Parameters of a catenary

Question

I am trying to remind myself of variational calculus and wanted to try and derive the catenary equation, but the final equation, although corresponding to what I can find on the internet, seems incomplete. Am I misunderstanding what I am actually asking the math to do?

I started with writing the total energy of the hanging wire as

$$ E = \int_{x_1}^{x_2}{\gamma gy\sqrt{1+y'^2}dx}, $$

where $\gamma$ is the linear density.

$\gamma g$ being just a scaling factor, I need to minimize

$$ \int_{x_1}^{x_2}{y\sqrt{1+y'^2}dx} = \int_{x_1}^{x_2}{L(y,y')dx}. $$

This leads to the usual formulation of

$$ \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} = 0. $$

In this case this has yielded $1 + y'^2 - yy''= 0$. Using substitution $y'= u$ I have arrived to the solution

$$ u = \frac{dy}{dx} = \sqrt{C_1^2y^2 - 1}, $$ $$ y = \frac{1}{C_1}\cosh(C_1x + C_2), $$

which, if $C_2 = 0$, apart from $C_1$ being upside down, seems to agree with the common one. The question I have now is where is the constant which would allow me to shift the whole shape vertically. It feels to me that the final equation should look like $y = \frac{1}{C_1}\cosh(C_1x + C_2) + C_3$. Otherwise, the catenary can't go below $0$, which it should be able to do, at least in my mind, due to the positions of the edge points being arbitrary. The problem is, this equation does not fit into the initial differential equation unless $C_3 = 0$.

Is this just a case of potential energy only being defined relative to a certain "floor" height? If I write the initial equation as

$$ E = \int_{x_1}^{x_2}{\gamma g(y - y_0)\sqrt{1+y'^2}dx}, $$

mathematically I get what I want, but it feels as if something is lacking in my understanding.

If possible, please point me in the direction of a ground-up explanation. My field being electromagnetics, I haven't seen rigorous mechanics in about 5 years and it is currently making my brain generate smoke when I think about it.

Answer 1

Is this just a case of potential energy only being defined relative to a certain "floor" height?

Yes. Energy has no absolute reference zero, so every time you write an energy expression without an unknown additive constant or an explicit reference zero, you’ve implicitly assumed a reference zero.

Here, your first energy expression assumes that the catenary hangs vertically near your (arbitrary) personal reference zero for gravitational potential energy. Someone else 1 km below might wonder why you omitted the potential energy corresponding to all that height. Your insertion of $y_0$ accommodates the observer at arbitrary height and provides an additional constant for the final shape equation.

Answer 2

Your intuition is correct in that the solution for a chain hanging under its own weight should have a constant term representing vertical offset. Moreover, it should not depend on where you choose to set your reference point for potential. However, the problem that you are actually solving using your method is slightly different from the one you think.

As nothing in your formulation constrains the length of the chain, the formula represents the minimum energy for a chain of arbitrary length. The solution to this problem does depend on the reference point. When the chains are further from that point additional length contributes more to the potential energy. Therefore, the solutions must curve less.

However, the typical real-world situation is a weighted chain of fixed length. To find the general equation for such a chain, you should use a Lagrange multiplier to constrain the length of the chain. We need to set the function $$F \left ( y , y' \right ) = \int_{x_1}^{x_2} \sqrt{1 + y'^2} dx - l$$ equal to zero (here $l$ is the total length of the chain). This can be done by minimising $$E' \left ( y , y' \right ) = E \left ( y , y' \right ) - \lambda F \left ( y , y' \right ) $$ where $\lambda$ is the Lagrange multiplier. Doing so yields a similar functional to your final expression, except with $\frac{\lambda}{\gamma g}$ in place of $y+0$.