Phase transitions graph [closed]

I have temperature T in terms of r as

T = 1/(4 Pi r) + 2 P r^2


and G in terms of r as

G = r/4 - 2(Pi P r^3)/3


How can I plot G vs. T in Mathematica? Assume P = 1/(96π) or P = 0.01.

• What have you tried so far and at which specific step did you get stuck? (Please edit the question and add this information.) – Szabolcs Jan 19 at 12:02
• For P=1. Try: ContourPlot[ Evaluate@(Eliminate[{T == 1/(4*Pi*r) + 2*P*r^2, G == r/4 - 2 (Pi*P*r^3)/3}, r] /. P -> 1), {T, -1, 2}, {G, -1, 2}, FrameLabel -> Automatic, PlotPoints -> 50] – Mariusz Iwaniuk Jan 19 at 12:36
• Can you please make it for given values. i need the same as above given in figure. – Ahmed 1 Jan 19 at 12:44
• Szabolcs, kindly check now. – Ahmed 1 Jan 19 at 13:02

You want to use ParametricPlot and I recommend defining G and T as functions. Like so:

T[P_, r_] := 1/(4 π r) + 2 P r^2
G[P_, r_] := r/4 - 2 (π P r^3)/3
With[{p = 1./(96 π)},
ParametricPlot[{T[p, r], G[p, r]}, {r, 0.00001, 2. π},
AxesLabel -> (Style[#, 14, Bold] & /@ {"T", "G"}),
AspectRatio -> 1]]


• How you give the value of {r, 0.00001, 2. π}? – Ahmed 1 Jan 19 at 16:18
• @Ahmed1. There is a singularity at r = 0, so I offset the lower bound at little. The upper was chosen arbitrarily. – m_goldberg Jan 19 at 17:11