LOG#243. Elliptic trigonometry.

Jacobi elliptic functions allow to solve many physical problems. Today I will review briefly some features. Let me first highlight that the simple pendulum, Euler asymmetric top, the heavy top, the Duffing oscillator, the Seiffert spiral motion, and the Ginzburg-Landau theory of superconductivity are places where you can find Jacobi functions to arise.

Firstly, you can know there are three special Jacobi functions, named \mbox{sn}, \mbox{cn} and \mbox{dn}. The addition formulae for these 3 functions resembles those from euclidean or hyperbolic geometry:

(1)   \begin{equation*} \tcboxmath{\mbox{sn}(\alpha+\beta)=\dfrac{\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)+\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha)}{1-k^2\mbox{sn}^2(\alpha)\mbox{sn}^2(\beta)}} \end{equation*}

(2)   \begin{equation*} \tcboxmath{\mbox{cn}(\alpha+\beta)=\dfrac{\mbox{cn}(\alpha)\mbox{cn}(\beta)-\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)}{1-k^2\mbox{sn}^2(\alpha)\mbox{sn}^2(\beta)}} \end{equation*}

(3)   \begin{equation*} \tcboxmath{\mbox{dn}(\alpha+\beta)=\dfrac{\mbox{dn}(\alpha)\mbox{dn}(\beta)-k^2\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{cn}(\beta)}{1-k^2\mbox{sn}^2(\alpha)\mbox{sn}^2(\beta)}} \end{equation*}

and where k^2=m is the modulus fo the Jacobi elliptic function. To prove these addition theorems, we can take some hard paths. Let me define the derivatives:

(4)   \begin{equation*} \dfrac{d\mbox{dn(\gamma)}}{d\gamma}=\mbox{cn}(\gamma)\mbox{dn}(\gamma) \end{equation*}

(5)   \begin{equation*} \dfrac{d\mbox{cn(\gamma)}}{d\gamma}=-\mbox{sn}(\gamma)\mbox{dn}(\gamma) \end{equation*}

(6)   \begin{equation*} \dfrac{d\mbox{dn(\gamma)}}{d\gamma}=-k^2\mbox{sn}(\gamma)\mbox{cn}(\gamma) \end{equation*}

and where

(7)   \begin{align*} \mbox{sn}^2(\alpha)+\mbox{cn}^(\gamma)=1\\ k^2\mbox{sn}^2(\gamma)+\mbox{dn}^2(\gamma)=1\\ \mbox{dn}^2(\gamma)-k^2\mbox{cn}^2(\gamma)=1-k^2 \end{align*}

and where the initial conditions \mbox{sn}(0)=0, \mbox{cn}(0)=1, \mbox{dn}(0)=1 are often assumed. A more symmetric form of these equations can be deduced (exercise!):

(8)   \begin{equation*} \tcboxmath{\mbox{sn}(\alpha+\beta)=\dfrac{\mbox{sn}^2(\beta)\mbox{dn}^2(\alpha)-\mbox{sn}^2(\alpha)\mbox{dn}^2(\beta)}{\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha)-\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)}} \end{equation*}

(9)   \begin{equation*} \tcboxmath{\mbox{cn}(\alpha+\beta)=\dfrac{\mbox{sn}(\beta)\mbox{cn}(\beta)\mbox{dn}(\alpha)-\mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{dn}(\beta)}{\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha)-\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)}} \end{equation*}

(10)   \begin{equation*} \tcboxmath{\mbox{dn}(\alpha+\beta)=\dfrac{\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\beta)-\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)}{\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha)-\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)}} \end{equation*}

Using that

(11)   \begin{align*} \mbox{dn}^2(\gamma)-\mbox{cn}^2(\gamma)=(1-k^2)\mbox{sn}^2(\gamma)\\ \dfrac{\mbox{dn}(\gamma)+\mbox{cn}(\gamma)}{\mbox{sn}(\gamma)}=(1-k^2)\dfrac{\mbox{sn}(\gamma)}{\mbox{dn}(\gamma)-\mbox{cn}(\gamma)} \end{align*}

you can derive the third form of the addition theorem for Jacobi elliptic functions:

(12)   \begin{equation*} \tcboxmath{\mbox{sn}(\alpha+\beta)=\dfrac{\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\beta)+\mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)}{\mbox{dn}(\alpha)\mbox{dn}(\beta)+k^2\mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\beta)}} \end{equation*}

(13)   \begin{equation*} \tcboxmath{\mbox{cn}(\alpha+\beta)=\dfrac{\mbox{cn}(\alpha)\mbox{dn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)-(1-k^2)\mbox{sn}(\alpha)\mbox{sn}(\beta)}{\mbox{dn}(\alpha)\mbox{dn}(\beta)+k^2\mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\beta)}} \end{equation*}

(14)   \begin{equation*} \tcboxmath{\mbox{dn}(\alpha+\beta)=\dfrac{(1-k^2)\mbox{sn}^2(\beta)+\mbox{cn}^2(\beta)\mbox{dn}^2(\alpha)}{\mbox{dn}(\alpha)\mbox{dn}(\beta)+k^2\mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\beta)}} \end{equation*}

