Expansions in thermodynamics using differentials

[CAF123]

Because the non-approximated X and Y are functions of two variables, and to integrate them you need a path. Once a path a selected, the 2D-functions are reduced to 1D, which is what happens above, too, albeit via a different approach.

So why, if I express F = F(L,T) and then write $dF = \frac{\partial F}{\partial T}dT + \frac{\partial F}{\partial L}dL$ I can then say $$F_2 - F_1 = \int_{T_1}^{T_1} \frac{\partial F}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F}{\partial L}dL$$ replace each integrand with the approximation and then integrate? (as in part A) for example)

I recall you mentioning that this was because dF was a total differential in this case and so this implies the change in tension was path independent. What is it about dF being a total differential implies path independency?

 

[voko]

So why, if I express F = F(L,T) and then write $dF = \frac{\partial F}{\partial T}dT + \frac{\partial F}{\partial L}dL$ I can then say $$F_2 - F_1 = \int_{T_1}^{T_1} \frac{\partial F}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F}{\partial L}dL$$

More verbosely, this is: $F_2 - F_1 = \int_{T_1}^{T_2} \frac{\partial F(L, T)}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F(L, T)}{\partial L}dL$. What is the value of L in the first term on the right, and the value of T in the second term? Observe L and T are not variables on the left.

 

[CAF123]

More verbosely, this is: $F_2 - F_1 = \int_{T_1}^{T_2} \frac{\partial F(L, T)}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F(L, T)}{\partial L}dL$. What is the value of L in the first term on the right, and the value of T in the second term?

In the first term on the RHS, we consider how F varies with T only, so the value of L in first term: L₁. Similarly for the second term on the RHS, we consider only changes in L so the value of T in the second term is T₁.

Edit: So this means each integral on the RHS is an integral over a single variable but how does that help me? For example, I can express $\frac{\partial F}{\partial T} dT = - \frac{\partial F}{\partial L}dL$ as $\frac{\partial F}{\partial T} dT + \frac{\partial F}{\partial L}dL = 0$ and each integral on the LHS are also integrals over a single variable.

Observe L and T are not variables on the left.

Yes, $F_2$ is the value of F at the state $(T_2, L_2)$ and $F_1$ the value of F at the state $(T_1, L_1)$

 

[Chestermiller]

I get to here by rearranging the previous equation: $$\frac{dL}{dT} = -\frac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial L}}$$

Did you convert the LHS to a partial because given F = F(L,T) we can express L = L(F,T), and so L also depends on F?

This is a standard result in working with partial differential equations. Check your math textbook on pde's. The left hand side is a partial derivative because F is being held constant (recall, dF = 0). So we are getting the rate of change of L with respect to T at constant F (i.e., the partial derivative of L with respect to T at constant F).

 

[voko]

In the first term on the RHS, we consider how F varies with T only, so the value of L in first term: L₁. Similarly for the second term on the RHS, we consider only changes in L so the value of T in the second term is T₁.

Then the RHS is not an integral over a single path such as L = L(T), so the equation cannot hold unless you show it results from a single-path integral and some other assumptions.

Edit: So this means each integral on the RHS is an integral over a single variable but how does that help me?

You cannot just integrate things independently over different paths and claim the equation still holds. In this case, where the rail is allowed to expand freely, you have some unknown path L = L(T), and you have to integrate the whole thing over this path - or some other path, but then you have to show the equivalence.

Because $dF$ is a full differential, you can say that you can integrate over any path and the result will be the same, provided the endpoint are the same. One such path could be made of two straight-line segments: $(L_1, T_1) \rightarrow (L_1, T_2) \rightarrow (L_2, T_2)$. Physically, this corresponds to heating the rail while holding it in place, then to letting it expand at constant temperature - we can expect that this should have the same end result. But this corresponds to $$ F_2 - F_1 = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{T_1}^{T_2} F_L(L_1, T) L'(T) dT + \int_{L_1}^{L_2} F_T(L, T_2) T'(L) dL + \int_{L_1}^{L_2} F_L(L, T_2) dL \\ = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{L_1}^{L_2} F_L(L, T_2) dL $$ Observe this is different from what you had earlier.

 

[CAF123]

Then the RHS is not an integral over a single path such as L = L(T).

So, for example, in the other equation I had when I rearranged dF = X dT + YdL for dL, I got $$\frac{\partial F}{\partial L} dL = - \frac{\partial F}{\partial L} dL,$$ where in this case L = L(T) so that this becomes $$\frac{\partial F(L,T)}{\partial L(T)} L'(T) dT = -\frac{\partial F(L,T)}{\partial T(L)} T'(L) dL$$ Provided L=L(T), this eqn then makes sense. Is this what you meant?

