Pendulum in imaginary time

Posted on by arkajad

Today’s post needs mathematical imagination and practice. Practice is important. For instance lot of practice is needed for a successful target slinging. Like in this video below

There are many stiles of slinging. For us probably the best is is Apache style, when the stone (representing our mathematical pendulum) is making vertical circles. We will practice the math behind such slings right below. But after we are done with that, we will check what happens with the sling when time is imaginary rather than than real. The problem of the pendulum with imaginary time has been posed by John Baez as an exercise in his notes on elliptic functions. It seems that the solution of this exercise has not been published yet.

But first let us considered the stone (or a bucket of water, if you wish) making vertical circles on a string. It is clear that there are certain cases (velocities) that are “forbidden”, for instance if you do not give the bucket on the string sufficient initial velocity, the water will fall on your head from the bucket.

The same with certain amusement park rides – not all speed are safe:

We have started looking into this problem yesterday in Cosmoplanetary pendulum, and Bjab, in his comments, proposed an answer, but without providing details. I promised to look into it and to post my own solution (in fact yesterday I have made an error in my calculations). Here is what I came with today. In fact, just a while ago, I have found this problem discussed, though not from the angle we need, in CBSE Class XI Supplementary Textual Material in POhysics, Unit IV, Motion in a Vertical Circle. I am borrowing a picture from this site:

It fits perfectly our needs, except that we use letter \mu for the mass. Letter m we use for the second argument of Jacobi elliptic functions discussed in previous posts. For a pendulum on a string to work the centrifugal force \mu\dot{\theta}^2 l experienced by the rotating mass must be greater or equal to the radial component of the gravitational force:

    \[\mu\dot{\theta}^2l\geq -\mu g \cos \theta.\]

The mass \mu cancels out, and g/l = \omega^2, thus

(1)   \begin{equation*}\dot{\theta}^2\geq -\omega^2 \cos \theta.\end{equation*}

For \dot{\theta} we use Eq. (1) from Cosmoplanetary pendulum, but without fixing the value of \omega:

(2)   \begin{equation*}\dot{\theta}^2=4\omega^2(\frac{1}{m}-\sin^2\frac{\theta}{2}).\end{equation*}

Using the last two equations we arrive at the inequality

    \[4(\frac{1}{m}-\sin^2\frac{\theta}{2})\geq-\cos \theta.\]

Now

    \[\cos \theta=1-2\sin^2\frac{\theta}{2},\]

therefore

    \[4(\frac{1}{m}-\sin^2\frac{\theta}{2})\geq 2\sin^2\frac{\theta}{2}-1\]

or

(3)   \begin{equation*}6 \sin^2\frac{\theta}{2}\leq \frac{4}{m}+1.\end{equation*}

Suppose first that m<1. Then \theta varies from 0 to 2\pi and the left hand side has the largest value for \sin^2\frac{\theta}{2}=1. Therefore

    \[\frac{4}{m}+1\geq 6,\]

or

(4)   \begin{equation*}m\leq\frac{4}{5}.\end{equation*}

We are done with m<1. When m>1, we already know from the discussion in previous posts that for the highest point of the pendulum we have \sin^2\frac{\theta}{2}=\frac{1}{m}.
Therefore

    \[ \frac{6}{m}\leq \frac{4}{m}+1,\]

or

(5)   \begin{equation*}m\geq 2.\end{equation*}

It follows that the forbidden region is for m between 4/5 and 2 – as correctly predicted by Bjab in his comment yesterday.

We now come to the second issue – that of imaginary time. Here is the exercise form notes by John Baez: The Pendulum, Elliptic Functions and Imaginary Time. There we find the following “exercise”:

11. Show that making the replacement

    \[t\mapsto it\]

in Newton’s law is equivalent to reversing the sign of all forces.

In the present problem, this amounts to reversing the force of gravity, making it pull the pendulum up. But an upside-down pendulum is just another pendulum. Therefore the function \mathrm{sn}(it, k) must also be periodic as a function of t. This suggests that \mathrm{sn}(z, k), as a function of z\in\mathbb{C}, is periodic in both the real and imaginary directions. And it’s true!

So, the pendulum gives a physical explanation of the fact that elliptic functions are periodic in two directions on the complex plane!

12. Prove, as rigorously as you can, that \mathtm{sn}(z, k) is periodic in two directions. You can do this either by fleshing out the above argument, or by studying the integral in equation (3) and worrying about those branch points. In fact we have

    \[\mathrm{sn}(z + 4K, k) = \mathrm{sn}(z, k),\quad \mathrm{sn}(z + 2iK',k) = \mathrm{sn}(z, k)\]

where for0 < k < 1

    \[K=\int_0^1\frac{dy}{\sqrt{(1-y^2)(1-k^2y^2}}\]

and

    \[K'=\int_0^{1/k}\frac{dy}{\sqrt{(y^2-1)(1-k^2y^2}}.\]

It seems that the first nine exercises from these notes (for MATH 241 course) has been solved by Toby Barlets and posted. But not 11 and 12. Can we make sense of Baez’idea of “upside-down pendulum is just another pendulum” and use it to derive the formula with 2K’ imaginary period?

Who can do it?

7 thoughts on “Pendulum in imaginary time”

  1. Centripetal force has to be formed by gravity and positive (or zero) string tension.

    The critical state is when the pendulum highest position is vertical or horizontal.

    In highest/vertical position we have

    \frac{\mu v^2}{l} = \mu g

    therefore

    v^2 = gl

    and

    E_{k,max}-E_{p,max}=\frac{\mu v^2}{2}
    E_{k,max}-E_{p,max}=\frac{\mu gl}{2}

    dividing by

    E_{p,max} which is equal to 2\mu gl

    we get

    \frac{1}{m} - 1 = \frac{1}{4}

    so

    m=\frac{4}{5}.

    In highest/horizontal positon we have

    E_{k}=0
    E_{p}=\frac{E_{p,max}}{2}
    E_{k,max} = \frac{E_{p,max}}{2}

    so

    m=2

  3. From my working table:

    I am not able to give meaning to the idea from John Baez exercise. I fail to see how pendulum in imaginary time can be “equivalent” to a pendulum with gravity upside down.

    Moreover Jacobi amplitude function of imaginary argument has imaginary values….

  6. While looking for something that talks about a symmetry group for the pendulum, I found this:

    https://arxiv.org/pdf/1601.07891.pdf

    IT also discusses “doubly periodic” and imaginary time. Don’t know if it’s any clearer or not; this is all way over my head.

    Here’s the abstract:

    The solutions that describe the motion of the classical simple pendulum have been known for very long time and are given in terms of elliptic functions, which are doubly periodic functions in the complex plane. The independent variable of the solutions is time and it can be considered either as a real variable or as a purely imaginary one, which introduces a rich symmetry structure in the space of solutions. When solutions are written in terms of the Jacobi elliptic functions the symmetry is codified in the functional form of its modulus, and is described mathematically by the six dimensional coset group Γ/Γ(2) where Γ is the modular group and Γ(2) is its congruence subgroup of second level. In this paper we discuss the physical consequences this symmetry has on the pendulum motions and it is argued they have similar properties to the ones termed as duality symmetries in other areas of physics, such as field theory and string theory. In particular a single solution of pure imaginary time for all allowed value of the total mechanical energy is given and obtained as the S-dual of a single solution of real time, where S stands for the S generator of the modular group.

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