Why do the two mouths of a wormhole have opposite charges?

[etotheipi]

It's about problem 106 and 107 in Gauge Fields, Knots & Gravity. There's a wormhole of topology $\mathbf{R} \times S^2$ on which has been defined a spherical metric $g = \mathrm{d}r^2 + f(r)^2 (\mathrm{d} \phi^2 + \sin^2{\phi} \mathrm{d} \theta^2)$ as well as a 1-form $E = (q\mathrm{d}r)/(4\pi f(r)^2)$. The first part is to figure out the "charge" of each mouth, which we can do by defining an orthonormal basis of 1-forms $\{\mathrm{d}r, f(r)\sin{\phi} \mathrm{d}\theta, f(r)\mathrm{d}\phi \}$ and an orientation $\omega = \mathrm{d}r \wedge f(r)\sin{\phi} \mathrm{d}\theta \wedge f(r)\mathrm{d}\phi$. Then $\mathrm{d}r \wedge *\mathrm{d}r = g(\mathrm{d}r,\mathrm{d}r) \mathrm{d}r \wedge f(r)\sin{\phi} \mathrm{d}\theta \wedge f(r)\mathrm{d}\phi$ implies that $*\mathrm{d}r = f(r)^2 \sin{\phi} \mathrm{d}\theta \wedge \mathrm{d} \phi$, so\begin{align*}

\int_{S^2} *E = \frac{q}{4\pi} \int_{S^2} \frac{*dr}{f(r)^2} = \frac{q}{4\pi} \int_{S^2} \sin{\phi} \mathrm{d} \theta \wedge d\phi = \frac{q}{4\pi} \int_0^{2\pi} \mathrm{d}\theta \int_0^{\pi} \sin{\phi} \mathrm{d} \phi = q

\end{align*}The confusing bit is problem 107 where he asks for a careful explanation of why, since electric field lines are flowing from one mouth to the other we'd expect one mouth to be positively $(q)$ charged and the other to be negatively $(-q)$ charged, whilst on the other hand we just showed the integral of $*E$ over any 2-sphere of constant radius is $q$.

I get that if we'd chosen the opposite orientation $f(r)^2 \sin{\theta} d\phi \wedge d\theta$ on $S^2$ then the sign of the integral would change, but there's got to be something more than that, surely? Can somebody explain the question? Thanks!

 

[PeterDonis]

I get that if we'd chosen the opposite orientation $f(r)^2 \sin{\theta} d\phi \wedge d\theta$ on $S^2$ then the sign of the integral would change

Yes. Which orientation would be the "natural" one for observers looking into each mouth from the outside? Would they be the same, or opposite?

 

[etotheipi]

Yes. Which orientation would be the "natural" one for observers looking into each mouth from the outside? Would they be the same, or opposite?

I reckon the mouth through which the field lines (flows of the vector field which is dual to $E$) enters naturally needs to be the one with the orientation which gives $-q$. But even then, due to the symmetry, there would be two possibilities for the direction of the vector field dual to $E$ in the first place, so it would still amount to an arbitrary choice right?

 

[PeterDonis]

there would be two possibilities for the direction of the vector field dual to $E$ in the first place, so it would still amount to an arbitrary choice right?

The arbitrary choice is which mouth we assign positive charge and which mouth we assign negative charge. But there is nothing arbitrary about the charges of the two mouths, as seen by observers outside their respective mouths, being of opposite sign. The question I asked in my last post was intended to indicate why the latter must be the case.

 

[etotheipi]

The arbitrary choice is which mouth we assign positive charge and which mouth we assign negative charge. But there is nothing arbitrary about the charges of the two mouths, as seen by observers outside their respective mouths, being of opposite sign. The question I asked in my last post was intended to indicate why the latter must be the case.

Thanks, I think I get it now! So I guess the point of that exercise is just to demonstrate that nothing physically changes if you re-label $q \mapsto -q$, the only important thing is that the two mouths need opposite orientations imposed on them.