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I've been reading Sean Carroll's book on GR and I stumbled upon an exercise on EM using $p$-forms. I think I've solved the problem correctly but I am having problems with my answers. I'll provide the question, my answer and what is troubling me, if anyone could help me it would be much appreciated.

Assuming we are in Euclidean three-space and $$\star F=q \sin{\theta}\;\; d\theta \wedge d\phi,$$ $\star$ is the Hodge dual. We must:

(a) Evaluate $\mathrm{d} \star F=\star J$:

My calculations so that $$\mathrm{d} \star F=q \cos\theta \mathrm{d}\theta \wedge \mathrm{d}\theta \wedge \mathrm{d}\phi=0.$$ This means $\star J=0$ and also $\star \star J=J=0$ which is my main problem as I will discuss a little bit later.

(b) What is the 2-from $F$ equal to.

Here applying the dual one more time we have $\star \star F=- F$ and so $F_{\mu\nu}=-\star \star F= -\frac{1}{\sqrt{-|g|}} \epsilon^{23}\phantom{}_{\mu\nu} q\sin\theta$. This is non-zero only in the $tr$ component so $F_{tr}=-\frac{q}{r^2}$. But this is equal to the Electric field in the $r$ direction so $\vec{E}= -\frac{q}{r^2} \hat{e}_r $.

My problem is that this is the field of a point charge at the origin, so I should have found $J^0=\rho=q\;\delta(r-0)$. My answer shows that there is a field but no source, and I can't pinpoint where my mistake lies.

For completion (c) said to find the Electric and Magnetic fields $B$ is zero, E is mentioned above.

(d) Calculate $\int_V \mathrm{d} \star F$. Using Stokes theorem this is $\int_S \star F=\int\limits_0^{2\pi}\int\limits_0^\pi q\sin\theta d\theta d\phi=4\pi\;q$. As expected by a point charge.

But this shows $\int \star J=4\pi\;q$, so $J$ isn't actually 0...This leads me to believe I did a mistake in the calculation of $\mathrm{d} \star F$.

Edit: I think my mistake is that $dr,\;d\theta,\;d\phi$ is not an orthonormal basis. So the exterior derivative I calculated is wrong.

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  • $\begingroup$ Which exercise? Which page? $\endgroup$
    – Qmechanic
    Commented Mar 16 at 16:02
  • $\begingroup$ @Qmechanic Sorry about that I should have mentioned it. It's Chapter 2, exercise 9, page 92. $\endgroup$ Commented Mar 16 at 16:11
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    $\begingroup$ You have just proved that punctured space has nontrivial second cohomology group. Congratulations. $\endgroup$ Commented Mar 17 at 2:39

2 Answers 2

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The first thing you need to note is that your $\star F$ is only defined on a specific subset, $\Omega_0$, of $\Bbb{R}\times\Bbb{R}^3$, where $\Omega_0$ is the cartesian product of $\Bbb{R}$ with the domain of the spherical coordinate chart $(r,\theta,\phi)$. This domain excludes quite a bit of stuff (e.g half a half a meridian from each sphere of radius $r>0$). In particular, the origin of $\Bbb{R}^3$ is not a part of this domain, so when you calculated $d(\star F)$, you should be mindful that you have only calculated it a-priori on $\Omega_0$.

Ok, this still isn’t the main issue, because the form $\star F$ as you have written can be written more globally as $q\cdot \frac{x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy}{r^2}$, where $r^2:=x^2+y^2+z^2$. Now, with this formula we see that $\star F$ is defined and smooth on $\Omega_1:=\Bbb{R}\times (\Bbb{R}^3\setminus\{0\})$, and since $\Omega_0$ is dense in $\Omega_1$, it follows that $d(\star F)$ vanishing on $\Omega_0$ implies it vanishes on $\Omega_1$. So, indeed, $\star J$, and thus $J$, vanishes on $\Omega_1$.

Next, you have indeed correctly computed the Hodge dual $F=-\star\star F=-\frac{q}{r^2}\,dt\wedge dr$; again in deriving this formula you a-priori assumed yourself to be working on the domain $\Omega_0$, but again you can convince yourself (by smoothness of everything) this holds on the larger domain $\Omega_1$.

