Twiss Matrix and stability criterion

tricks

Unprotect[Csc,Sec];
Format[Csc[x_]]:=HoldForm[1/Sin[x]];
Format[Sec[x_]]:=HoldForm[1/Cos[x]];
Protect[Csc,Sec];

here we go

mp = {{c,s},{c',s'}};MatrixForm[mp]

mJ = {{α,β},{-γ,-α}};MatrixForm[mJ]

mI=IdentityMatrix[2];MatrixForm[mI]

mTwiss = Cos[μ] *mI + mJ*Sin[μ];MatrixForm[mTwiss]

γrule = Solve[Det[mTwiss]==1,γ]//Simplify//Flatten

Step by Step to ((Subscript[M, p])^N)

mTwiss//MatrixForm

m2 = Simplify[TrigReduce[mTwiss.mTwiss],Det[mTwiss]==1];MatrixForm[m2]

m3 = FullSimplify[TrigReduce[m2.mTwiss] /. γrule,{Det[m2]==1, Det[mJ]==1}];MatrixForm[m3]

We make a proof by complete induction

(Subscript[mn, n_]) = {{Cos[n μ]+α Sin[n μ],β Sin[n μ]},
	{-γ Sin[n μ],Cos[n μ]-α Sin[n μ]}};MatrixForm[(Subscript[mn, n])]

mnext = FullSimplify[TrigReduce[mTwiss.(Subscript[mn, n])],{Det[mn]==1, -(α^2)+β γ ==1}];MatrixForm[mnext]

Simplify[mnext==(Subscript[mn, n+1])]

Tr[mTwiss]

Show that Tr((Subscript[M, p])(s)) is independent of s

This is the statement, that the Trace of the matrix is conserved under similarity transformations..

Simplify[Trace[Inverse[ma].mb.ma]==Trace[mb]]

Note : the Trace of (Subscript[M, p]) and thus μ  is independent of s, the matrix elements are NOT. 
The coefficients β(s), α(s) and γ(s) and μ  are called “Beta functions” or “optical functions” of the ring. 

The two transverse planes are (normally) independent, that is we have (Subscript[μ, x]), (Subscript[β, x]), (Subscript[α, x]), (Subscript[γ, x])  and  (Subscript[μ, y]), (Subscript[β, y]), (Subscript[α, y]), (Subscript[γ, y])