Why does the curve of a hanging chain not minimize the area below it?

Question

If we have a chain of fixed length hanging from two points we know that it will form a curve that minimizes the chain's potential energy.

If we imagine the chain as having many small segments, then the potential energy of each segment is $E_p = mgh$. As the number of small segments approaches infinity, their masses equalize because the difference in mass between any two segments goes to $0$ as the number of segments goes to infinity.

So if we want to minimize the total potential energy of the chain, we need to minimize the sum of potential energies of every segment as the number of segments goes to infinity. However, since we know that in the limiting case all small segments have the same mass, minimizing the sum of their potential energies is equivalent to minimizing the sum of their heights (as the number of segments goes to infinity).

This limit of a sum is by definition the integral of the chain curve. Therefore, to minimize the potential energy of a hanging chain, the area under the chain must be minimal.

However, this is not the case because the curve that minimizes that area is known to be a semicircle. This would mean that the shape of a hanging chain is a semicircle which is obviously false.

What is wrong with my argument? I don't see a reason why minimizing the area doesn't minimize the potential energy. By the way, I don't know physics above high school level, so I would be very thankful if someone can answer this without super heavy math. Calculus is fine.

Answer 1

What is wrong with your argument is this paragraph:

If we imagine the chain as having many small segments, then the potential energy of each segment is $E_p=mgh$. As the number of small segments approaches infinity, their masses equalize because the difference in mass between any two segments goes to $0$ as the number of segments goes to infinity.

Especially the part "their masses equalize".

The shape of the curve is $y=\cosh(x)$, but for a segment of chain with a small change in $x$ of $\mathrm{d}x$ in the middle, where the gradient is zero, the length of the segment is shorter by a significant fraction than for a part at the edge with the same $\mathrm{d}x$.

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Due to the comments

You said that the difference in the $\mathrm{d}x$ values would go to zero, that's true but it's the ratio, or fraction, of the $\mathrm{d}x$ values at different parts of the chain that matter as follows...

1cm of chain (with the 1cm being a change in $x$) near the edge might be made up of a million small $\mathrm{d}x$ parts, but these million would add up to a greater mass of chain than a million similar parts that make up a a 1cm section (in the $x$ direction) in the middle.

Answer 2

To put the accepted answer in mathematical terms, if you have a curve $y(x)$, hanging fixed at $x_0$ and $x_L$ at an height $h=y(x_0)=y(x_L)$, of total length $L$ and mass $M$ then then linear mass density is going to be $\lambda = M/L$.

The length of the curve is given by the integral of the arc-length

$$L=\int_{x_0}^{x_L} \sqrt{ 1+\left({dy \over dx}\right)^2} dx$$

i.e. for each "$dx$" (variation of the $x$ coordinate) the local length of the curve is not given by $dx$ but by the length "along" the curve $$dL=\sqrt{1+\left({dy \over dx}\right)^2} dx$$ which as you can see depends on the derivative of the curve $y'(x)={dy\over dx}$.

This means that the length of each segment along $x$ is not the same and thus the infinitesimal mass is going to be a function of $x$ given by

$$dm(x)=\lambda dL(x) = \lambda \sqrt{1+y'(x)^2} dx $$

So the infinitesimal potential energy (of a small element of mass $dm(x)$ at height $h=y(x)$) is going to be $$dU(x)=g dm(x)y(x)= g \lambda dL y(x)= g \lambda \sqrt{1+y'(x)^2} y(x)dx $$ and the intgral we want to minimise is not the area under the curve (which would be $A=\int_{x_0}^{x_L} y(x)dx$ ) but rather the energy integral

$$U=\int dU(x) = g \lambda \int_{x_0}^{x_L} \sqrt{1+y'(x)^2} y(x)dx $$

where the only part we want to minimise is actually

$$A_U=\int_{x_0}^{x_L} \sqrt{1+y'(x)^2} y(x) dx \ne A$$

(Unless of course the curve has $y'(x)=0$)

Answer 3

This limit of a sum is by definition the integral of the chain curve.

There's no such thing as "the integral of the chain curve". Curves don't have integrals. For an integral, you need three things: an integrand, measure function, and a set over which those are defined (in basic integrals, these are the function you're integrating, the differential, and the limits of integration, respectively). The set is split into subsets, the measure of each subset is multiplied by the value of the function in that subset, and all of the resulting values are added together. Of course, for subsets of finite size, the function doesn't have a unique value, so one might take the minimum, the maximum, or some other value. If all of the options converge to the same value when the measures of the subsets go to zero (for a precise definition of convergence that I won't get into), then the function is integrable, and that value is its integral.

You started out with your set being the set of all points along the chain, and each subset is a segment of chain. You take the length of the chain to be the measure of the segment, and then the potential energy of that segment is the height (integrand) times the length of the segment (measure), and you add up all the potential energies across the length of chain (set over which you're integrating).

However, you then switched over to talking about area. The area under a curve is the height of the curve (integrand) times the width of the curve (measure). The width of a curve is different from the length; the width is the distance along the x-axis, while the length is the distance along the curve itself.

It's often obvious from context what one is integrating with respect to, and so you've gotten used to it not being explicitly stated. But this is a case where there are two different things that you can integrate with respect to: you can integrate with respect to the width of the chain, or with respect to the length.

You're also missing the constraint that the ends of the chain are attached to the ends. The length of the chain will not in general match the circumference of a semi circle that matches the distance between the two points. I'm pretty sure that if the length is too short, the curve that minimizes area under it will be an arc of circle, rather than a semicircle, and if it's longer than what's necessary, it will be vertical drops on each side and a semicircle in between.

Answer 4

Minimizing the area underneath the chain would work if the chain-mass was equal in each equally-widthed vertical cross-section. If this were the case then the area under the chain would then be proportional to potential energy of the chain.

However, we know that there is more mass in those equally-widthed vertical cross-sections where the chain is slanted than those sections where the chain is flat. Therefore, this will "weight" sections with a more-slanted chain length more than those with a less-slanted chain length, causing more potential energy for a more-slanted chain length than one with a less-slanted chain length for equivalent heights+widths. Thus, aside special cases (e.g. where the the chain is a straight line), the area underneath the chain for a given width will not be proportional to the chain's potential energy for that width.