Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The curve for a chain hanging between two poles in a uniform gravitational field is known as the catenary.

Is there known an expression for the curve of a hanging chain on a planet of mass $M$ which has a gravitational field of $M/r^2$ ?

Let the rope have length $L$, uniform mass-density $\rho$ and negligible thickness. The poles are a distance $D$ apart, each a distance $R$ from the planets centre.

I am interested in solutions where the gravitational field may be taken to point straight down as well as when it point towards the center of the planet.

share|cite|improve this question
    
Note that near the surface of the Earth $r \approx 6400\text{ km}$. How long and high do you intend to have the arc be? How big a deviation from constant gravity will be observed? Do you think the effect will be larger or small then the size of the links? Then imperfections in the links? – dmckee Mar 29 '11 at 2:10
    
@dmckee There is no links or anything, its an ideal rope, this is a problem in theoretical physics – TROLLKILLER Mar 29 '11 at 3:47
    
Interesting solution when $L=2R$. – Carl Brannen Mar 29 '11 at 3:47
up vote 7 down vote accepted

It took a while, but I found a good reference on this problem: A study of the generalized catenary problem (PDF), an undergraduate mathematical physics thesis by Hai Xuan Nguyen. It explores the equations governing the equilibrium shapes of chains and arches in radial potentials $V(r)$. I've extracted and adapted the relevant parts for this answer, and referenced the corresponding equations in the thesis using tags.

Anyway, our situation is as follows: the chain is suspended from two poles of radius $R$ with angular separation $\Delta = \frac{D}{R}$ if I understand the problem correctly. Let the shape of the rope be given by a function $r(\phi)$ with boundary conditions $r(\phi_1) = r(\phi_2) = R$, $\phi_2 - \phi_1 = \Delta$. Then the potential energy of the rope is $$U_G = -\int_{\phi_1}^{\phi_2} \frac{GM}{r}\rho\sqrt{r'^2 + r^2}\mathrm{d}\phi\tag{cf. eq. 9 of the thesis}$$ with $\rho$ being the linear mass density of the rope, and I've left the $\phi$ dependence of $r(\phi)$ implicit.

We need a constraint that enforces the total length of the rope remain fixed at $L$. That constraint is going to be $$\int_{\phi_1}^{\phi_2} \sqrt{r'^2 + r^2}\mathrm{d}\phi = L\tag{cf. 10}$$ Then we can minimize the potential energy subject to the constraint by adding a term proportional to the constraint, $U_L = \int_0^\Delta \lambda\sqrt{r'^2 + r^2}\mathrm{d}\phi$, for a total of $$U = \int_{\phi_1}^{\phi_2} \biggl(-\frac{GM}{r}\rho + \lambda\biggr)\sqrt{r'^2 + r^2}\mathrm{d}\phi$$ where $\lambda$ is the Lagrange multiplier. This can be solved by the Euler-Lagrange equation, $$\frac{\mathrm{d}}{\mathrm{d}\phi}\frac{\partial}{\partial r'}\biggl[\biggl(\frac{GM}{r}\rho - \lambda\biggr)\sqrt{r'^2 + r^2}\biggr] = \frac{\partial}{\partial r}\biggl[\biggl(\frac{GM}{r}\rho - \lambda\biggr)\sqrt{r'^2 + r^2}\biggr]\tag{cf. 11}$$ If you multiply both sides by $r'$ and do a little manipulation, you can get this in the form $$\underbrace{\biggl(\frac{\mathrm{d}r}{\mathrm{d}\phi}\frac{\partial}{\partial r} + \frac{\mathrm{d}r'}{\mathrm{d}\phi}\frac{\partial}{\partial r'}\biggr)}_{\mathrm{d}/\mathrm{d}\phi}\biggl[\biggl(\frac{GM}{r}\rho - \lambda\biggr)\sqrt{r'^2 + r^2}\biggr] = \frac{\mathrm{d}}{\mathrm{d}\phi}\biggl[\biggl(\frac{GM}{r}\rho - \lambda\biggr)\frac{r'^2}{\sqrt{r^2 + r'^2}}\biggr]$$ which integrates to $$\frac{GM\rho}{\alpha} = \biggl(\frac{GM\rho}{r} - \lambda\biggr)\frac{r^2}{\sqrt{r^2 + r'^2}}\tag{cf. 12, 18}$$ Here $\frac{GM\rho}{\alpha}$ is a constant of integration. Actually, $\alpha$ is the arbitrary constant, but I write it in this way because, with the substitution $u = \frac{GM\rho}{\lambda r}$, this becomes $$\frac{1}{\alpha} = \frac{u - 1}{\sqrt{u^2 + u'^2}}$$ Hanging chains correspond to solutions of this equation for $\alpha > 1$ (see section 3 of the thesis, also note that the right side is always less than $1$ given that $u > 0$, $u' > 0$). Solving the equation and reversing the variable substitution gives $$r(\phi) = \frac{GM\rho}{\lambda}\frac{\alpha^2 - 1}{\alpha\bigl(\alpha - \cosh\bigl[\sqrt{\alpha^2 - 1}(\phi - \phi_0)\bigr]\bigr)}\tag{cf. 23}$$ Note: I'm not convinced the most general solution can be written in this form, but for simplicity and consistency with the paper I'll stick to this unless shown otherwise.
The three undetermined constants $\phi_0$, $\alpha$, and $\lambda$ can be used to fit the boundary conditions and the length constraint.

