TSny said:
Ok. But, what is the value of ##\frac{\partial \sin \theta }{\partial t}##?
The exponents on the ##(-1)## look good. And the occurrence of ##dt \wedge dr## also looks good. But the ##\sin \theta## is not correct. Things get a little complicated because we're working in spherical coordinates. From Carroll's equation (2.82) $$\left(\ast \ast F\right)_{\alpha \beta} = \frac{1}{2!}{\epsilon^{\mu \nu}}_{\alpha \beta} (\ast F)_{\mu \nu}$$
Since you know that the only nonzero components of ##\ast F## are ##(\ast F)_{\theta \phi}## and ##(\ast F)_{\phi \theta}##, convince yourself that the above equation reduces to $$\left(\ast \ast F\right)_{\alpha \beta} = {\epsilon^{\theta \phi}}_{\alpha \beta} (\ast F)_{\theta \phi}$$ So, now we have to work out ##{\epsilon^{\theta \phi}}_{\alpha \beta} ##. Write it as
$${\epsilon^{\theta \phi}}_{\alpha \beta} = g^{\theta \rho} g^{\phi \sigma} \epsilon_{\rho \sigma \alpha\beta} $$ Work this out using what you know about the metric components in spherical components and using Carroll's equation (2.69), $$\large \epsilon_{\rho \sigma \alpha\beta} = \sqrt{|g|} \, \tilde \epsilon_{\rho \sigma \alpha\beta}$$ Here, ##\large \tilde \epsilon_{\rho \sigma \alpha\beta}## is the Levi-Civita symbol defined in Carroll's equation (2.65). See what you get.Thank you. But you might want to hold that thought.
Daer TSny;
Sorry for delay. I had decided to review chapter 2 of Carroll again and it had took time.
(a) $$
d(\ast F) = \partial_t (q \sin \theta) dt \wedge d \theta \wedge d \phi + \partial_r (q \sin \theta) dr \wedge d \theta \wedge d \phi = 0
$$
heave a sigh of relief?
(b)
I took ##\ast F = A## so:
$$
(\ast A)_{\rho \sigma} = \frac{1}{p!} {\epsilon^{\mu \nu}}_{\rho \sigma} A_{\mu \nu}
$$
now I should lower the indices of ##\epsilon##, that is:
$$
(\ast A)_{\rho \sigma} = \frac{1}{2!} g^{\mu \alpha} g^{\nu \beta}\epsilon _{\mu \nu \rho \sigma} A_{\mu \nu}
$$
changinf Levi-Civita tensor with its symbol gets:
$$
(\ast A)_{\rho \sigma} = \frac{1}{2!} g^{\mu \alpha} g^{\nu \beta} \sqrt{|g|} \tilde\epsilon _{\mu \nu\rho \sigma} A_{\mu \nu}
$$
because we are working in spherical coordinate:
$$
g_{\mu \nu} = diag (1,1,r^2 , r^2 \sin^2 \theta) ; \sqrt{|g|} = r^2 \sin \theta
$$
I really do'nt know how to evaluate ##g^{\mu \alpha} g^{\nu \beta}##, formally it should raise the indices of ##A_{\mu \nu}##, but my question is still remaining...
$$
(\ast A)_{tr} = \frac{1}{2!} g^{\mu \alpha} g^{\nu \beta} (r^2 \sin \theta)(q \sin \theta)
A _{\theta \phi}= \frac{1}{2!} g^{\mu \alpha} g^{\nu \beta} (qr^2 \sin^2 \theta) dt \wedge dr
$$
That minus sign we evaluated before will be added finally.
(c) Here I can easily read E and B from elements of matrix #F_{\mu \nu}#
(d) Integration in Euclidean 3 space:
$$
\int_D \omega = \int \frac{1}{n!} \omega_{\mu_1 ... \mu_n} dx^{{\mu_1} ...{\mu_n}} = \int \omega_{1...n} dx^1 ... dx^n
$$
it means I should take integral like this:
$$
\int_V d \ast F = q\int \sin \theta d\theta d\phi = -q \phi \cos \theta
$$
I don't know how to insert the invariant measure!
but some how I am sure that it cannot be true, I suppose the answer is :
using ##I = \int \phi (x) \sqrt{|g|} d^n x##
$$
I = \int d \theta d \phi (r^2 \sin \theta)(q \sin \theta)= qr^2 \int d \theta d \phi \sin^2 \theta = \frac{qr^2\phi}{2} ( cos^2 \theta)
$$
yes?