But now is where you need to be extremely mindful. Note carefully that $\Omega_1\neq\Bbb{R}\times\Bbb{R}^3$, and that all the forms $F,\star F, J,\star J$ have singularities at the set of points with $r=0$. So, none of the above calculations can be extended beyond the domain $\Omega_1$. Therefore, you will NOT be able to recover a Dirac delta expression simply by fiddling around with $d(\star F)$ in this manner. If you want to properly treat this, then you need to treat it distributionally right from the beginning, then you need to compute exterior derivatives in the distributional sense (which is essentially defined by ensuring that Stokes’ theorem holds). Once you calculate things (i.e introduce your test-forms, and define a functional by integrating against them etc), you’ll recover your beloved Dirac delta.


Further Remarks.

At this point I should point out that Carroll is really being quite sloppy (I guess intentionally so) with his prompt in (d), because he’s applying Stokes’ theorem in the classical sense on a domain where it is not applicable. This is one of the biggest reasons why confusion arises (for students especially) in Physics. What he really wants to do is treat the exterior derivative distributionally, but he’s not being explicit about it (because obviously he doesn’t introduce distributions formally).

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  • $\begingroup$ Thank you very much for this insightful answer. I see that I couldn't have gotten the result at r=0 cause it was never part of the domain I could differentiate. You are correct the main reason I jumped to the assumption I forgot some term in the derivative is that the integral was non-zero. Again thanks for your time and the clarification. $\endgroup$ Commented Mar 16 at 22:24
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There are no big mistakes. Even without the differential form/Hodge dual formalism we always knew that the Coulomb potential leads to an electric field that is divergence free except in the singularity at zero. But since Gauss' theorem should hold there is a Dirac delta at zero to fix that. From a rigorous mathematical stand point nothing can be said about $r=0$ because it has no polar coordinates. In particular $J=0$ except in the origin.

Few things I want to highlight:

  • Sean Carroll writes that we are in Minkowski space (not Euclidean space) which determines how we calculate Hodge duals. In spherical polar coordinates for the space components we therefore have the metric $$g =\begin{pmatrix}-1&0&0&0\\ 0&1&0&0\\ 0&0&r^2&0\\ 0&0&0&r^2\sin^{2}\theta \end{pmatrix}\,. $$

  • A Hodge dual requires an orthonormal basis and to get that we will inevitably introduce a singularity at $r=0\,.$ The orthonormal coordinate basis is $\{\partial_t,\partial_r,\frac1r\partial_\theta,\frac1{r\sin\theta}\partial_\phi\}$ and the orthonormal dual basis is $\{dt,dr,r\,d\theta,r\sin\theta\,d\phi\}\,.$ Therefore, $$\tag{1} \star\Big(r\,d\theta\,\wedge\, (r\sin\theta\,d\phi)\Big)=dt\wedge dr\,. $$ Writing $\star F=q\sin\theta\,d\theta\wedge d\phi$ in the dual orthonormal basis gives $$ \star F=\frac q{r^2}\Big(r\,d\theta\,\wedge\, (r\sin\theta\,d\phi)\Big)\,. $$ By (1) $$ \star\star F=\frac q{r^2}\,dt\wedge dr\,. $$ Therefore, $F$ is the two-form $$ F=-\frac q{r^2}\,dt\wedge dr\,. $$ This is an electric field with Coulomb potential. The magnetic field is zero.

  • It is also quite obviously true that $d\star F=0$ so that, by Maxwell, $\star J=0\,.$ But, as I said above, $\star\star J=-J=0$ except at $r=0$ since we work with polar coordinates.

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  • $\begingroup$ Thank you for taking the time to answer. I am quite new to this formalism and my first reaction was to think i missed something somewhere and completely forgot about the singularity of the coordinates. I have one more question, I see that taking the Hodge dual on the orthonormal basis is more convenient. But can I always use the (2.82) relation for the components $\star A_{\mu_1 \cdots \mu_{n-p}}=\frac{1}{p!} \epsilon^{\nu_1\cdots\nu_p}_{\mu_1\cdots\mu_{n-p}}A_{\nu_1\cdots\nu_p}$ in some general $dx^\mu$ basis and get the correct answer? Or I happened to get the correct result this time? $\endgroup$ Commented Mar 16 at 22:14
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    $\begingroup$ @user20046481 Carroll mentions very briefly right after (2.82) that that this expression depends on the metric since we had to raise indices of the Levi-Civita symbol. Working this out leads to the formula that you have used in OP with $1/\sqrt{-|g|}\,.$ It was not an accident that it worked this time. :) $\endgroup$
    – Kurt G.
    Commented Mar 17 at 7:48

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