From the form of the solution, you can tell that $\lambda$ and $\phi_0$ have simple physical interpretations as a scaling factor for $r$ and a shift for $\phi$, respectively. The only parameter that has a nontrivial effect on the shape of the solution is $\alpha$. So suppose we define a "reduced $r$" as $$\bar{r}(\varphi) = \frac{\lambda}{GM\rho}r(\varphi + \phi_0) = \frac{\alpha^2 - 1}{\alpha\bigl[\alpha - \cosh\bigl(\varphi\sqrt{\alpha^2 - 1}\bigr)\bigr]}$$ This function is symmetric, with its minimum at $\varphi = 0$ and vertical asymptotes at $\varphi = \pm \frac{\cosh^{-1}\alpha}{\sqrt{\alpha^2 - 1}}$. Plotted for various values of $\alpha$, it looks like this:

curve shapes for various values of alpha

If you pick a value of $\alpha$, that sets the shape of the curve, and then you can rotate it (by choosing $\phi_0$) and scale it (by choosing $\lambda$) to match any boundary conditions you want.

The value of $\alpha$ you pick will have to be the one that satisfies the length constraint. It helps that we can actually compute the length of the curve analytically. One can do the computation in general, but I'm going to use the fact that the endpoints are at the same radius to simplify matters a bit. It works out to $$\begin{align} \int_{-\Delta/2}^{\Delta/2} \sqrt{\bar{r}(\varphi)^2 + \bar{r}'(\varphi)^2}\mathrm{d}\varphi &= \frac{2\sqrt{\alpha^2 - 1}}{\alpha}\frac{\sinh\bigl(\frac{\Delta}{2}\sqrt{\alpha^2 - 1}\bigr)}{\alpha - \cosh\bigl(\frac{\Delta}{2}\sqrt{\alpha^2 - 1}\bigr)} \\ &= \frac{2}{\sqrt{\alpha^2 - 1}}\sinh\biggl(\frac{\Delta}{2}\sqrt{\alpha^2 - 1}\biggr)\bar{r}\biggl(\frac{\Delta}{2}\biggr) = \frac{2\bar{r}'(\Delta/2)}{\alpha\bar{r}(\Delta/2)} \end{align}$$ and putting that in terms of $r(\phi)$, you get $$\begin{align} \int_{\phi_0 - \Delta/2}^{\phi_0 + \Delta/2} \sqrt{r(\phi)^2 + r'(\phi)^2}\mathrm{d}\phi &= \frac{GM\rho}{\lambda}\int_{-\Delta/2}^{\Delta/2} \sqrt{\bar{r}(\varphi)^2 + \bar{r}'(\varphi)^2}\mathrm{d}\varphi \\ L &= \underbrace{\frac{GM\rho}{\lambda}\bar{r}\biggl(\frac{\Delta}{2}\biggr)}_{R}\frac{2}{\sqrt{\alpha^2 - 1}}\sinh\biggl(\frac{\Delta}{2}\sqrt{\alpha^2 - 1}\biggr) \end{align}$$

So in the end, here's how to get the (or at least a) solution given endpoint radius $R$, surface separation $D$, and chain length $L$, as long as $L < 2R$.

