Mathematica Black hole Joule-Thomson expansion in Mathematica

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To recreate the T-P plane from the article on black hole Joule-Thomson in Mathematica, it is suggested to solve the function f(r) directly for pressure P instead of finding the roots numerically. The equations for P and temperature T can be derived from the original function, allowing for a direct plot of T versus P using ParametricPlot. The resulting graph resembles the first curve in Fig. 3a of the reference paper, but discrepancies in peak amplitude and pressure values should be cross-checked against the original source. This method simplifies the process and avoids the complexity of solving a quintic polynomial. The discussion emphasizes the importance of verifying calculations to ensure accuracy in the graphical representation.
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I am trying to recreate the ##T-P## plane of figure 3 in the article black hole Joule-Thomson in Mathematica. I am following the procedure but cannot obtain the graph. I don't know exactly where am i doing wrong. Can any one tell me how to fix it? My notebook is attached here.
 
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djymndl07 said:
I am trying to recreate the ##T-P## plane of figure 3 in the article black hole Joule-Thomson in Mathematica. I am following the procedure but cannot obtain the graph. I don't know exactly where am i doing wrong. Can any one tell me how to fix it? My notebook is attached here.
Finding the zeros of your function:$$f\left(r\right)=1-\frac{2Mr^{2}}{\left(r^{2}+q^{2}\right)^{3/2}}+\frac{8\pi Pr^{2}}{3}\tag{1}$$requires solving a quintic (5th-order) polynomial in ##r^2##. This is basically impossible in Mathematica using Solve. One alternative is to proceed numerically by using FindRoot to solve ##f(r_H)=0##.
But for the ##T## vs. ##P## graph you're interested in, why bother trying to find ##r_H## at all? Just directly solve ##f(r_H)=0## for the pressure ##P## instead:$$P\left(r_{H},q,M\right)=-\frac{3}{8\pi r_{H}^{2}}+\frac{3M}{4\pi\left(r_{H}^{2}+q^{2}\right)^{3/2}}\tag{2}$$From your notebook, the temperature ##T## is then:$$T\equiv\frac{1}{4\pi}\frac{\partial f\left(r\right)}{\partial r}\left|_{r=r_{H}}\right.=\frac{4Pr_{H}}{3}-\frac{Mr_{H}\left(2q^{2}-r_{H}^{2}\right)}{2\pi\left(r_{H}^{2}+q^{2}\right)^{5/2}}\tag{3}$$or:$$T\left(r_{H},q,M\right)=-\frac{1}{2\pi r_{H}}+\frac{3Mr_{H}^{3}}{2\pi\left(r_{H}^{2}+q^{2}\right)^{5/2}}\tag{4}$$after substituting eq.(2) into eq.(3). With the specific forms of ##P(r_H),T(r_H)## given by eqs.(2,4) you can directly plot ##T## vs. ##P## by varying ##r_H## with ParametricPlot:
1722832604856.png

This more or less looks like the first curve in Fig. 3a of your reference paper, except that the peak amplitude is significantly smaller here and the peak occurs at a smaller value of ##P##. Since I just used the equations from your notebook, you'll have to crosscheck your notebook with the paper to see why there's a discrepancy.
 

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