading. Could you elaborate on this situation? I don't get the issue you're trying put forward.

correction: I shouldn't say the ideal gas law is derived from the second law. It just corresponds to the definition of temperature that I gave.

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- #18

Gerenuk

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Imagine you are

For the second part:

1) you've somehow managed to build a thermometer

2) you stick it in body A and read off T_a

3) you remove the thermometer

4) you bring body A and body B in contact

5) you stick the thermometer in body B and read off T_b

6) can you guarantee with your definition of temperature that T_a=T_b as it should be?

- #19

Mr.Miyagi

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I look for a device that, when put in the mini-chamber, reacts predictably to the different pressures (and thus temperatures, by the ideal gas law). I then assign values to the states of the device.

Then, since the ideal gas law is compatible with the definition of temperature as given before, the device could now be used to measure other things. These other things don't have to follow the ideal gas law, just the second law of entropy.

I still don't see why measurement of body B needs to give the same reading as body A. It seems to me that this should only true if body B does actually have the same temperature as body A. If this is not the case, there is no thermal equilibrium when the two are put in contact and everything changes. I can guarantee that when the bodies are in contact and thermal equilibrium has been achieved, the readings will be the same (when both measurements are made after equilibrium). I can also guarantee that different readings will be made when the bodies are not in thermal equilibrium.

- #20

Gerenuk

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How do you pick which number exactly you want to assign? Who is telling you which test chamber pressure corresponds to which temperature?I then assign values to the states of the device.

Why do you think that if you take two different devices, you won't get contradictory results when cross-checking the temperature of the same test object?Then, since the ideal gas law is compatible with the definition of temperature as given before, the device could now be used to measure other things.

When do you ever need your quoted definition of temperature? You might as well have said [itex]\partial \sqrt{S}/\partial E=T[/itex] and nothing would be different?These other things don't have to follow the ideal gas law, just the second law of entropy.

Step 4) was to bring both bodies in thermal contact (for a long time)I still don't see why measurement of body B needs to give the same reading as body A. It seems to me that this should only true if body B does actually have the same temperature as body A.

- #21

Holmz

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0k ?

- #22

SpectraCat

Science Advisor

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Now say you measure temperature of body A. Then you bring body A in contact with body B. Now you measure temperature of body B. Why are you sure that temperature B will give the same reading?

Because that is the zeroth law of thermodynamics. If A is in thermal equilibrium with B, and B is in thermal equilibrium with C (the thermometer in this case), then A is also in thermal equilibrium with C. That is the thermodynamic definition of what it means for things to be in thermal equilibrium ... it provides the basis for what temperature *means*, although it does not provide a quantitative scale for measurement.

EDIT: Note that I am assuming you are working with ideal cases here, and not worrying about thermal losses to the surroundings that might be present in a "real" measurement.

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- #23

Gerenuk

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Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.Because that is the zeroth law of thermodynamics. If A is in thermal equilibrium with B, and B is in thermal equilibrium with C (the thermometer in this case), then A is also in thermal equilibrium with C. That is the thermodynamic definition of what it means for things to be in thermal equilibrium ... it provides the basis for what temperature *means*, although it does not provide a quantitative scale for measurement.

I think the definition

[tex]\frac{1}{T}=\frac{\partial \ln\Omega}{\partial E}[/tex]

with knowledge about microscopic partition functions is best. I just was trying to point out to Miyagi that he uses many hidden assumptions that people have derived before. They can all be resolved, but one has to think about it.

- #24

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It's not complete- one or two simple examples should suffice to show that.

Blackbody radiation is defined as a photon gas in thermal equilibrium with a reservoir at some temperature T. Say blackbody radiation is emitted, and then passes through a spectral filter- the spectral filtering can be as broad or as narrow as you wish. In fact, we could simply pass it through a polarizer and separate out the polarization components. The energy is well-defined, the change in entropy is well-defined, but the photon gas no longer has a temperature- even if all the optical components are at the same (initial) temperature.

Same for monochromatic radiation- a well defined energy, a well-defined entropy, but it cannot be assigned a temperature.

- #25

SpectraCat

Science Advisor

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Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.

I don't think I agree with that ... I would rather say that one can have as many independent definitions of temperature as one cares to,

In the latter case, you take the temperature of a universal physical system with a constant temperature, such as the triple point of water, and use it as the reference point for the absolute temperature (e.g. make it T=100). Now take the system for which you want to measure the temperature and set it as the heat source for your ideal Carnot engine, with the cold sink in equilibrium with the triple point of water. Measure the efficiency of the engine, and use the Carnot formula to calculate the temperature of the hot source:

[tex]efficiency=\frac{W}{Q_H}=1-\frac{T_C}{T_H}[/tex]

where Q

I think the definition

[tex]\frac{1}{T}=\frac{\partial \ln\Omega}{\partial E}[/tex]

with knowledge about microscopic partition functions is best.

Yeah .. but that doesn't work for your block of wood example, right?

- #26

Gerenuk

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Maybe for some purposes. But wouldn't that screw up all nice physical equations with T in them?I don't think I agree with that ... I would rather say that one can have as many independent definitions of temperature as one cares to,provided that they all obey the zeroth law.

