Why Must Constants in Elastic String Equations Be Equal?

[CAF123]

Homework Statement

Consider a constant volume elastic string. A change in internal energy of the string is given by dU = TdS + fdL.

The elastic string obeys the following two equations; $$\left(\frac{\partial f}{\partial L}\right)_L = \gamma_1 T\,\,\,\,\,\,\,\,\,\,\,\left(\frac{\partial f}{\partial T}\right)_L = \gamma_2L,$$ $\gamma_1, \gamma_2$ constants.

A) Why do we require $\gamma_1 = \gamma_2$
B)Derive an expression for f(T,L). Explain whether this expression gives a complete thermodynamic description of the system.

Homework Equations

Total differential

The Attempt at a Solution

A) It is not exactly clear to me why the constants have to be the same. I was thinking that since L and T are independent variables that $\gamma_1, \gamma_2$ may be interpreted as separation constants, but I am unsure.

B)If $df = \gamma_1 T dL + \gamma_2 L dT$, then I am tempted to write $\Delta f = \gamma_2 T \Delta L + \gamma_2 L \Delta T$, but I don't think this is correct since L ( and T) are not state variables.

I think the resulting expression will give a complete thermodynamic description since we have expressed f as a function of two independent variables, and we have specified what is being held constant in each case.

Thanks.

 

[Simon Bridge]

A) What property of the string do the gammas represent?

B) Are you no longer requiring that $\gamma_1 = \gamma_2$ ?

What does that subscript L after the partials, in your equations, mean?

Aren't you looking for f(L,T) rather than delta-f?

If L (and T) are not state variables, then does the expression give "a complete thermodynamic description of the system"?

 

[Chestermiller]

If you take the partial of the first equation with respect to T, you get \frac{\partial ^2 f}{\partial T \partial L}. What do you get if you take the partial of the second equation with respect to L?

 

[CAF123]

There is a typo in the OP, the first partial derivative should be at constant T, not L. I figured that if I can find delta f, then this can be replaced by f - f_o and so when I rearrange I will have an expression for f. But since L and T are not state variables, the integral will be path dependent and so what I did (I think) is not correct. Do you have a pointer in the right direction?
Thanks.

 

[Chestermiller]

There is a typo in the OP, the first partial derivative should be at constant T, not L. I figured that if I can find delta f, then this can be replaced by f - f_o and so when I rearrange I will have an expression for f. But since L and T are not state variables, the integral will be path dependent and so what I did (I think) is not correct. Do you have a pointer in the right direction?
Thanks.

Good. So if you follow what I said in my previous post, you will find that γ₁=γ₂=γ. You also know that:

df=\frac{\partial f}{\partial L}dL+\frac{\partial f}{\partial T}dT
Substitute your original two equations into this relationship and see what you get.

 

[CAF123]

Yes, I got to $df = \gamma[TdL + LdT]$ and then said $f= f_0 + \gamma[T(L_2-L_1) + L(T_2-T_1)]$. I Think this is incorrect because L and T are not state variables.

 

[haruspex]

Yes, I got to $df = \gamma[TdL + LdT]$

Seems reasonable.

and then said $f= f_0 + \gamma[T(L_2-L_1) + L(T_2-T_1)]$.

That would only be an approximation for small changes, $L_2-L_1$ and $T_2-T_1$. Why not the more obvious solution

Spoiler

f = γLT+c

?

I Think this is incorrect because L and T are not state variables.

That would be a concern if you were deriving an expression for U, but it's not obvious to me that it matters in finding expression for f. After all, finding an equation for f as a function of L and T is exactly what you were asked to do.

 

[CAF123]

Hi haruspex,

That would only be an approximation for small changes, $L_2-L_1$ and $T_2-T_1$.

Could you explain a bit more why this is only valid for small changes?

That would be a concern if you were deriving an expression for U, but it's not obvious to me that it matters in finding expression for f. After all, finding an equation for f as a function of L and T is exactly what you were asked to do.

The reason I thought I did something wrong was because the integral I computed was $$\int_{f_0}^{f'} df = \gamma \left( T \int_{L_1}^{L_2} dL + L \int_{T_1}^{T_2} dT\right)$$ L and T are not state variables, so I think I need to specify the path. But I have only specified the end points here, and not the actual path.

 

[haruspex]

Hi haruspex,
Could you explain a bit more why this is only valid for small changes?

Looking at it again, I'm not even sure what your equation means. I assume T1, L1 refer to some initial state and T2, L2 to a final state, but you also have T and L. What do they represent?
Maybe you meant f2 =f1+γ[T2 L2 - T1 L1)], which comes to the same as I posted. E.g. consider the case where one changes then the other:
L: L1 L2 L2
T: T1 T1 T2
So f first increases by (L2-L1)T1, then by (T2-T1)L2. Total increase T2L2-T1L1.

Going back to your OP equations, I guess you meant
$\left(\frac{\partial f}{\partial L}\right)_T = \gamma_1T$ (the subscript denoting which variable stays constant, yes?).
You can integrate that straight away to obtain $f = \gamma_1 T L + g(T)$. With the other differential, we likewise get $f = \gamma_2 T L + h(L)$.

L and T are not state variables, so I think I need to specify the path. But I have only specified the end points here, and not the actual path.

But if that's a problem, would it not apply to any solution f = f(L, T)? In which case, how can you answer the question? The equations dictate this result - how can it be wrong?
Maybe the path matters for U but not for f?

 

[Chestermiller]

Have you learned the rule about differentiation of a product? d(uv) = udv+vdu

Chet