Moving to imaginary time

Posted on by arkajad

Before moving to imaginary time let us warm ourselves up first doing a small exercise while still in real time. We did not quite finish our business with periods of elliptic functions in the last post “Periods of Jacobi elliptic functions – Part 1.” We did not answer the question: what about the periods for m>1?

We know (see The case of inverted modulus – Treading on Tiger’s tail and Jacobi elliptic cn and dn) that for m>1 we have the formulas

(1)   \begin{eqnarray*} \sn(u,m)&=&\frac{1}{k}\sn(ku,\frac{1}{m}),\\ \cn(u,m)&=&\dn(ku,1/m),\\ \dn(u,m)&=&\cn(ku,1/m). \end{eqnarray*}

We know, from Periods of Jacobi elliptic functions – Part 1, that the period of \sn(u,m) for m<1 is 4K(m). Therefore, from Eq. (1) we deduce that the period of \sn(u,m) for m>1 is 4K(1/m)/k. For instance, we have \sn(4K(1/m)/k,m)=\frac{1}{k}\sn(4K(1/m),1/m)=0. But in the previous post we have shown that for m>1 we have

(2)   \begin{equation*}K(m)=\frac{K(1/m)-iK(1-1/m)}{k}.\end{equation*}

Therefore, for m>1 the period of \sn(u,m) is 4Re(K(m)). From two last equations in (1) we see that, for m>1, the period for \dn(u,m) is the same as that of \sn(u,m), and the period of \cn(u,m) is half of that.
Below is the plot of the function \dn(u,m) that shows its periodicity.

Plot of \dn(u,m) for |u|<10 and 0<m<4.

After these warm-up exercises we now open the door to the universe with the imaginary time. In fact there are several different doors leading there, and we will choose one that has been used by Alfred Cardew Dixon, MA, in his 1894 book “The Elementary Properties of the Elliptic Functions with Examples“.

Namely, we take the Eqs. (1)-(4) from Periods of Jacobi elliptic functions – Part 1 as the defining relations of the Jacobi elliptic functions.
In other words: whenever we have three functions S(v),C(v),D(v) and a constant \lambda satisfying

(3)   \begin{equation*} \frac{d}{dv}\,S= CD,\end{equation*}

(4)   \begin{equation*} S^2 +C^2=1,\end{equation*}

(5)   \begin{equation*} D^2+\lambda^2S^2=1,\end{equation*}

(6)   \begin{equation*}S(0)=0,\quad C(0)=D(0)=1,\end{equation*}

then

(7)   \begin{equation*}S(v)=\sn(v,\lambda),\,C(v)=\cn(v,\lambda),\,D(v)=\dn(v,\lambda).\end{equation*}

We are now ready for the trick invented by Jacobi, called “Jacobi Imaginary Transformation“. But before that, once we are in the land of elliptic functions, let us familiarize ourselves with the language used by some of the native population of this country. Following one Dr. Glaisher, the following notation is sometimes used in conversations in local pubs:

    \[\frac{\cn\, u}{\dn\, u}=\mathrm{cd}\, u,\quad \frac{\sn\, u}{\cn\, u}=\mathrm{sc}\, u\]

    \[\frac{\dn\, u}{\cn\, u}=\mathrm{dc}\, u,\quad \frac{1}{\sn\, u}=\mathrm{ns}\,u\]

    \[\frac{1}{\cn\, u}=\mathrm{nc}\,u,\quad \rm{etc.}\]

Now we are really ready.
It is a question of simple rules about derivatives of fractions and a simple algebra (we can use REDUCE to this end) to establish the following properties:

(8)   \begin{equation*} \frac{d}{du} \mathrm{sc}\,u =\mathrm{dc}\,u\,\mathrm{nc}\,u,\end{equation*}

(9)   \begin{equation*}\mathrm{nc}^2\,u-\mathrm{sc}^2\,u=1,\end{equation*}

(10)   \begin{equation*}\mathrm{dc}^2\,u-k'^2\mathrm{sc}^2\,u=1, \quad k'=\sqrt{1-k^2}.\end{equation*}


Here is the REDUCE code checking these relations – it produces, on output, three zeros.:

OPERATOR cn,sn,dn,sc,dc,nc;
For all u let sc(u)=sn(u)/cn(u);
For all u let dc(u)=dn(u)/cn(u);
For all u let nc(u)=1/cn(u);
For all u let DF(sn(u),u)=cn(u)*dn(u);
For all v let DF(cn(v),v)=-sn(v)*dn(v);
For all v let DF(dn(v),v)=-k2*sn(v)*cn(v);
For all u let cn(u)^2=1-sn(u)^2;
For all u let dn(u)^2=1-k2*sn(u)^2;
kp2:=1-k2;
l1:=DF(sc(u),u)-dc(u)*nc(u);
l2:=nc(u)^2-sc(u)^2-1;
l3:=dc(u)^2-kp2*sc(u)^2-1;
END;

Comparing Eqs. (8–10) with (3–5) we notice that if we set

    \[S(v)=i\mathrm{sc}\,\,u,\quad C(v)=\mathrm{nc}\, u,\quad D(v)=\mathrm{dc}\, u,\quad v=iu,\quad \lambda=ik'\]

,
then, since S(0)=0,C(0)=D(0)=1, we must have S(v)=\sn(v,k'),\,C(v)=\cn(v,k'),\,D(v)=\dn(v,k'). Therefore
\sn(iu,k')=i\mathrm{sc}(u,k), \cn(iu,k')=\mathrm{nc}(u,k), \mathrm{dn}(iu,k')=\mathrm{dc}(u,k). Bu since the relation between k and k' is symmetric, we can as well write

(11)   \begin{eqnarray*} \sn(iu,k)&=&i\mathrm{sc}(u,k'),\\ \cn(iu,k)&=&\mathrm{nc}(u,k'),\\ \mathrm{dn}(iu,k)&=&\mathrm{dc}(u,k'). \end{eqnarray*}

The functions on the right hand side have the period 4K(k'), denoted simply as 4K'. Therefore the function on the left hand side have the period 4iK'. We can use now addition formula and
calculate the functions sn\,z, \cn\,z,\dn\, z for any complex z. The period along the real direction is 4K, the period along the imaginary direction is 4iK'.

Plot of of the absolute value of \cn(x+iy,m) for m=0.8. We have 4K=4K(0.8)=9.02882, 4K'=4K(0.2)=6.63849.

Now we can navigate with ease in our complex times. In real time and in imaginary time.

15 thoughts on “Moving to imaginary time”

    1. For real argument they are all real. For imaginary argument sn is imaginary, cn and dn are real. So, whenever the two letter function has s, it is imaginary, otherwise it is real.
      For a complex argument they will all be, in general, complex.

      1. Ark,
        cn(z) is a real valued function, I suppose.
        So please present the graph not of absolute value of this function but the value itself.
        (Graph with the same domain and m as in the post).

          1. So formula (5) also relates to not real arguments?
            I didn’t think that way.

          3. Perhaps, but it will not bite you. The main points are:

            1) how to define these functions for complex z.
            2) How to prove the addition formulas for complex z

            Here is the possible answer: we use the addition formulas to define the functions for a complex argument. Then we verify that they still hold for a general complex argument.

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