The internal energy, $U$, of a given thermodynamic state, $\boldsymbol{R}$, as introduced in thermodynamics textbooks like Callen seems to be defined as:

$U (\boldsymbol{R}) = \Delta U (\boldsymbol{R}_{Ref} \rightarrow \boldsymbol{R})$

where $\boldsymbol{R}_{Ref}$ is some other thermodynamic state of the system that I choose as a reference. Is that correct?

Under this formalism $U (\boldsymbol{R}_{Ref}) = 0$ always, so I cannot actually assign an arbitrary value to the function $U (\boldsymbol{R}_{Ref})$. If I want to assign an arbitrary value $k$ to the internal energy of state $\boldsymbol{R}_{Ref}$, I have to define a new internal energy function $U'(\boldsymbol{R}_{Ref}) = k$. However, I am really just picking a different reference state $\boldsymbol{R}_{Ref}'$ such that:

$U' (\boldsymbol{R}_{Ref}) = \Delta U (\boldsymbol{R}_{Ref}' \rightarrow \boldsymbol{R}_{Ref}) = k$

So the energy of my true reference state, the one actually included in the internal energy function ($\boldsymbol{R}_{Ref}'$ in this case), is still zero.

Is my understanding correct? If so, why is it so hard to find this stated explicitly? And why is it often said that "you can assign an arbitrary value of energy to your reference state." This is technically true, but I find it to be misleading. As soon you assign a non-zero internal energy to a given state, it is no longer the "true" reference state. Maybe a better statement would be "you can assign an arbitrary value of internal energy to a given state by selecting an appropriate reference state."


Your definition is a bit funky. I'll explain.


As much as I like that you explicitly express internal energy $U$ as a function of state, for clarity I'm going to drop $\boldsymbol{R}$ for now.

We can't directly measure the absolute internal energy of a system, we can only infer its change by understanding some defined process. Consider the change in internal energy between two arbitrary states. By definition this is:

$\Delta U_{1 \rightarrow 2} = U_2 - U_1$

This applies equally to the case where we are considering internal energy relative to some reference state. We can express this as

$\Delta U = U - U_\text{ref}$

It doesn't matter what our reference state is, we can set $U_\text{ref} = 0$ so that

$\Delta U = U$

For a given state, internal energy is always relative to the internal energy of some other (reference) state. In practice, analysis of a non-trivial system will involve multiple states, we could choose any as the reference or pick an appropriate standard state; in any case setting the value to zero or dealing exclusively with $\Delta U$ has a similar effect.

For some arbitrary state of interest (rather than processes), I'd argue setting $U_\text{ref} = 0$ is not only convenient, it's semantically the most appropriate thing to do. We make it zero by definition rather than making it, say $12.7$ (which would mean $U = \Delta U + 12.7$). There's no such thing as the "true" reference state, only the one you define.


Here's an illustration to underline why the absolute value of the reference state is unimportant.

Internal energy and reference states

This is a generic visualisation relevant to two arbitrary states of interest.

  1. The relative positioning of $\boldsymbol{R}_\text{ref}$ doesn't affect $\Delta U_{1 \rightarrow 2}$.

  2. If we set $\Delta U_1 = 0$, we're defining $\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$

  3. $\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$ implies $\Delta U_{1 \rightarrow 2} = \Delta U_2 = U_2$

  • $\begingroup$ Yes, everything you have said is consistent with my understanding, but again I think the language you use is confusing. I think that the fundamental definition is $U(\boldsymbol{R}) \equiv \Delta (\boldsymbol{R}_{Ref} \rightarrow \boldsymbol{R})$. From this it follows that $U(\boldsymbol{R}_{Ref}) = 0$. This is not arbitrary. I cannot have $U(\boldsymbol{R}_{Ref}) = 12.7$ unless I change my definition of $U$. $\endgroup$ – CraftyVisage Oct 8 '14 at 19:40
  • $\begingroup$ I agree. Problem is, that notation is unfamiliar and, frankly, odd. If that's an identity it makes no sense if the RHS is shorthand for $U(\boldsymbol{R}) - U(\boldsymbol{R_{Ref}})$. It can't appear on both sides. If it's an equality, it can, but then you are right it implies $U(\boldsymbol{R}) = 0$ which is not arbitrary. The thing is, it is arbitrary, and not a single one of the 7 texts I just checked defines it anything like this. Let me get back to you... $\endgroup$ – Electric Head Oct 8 '14 at 20:26
  • $\begingroup$ Sorry, I was missing a $U$. I'm arguing the fundamental definition should be: $U(\boldsymbol{R}) \equiv \Delta U (\boldsymbol{R}_{Ref} \rightarrow \boldsymbol{R})$ $\endgroup$ – CraftyVisage Oct 8 '14 at 20:52
  • $\begingroup$ Ahhh. Callen doesn't define $U$ this way either; pretty sure the problem is your notation/definition. What you are effectively arguing is $U \equiv \Delta U \equiv U - U_\text{ref}$ which is why we have this fixed value of $U_\text{ref} = 0$. It should strictly be $\Delta U = U - U_\text{ref}$. No worries about the typo, looks like I made one in the last comment too ;) I've updated the answer, hopefully addressed it now! $\endgroup$ – Electric Head Oct 8 '14 at 22:53
  • $\begingroup$ I see, so the precise definition of a reference state is: any state of my choosing that I assign an arbitrary $\ U$ value to. This $U$ value doesn't have to be zero, but in practice this choice often makes sense. $\endgroup$ – CraftyVisage Oct 15 '14 at 16:01

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