General relativity, integration over a manifold exercise

[kyp4]

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 each part a.) I arrived at the lengthy result of

$$\frac{1}{2}[\partial_\mu(*F)_{\mu\rho} - \partial_\nu(*F)_{\mu\rho} + \partial_\nu(*F)_{\rho\mu} - \partial_\rho(*F)_{\nu\rmu} + \partial_\rho(*F)_{\mu\nu} - \partial_\mu(*F)_{\rho\nu} = \epsilon^\sigma_{\mu\nu\rho}J_\sigma$$

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

$$F = -(**F) = \left[ \begin{array}{cccc} 0 & \frac{q}{r^4 \sin{\theta}} & 0 & 0\\ \frac{-q}{r^4 \sin{\theta}} & 0 & 0 &0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array} \right]$$

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.

 

[CFDFEAGURU]

This needs to be moved to the Special and General Relativity forum.

Perhaps there is will get some attention.

Thanks
Matt

 

[kyp4]

Thanks, I will try my luck over there.