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.
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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