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Vertical spring statistical mechanics

  1. Jul 13, 2014 #1

    WannabeNewton

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    Consider the following problem: A spring with spring constant ##k## is suspended vertically from a fixed support and the spring is in equilibrium with a heat bath of temperature ##T## and resides in a gravitational field ##g##. A mass ##m## is hung from the other end of the spring. Find the average elongation of the spring.

    This isn't a homework problem as I know how to get the solution; rather, I've got a question about the way the book goes about getting the solution. Let ##x = 0## denote the location of the unstretched length of the spring as well as the zero of the gravitational potential so that ##x_{eq} \equiv -\frac{mg}{k}## is the equilibrium position of the spring. My first issue is in the interpretation of the phrase "average elongation". Presumably when the book uses this phrase, it really means the average displacement of the spring about its equilibrium position ##x_{eq}## and not the average equilibrium position itself, which is in fact the elongation of the spring from its unstretched length due to the weight ##mg##, for then it wouldn't make any sense as it would make the problem trivial since given an ensemble of such springs, every single one of them has the same equilibrium position ##x_{eq} = \frac{mg}{k}## as determined to exactness by Newton's 2nd law for each spring.

    The average displacement given an ensemble of vertical springs in simple harmonic motion is just ##\langle x \rangle = \int dx x e^{-\frac{1}{2}\beta kx^2 + mgx}/\int dx e^{-\frac{1}{2}\beta kx^2 + mgx} \\= \int dx x e^{-\frac{\beta k}{2}(x - \frac{mg}{k})^2}/\int dx e^{-\frac{\beta k}{2}(x - \frac{mg}{k})^2} \\= \int dx' x' e^{-\frac{\beta k}{2}x'^2}/\int dx' e^{-\frac{\beta k}{2}x'^2} + \frac{mg}{k} = \frac{mg}{k} = x_{eq}##

    which certainly makes sense after the fact because each spring in the ensemble spends most of its time around the equilibrium position so a statistical averaging over displacements about equilibrium across the entire ensemble would naturally yield ##x_{eq}##.

    But what the book says is "The mean elongation is found by equating the gravitational and restoring forces" so that ##k\langle x \rangle = mg## thus ##\langle x \rangle = \frac{mg}{k}##. But this makes no sense to me. First of all it is the equilibrium position that one obtains by equating the gravitational and restoring forces i.e. ##k x_{eq} = mg## and as noted above the equilibrium position is the same across all springs in the ensemble so there is no need for a statistical averaging over the ensemble if one just seeks the equilibrium position. So the book's solution would only work if one knew a priori that ##x_{eq} = \langle x\rangle##. While this is certainly a reasonable intuitive guess as mentioned above, I see no reason why this would have to hold a priori, that is before actually calculating the average displacement about equilibrium ##\langle x \rangle## from the partition function and consequently verifying this guess. Could anyone shed light on this? Thanks in advance.
     
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  3. Jul 14, 2014 #2

    Matterwave

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    I think by symmetry you can conclude the result sort of a priori. There is no such other possibility for <x> as by symmetry the spring exerts the same force in opposite directions as displacements from <x>. Or in other words, the potential energy function is symmetric about x_eq.

    It seems to me similar to the conclusion that in a random walk the average displacement from 0 is 0, since the problem is symmetric about this point.

    By the way, are you missing a kinetic energy term in your integrals?
     
  4. Jul 14, 2014 #3

    WannabeNewton

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    That certainly makes sense, thanks! I do have a follow-up question however:

    The problem in the book then asks for, and I quote, "the magnitude ##\langle (x - \langle x \rangle )^2 \rangle## of the thermal fluctuations of the object about its equilibrium position". The problem basically gives away the fact that ##\langle x \rangle = x_{eq}## but with that humorous tidbit aside, the dispersion ##\sigma^2 =\langle (x - \langle x \rangle )^2 \rangle=\langle x^2 \rangle - \langle x \rangle ^2## gives the thermal fluctuations about the spring's average position, which in this case happens to be its equilibrium position but say we didn't know that. Then wouldn't "the magnitude ##\langle (x - \langle x \rangle )^2 \rangle## of the thermal fluctuations of the object about its equilibrium position" necessarily be zero? All the springs in the ensemble have the same equilibrium position so if we make measurements of the equilibrium position across the entire ensemble, we will always get the same result and so there won't be any dispersion from this value.

    I know I'm being pedantic here but surely there's a crucial difference in asking for "the magnitude ##\langle (x - \langle x \rangle )^2 \rangle## of the thermal fluctuations of the object about its equilibrium position" and "the magnitude ##\langle (x - \langle x \rangle )^2 \rangle## of the thermal fluctuations of the object about its average position which just happens to be its equilibrium position", is there not? The latter refers to a measurement of the position of each spring in the ensemble, not necessarily the equilibrium position, and on average this measurement will yield the equilibrium position but we will have some dispersion ##\sigma^2## due to thermal fluctuations which in this case can be determined from the equipartition theorem. But this is definitely different from a measurement of the equilibrium position of each spring in the ensemble, which will all be the same thus yielding zero dispersion.

    The same term appears in the numerator and denominator of ##\langle x \rangle## so they cancel.

    Thanks again.
     
  5. Jul 14, 2014 #4

    Matterwave

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    Let me just make sure I got your question right. Is your problem with the text specifically its use of the word "equilibrium position" where you would use the word "average position"? You don't have a problem with the mathematical expression ##\sigma^2=\left<(x-\left<x\right>)^2\right>## do you? In other words, if the book omitted any words at all and just asked you to find ##\sigma^2## you'd have no problems right?
     
  6. Jul 14, 2014 #5

    WannabeNewton

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    Yes exactly. It may seem pedantic but it did confuse me at first and I just want to make sure my confusion isn't ill founded.
     
  7. Jul 14, 2014 #6

    Matterwave

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    Well, my two cents on this issue is that it is purely a semantic one. When in doubt, use the mathematical expression to figure out what they want.

    I think in this case, since they said "thermal fluctuations of THE object ABOUT it's equilibrium position", even though they used "equilibrium position", it's still reasonable to assume they mean ##\sigma^2##. In other words, I think it's reasonable to think they are just interchanging the words equilibrium and average without changing any meaning in the problem.

    If they wanted you to find the fluctuations of the equilibrium position of an ensemble of such springs, I think they would have said as such "find the fluctuations of THE equilibrium position of an ensemble of such springs".
     
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