Carroll chapter 2 questions 9 and 10 on Manifolds

[Pouramat]

Summary:: hodge Duality...

Does anybody has idea for solving these 2 problems?

 

[PeterDonis]

Moderator's note: Thread moved to homework forum. @Pouramat please provide some attempt at a solution.

 

[Pouramat]

9.a. I did this part and get the answer as q sin theta dr d theta d phi

 

[PeterDonis]

For equations, please use the PF LaTeX feature; a link to the LaTeX Guide is at the bottom left of the post window.

 

[TSny]

See the errata for Carroll's textbook corresponding to page 92 as given below. It fixes the wording of problem 9.

 

[Pouramat]

tnx, I

See the errata for Carroll's textbook corresponding to page 92 as given below. It fixes the wording of problem 9.

View attachment 274846
got this in my solution.

 

[Pouramat]

See the errata for Carroll's textbook corresponding to page 92 as given below. It fixes the wording of problem 9.

View attachment 274846

And did you solve it in 4-dim Minkowski space? I want to know the coorect solution and answer. tnx

 

[TSny]

And did you solve it in 4-dim Minkowski space? I want to know the coorect solution and answer. tnx

In the homework forum, we require that you show your work. We don't provide answers, but we can try to provide some guidance.

In post #3 you wrote

9.a. I did this part and get the answer as q sin theta dr d theta d phi

This is not the correct answer. Can you explain how you got this result? This will help us see how you are thinking about the problem.

Also, as @PeterDonis said, please use Latex to format your mathematical expressions, if possible.

 

[Pouramat]

Dear TSny
In part (a) I should take exterior derivatives from *F so:
$$
d\star F = (q \sin \theta) dr \wedge d \theta \wedge d \phi + (\dot q \cos \theta +q (d/dt (cos \theta) )) dt \wedge d \theta \wedge d \phi
$$
but I 'm not sure about my solution.

In part (b) I have no idea how to read F form *F. can you help me?

 

[TSny]

$$
d\star F = (q \sin \theta) dr \wedge d \theta \wedge d \phi + (\dot q \cos \theta +q (d/dt (cos \theta) )) dt \wedge d \theta \wedge d \phi
$$
but I 'm not sure about my solution.

$q$ is just a constant. So, $\dot q = 0$.

The time derivative $d/dt (\cos \theta)$ should be a partial derivative $\large \frac{\partial \cos \theta}{\partial t}$. What does this evaluate to? How did $\sin \theta$ become $\cos \theta$ in the last part of your answer?

Shouldn't there be a partial derivative with respect to $r$ in the first part of your answer?

See Example 1 here.

In part (b) I have no idea how to read F form *F. can you help me?

$F$ is related to $\ast (\ast F)$. See equation (2.83) in Carroll: $\ast \ast A \equiv \ast (\ast A) = (-1)^{s+p(n-p)}A$

So, try to work out $\ast\ast F$ by applying the Hodge star operator to $\ast F$.

 

[Pouramat]

$q$ is just a constant. So, $\dot q = 0$.

The time derivative $d/dt (\cos \theta)$ should be a partial derivative $\large \frac{\partial \cos \theta}{\partial t}$. What does this evaluate to? How did $\sin \theta$ become $\cos \theta$ in the last part of your answer?

Shouldn't there be a partial derivative with respect to $r$ in the first part of your answer?

See Example 1 here.

$F$ is related to $\ast (\ast F)$. See equation (2.83) in Carroll: $\ast \ast A \equiv \ast (\ast A) = (-1)^{s+p(n-p)}A$

So, try to work out $\ast\ast F$ by applying the Hodge star operator to $\ast F$.

Dear TSny
so the correct answer is :
$$
d \ast F = q \frac{\partial \sin \theta }{\partial t} dt \wedge d \theta \wedge d \phi
$$
and for part (b),
$$
\ast \ast F = (-1)^{1+2(2)}(q sin \theta) dt \wedge dr
$$
You are a good teacher...

 

[TSny]

Dear TSny
so the correct answer is :
$$
d \ast F = q \frac{\partial \sin \theta }{\partial t} dt \wedge d \theta \wedge d \phi
$$

Ok. But, what is the value of $\frac{\partial \sin \theta }{\partial t}$?

and for part (b),
$$
\ast \ast F = (-1)^{1+2(2)}(q sin \theta) dt \wedge dr
$$

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.

