0
$\begingroup$

I have an equation for the effective potential for a neutral particle (non-spinning), i.e.

VeffNonSPIN[r_, θ_] := (1 - 2/r)*(1 + LL^2/(r^2 Sin[θ]^2));

If I take double derivative with respect to θ, plug in a value, and simplify

Simplify[1/(r (r - 2))*D[D[VeffNonSPIN[r, θ], θ], θ] /. θ -> Pi/2]

I found the output

(2 LL^2)/r^4

This is OK.

Now, I have an equation for a spinning particle, which is a generalization of the above mentioned case, i.e.

VeffSpinP[r_, θ_] := ((((-2 + r) (-LL^2 (-3 + r) + r^3 - SS^2))/(r^3 - (-2 + r) SS^2) + (
    LL^2 (-2 + r) (r^3 (-3 + 2 r) + (-6 + r) SS^2) Sin[θ]^2)/(r^3 - (-2 + r) SS^2)^2 + (
    LL (r^3 (-5 + 2 r) - (-2 + r) SS^2) Sin[θ] Sqrt[((-2 + 
        r) ((LL - SS) (LL + SS) (-r^3 + (-2 + r) SS^2) + 
        LL^2 (r^3 + 2 SS^2) Sin[θ]^2))/(r^3 - (-2 + 
          r) SS^2)^2])/(r^4 - (-2 + r) r SS^2))/(
   Sqrt[-2 + r] Sqrt[r] Sqrt[
    1 + ((LL r^2 Sin[θ])/(r^3 - (-2 + r) SS^2) + 
       Sqrt[((-2 + r) ((LL - SS) (LL + SS) (-r^3 + (-2 + r) SS^2) + 
          LL^2 (r^3 + 2 SS^2) Sin[θ]^2))/(r^3 - (-2 + 
            r) SS^2)^2])^2]))^2;

For

AA = Simplify[1/(r (r - 2))*D[D[VeffSpinP[r, θ], θ], θ] /. θ -> Pi/2]

After simplification, in the limiting case (SS->0) my results should be for the same as for the neutral (non-spinning case) above mentioned case. But I'm not able to find the correct results; instead I'm getting infinity.

I have tried using

Limit[AA, ss -> 0]

I got the following output

DirectedInfinity[-((Sign[LL]^3 Sign[-3 + r] Sign[-2 + r]^2)/(Sign[r]^4 Sqrt[Sign[(-2 + r) (LL^2 + r^2)]/Sign[r]^5]))]/((-2 +  r)^2 r^2) 

Here I want to remove DirectedInfinity and Sign. Can anyone help me please?

I also tried using

 SS=0; Simplify[AA]

but again found indeterminate form (infinity).

Can anyone please help me how can I remove this indeterminate form.

$\endgroup$
8
  • 1
    $\begingroup$ Have you double-checked your spinning model equations? There is nothing obviously wrong with your MMA code that I can see. As a note, the double derivative can be written as D[VeffNonSPIN[r, θ], {θ, 2}] which is more compact. $\endgroup$ – MarcoB Jan 3 at 20:10
  • $\begingroup$ Yes, I have checked, I'm getting infinity answer for Limit[AA, ss -> 0], instead answer should be (2 LL^2)/r^4 as I mentioned in the question. Actually after calculating double derivatives I want to take limit for SS->0 $\endgroup$ – MMS Jan 4 at 5:14
  • $\begingroup$ For Limit[A, SS -> 0], I'm getting this answer DirectedInfinity[-((Sign[LL]^3 Sign[-3 + r] Sign[-2 + r]^2)/(Sign[r]^4 Sqrt[Sign[(-2 + r) (LL^2 + r^2)]/Sign[r]^5]))]/((-2 + r)^2 r^2) , If I use the command Simplify[Limit[A, SS -> 0], r > 3 && LL > 0] then I have - infinity. but my answer should be (2 LL^2)/r^4 $\endgroup$ – MMS Jan 4 at 5:29
  • $\begingroup$ In the general case $\lim_{x\to a} f(x) \neq f(a)$ (see textbooks on calculus). $\endgroup$ – user64494 Jan 4 at 10:00
  • 4
    $\begingroup$ I’m voting to close this question because the issue seems to be with the underlying math, rather than with the MMA code. $\endgroup$ – MarcoB Jan 4 at 22:04
1
$\begingroup$

If you try

Normal[Series[
D[VeffSpinP[r, \[Theta]] /. SS -> 0, {\[Theta], 2}] /. \[Theta] -> 
Pi/2 - \[Epsilon], {\[Epsilon], 0, 0}]]

the result is an expression without spin. If your spinning model is correct this should equal the non spinning result. Does it?