But the more serious point is that you actually have to define some kind of exact experimental procedure which gives consistent temperatures. And for no physical process you can be sure it would obey the 0th law?

A theoretical microscopic definition however makes some calculation and proof possible. What's left is the task to find an ideal object which behaves according to theory :)

I thought about this definition for a while, but it's more tricky than that. It is crucial that you find aNow take the system for which you want to measure the temperature and set it as the heat source for your ideal Carnot engine, with the cold sink in equilibrium with the triple point of water. Measure the efficiency of the engine, and use the Carnot formula to calculate the temperature of the hot source:

That's not the whole story. Of course I need to calculate [itex]\Omega[/itex] which I can only do for something idealized like an ideal gas. So in the end you can derive all of thermodynamics.Yeah .. but that doesn't work for your block of wood example, right?

The little dilemma is, that I can never be sure that I'm really dealing with a perfect gas.

- #27

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Sure, but but cannot have two different definitions/properties of temperature. You have to derive all properties from one definition. Other consistency isn't guaranteed.

I think the definition

[tex]\frac{1}{T}=\frac{\partial \ln\Omega}{\partial E}[/tex]

with knowledge about microscopic partition functions is best. I just was trying to point out to Miyagi that he uses many hidden assumptions that people have derived before. They can all be resolved, but one has to think about it.

They *can't* all be resolved. Your definition works fine for a very limited set of processes (thermostatics), but there is not yet a sound foundation for thermodynamics. Experiments do not use ideal gases; why should you require such a restricted basis for the measurement of temperature? Dissipative processes are omitted from the kinetic theory formulation. And what happens if you do not know the "microscopic partition functions"- which is the case for almost all real objects?

- #28

Gerenuk

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They can be resolved and work for all of thermdynamics, but I'm going to write down the derivation again. If you insist I can try to dig out a part of the derivation that I wrote in another thread.They *can't* all be resolved. Your definition works fine for a very limited set of processes (thermostatics), but there is not yet a sound foundation for thermodynamics.

Because no other way is possible. In another thread I asked before, and the only answer was the ideal gas method (link that I gave earlier in this thread).Experiments do not use ideal gases; why should you require such a restricted basis for the measurement of temperature?

The Carnot definition requires hypothetical engines which do not necessarily exist and no physical device can be confirmed to be strictly Carnot-like.

Then you cannot make any statement about the system. It's just some arbitrary system with completely arbitrary dynamics. Of course then you cannot say anything about it.And what happens if you do not know the "microscopic partition functions"- which is the case for almost all real objects?

- #29

SpectraCat

Science Advisor

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The Carnot definition requires hypothetical engines which do not necessarily exist and no physical device can be confirmed to be strictly Carnot-like.

*Ahem* ... please mail me a vial of ideal gas. I will send you my mailing address via PM.

How is the ideal gas version any more achievable experimentally than the Carnot engine version?

- #30

Gerenuk

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- #31

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They can be resolved and work for all of thermdynamics, but I'm going to write down the derivation again. If you insist I can try to dig out a part of the derivation that I wrote in another thread.

Because no other way is possible. In another thread I asked before, and the only answer was the ideal gas method (link that I gave earlier in this thread).

The Carnot definition requires hypothetical engines which do not necessarily exist and no physical device can be confirmed to be strictly Carnot-like.

Then you cannot make any statement about the system. It's just some arbitrary system with completely arbitrary dynamics. Of course then you cannot say anything about it.

This is really disappointing- you have unilaterally declared 99% of reality to be off-limits of physics.

- #32

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Let me back up a bit here: "..no better method than ideal gas"

I don't understand what you are saying- no better method *for what*? thermodynamics? statistical mechanics? Something else?

- #33

Gerenuk

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Or maybe it's rather 1% and you don't know how to write the partition function for 98% of physics.This is really disappointing- you have unilaterally declared 99% of reality to be off-limits of physics.

- #34

Gerenuk

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A better method for measuring the temperature. Just describe one. But one which doesn't use ideal gases and no Carnot engine. Especially Carnot engines don't exist.Let me back up a bit here: "..no better method than ideal gas"

I don't understand what you are saying- no better method *for what*? thermodynamics? statistical mechanics? Something else?

- #35

Mr.Miyagi

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It's not complete- one or two simple examples should suffice to show that.

Blackbody radiation is defined as a photon gas in thermal equilibrium with a reservoir at some temperature T. Say blackbody radiation is emitted, and then passes through a spectral filter- the spectral filtering can be as broad or as narrow as you wish. In fact, we could simply pass it through a polarizer and separate out the polarization components. The energy is well-defined, the change in entropy is well-defined, but the photon gas no longer has a temperature- even if all the optical components are at the same (initial) temperature.

Same for monochromatic radiation- a well defined energy, a well-defined entropy, but it cannot be assigned a temperature.

Thank you for the counter-examples. But when you say the entropy and the energy are known, are they not related? Is that the problem?

I'll try to read up a bit more.

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