You are a good teacher...

Thank you. But you might want to hold that thought.

 

[Pouramat]

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?

 

[TSny]

(a) $$
d(\ast F) = \partial_t (q \sin \theta) dt \wedge d \theta \wedge d \phi + \partial_r (q \sin \theta) dr \wedge d \theta \wedge d \phi = 0$$

Yes. So, $d(\ast F) = 0$ at all spacetime points except possibly at the origin of the spherical coordinate system. $\sin \theta$ is not well-defined at the origin, so I don't think we can necessarily say that $d(\ast F) = 0$ at the origin. This might be relevant to the result that you will get for the electric field in part (d).

(b)
I took $\ast F = A$ so:
$$
(\ast A)_{\rho \sigma} = \frac{1}{p!} {\epsilon^{\mu \nu}}_{\rho \sigma} A_{\mu \nu}
$$

Ok

now I should lower the indices of $\epsilon$, that is:
$$
(\ast A)_{\rho \sigma} = \frac{1}{2!} g^{\mu \alpha} g^{\nu \beta}\epsilon _{\mu \nu \rho \sigma} A_{\mu \nu}
$$

This is not correct. Note that you have the index $\mu$ appearing three times on the right hand side. Likewise for $\nu$. Also, $\alpha$ and $\beta$ do not appear on the left. So, they each need to appear twice as dummy summation indices on the right side. Consider the example $T^{\mu \nu} = g^{\mu \alpha} g^{\nu \beta} T_{\alpha \beta}$. See if you can correct your expression for $(\ast A)_{\rho \sigma}$

because we are working in spherical coordinate:
$$
g_{\mu \nu} = diag (1,1,r^2 , r^2 \sin^2 \theta) ; \sqrt{|g|} = r^2 \sin \theta
$$

Yes, except for the sign of the first "1" in $diag (1,1,r^2 , r^2 \sin^2 \theta)$.

I really do'nt know how to evaluate $g^{\mu \alpha} g^{\nu \beta}$

The matrix with elements $g^{\mu \nu}$ is the inverse of the matrix with elements $g_{\mu \nu}$. Since the matrix with elements $g_{\mu \nu}$ is diagonal, the matrix with elements $g^{\mu \nu}$ will also be diagonal and the diagonal elements are $g^{\mu \mu} = 1/g_{\mu \mu}$.

 

[Pouramat]

Yes. So, $d(\ast F) = 0$ at all spacetime points except possibly at the origin of the spherical coordinate system. $\sin \theta$ is not well-defined at the origin, so I don't think we can necessarily say that $d(\ast F) = 0$ at the origin. This might be relevant to the result that you will get for the electric field in part (d).

Ok

This is not correct. Note that you have the index $\mu$ appearing three times on the right hand side. Likewise for $\nu$. Also, $\alpha$ and $\beta$ do not appear on the left. So, they each need to appear twice as dummy summation indices on the right side. Consider the example $T^{\mu \nu} = g^{\mu \alpha} g^{\nu \beta} T_{\alpha \beta}$. See if you can correct your expression for $(\ast A)_{\rho \sigma}$

Yes, except for the sign of the first "1" in $diag (1,1,r^2 , r^2 \sin^2 \theta)$.The matrix with elements $g^{\mu \nu}$ is the inverse of the matrix with elements $g_{\mu \nu}$. Since the matrix with elements $g_{\mu \nu}$ is diagonal, the matrix with elements $g^{\mu \nu}$ will also be diagonal and the diagonal elements are $g^{\mu \mu} = 1/g_{\mu \mu}$.