  1. Calculate $\Delta = \frac{D}{R}$.
  2. Solve $\frac{L}{R} = \frac{2}{\sqrt{\alpha^2 - 1}}\sinh\bigl(\frac{\Delta}{2}\sqrt{\alpha^2 - 1}\bigr)$ for $\alpha$.
  3. Plug $\alpha$ and $\Delta$ in to $r(\phi_0 + \Delta/2) = R$ (remember $\phi_0$ cancels out) and solve for $\lambda$.
  4. Pick $\phi_0$ to be whatever it needs to be, given your coordinate system. Probably $\phi_0 = 0$, unless you have a particular reason to choose otherwise.

Here are some sample curves for a situation representative of the Earth. The circle has a radius of $6400\ \mathrm{km}$, the endpoints are separated by $1000\ \mathrm{km}$ along the surface, and the chain lengths range from $1000\ \mathrm{km}$ to $9000\ \mathrm{km}$.

hanging chains

share|cite|improve this answer

We have had a related problem with the brachistochrone problem in Schwarzschild coordinates here. This problem here has struck me as rather odd, for this seems to refer to geodesics, where to me the . However, it has occurred to me there might be a problem here of some relevancy. Suppose we have a string in some orbit around a black hole. We then impose some condition that this string has end points attached to a fixed surface of some kind --- a Dp-brane? This string is then in an orbit around the black hole and will assume some configuration in the curved spacetime.

So to set this up we start with the line element, which gives the unit 4-velocity, $$ 1~=~\Big(1~-~\frac{R}{r}\Big) \Big(\frac{dt}{ds}\Big)^2~-~ \Big(1~-~\frac{R}{r}\Big)^{-1} \Big(\frac{dr}{ds}\Big)^2~-~r^2\Big(\frac{d\theta}{ds}\Big)^2~-~r^2sin^2\theta\Big(\frac{d\phi}{ds}\Big)^2, $$ where $R~=~2GM/r$. We identify the $dt/ds$ with the energy $E$ and do some algebra to obtain an equation of motion $$ \Big(\frac{dr}{ds}\Big)^2~=~\frac{E}{m}~-~1~-~V(r) $$ for the potential function $$ V(r)~\simeq~\frac{1}{2}\Big(\frac{R}{r}~-~\frac{L^2}{2mr^2}~+~\frac{R}{mr^2}\Big) $$ The $L$ is the angular momentum of the object in orbit.

We now use this in the potential for the computation of the shape of the string $$ U_G~=~\int_0^\Delta V(r)\lambda\sqrt{r'^2~+~r^2}d\phi $$ $\Delta~=~D/r$ and $r’~=~dr/d\phi$ and employ the Lagrange multiplier $U_L$. The Euler-Lagrange equations are applied to the $$ {\cal L}~=~(V(r)~+~L)\sqrt{r’^2~+~r^2} $$

This is somewhat artificial for it assumes the string endpoints are held apart by some fixed distance. This could be relaxed by treating the endpoints are particle geodesic or orbits. In that case the angular separation of the fixed points varies and the problem is time dependent. This might be an interesting tool to use in some problems with strings and black holes.

share|cite|improve this answer

The explicit form of catenaries in a central force field proportional to a power of $r$ has been known since the 19th centruay. They are the so-called MacLaurin spirals which were discovered much earlier by this Scottish mathematician who showed that they were orbits of planets moving in a $r^\alpha$ field. They are the curves with polar equations $r^n \cos (n \theta) =1$ and their dilations and rotations (the index $n$ is related in a simple manner to the powers of $r$ involved). This is contained in the classic treatise on special curves by Teixeira Gomes. Unfortunately, I do not have access to a copy at the moment and so cannot give a more precise reference. A modern treatment can be found in the arXiv article 1102.1579 where it is shown that these and many other properties have their origin in the fact that the curves have equations of the form $rf(\theta)=1$ where $f$ is such that $f^2+f'^2$ is proportional to a power of $f$. In this paper, a further class of curves is introduced---the MacLaurin catenaries. They are characterised by the fact that they have parametrisations of the form $(F(t),f(t))$ where $f$ is as above and $F$ is a primitive of $f$. These have similar properties, now with respect to parallel forces proportional to a power of $y$. They thus generalise the catenaries---hence the name.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.