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Re: Quaternion problem--> Jacobian derivation/ Metric

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68137] Re: Quaternion problem--> Jacobian derivation/ Metric
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Tue, 25 Jul 2006 04:01:43 -0400 (EDT)
  • References: <e7ap5q$952$1@smc.vnet.net> <e7lf1d$3l9$1@smc.vnet.net> <e7no3j$97f$1@smc.vnet.net> <e9ri0i$pa1$1@smc.vnet.net> <e9sukq$rl6$1@smc.vnet.net> <ea25uv$1cn$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

 SU(3) Rodrigues Jacobian using Paul's better version:

jacobian = D[Exp[r0]*{Cos[
    r], Sin[r]*Sin[p]*Cos[t], Sin[r]*
        Sin[p]*Sin[t], Sin[r]*Cos[p], Cos[r1], Sin[
        r1]*Sin[p1]*Cos[t1], Sin[r1]*Sin[p1]*Sin[t1]}, {{r0, t, p, r,
           t1, p1, r1,}}]
MatrixForm[jacobian]
 Tr[ Transpose[jacobian] . jacobian, List] // Simplify

I
Paul Abbott wrote:

>In article <e9sukq$rl6$1 at smc.vnet.net>,
> Roger Bagula <rlbagula at sbcglobal.net> wrote:
>
>  
>
>>I did Jacobians and metrics way back in Mathematica using
>>a Jacobian matrix form from Theoretical Mechanics ,Ames and Murnagham, 
>>Dover Books.
>>    
>>
>
>D was extended to compute the Jacobian directly in version 5.1. 
>Alternatively, you can compute the Jacobian using Outer:
>
>  Outer[D, f, vars]
>
>where f is list of functions and vars is a list of variables.
>
>  
>
>>(*The hard way:  there has got to be a more compact way of doing this!*)
>>    
>>
>
>Indeed there is.
>
>  
>
>>Exp[r]*{Cos[d], Sin[d]*Sin[p]*Cos[t], Sin[d]*Sin[p]*Sin[t], Sin[d]*Cos[p]}
>>x = Exp[r]*Sin[d]*Sin[p]*Cos[t]
>>y = Exp[r]*Sin[d]*Sin[p]*Sin[t]
>>z = Exp[r]*Sin[d]*Cos[p]
>>t0 = Exp[r]*Cos[d]
>>xr = D[x, r]
>>yr = D[y, r]
>>zr = D[z, r]
>>tr = D[t0, r]
>>xt = D[x, t]
>>yt = D[y, t]
>>zt = D[z, t]
>>tt = D[t0, t]
>>xp = D[x, p]
>>yp = D[y, p]
>>zp = D[z, p]
>>tp = D[t0, p]
>>xd = D[x, d]
>>yd = D[y, d]
>>td = D[t0, d]
>>    
>>
>
>No need to explicitly define x, y, z, and t0. The Jacobian is just
>
>  jacobian = D[Exp[r] {Cos[d], Sin[d] Sin[p] Cos[t], 
>    Sin[d] Sin[p] Sin[t], Sin[d] Cos[p]}, {{r,t,p,d}}];
>
>Alternatively in Mathematica pre 5.1,
>
>  jacobian = Outer[D, Exp[r] {Cos[d], Sin[d] Sin[p] Cos[t], 
>    Sin[d] Sin[p] Sin[t], Sin[d] Cos[p]}, {r,t,p,d}];
>
>  
>
>>h1 = FullSimplify[1/Sqrt[xr^2 + yr^2 + zr^2 + tr^2]]
>>h2 = FullSimplify[1/Sqrt[xt^2 + yt^2 + zt^2 + tt^2]]
>>h3 = FullSimplify[1/Sqrt[xp^2 + yp^2 + zp^2 + tp^2]]
>>h4 = FullSimplify[1/Sqrt[xd^2 + yd^2 + zd^2 + td^2]]
>>
>>rra = Simplify[Expand[1/h1^2]]
>>tta = Simplify[Expand[1/h2^2]]
>>ppa = Simplify[Expand[1/h3^2]]
>>dda = Simplify[Expand[1/h4^2]]
>>
>>    
>>
>The square of the scale factors can be obtained directly as
>
>  Tr[ Transpose[jacobian] . jacobian, List] // Simplify
>
>Cheers,
>Paul
>
>_______________________________________________________________________
>Paul Abbott                                      Phone:  61 8 6488 2734
>School of Physics, M013                            Fax: +61 8 6488 1014
>The University of Western Australia         (CRICOS Provider No 00126G)    
>AUSTRALIA                               http://physics.uwa.edu.au/~paul
>
>  
>


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