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I'm changing a basis from one metric to another. When simplifying the result Mathematica doesn't seem to grasp all the possible simplification:

r1 = (R*Cos[η])/(Cosh[ρ] - Sinh[ρ]*Cos[θ]);
r2 = (R*Sinh[ρ]*Sin[θ])/(Cosh[ρ] - Sinh[ρ]*Cos[θ]);
y = (R*Sin[η])/(Cosh[ρ] - Sinh[ρ]*Cos[θ]);
difx = {dη, dρ, dθ};
dr1 = Dot[D[r1, {{η, ρ, θ}}], difx];
dr2 = Dot[D[r2, {{η, ρ, θ}}], difx];
dy = Dot[D[y, {{η, ρ, θ}}], difx];
ds = L^2/y^2 (dy^2 + dr1^2 + dr2^2 + r1^2*dψ^2 + r2^2*dϕ^2) // FullSimplify

The answer I get is:

$\frac{1}{4} L^2 \csc ^2(\eta ) \left(4 \text{d$\eta $}^2+\left(2 \text{d$\theta $}^2+\text{d$\phi $}^2\right) \cosh (2 \rho )-2 \text{d$\theta $}^2+4 \text{d$\rho $}^2+2 \text{d$\psi $}^2 \cos (2 \eta )+2 \text{d$\psi $}^2-2 \text{d$\phi $}^2 \cos (2 \theta ) \sinh ^2(\rho )-\text{d$\phi $}^2\right)$

I know that this can be simplified more by grouping all the differentials together. Why isn't Mathematica doing that?

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  • $\begingroup$ Could you show us what you would like your further simplification to look like? Would Collect[ds, difx] help? $\endgroup$
    – MarcoB
    Commented Apr 26, 2016 at 23:19
  • $\begingroup$ For example: $2d\psi^2 + 2d\psi^2 cos(2\eta) = 4 cos^2(\eta)$ $\endgroup$
    – Jasimud
    Commented Apr 26, 2016 at 23:33
  • $\begingroup$ Jasimud, perhaps I'm slow this morning, but I don't understand why that is a valid substitution. At first sight it doesn't seem to be an obvious consequence of your definitions, so how should the system know about it? $\endgroup$
    – MarcoB
    Commented Apr 27, 2016 at 12:29
  • $\begingroup$ I've missed the $d\psi^2$, so: $2d\psi^2 + 2d\psi^2cos(2\eta) = 2d\psi^2(1+cos(2\eta)) = 4cos^2(\eta) d\psi^2$, wich is only grouping terms and making some trigonometric simplification $\endgroup$
    – Jasimud
    Commented Apr 27, 2016 at 14:41
  • $\begingroup$ Simplify /@ Collect[ds, {d\[Eta], d\[Rho], d\[Theta], d\[Psi], d\[Phi]}]? $\endgroup$
    – march
    Commented Apr 27, 2016 at 15:22

1 Answer 1

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I am not sure, if this is what you are after, but try something in this direction. Provided all your definitions are evaluated:

 ds = (
   L^2/y^2 (dy^2 + dr1^2 + dr2^2 + r1^2*dψ^2 + r2^2*dϕ^2
 ) //  Simplify[#, 
     {r1 > 0, L > 0, ρ > 0}, 
     ComplexityFunction -> (Count[{#}, _[2 η]] &)
  ] &) /.   Sin[η]^2 -> 1 - Cos[η]^2;

Collect[ds, {dη^2, dρ^2, dϕ^2, dψ^2, dθ^2}]

returning

 dη^2 L^2 Csc[η]^2 + dρ^2 L^2 Csc[η]^2 - 
 1/4 dψ^2 L^2 (-2 - 2 (-1 + 2 Cos[η]^2)) Csc[η]^2 - 
 1/4 dθ^2 L^2 (2 - 2 Cosh[2 ρ]) Csc[η]^2 - 
 1/4 dϕ^2 L^2 (
     1 - Cos[2 θ] + 1/2 Cos[2 θ - 2 I ρ] + 1/2 Cos[2 θ + 2 I ρ] - Cosh[2 ρ]
 ) Csc[η]^2  

Have fun!

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