0
$\begingroup$

I have a system of five equations with ten variables, i.e., the radial distance $r$, polar angle $\theta$; four spin vector components ($St, Sr, S\theta, S\phi$) and four momentum components ($Pt, Pr, P\theta, P\phi$). We can set $r=15$ and $\theta = \pi/2$. $EE$ and $LL$ are energy and angular momentum. We can also give numerical value to these objects, i.e., $EE=0.95$, and $LL=1.5$

Now, we have five equations with eight variables. So, we can give numerical values to any three, i.e., $Pr, Sr$, and $S\theta$.

I want to find the initial data using these five equations using Fixed Point method or Newton Raphson method. I am finding difficulties. Can anyone help me please to find the data which satisfy these equations.

    EQ1 = EE + Pt + 
       1/2 (-((2 Pϕ r^2 Sθ Sin[θ]^2)/Sqrt[
           r^4 Sin[θ]^2]) + (
          2 Pθ r^2 Sϕ Sin[θ]^2)/Sqrt[
          r^4 Sin[θ]^2]);
    EQ2 = -LL + Pϕ + 
       1/2 ((2 Pt (-1 + 2/r) r^2 Sr Cos[θ] Sin[θ])/((1 - 2/
             r) Sqrt[r^4 Sin[θ]^2]) - (
          2 Pr (-1 + 2/r) r^2 St Cos[θ] Sin[θ])/((1 - 2/
             r) Sqrt[r^4 Sin[θ]^2]) + (
          2 Pθ (-1 + 2/r) r^3 St Sin[θ]^2)/Sqrt[
          r^4 Sin[θ]^2] - (
          2 Pt (-1 + 2/r) r^3 Sθ Sin[θ]^2)/Sqrt[
          r^4 Sin[θ]^2]);
    EQ3 = 1 + (Pr^2 (-2 + r))/((1 - 2/r)^2 r) + (Pt^2 (-1 + 2/r)^2 r)/(
       2 - r) + Pθ^2 r^2 + Pϕ^2 r^2 Sin[θ]^2;
    EQ4 = (Pr (-2 + r) Sr)/((1 - 2/r)^2 r) + (Pt (-1 + 2/r)^2 r St)/(
       2 - r) + Pθ r^2 Sθ + 
       Pϕ r^2 Sϕ Sin[θ]^2;
    EQ5 = -(((-2 + r) Sr^2)/((1 - 2/r)^2 r)) + 
       SS^2 - ((-1 + 2/r)^2 r St^2)/(2 - r) - r^2 Sθ^2 - 
       r^2 Sϕ^2 Sin[θ]^2;

Using $EE = 0.95; LL = 1.5; r = 9; \theta = Pi/2; Pr = 0; Sr = 0.01; S\theta = 0.001; SS = 1;$ I simpilifed the equations and get

EQ1[Pt_, Pθ_, Pϕ_, Sϕ_] := 
  0.95` + Pt + 1/2 (-0.002` Pϕ + 2 Pθ Sϕ);
EQ2[Pt_, Pθ_, Pϕ_, St_] := -1.5` + Pϕ + 
   1/2 (0.` + 0.013999999999999999` Pt - 14 Pθ St);
EQ3[Pt_, Pθ_, Pϕ_] := 
  1 - (7 Pt^2)/9 + 81 Pθ^2 + 81 Pϕ^2;
EQ4[Pt_, Pθ_, Pϕ_ , St_, Sϕ_] := 
  0.` + 0.081` Pθ - (7 Pt St)/9 + 81 Pϕ Sϕ;
EQ5[St_, Sϕ_] := 0.9997904285714285` + (7 St^2)/9 - 81 Sϕ^2;

I was taking help from the help section, where FixedPoint[f, expr] command was used for one equation. That's why, first I was trying to to convert the system in such a way that one equation contains one variable using

Solve[{EQ1[Pt, Pθ, Pϕ, Sϕ] == 0, 
  EQ2[Pt, Pθ, Pϕ, St] == 0, 
  EQ3[Pt, Pθ, Pϕ] == 0, 
  EQ4[Pt, Pθ, Pϕ , St, Sϕ] == 0, 
  EQ5[St, Sϕ] == 0}, {Pt, Pθ, Pϕ, St, Sϕ}] 

But, there was some error. Actually I want to find the initial data which satisfy EQ1, EQ2,..., EQ5.

I shall be highly thankful if someone ca help me.

$\endgroup$
4
  • $\begingroup$ I see undefined objects EE~ and LL` there. $\endgroup$ – murray Feb 3 '20 at 15:44
  • $\begingroup$ How have you tried to use the fixed point method? Can you provide the code you have tried? $\endgroup$ – CA Trevillian Feb 3 '20 at 17:26
  • $\begingroup$ What are FF1[] etc.? $\endgroup$ – Michael E2 Feb 4 '20 at 14:16
  • $\begingroup$ These are the equations EQ1, EQ2, ... I just rename. EQ1 to FF1[ ]. $\endgroup$ – MMS Feb 4 '20 at 15:13
1
$\begingroup$

I take your preset values and turn them into rules

presets = {EE -> 0.95, LL -> 1.5, r -> 9, θ -> Pi/2, Pr -> 0, Sr -> 0.01, Sθ -> 0.001, SS -> 1}

Then

eqs = {EQ1, EQ2, EQ3, EQ4, EQ5} /. presets

Solve[eqs == {0, 0, 0, 0, 0}, {Pt, Pθ, Pϕ, Sϕ, St}]

Just one answer shown...

{{Pt -> -0.963824 - 0.0047205 I,

Pθ -> -0.0337024 - 0.132886 I,

Pϕ -> 0.120239 - 0.036884 I,

Sϕ -> -0.0581202 + 0.0901944 I,

St -> 0.392471 - 1.391 I},

...

}

$\endgroup$
1
  • $\begingroup$ I was also trying in this way, but I found these imaginary values and some error in mathematica. So, I wanted to apply Newton Raphson or fixed point method. $\endgroup$ – MMS Feb 4 '20 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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