Ooooh yes I Know:
$$
(\ast A)_{\rho \sigma} = \frac{1}{p!} g^{\mu \alpha} g^{\nu \beta} \epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \alpha} g^{\nu \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$
so here I should put the exact elemet in $g^{\mu \alpha}$ and $g^{\nu \beta}$ and sum over?
because $g^{\mu \nu}$ is diagonal so:
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \mu} g^{\nu \nu} \tilde\epsilon_{\mu \nu \rho \sigma} A_{\mu \nu}
$$
ok, now by replacing the indices we have:
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} (r^2 \sin \theta) g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi t r} A_{\theta \phi}
$$
$$
(\ast A)_{\rho \sigma} =\frac{2}{2!} (r^2 \sin \theta) (\frac{1}{r^2}) (\frac{1}{r^2 \sin^2}) (q \sin \theta)
$$
the factor 2 comes from the change of indices $\theta$ with $\phi$ and r with t.

 

[TSny]

Ooooh yes I Know:
$$
(\ast A)_{\rho \sigma} = \frac{1}{p!} g^{\mu \alpha} g^{\nu \beta} \epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \alpha} g^{\nu \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$
so here I should put the exact elemet in $g^{\mu \alpha}$ and $g^{\nu \beta}$ and sum over?
because $g^{\mu \nu}$ is diagonal so:
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \mu} g^{\nu \nu} \tilde\epsilon_{\mu \nu \rho \sigma} A_{\mu \nu}
$$
ok, now by replacing the indices we have:
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} (r^2 \sin \theta) g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi t r} A_{\theta \phi}
$$
$$
(\ast A)_{\rho \sigma} =\frac{2}{2!} (r^2 \sin \theta) (\frac{1}{r^2}) (\frac{1}{r^2 \sin^2}) (q \sin \theta)
$$
the factor 2 comes from the change of indices $\theta$ with $\phi$ and r with t.

I believe your final result is correct even though I don't quite follow your individual steps. The subscripts $\rho \sigma$ should be $t r$ on the left side of your last two equations. Your final result will simplify greatly.

Here's another way to get this. Start with
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \alpha} g^{\nu \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$ We know that all $A_{\mu \nu} = 0$ except $A_{\theta \phi}$ and $A_{\phi \theta}$. So, we have $$(\ast A)_{\rho \sigma} =\frac{1}{2!}\left[ \sqrt{|g|} g^{\theta \alpha} g^{\phi \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\theta \phi} + \sqrt{|g|} g^{\phi \alpha} g^{\theta \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\phi \theta} \right ]$$ Using the antisymmetry properties of $A_{\mu \nu}$ and $\tilde\epsilon_{\alpha \beta \rho \sigma}$, the two terms in the bracket are equal. So, $$(\ast A)_{\rho \sigma} = \sqrt{|g|} g^{\theta \alpha} g^{\phi \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\theta \phi} $$ Since the metric tensor is diagonal, this reduces to $$(\ast A)_{\rho \sigma} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi \rho \sigma} A_{\theta \phi} $$

Therefore, the only nonzero elements of $(\ast A)_{\rho \sigma}$ are $(\ast A)_{t r} = -(\ast A)_{r t}$, where $$(\ast A)_{t r} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi t r} A_{\theta \phi} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} A_{\theta \phi} $$ which agrees with your result.

So, after simplification, what do you get for the components of the two-form $F$?

 

[Pouramat]

I believe your final result is correct even though I don't quite follow your individual steps. The subscripts $\rho \sigma$ should be $t r$ on the left side of your last two equations. Your final result will simplify greatly.

Here's another way to get this. Start with
$$
(\ast A)_{\rho \sigma} =\frac{1}{2!} \sqrt{|g|} g^{\mu \alpha} g^{\nu \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\mu \nu}
$$ We know that all $A_{\mu \nu} = 0$ except $A_{\theta \phi}$ and $A_{\phi \theta}$. So, we have $$(\ast A)_{\rho \sigma} =\frac{1}{2!}\left[ \sqrt{|g|} g^{\theta \alpha} g^{\phi \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\theta \phi} + \sqrt{|g|} g^{\phi \alpha} g^{\theta \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\phi \theta} \right ]$$ Using the antisymmetry properties of $A_{\mu \nu}$ and $\tilde\epsilon_{\alpha \beta \rho \sigma}$, the two terms in the bracket are equal. So, $$(\ast A)_{\rho \sigma} = \sqrt{|g|} g^{\theta \alpha} g^{\phi \beta} \tilde\epsilon_{\alpha \beta \rho \sigma} A_{\theta \phi} $$ Since the metric tensor is diagonal, this reduces to $$(\ast A)_{\rho \sigma} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi \rho \sigma} A_{\theta \phi} $$

Therefore, the only nonzero elements of $(\ast A)_{\rho \sigma}$ are $(\ast A)_{t r} = -(\ast A)_{r t}$, where $$(\ast A)_{t r} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} \tilde\epsilon_{\theta \phi t r} A_{\theta \phi} = \sqrt{|g|} g^{\theta \theta} g^{\phi \phi} A_{\theta \phi} $$ which agrees with your result.

