Inverse of Curved Quaternions

doug  <sweetser@alum.mit.edu>

Define the curved quaternion:

qCurved[t_, x_, y_, z_,M_:M,G_:G,C_:C] :=  (({{t Exp[-((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -x Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -y Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -z Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))]}, {x Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], t Exp[-((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -z Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], y Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))]}, {y Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], z Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], t Exp[-((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -x Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))]}, {z Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], -y Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], x Exp[((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))], t Exp[-((G M)/((C^2)(Sqrt[(x^2)+(y^2)+(z^2)])))]}}))

Show the inverse is the same as the Hamilton quaernion should M go to zero

Simplify[Inverse[qCurved[t,x,y,z]/.G->0].{1,0,0,0}]

Get the inverse in the general case:

Simplify[Inverse[qCurved[t,x,y,z]].{1,0,0,0}]

Look at a specific example:

Simplify[Inverse[qCurved[t,x,0,0]].{1,0,0,0}]

Simplify[Inverse[qCurved[t,x,0,0]/.{t->1,x->1}].{1,0,0,0}]

Trivial proof.

Simplify[qCurved[t,x,0,0].Inverse[qCurved[t,x,0,0]].{1,0,0,0}]

Look at the product of t with its inverse:

((E^((G M)/(C^2)))/(1+(E^((4 G M)/(C^2)))))(E^(-((G M)/(C^2))))

One over the norm.

Look at the product of x with its inverse:

-((E^((3 G M)/(C^2)))/(1+(E^((4 G M)/(C^2)))))(E^((G M)/(C^2)))

This is minus an exponential factor to the fourth.

Subtract these two results due to the sign of x*x

Simplify[((E^((G M)/(C^2)))/(1+(E^((4 G M)/(C^2)))))(E^(-((G M)/(C^2))))+((E^((3 G M)/(C^2)))/(1+(E^((4 G M)/(C^2)))))(E^((G M)/(C^2)))]

Bingo, bing.