Differential Forms Integration Exercise

[kyp4]

Hi all, I posted this awhile back in the homework sections of the forums and received only one reply, which suggested that I post it here instead, though I understand that it belongs in the homework section. The fundamental problem is not this particular exercise but about integration of differential forms in general. But this is best illustrated in an example and, for simplicity, I just copied the original post:

Homework Statement

The problem I am facing is 2.9 in Sean Carroll's book on general relativity (Geometry and Spacetime) I should note that I am not studying this formally and so a full solution would not be unwelcome, though I understand that forum policy understandably prohibits it. The full problem is stated as

In Minkowski space, suppose that *F = q \sin{\theta} d\theta\wedge d\phi.

a.) Evaluate d*F=*J
b.) What is the two-form F equal to?
c.) What are the electric and magnetic fields equal to for this solution?
d.) Evaluate \int_V d*F where V is a ball of radius R in Euclidean three-space at a fixed moment of time.

Homework Equations

In the above the asterisk denotes the Hodge dual and the d denotes the exterior derivative. The definitions of these operators should be well known.

The Attempt at a Solution

I think I have the first three parts solved. For part a.) I arrived at the lengthy result of

<br /> \frac{1}{2}[\partial_\mu(*F)_{\nu\rho} - \partial_\nu(*F)_{\mu\rho} + \partial_\nu(*F)_{\rho\mu} - \partial_\rho(*F)_{\nu\mu} + \partial_\rho(*F)_{\mu\nu} - \partial_\mu(*F)_{\rho\nu}] = \epsilon^\sigma_{\,\,\mu\nu\rho}J_\sigma<br />

For part b.) I got (with the help of my TI-89) the two-form, in matrix form,

<br /> F = -(**F) = \left[<br /> \begin{array}{cccc}<br /> 0 &amp; \frac{q}{r^4 \sin{\theta}} &amp; 0 &amp; 0\\<br /> \frac{-q}{r^4 \sin{\theta}} &amp; 0 &amp; 0 &amp;0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right]<br />

and for c.) I arrived at, directly from part b.),

E_r = \frac{-q}{r^4 \sin{\theta}}
E_\theta = 0
E_\phi = 0

B_\mu = 0

For \mu=1,2,3.

Verification of these would be appreciated but I am especially confused about the last part d.). In particular the integrand is a three-form and I have no idea how to integrate this and how to transform the integral into a familiar volume integral that can be computed by standard multivariable calculus.

 

[Orodruin]

For (a) I would answer in terms of $F$ rather than in terms of $*F$, which makes things much simpler. Note that differential forms by definition are anti-symmetric so the six terms you have in the left-hand side reduce to three terms. Anyway, taking the hodge of the equation, you should find the remaining set of Maxwell's equations that are not covered by $dF = 0$, i.e., $\partial_\mu F^{\mu\nu} = J^\nu$ (with possibly a sign depending on some conventions).

In terms of the integration in (d), your region of spacetime is a space-like 3-surface and so what can be integrated over it are 3-forms. In particular $d*F$ is a 3-form and can be integrated over it. In general, the integral of a $p$-form $\omega$ over a $p$-dimensional region $S$ parametrised by some parameters $t_a$ is given by
$$
\int_{S} \omega = \int_{S_t} \omega(\dot \gamma_1,\ldots,\dot\gamma_p) dt_1\ldots dt_p,
$$
where $S_t$ is the region in the parameters $t_a$ that describe the region $S$ and $\dot\gamma_k$ is the tangent vector to the curve parametrised with $t_k$ (while keeping all other $t_a$ constant).

You would go about doing this integral much as you do any surface integral: First parametrise the surface. In your case the surface is given by $t = t_0$ and you can parametrise the rest using your spherical coordinates with the region being defined by $r < R$. However, it will be easier to perform the integral after applying Stokes' theorem
$$
\int_\Omega d\omega = \int_{\partial\Omega} \omega.
$$
In your case, you would use $\omega = *F$ and find that
$$
\int_V d*F = \int_S *F,
$$
where $S$ is the boundary of your ball, i.e., it is described by $t = t_0$ and $r = R$ and parametrised by $\theta$ and $\varphi$ such that
$$
0 < \theta < \pi, \quad 0 < \varphi < 2\pi.
$$
This leads to
$$
\int_V d*F = \int_{\theta = 0}^\pi \int_{\varphi = 0}^{2\pi} *F(\partial_\theta,\partial_\varphi) d\theta\,d\varphi.
$$
Since your particular form of $*F$ is $*F = q\sin(\theta)d\theta \wedge d\varphi$, it holds that
$$
*F(\partial_\theta,\partial_\varphi) = q\sin(\theta)
$$
and therefore
$$
\int_V d*F = q \left[\int_0^\pi \sin(\theta) d\theta\right] \left[\int_0^{2\pi} d\varphi\right]
= q[2][2\pi] = 4\pi q.
$$

Note that, since $d*F = *J$, it generally holds that
$$
\int_{\partial V} *F = \int_V *J.
$$
If $V$ is in a single moment in time, then the right-hand side essentially reduces to the integral of $J_0$ over the volume. This is just Gauss' law in electrodynamics on integral form (which should not come as a surprise since $d*F = *J$ includes Gauss' law on differential form ...).

 

[axolotl]

The phrasing of the exercise is a bit strange for parts (a) and (d). If you take the derivative of
$*F = q \sin\theta d\theta \wedge d\phi$, you get $d*F = q \cos\theta d\theta \wedge d\theta \wedge d\phi = 0$. So (a) and (d) become trivial.

For (b), you made a mistake. $F_{tr}$ should be $-\frac{q}{r^2}$. I guess that you used the Levi-Civita symbol, instead of the Levi-Civita tensor, to calculate $**F$. The latter would give an extra factor of $\sqrt{|g|} = r^2 \sin \theta$. So the correct answer to (c) is that we are simply dealing with a point charge $q$ at the origin.

 

[PeroK]

The phrasing of the exercise is a bit strange for parts (a) and (d). If you take the derivative of
$*F = q \sin\theta d\theta \wedge d\phi$, you get $d*F = q \cos\theta d\theta \wedge d\theta \wedge d\phi = 0$. So (a) and (d) become trivial.

For (b), you made a mistake. $F_{tr}$ should be $-\frac{q}{r^2}$. I guess that you used the Levi-Civita symbol, instead of the Levi-Civita tensor, to calculate $**F$. The latter would give an extra factor of $\sqrt{|g|} = r^2 \sin \theta$. So the correct answer to (c) is that we are simply dealing with a point charge $q$ at the origin.

The original post is nearly 13 years old and the original poster long gone!