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Thermal Expansion Question

  1. Dec 12, 2011 #1
    Hi Physicsforums,

    I was trying to derive the formula that the coefficient for volume expansion, β, is 3 times the coefficient for length expansion, [itex]\alpha[/itex].

    As a reminder, the formulas are:

    [itex]\Delta[/itex]L = [itex]\alpha[/itex]Lo[itex]\Delta[/itex]T

    and [itex]\Delta[/itex]V = [itex]\beta[/itex]Vo[itex]\Delta[/itex]T

    where, supposedly, [itex]\beta[/itex] = 3[itex]\alpha[/itex].

    My proof actually gave me a completely different result. I double-checked it a few times, and I can't find the problem. Does anyone know what's wrong?

    here it is: http://imageshack.us/photo/my-images/267/img20111212204020.jpg/

    thanks in advance,

  2. jcsd
  3. Dec 12, 2011 #2

    Doc Al

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    Staff: Mentor

    Get rid of all the terms with higher powers of α. α is usually quite small.
  4. Dec 12, 2011 #3
    Well, then how come in this proof it works out EXACTLY?

    V = L3
    dV/dL = 3L2

    dL = [itex]\alpha[/itex]LodT
    dL/dT = [itex]\alpha[/itex]Lo

    dV/dT = (dV/dL)(dL/dT) = 3L2 * [itex]\alpha[/itex]Lo = 3[itex]\alpha[/itex]Vo. Now just turn dV/dT into [itex]\Delta[/itex]V/[itex]\Delta[/itex]T (since dV/dT is constant, you can do that), and you have EXACTLY the equation.

    It should come out to the same result, regardless of which proof you use, if you do no approximations.

    Edit: I'm guessing the error in this derivation is that L is a function of T, so (L^2)*Lo is not Vo. Checking now if that results in the same equation as the original.
    Last edited: Dec 12, 2011
  5. Dec 12, 2011 #4

    β≈3α. It's only an approximation. That's why your proof went wrong.
    If you neglect any higher order terms of α, then you will find β=3α.
  6. Dec 12, 2011 #5
    Okay. I'm glad that I know it's an approximation. It bothers me, because my textbook (Y&F University Physics) didn't mention that detail WHATSOEVER, and in fact their proof is just plain incorrect; they made an illegal substitution that results in the coefficients being exact multiples.

    Anyway - I still have a question. The two ways I'm deriving the equation don't agree. The first method is using V + [itex]\Delta[/itex]V = (LO+[itex]\Delta[/itex]L)3, and plugging in. The second way is to derive the equation starting from V = L^3 and the length expansion equation. I have both of my proofs in the following image.

    Why do they not agree?? I re-did them multiple times. What step was illegal?


    Also - ALL of these calculations were just for cubes. Nothing is a perfect cube in real life. How do I know that other shapes follow similar rules?
    Last edited: Dec 12, 2011
  7. Dec 12, 2011 #6

    Doc Al

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    Staff: Mentor

    No reason they should agree. The first involves no approximations at all. In the second method, as soon as you introduce a derivative you've introduced an approximation.

    dV = 3L2dL is only true for small changes, yet you treat it as equivalent to ΔV = 3L2ΔL. But that's not quite accurate, as the first method demonstrates. (ΔV really equals (L + ΔL)3 - L3)

    As long as you get rid of all higher powers, the two derivations give the same approximate answer.
  8. Dec 12, 2011 #7
    Nice, that makes perfect sense. Thank you so much. Any idea on the other shapes question? Or is it just true that regardless of the shape, it's always approximately true that [itex]\Delta[/itex]V = 3[itex]\alpha[/itex]VO[itex]\Delta[/itex]T, as long as the equation for linear expansion holds equally in every dimension.
  9. Dec 13, 2011 #8

    Doc Al

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    Staff: Mentor

    Exactly. (You can think of other shapes as composed of small cubes, if you like.)
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