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Maybe it's a very basic physics question but I'm confused and not sure if I should ask my question here. From this paper, I have the potential vector

$$ A=A_t d t+A_\phi(d \phi-\omega d t) $$

where

$$ \begin{aligned} & A_t=-\frac{Q}{r}+\frac{3}{4} Q B^2 r\left(1+f \cos ^2 \theta\right), \\ & A_\phi=\frac{2}{B}-H^{-1}\left[\frac{2}{B}+\frac{1}{2} B\left(r^2 \sin ^2 \theta+3 Q^2 \cos ^2 \theta\right)\right] \end{aligned} $$

and I want to calculate the components of the Faraday tensor. To do this, I took the potential vector in the following way

$$ A_{\mu} = (A_t - \omega A_{\phi}, 0, 0, A_{\phi}) $$

$$ f = 1 - \frac{2 M}{r} + \frac{Q^{2}}{r^{2}} $$

and in Mathematica I wrote

Clear["Global`*"]
coords = {t, r, \[Theta], \[Phi]};
At = -(Q/r) + 
    3/4*Q*B^2*r*(1 + f*(Cos[\[Theta]])^2) /. {f -> 
     1 - (2*M)/r + Q^2/r^2};
Aphi = 2/B - (H)^-1*(2/B + 
      1/2*B*(r^2*(Sin[\[Theta]])^2 + 3*Q^2*(Cos[\[Theta]])^2));
A = {At - Aphi *\[Omega], 0, 0, Aphi};
F = Table[
   D[A[[j]], coords[[k]]] - D[A[[k]], coords[[j]]], {j, 1, 4}, {k, 1, 
    4}];

And the items I get are not what appears in the article. For example, according to the article, it should have the following components (Eq. 8, 9, 10 and 11)

$$ \begin{aligned} & F_{r t}=-\frac{Q}{r^2}+\frac{3}{4} Q B^2\left(1+\left(1-\frac{Q^2}{r^2}\right) \cos ^2 \theta\right) \\ & F_{\theta t}=-\frac{3}{2} Q B^2 r\left(1-\frac{2 M}{r}+\frac{Q^2}{r^2}\right) \sin \theta \cos \theta \end{aligned} $$

I do not write the $F_{r \phi}$ and $F_{\theta \phi}$ components because they are very extensive.

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1 Answer 1

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You can do it mimicking exterior calculus with a 1-form for A

$$A =\text{dt}\ \left(\frac{3}{4} B^2\ Q \ r *\left(f \cos ^2(\theta )+1\right)-\frac{Q}{r}\right)\ + \ \text{d$\phi $} \left(\frac{2}{B}-\frac{1}{H}* \left(\frac{1}{2} B \left(3 Q^2 \cos ^2(\theta )+r^2 \sin ^2(\theta )\right)+\frac{2}{B}\right)\right)$$

Differentiate by Dt in Mathematica and replace products of d's by wedge products

F = Expand[
Dt[A,  Constants -> {Q, B, H, f, dt, d\[Phi]}] ] /. 
 {u___*  x_?(MemberQ[ {dt, dr, d\[Theta], d\[Phi]}, #] &)*
  HoldPattern[Dt[r_, _]] :> u x\[Wedge]\[DifferentialD]r } // 
 Expand

The result can be ordered by

  FullSimplify /@ Collect[ F, Union@Cases[F, _Wedge, \[Infinity]]] 

$$F=\frac{1}{4} Q\ \left(3 B^2 f \cos ^2(\theta )+3 B^2+\frac{4}{r^2}\right) \ \text{dt}\wedge dr -\frac{1}{2} 3 B^2 f Q r \sin (\theta ) \cos (\theta )\ \text{dt}\wedge d\theta $$ $$+\frac{B}{H} \sin (\theta ) \cos (\theta ) \left(3 Q^2-r^2\right) \ \text{d$\phi $}\wedge d\theta -\frac{B}{H} r \sin ^2(\theta ) \ \text{d$\phi $}\wedge dr$$

The signs are governed by the rule, that the coefficients are differentiates leftmost, setting the the original italics d's on right and the diffential straight d's on the left.

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