Derivation from the 0th Law of Thermodynamics

In summary, the author is trying to solve an equation for C1 by cancelling the coordinates of C. However, this is already done for them by invoking the 0th law.
  • #1
PhiJ
44
0

Homework Statement



I really don't know if I'm in the right subforum...
I've started reading this text on statistical mechanics from MIT, but I'm stuck on page 2. Here's the statement:

Let the equilibrium state of systems A, B, and C be described by the coordinates
{A1, A2,· · ·}, {B1, B2,· · ·}, and {C1, C2,· · ·} respectively. The assumption that A and C are in equilibrium implies a constraint between the coordinates of A and C, i.e. a change in A1 must be accompanied by some changes in {A2,···;C1, C2,···} to maintain equilibrium of A and C. Denote this constraint by

fAC(A1, A2,···;C1, C2,···) = 0. (I.1)

The equilibrium of B and C implies a similar constraint
fBC(B1, B2,···;C1, C2,···) = 0. (I.2)

Each of the above equations can be solved for
C1 to yield
C1 = FAC(A1, A2,···; C2,···) (I.3a)
C1 = FBC(B1, B2,···; C2,···) (I.3b)​

Thus if C is separately in equilibrium with A and B we must have
FAC(A1, A2,···; C2,···) = FBC(B1, B2,···; C2,···) (I.4)

However, according to the zeroth law there is also equilibrium between A and B, implying
the constraint
fAB(A1, A2,···;B1, B2,···) = 0. (I.5)
Therefore it must be possible to simplify eq.(I.4) by cancelling the coordinates of C.​

Homework Equations


(I.1) through (I.5) above.

The Attempt at a Solution



I understand up to I.2. For I.3 I'm going to assume there is a unique solution - I'm sure the solution isn't always unique, but it means their argument is correct and I don't have to play mind twister to agree. So given that assumption I'm happy with I.3. I.4 is then obviously true, and I.5 is obviously true by the zeroth law.

I suppose 'cancel the Cs' means substitute some constants into the values for CX, so define:

FA(A1, A2,···) = FAC(A1, A2,···; c2,···)
FB(B1, B2,···) = FBC(B1, B2,···; c2,···)
where cx is a constant.

But then:

FA(A1, A2,···) - FB(B1, B2,···) = 0

And we can define fAB = FA - FB

i.e. we already have a constraint in terms of A and B. Therefore the zeroth law is not a foundational assumption, it's something you can prove. I'm sure I'm wrong on that count, but I can't see why. I realized I've made an assumption to accept I.3, but is I.3 wrong? It's stated in the text.
 
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  • #2
PhiJ said:
FA(A1, A2,···) - FB(B1, B2,···) = 0

And we can define fAB = FA - FB

i.e. we already have a constraint in terms of A and B. Therefore the zeroth law is not a foundational assumption, it's something you can prove.
You arrived at that equation because you could cancel out the coordinates of C, which was possible by invoking the 0th law. Proving the 0th law using this would be circular reasoning.
 
  • #3
I think I 'cancelled out' the coordinates of C without using the zeroth law.

Substituting constants into the equation is standard mathematical practise. That gives me:

FAC(A1, A2,···; c2,···) = FBC(B1, B2,···; c2,···)

Then I defined two functions:
PhiJ said:
FA(A1, A2,···) = FAC(A1, A2,···; c2,···)
FB(B1, B2,···) = FBC(B1, B2,···; c2,···)
where cx is a constant.

then substituted those defined functions in:

FA(A1, A2,···) = FB(B1, B2,···)

Then subtracted:

FA(A1, A2,···) - FB(B1, B2,···) = 0

Then defined again fAB = FA - FB

And I'm done. Those steps require the normal rules for functions and arithmetic, and that there is a solution to fAC and fBC. The existence of a solution is implied by the question: if A and C are in equilibrium, then there is a constraint between A and C, but the implication is that this is a constraint with a solution. If A and C are in equilibrium, and B and C are also in equilibrium, then there are two constraints, and the implication is that there is a solution that fits both constraints.

I suppose without the zeroth law, we'd still have the constraint, but it wouldn't be a constraint that defines a thermal equilibrium between A and B?
 
  • #4
PhiJ said:
Then I defined two functions:
PhiJ said:
FA(A1, A2,···) = FAC(A1, A2,···; c2,···)
FB(B1, B2,···) = FBC(B1, B2,···; c2,···)
where cx is a constant.
I think that the crux is that this should work for any cx, not a particular one.
 
  • #5
DrClaude said:
I think that the crux is that this should work for any cx, not a particular one.

Of course! FA - FB has to have the same trace, whichever c you pick, or your constraint is only true under certain conditions. I think I was getting confused by the concept of cancelling variables within an arbitrary function, because it's obvious in hindsight.

Thanks for the help.
 

Related to Derivation from the 0th Law of Thermodynamics

What is the 0th Law of Thermodynamics?

The 0th Law of Thermodynamics states that if two systems are in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other. This means that if two objects have the same temperature as a third object, then they also have the same temperature as each other.

What is the significance of the 0th Law of Thermodynamics?

The 0th Law of Thermodynamics is significant because it allows us to define the concept of temperature. It states that temperature is a property that can be used to describe the thermal equilibrium between two or more systems. This law also forms the basis of the Celsius and Fahrenheit temperature scales.

How does the 0th Law of Thermodynamics relate to the other laws of thermodynamics?

The 0th Law of Thermodynamics is the most fundamental of the three laws of thermodynamics. It provides the basis for the measurement of temperature and the definition of thermal equilibrium, which are important concepts in understanding the other two laws.

Can the 0th Law of Thermodynamics be violated?

No, the 0th Law of Thermodynamics cannot be violated. It is a fundamental law of nature that has been experimentally verified and holds true in all known cases. If the 0th Law was violated, it would mean that temperature would not be a well-defined property and thermal equilibrium would not exist.

How does the 0th Law of Thermodynamics apply to everyday life?

The 0th Law of Thermodynamics has many practical applications in everyday life. For example, it is the basis for the use of thermometers to measure temperature, the functioning of refrigerators and air conditioners, and the distribution of heat in cooking and heating systems. It also helps us understand how heat is transferred and distributed in different systems.

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