Because $dF$ is a full differential, you can say that you can integrate over any path and the result will be the same, provided the endpoint are the same.

Can you point me to a proof of this?

But this corresponds to $$ F_2 - F_1 = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{T_1}^{T_2} F_L(L_1, T) L'(T) dT + \int_{L_1}^{L_2} F_T(L, T_2) T'(L) dL + \int_{L_1}^{L_2} F_L(L, T_2) dL \\ = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{L_1}^{L_2} F_L(L, T_2) dL $$

Is your use of notation $F_T$ standard, $F_T = \frac{\partial F}{\partial T}$? Is that top line, the general form of $F_2 - F_1$ and then by the choice of path, you were then able to say the middle terms were zero and so arrived at the bottom line?

Observe this is different from what you had earlier.

Different in the sense that I said the value of T in the second term was T₁ above?

 

[voko]

So, for example, in the other equation I had when I rearranged dF = X dT + YdL for dL, I got $$\frac{\partial F}{\partial L} dL = - \frac{\partial F}{\partial L} dL,$$ where in this case L = L(T) so that this becomes $$\frac{\partial F(L,T)}{\partial L(T)} L'(T) dT = -\frac{\partial F(L,T)}{\partial T(L)} T'(L) dL$$ Provided L=L(T), this eqn then makes sense. Is this what you meant?

The differential equation makes sense without regard to any path. What I meant is that in your previous integration attempt, you had the (L, T) point follow different paths in the two integrals, and it was not obvious that those are just parts of a bigger path, with some other segments of the path evaluating to zero.

Can you point me to a proof of this?

This is a corollary of Green's theorem in 2D, and Stoke's theorem in higher dimensions. This is also what makes "potential energy" possible as a concept.

Is your use of notation $F_T$ standard, $F_T = \frac{\partial F}{\partial T}$? Is that top line, the general form of $F_2 - F_1$ and then by the choice of path, you were then able to say the middle terms were zero and so arrived at the bottom line?


Different in the sense that I said the value of T in the second term was T₁ above?

All correct.

 

[CAF123]

What I meant is that in your previous integration attempt, you had the (L, T) point follow different paths in the two integrals,

How so?

 

[voko]

Refer to #33. What paths are followed by (L, T) in the RHS terms?

 

[CAF123]

In #33, I wasn't referring to a path. I just knew that $F_i$ would be $(L_i, T_i)$ for i = 1,2. and integrated by just considering the end points since dF was path independent.

(I've noticed at the back of my book an appendix which gives a couple of arguments to illustrate when and when not a differential is exact)

Going back to the eqn at the top of #36, does it not need to be the case that L=L(T) for that eqn to make sense?

 

[voko]

In #33, I wasn't referring to a path. I just knew that $F_i$ would be $(L_i, T_i)$ for i = 1,2. and integrated by just considering the end points since dF was path independent.

When you integrate something whose dim > 1, you always have a path (or surface, or hypersurface) involved - that's how an integral is defined. In your original attempt you could say the paths were different simply because the end points were different.

Going back to the eqn at the top of #36, does it not need to be the case that L=L(T) for that eqn to make sense?

Yes and no. The differential of $f(x)$, where $x$ is a vector, is defined as $h(x, dx)$ such that $f(x + dx) - f(x) = h(x, dx) + r(x, dx)$, where $h(x, dx)$ is linear in the second argument, and $\lim r(x, dx)/|dx| = 0$. Because of the linearity in $dx$ it can be written in terms of its components, traditionally, as $f(X + dX, Y + dY) - f(X, Y) ≈ f_X dX + f_Y dY$, which looks as if $dX$ and $dY$ were completely independent; but they must in fact be components of the infinitesimal vector $dx$. So not only is $dY = dY(dX)$ here, it is even stronger: $dY = k dX$, so you do not really have to think about the path when you deal with the differential form. And when $x$ is itself a function of something else, the same applies to its differential, and we end end up with a product of two linear maps, which is again linear, so the result holds, which basically means we can use the chain rule. This is sometimes called the invariance of (first order) differentials.

 

[CAF123]

Thanks voko, I appreciate the help.

When we started with $dF = F_T dT + F_L dL$ and for part B), dF = 0 we then had $0 = F_T dT + F_L dL \Rightarrow F_T dT = -F_L dL$ Is it correct to say that this equation is true for any path taken between $(L_1, T_1)$ and $(L_2, T_2)$ since it is really the same as evaluating $0 = F_T dT + F_L dL$ which is path independent.

 

[voko]

Is it correct to say that this equation is true for any path taken between $(L_1, T_1)$ and $(L_2, T_2)$ since it is really the same as evaluating $0 = F_T dT + F_L dL$ which is path independent.

Correct.