Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am using the Package RGTC to do some calculations in Supergravity. It allows to define differential forms after specifying a co-frame. I am working in with the 10d coordinates $\{t,x,y,z,r,\phi_1,\ldots,\phi_5\}$ and the coframe $\{dt,dx,dy,dz,d\phi_1,\ldots,d\phi_5\}$ and have the 5-form $F_5=-\frac{4L^2}{(1+\frac{L^4}{r^4})^2 r^5}\sqrt{R^4+r_0^4}(1+\star)dt\wedge dx\wedge dy\wedge dz\wedge dr$
where $R$ and $r_0$ are constants. Computing the Hodge dual in the expression is not a problem, since there is a function Hstar[_] in the package that does the job. What I need to compute is $F_{MPQRS}F_N{}^{PQRS}$, i.e. I need to extract the components from the differential form in order to perform the contractions.

The actual code is a bit cryptic, because the space-time I am considering is a bit complicated, but I will include it anyways:

J is the Jacobian for changing from Cartesian to spherical coordinates, xSugra my set of coordinates and g3branes the metric tensor

<< EDCRGTCcode.m
J = D[{Cos[ϕ1], Sin[ϕ1] Cos[ϕ2], 
Sin[ϕ1] Sin[ϕ2] Cos[ϕ3], 
Sin[ϕ1] Sin[ϕ2] Sin[ϕ3] Cos[ϕ4], 
Sin[ϕ1] Sin[ϕ2] Sin[ϕ3] Sin[ϕ4] Cos[ϕ5], 
Sin[ϕ1] Sin[ϕ2] Sin[ϕ3] Sin[ϕ4]\
Sin[ϕ5]}, {{ϕ1, ϕ2, ϕ3, ϕ4, ϕ5}}];
H[r_] := 1 + L^4/r^4;
f[r_] := 1 - r0^4/r^4;
xSugra = {t, x, y, z, r, ϕ1, ϕ2, ϕ3, ϕ4, ϕ5};
coFrame = {d[t], d[x], d[y], d[z], d[r], d[ϕ1], d[ϕ2], d[ϕ3],
d[ϕ4], d[ϕ5]};
d[L] = 0; d[r0] = 0;
g3branes = 
ArrayFlatten[{{H[r]^(-(1/2)) DiagonalMatrix[{-f[r], 1, 1, 1}], 0, 
 0}, {0, H[r]^(1/2)/f[r], 0}, {0, 0, 
 FullSimplify[H[r]^(1/2) r^2 Transpose[J].J]}}];
simpRules = TrigRules;
RGtensors[g3branes, xSugra, coFrame]
F5 = Simplify[-((4 L^2)/(H[r]^2 r^5)) Sqrt[
L^4 + r0^4] (d[t]⋀d[x]⋀d[y]⋀d[z]⋀d[r] +
  HStar[d[t]⋀d[x]⋀d[y]⋀d[z]⋀d[r]])]

where I have not included output.

Maybe I am lucky and there is someone who has worked with this package or a similar one before.

Cheers, physicus

share|improve this question

the function FormCoef[x_,y_] defined in the package should give you the result you want.

This function gives the "left-coefficient" of the differential form y in the differential form expression x, i.e., writes x as Wedge[w,y] + terms not containing y and returns w.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.