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Force on an elastic string

  1. Nov 28, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    Total differential

    3. 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.
     
  2. jcsd
  3. Nov 28, 2013 #2

    Simon Bridge

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    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"?
     
  4. Nov 28, 2013 #3
    If you take the partial of the first equation with respect to T, you get [tex]\frac{\partial ^2 f}{\partial T \partial L}[/tex]. What do you get if you take the partial of the second equation with respect to L?
     
  5. Nov 29, 2013 #4

    CAF123

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    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 - fo 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.
     
  6. Nov 29, 2013 #5
    Good. So if you follow what I said in my previous post, you will find that γ12=γ. You also know that:

    [tex]df=\frac{\partial f}{\partial L}dL+\frac{\partial f}{\partial T}dT[/tex]
    Substitute your original two equations into this relationship and see what you get.
     
  7. Nov 29, 2013 #6

    CAF123

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    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.
     
  8. Nov 29, 2013 #7

    haruspex

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    Seems reasonable.
    That would only be an approximation for small changes, ##L_2-L_1## and ##T_2-T_1##. Why not the more obvious solution
    f = γLT+c
    ?
    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.
     
  9. Nov 30, 2013 #8

    CAF123

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

    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.
     
  10. Nov 30, 2013 #9

    haruspex

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    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)##.
    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?
     
  11. Nov 30, 2013 #10
    Have you learned the rule about differentiation of a product? d(uv) = udv+vdu

    Chet
     
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