rmin depends on p and q in a complex way, but for a given p and q it can be solved for numerically.

I have three functions

L(p,q,rmin)

Theta(p,q,rmin)

S(p,q,rmin)

What I want to do is to plot S against L for a set of the values p,q for which Theta is fixed. Can Mathematica do this efficiently?

My code is below:

K[r_] = 1 + b^2/r^2

h[r_] = 1 - 1/r^4

gtt[r_] = -h[r]*K[r]^(-1)*(r/R)^2*(1 + b^2*r^2)*rh^2

gyt[r_] = -2*b^2*r^2*(K[r]^(-1)*(r/R)^2)*h[r]*rh^2

gyy[r_] = K[r]^(-1)*(r/R)^2*(1 - b^2*r^2*h[r])*rh^2

gzz[r_] = (r/R)^2*rh^2

grr[r_] = (R/r)^2*h[r]^(-1)

gxx[r_] = (r/R)^2*rh^2

s = (c^2 - 1)^(1/2)

ggtt[r_] = FullSimplify[c^2*gtt[r] + s^2*gzz[r]]

ggyt[r_] = FullSimplify[c*gyt[r]]

ggzt[r_] = FullSimplify[-2*c*s*(gtt[r] + gzz[r])]

ggyz[r_] = FullSimplify[-s*gyt[r]]

ggzz[r_] = FullSimplify[c^2*gzz[r] + s^2*gtt[r]]

ggyy[r_] = FullSimplify[K[r]^(-1)*(r/R)^2*(b*R)^2*(1 - h[r])]

ggrr[r_] = FullSimplify[(R/r)^2*h[r]^(-1)]

ggxx[r_] = FullSimplify[gxx[r]]

(* Note dil[r]=e^(-2\[CapitalPhi])*)

dil[r] = K[r]

c = 1.0

b = 1.0

rh = 1

R = 1

\[CapitalLambda] = 100000

xsigma[r_] = p/q*(-gtt[r]*gzz[r] + gtx[r]^2)/(-gtt[r]*gxx[r])

rsigma[r_] =

1/q/(-gtt[r]*grr[r])^(0.5)*(-gtt[r]*gzz[r] +

gtz[r]^2)^(0.5)*((-gtt[r]*gzz[r] + gtz[r]^2) (dil[r] +

p^2/(gtt[r]*gxx[r])) - q^2)^(0.5)

rmin[p_, q_?NumericQ] :=

NSolve[(-gtt[r]*gzz[r] + gtz[r]^2) (dil[r] + p^2/(gtt[r]*gxx[r])) -

q^2 == 0, r]

Theta[rmin_, p_, q_?NumericQ] :=

NIntegrate[xsigma[r]/rsigma[r], {r, rmin, \[CapitalLambda]}]/

NIntegrate[1/rsigma[r], {r, rmin, \[CapitalLambda]}]

L[rmin_, p_, q_?NumericQ] :=

2*((NIntegrate[

1/rsigma[r], {r, rmin, \[CapitalLambda]}])^2 + (NIntegrate[

xsigma[r]/rsigma[r], {r, rmin, \[CapitalLambda]}])^2)^(0.5)

S[rmin_, p_, q_?NumericQ] :=

dil[r]/q*(gtz[r]^2 - gtt[r]*gzz[r])/rsigma[r]

ParametricPlot[{L[p_, q_], S[p_, q_]}, {p, 0, 10},

AspectRatio -> 1/GoldenRatio, Frame -> True, Axes -> False]

If I wanted to do a Log plot of this with a parametric graph would this be possible in Mathematica?