Black hole Joule-Thomson expansion in Mathematica

  • Mathematica
  • Thread starter djymndl07
  • Start date
  • Tags
    Hole
  • #1
djymndl07
19
0
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.
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes djymndl07

Similar threads

Replies
0
Views
649
Replies
3
Views
2K
Replies
4
Views
509
Replies
1
Views
1K
Replies
19
Views
890
Replies
2
Views
595
Replies
4
Views
2K
Replies
2
Views
2K
Replies
3
Views
7K
Back
Top