Finally the fourth form of the addition theorem for these functions can be found from algebra, to yield:

(15)   \begin{equation*} \tcboxmath{\mbox{sn}(\alpha+\beta)=\dfrac{\mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{dn}(\beta)+\mbox{sn}(\beta)\mbox{cn}(\beta)\mbox{dn}(\alpha)}{\mbox{cn}(\alpha)\mbox{cn}(\beta)+\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)}} \end{equation*}

(16)   \begin{equation*} \tcboxmath{\mbox{cn}(\alpha+\beta)=\dfrac{\mbox{cn}^2(\beta)\mbox{dn}^2(\alpha)-(1-k^2)\mbox{sn}^2(\alpha)}{\mbox{cn}(\alpha)\mbox{cn}(\beta)+\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)}} \end{equation*}

(17)   \begin{equation*} \tcboxmath{\mbox{dn}(\alpha+\beta)=\dfrac{\mbox{cn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)+(1-k^2)\mbox{sn}(\alpha)\mbox{sn}(\beta)}{\mbox{cn}(\alpha)\mbox{cn}(\beta)+\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)}} \end{equation*}

Du Val showed long ago that these 4 forms can be derived from a language of five 4d vectors that are parallel to each other. The vectors are

(18)   \begin{equation*} V_1=\begin{pmatrix} \mbox{sn}(\alpha+\beta)\\ \mbox{cn}(\alpha+\beta)\\ \mbox{dn}(\alpha+\beta)\\ 1 \end{pmatrix} \end{equation*}

(19)   \begin{equation*} V_2=\begin{pmatrix} \mbox{sn}^2(\alpha)-\mbox{sn}^2(\beta)\\ \mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{dn}(\beta)-\mbox{sn}(\beta)\mbox{cn}(\beta)\mbox{dn}(\alpha)\\ \mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)-\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\beta)\\ \mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)-\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha) \end{pmatrix} \end{equation*}

(20)   \begin{equation*} V_3=\begin{pmatrix} \mbox{sn}(\alpha)\mbox{cn}(\alpha)\mbox{dn}(\beta)+\mbox{sn}(\beta)\mbox{cn}(\beta)\mbox{dn}(\alpha)\\ \mbox{cn}^2(\beta)\mbox{dn}^2(\alpha)-(1-k^2)\mbox{sn}^2(\alpha)\\ \mbox{cn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)-(1-k^2)\mbox{sn}(\alpha)\mbox{sn}(\beta)\\ \mbox{cn}(\alpha)\mbox{cn}(\beta)+\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta) \end{pmatrix} \end{equation*}

(21)   \begin{equation*} V_4=\begin{pmatrix} \mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)+\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\beta)\\ \mbox{cn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)-(1-k^2)\mbox{sn}(\alpha)\mbox{sn}(\beta)\\ (1-k^2)\mbox{sn}^2(\beta)+\mbox{cn}^2(\beta)\mbox{dn}^2(\alpha)\\ \mbox{dn}(\alpha)\mbox{dn}(\beta)+k^2\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{cn}(\beta9 \end{pmatrix} \end{equation*}

(22)   \begin{equation*} V_5=\begin{pmatrix} \mbox{sn}(\alpha)\mbox{cn}(\beta)\mbox{dn}(\beta)+\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{dn}(\alpha)\\ \mbox{cn}(\alpha)\mbox{cn}(\beta)-\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{dn}(\alpha)\mbox{dn}(\beta)\\ \mbox{dn}(\alpha)\mbox{dn}(\beta)-k^2\mbox{sn}(\alpha)\mbox{sn}(\beta)\mbox{cn}(\alpha)\mbox{cn}(\beta)\\ 1-k^2\mbox{sn}^2(\alpha)\mbox{sn}^2(\beta) \end{pmatrix} \end{equation*}

Du Val also grouped the vectors V_2 to V_5 in a compact matrix \mathcal{A} invented by Glaisher in 1881:

    \[ \begin{pmatrix} s_1^2-s_2^2 & s_1c_1d_2+s_2c_2d_1 & s_1c_2d_1+s_2c_1d_2 & s_1c_2d_2+s_2c_1d_1\\ s_1c_1d_2-s_2c_2d_1 & c_2^2d_1^2-(1-k^2)s_1^2 & c_1c_2d_1d_2-(1-k^2)s_1s_2 & c_1c_2-s_1s_2d_1d_2\\ s_1c_2d_1-s_2c_1d_2 & c_1c_2d_1d_2+(1-k^2)s_1s_2 & (1-k^2)s_2^2+c_2^2d_1^2 & d_1d_2-k^2s_1s_2c_1c_2\\ s_1c_2d_2-s_2c_1d_1 & c_1c_2+s_1s_2d_1d_2 & d_1d_2+k^2s_1s_2c_1c_2 & 1-k^2s_1^2s_2^2 \end{pmatrix} \]

This matrix has a very interesting symmetry \mathcal{A}^T(\alpha,\beta)=\mathcal{A}(\alpha,-\beta). You can also define the antisymmetric tensor F_{jk}=a_jb_k-a_kb_j from any vector pair a_i, b_j. In fact, you can prove that the tensor

(23)   \begin{equation*} F_{kl}=m\varepsilon_{klpq}\mathcal{A}_{pq} \end{equation*}

where m equals to 1, k^2, 1-k^2, and the \varepsilon tensor is the Levi-Civita tensor, holds as identity between the matrix \mathcal{A}, and the division in two couples the quartets of vectors.  It rocks!