Edit:

Well I looked at the paper and the correct Potential is:

VeffSpinP[r_, \[Theta]_] := (((LL*Sin[\[Theta]])/
     r - (LL*r*Sin[\[Theta]])/(r^2 - (1 - 2/r)*SS^2) - 
    Sqrt[(LL^2*r^2*
         Sin[\[Theta]]^2)/(r^2 - (1 - 2/r)*
           SS^2)^2 - ((1 - 2/r)*(LL^2 - SS^2) + (2*LL^2*
            Sin[\[Theta]]^2)/r)/
                 (r^2 - (1 - 2/r)*SS^2)])*((LL*r*
       Sin[\[Theta]])/(r^2 - (1 - 2/r)*SS^2) + 
    Sqrt[(LL^2*r^2*
         Sin[\[Theta]]^2)/(r^2 - (1 - 2/r)*
           SS^2)^2 - ((1 - 2/r)*(LL^2 - SS^2) + (2*LL^2*
            Sin[\[Theta]]^2)/r)/
                 (r^2 - (1 - 2/r)*SS^2)]))/(Sqrt[1 - 2/r]*r*
        Sqrt[1 + ((LL*r*Sin[\[Theta]])/(r^2 - (1 - 2/r)*SS^2) + 
       Sqrt[(LL^2*r^2*
            Sin[\[Theta]]^2)/(r^2 - (1 - 2/r)*
              SS^2)^2 - ((1 - 2/r)*(LL^2 - SS^2) + (2*LL^2*
               Sin[\[Theta]]^2)/r)/(r^2 - (1 - 2/r)*
             SS^2)])^2]) + 
   Sqrt[1 - 2/r]*
Sqrt[1 + ((LL*r*Sin[\[Theta]])/(r^2 - (1 - 2/r)*SS^2) + 
     Sqrt[(LL^2*r^2*
          Sin[\[Theta]]^2)/(r^2 - (1 - 2/r)*
            SS^2)^2 - ((1 - 2/r)*(LL^2 - SS^2) + (2*LL^2*
             Sin[\[Theta]]^2)/r)/
                    (r^2 - (1 - 2/r)*SS^2)])^2];

Possibly you may further simplify.

Then you have:

FullSimplify[VeffSpinP[r, \[Theta]]^2 /. SS -> 0 /. \[Theta] -> Pi/2]
(*((-2 + r)*(LL^2 + r^2))/r^3*)

Edit 2:

I found now that your VeffSpinP is just the square of the one I gave. Both coincide at \Theta=π/2 and SS=0 with the spinless case. Why do you bother with the second derivative? If the two potentials (with and without spin) are the same at a point ( \Theta=π/2 and SS=0) it doesn't mean that their derivatives wrt. \Theta are identical.

$\endgroup$
6
  • $\begingroup$ No I'm not getting this result. I'm sure, our spinning model is correct, I have taken this equation from already published articles. This equation is 2.31 from [arxiv.org/pdf/gr-qc/9604020.pdf]. $\endgroup$ – MMS Jan 5 at 13:33
  • $\begingroup$ @ Andreas thanks for the help, actually I'm trying to calculate the frequency of the particle and in formula I have to use double derivative w.r.t \theta, and after calculations my results of spinning case should be reduced in spin-less case. I have done the same for radial co-ordinate (i.e., derivative w.r.t r), my results were reducing to spin-less case. $\endgroup$ – MMS Jan 5 at 19:50
  • 1
    $\begingroup$ @MMS With FullSimplify[ PowerExpand[ Normal[Series[ Normal[Series[ D[VeffSpinP[r, \[Theta]], {\[Theta], 2}] /. \[Theta] -> Pi/2 - \[Epsilon], {\[Epsilon], 0, 0}]], {SS, 0, 0}]], {r, SS}]] you can see why there is a problem with the second derivative for SS->0 $\endgroup$ – Andreas Jan 5 at 20:10
  • $\begingroup$ @ Andreas thanks for the help, I have tried this, this command is giving some output. But how can I see the problem for second derivative with the command you have mentioned in the comment. I'm sorry, I'm beginner, I'm not able to understand. $\endgroup$ – MMS Jan 6 at 9:02
  • $\begingroup$ @MMS There appears a fraction with SS in the denominator which then will go to infinity with SS->0 $\endgroup$ – Andreas Jan 6 at 11:17

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.