How, a refresher of classical mechanics. The first order hamiltonian Mechanics reads

(24)   \begin{equation*} \begin{pmatrix} \dot{q}\\ \dot{p} \end{pmatrix}=\begin{pmatrix} 0 & +1\\ -1 & 0\end{pmatrix}\begin{pmatrix}\dfrac{\partial H}{\partial q}\\ \dfrac{\partial H}{\partial p}\end{pmatrix} \end{equation*}

From these equations, you get the celebrated Hamilton equations

(25)   \begin{equation*} \dot{p}^i=-\dfrac{\partial H}{\partial q_i} \end{equation*}

(26)   \begin{equation*} \dot{q}^i=+\dfrac{\partial H}{\partial p_i} \end{equation*}

Strikingly similar to F_i=-\nabla_i \varphi, or \dot{p}^i=-\nabla_i U. First order lagrangian theory provides

(27)   \begin{equation*} \dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot q}\right)=0 \end{equation*}

Also, it mimics classical newtonian mechanics if you allow

(28)   \begin{equation*} \dfrac{\partial L}{\partial q}=-\dfrac{d}{dt}p \end{equation*}

There is a relation between the lagrangian L and the hamiltonian function H via Legendre transforamations:

(29)   \begin{equation*} H(q,p,t)=\sum_i p_i\dot{q}_i-L \end{equation*}

where the generalized momentum is

(30)   \begin{equation*} p_i=\dfrac{\partial L}{\partial \dot{q}_i} \end{equation*}

There is also routhian mechanics, by Routh, where you have (n+s) degrees of freedom chosen to be n q_i and s p_j, such as

(31)   \begin{align*} R=R(q,\zeta, p,\dot{\zeta},t)=p_i\dot{q}_i-L(q,\zeta,p,\dot{\zeta},t)\\ \dot{q}_i=\dfrac{\partial R}{\partial p_i}\\ \dot{p}_i=-\dfrac{\partial R}{\partial q_i}\\ \dfrac{d}{dt}\left(\dfrac{\partial R}{\partial \dot{\zeta}_j}\right)=\dfrac{\partial R}{\partial \zeta_j} \end{align*}

and where there are 2n routhian-ham-equations, and s routhian-lag-equations. The routhian energy reads off easily

(32)   \begin{equation*} E_R=R-\sum_j^s\dot{\zeta}_j\dfrac{\partial R}{\partial\dot{\zeta}_j} \end{equation*}

and

(33)   \begin{equation*} \dfrac{\partial R}{\partial t}=\dfrac{d}{dt}\left(R-\sum_j^s \dot{\zeta}_j\dfrac{\partial R}{\partial\dot{\zeta}_j}\right) \end{equation*}

Finally, the mysterious Nambu mechanics. Yoichiru Nambu, trying to generalize quantum mechanics and Poisson brackets, introduced the triplet mechanics (and by generalization the N-tuplet) with two hamiltonians H,G as follows. For a single N=3 (triplets):

(34)   \begin{equation*} \dot{f}=\dfrac{\partial(f,G,H)}{\partial(x,y,z)}+\dfrac{\partial f }{\partial t} \end{equation*}

and for several triplets

(35)   \begin{equation*} \dot{f}=\displaystyle{\sum_{a=1}^N\dfrac{\partial(f,G,H)}{\partial(x_a,y_a,z_a)}+\dfrac{\partial f }{\partial t}} \end{equation*}

 and where f=f(r_1,r_2,\cdots,r_N,t). Sometimes it is written as \dot{f}=\nabla G\times \nabla H. In the case of N-n-plets, you have

(36)   \begin{equation*} \tcboxmath{\dfrac{df}{dt}=\dot{f}=\{f,H_1,H_2,\ldots,H_{N-1}\}} \end{equation*}

and also you get an invariant form for the triplet Nambu mechanics

(37)   \begin{equation*} \omega_3=dx_1^1\wedge dx_1^2\wedge dx_1^3+\cdots+dx_N^1\wedge dx_N^2\wedge x_N^3 \end{equation*}

This 3-form is the 3-plet analogue of the symplectic 2-form

(38)   \begin{equation*} \omega_2=\displaystyle{\sum_i dq_i\wedge dp_i} \end{equation*}

The analogue for N-n-plets can be easily derived:

(39)   \begin{equation*} \tcboxmath{\omega_n=\displaystyle{\sum_j dx^1_j\wedge dx^2_j\wedge\cdots\wedge dx^n_j}} \end{equation*}

 

The quantization of Nambu mechanics is a mystery, not to say what is its meaning or main applications. However, Nambu dynamics provides useful ways to solve some hard problems, turning them into superintegrable systems.

See you in other blog post!

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