So, after simplification, what do you get for the components of the two-form $F$?

Finally we get the expression for $F$ as following:
$$
(F)_{tr} = \frac{1}{r^2}
$$
We should not forget the minus sign we'd evaluated before:
$$
(\ast \ast F)_{\alpha \beta} = (-1)^{-1+2(2)} {\epsilon^{\mu \nu}}_{\alpha \beta}(\ast F)_{\mu \nu}
$$
$$
F = -\frac{1}{r^2} dt \wedge dr = \frac{1}{r^2}dr \wedge dt
$$
Can I ask your opinion about integration in part (d), you can see my attempt at solution in post #13

and a point about part (a):
$$
\partial_t(\sin \theta) = \frac{\partial \theta}{\partial t} \frac{\partial}{\partial \theta}(\sin \theta ) = \dot \theta (\sin \theta)
$$
We may write the final answer as:
$$
d(\ast F) = q \dot \theta (\sin \theta) dt \wedge d \theta
$$
Do you agree?

 

[TSny]

Finally we get the expression for $F$ as following:
$$
(F)_{tr} = \frac{1}{r^2}
$$
We should not forget the minus sign we'd evaluated before:
$$
(\ast \ast F)_{\alpha \beta} = (-1)^{-1+2(2)} {\epsilon^{\mu \nu}}_{\alpha \beta}(\ast F)_{\mu \nu}
$$
$$
F = -\frac{1}{r^2} dt \wedge dr = \frac{1}{r^2}dr \wedge dt
$$

Ok. Looks good except that the factor of $q$ is missing.

Using this result, what do you get for the electric field and magnetic field associated with $F$?

and a point about part (a):
$$
\partial_t(\sin \theta) = \frac{\partial \theta}{\partial t} \frac{\partial}{\partial \theta}(\sin \theta ) = \dot \theta (\sin \theta)
$$

In evaluating the partial derivative $\partial_t(\sin \theta)$, the variables $r$, $\theta$, and $\phi$ are held constant. So, $\partial_t(\sin \theta) = 0$.

In post #13 you had the correct result:

$$
d(\ast F) = \partial_t (q \sin \theta) dt \wedge d \theta \wedge d \phi + \partial_r (q \sin \theta) dr \wedge d \theta \wedge d \phi = 0
$$

As mentioned in post #14, the result $d(\ast F) = 0$ might not be valid at the origin ($r = 0$) of the spherical coordinate system.

Can I ask your opinion about integration in part (d), you can see my attempt at solution in post #13

In post #13 you wrote

(d) Integration in Euclidean 3 space:
$$
\int_V d \ast F = q\int \sin \theta d\theta d\phi = -q \phi \cos \theta
$$

The volume $V$ is a ball in 3-dimensional space. So, the integral on the left is over a 3-dimensional region. You then wrote that as an integral over a 2-dimensional region parameterized by $\theta$ and $\phi$. If the ball is centered on the origin, then I think that what you wrote is correct. But can you state the theorem that you used to go from a 3-dimensional integral to a 2-dimensional integral? And can you describe the 2-dimensional region that you are integrating over? The integration should be a definite integral with specific limits for the variables $\theta$ and $\phi$.

 

[Pouramat]

Using Field strength tensor we can read electrical and magnetic indices as following:
$$
F_{\mu \nu} =
\begin{pmatrix}
0 & -E_1 & -E_2 & -E_3 \\
E_1 & 0 & B_3 & -B_2 \\
E_2 & -B_3 & 0 & B_1\\
E_3 & B_2 & -B_1 & 0
\end{pmatrix}
$$
and now $F$ is :
$$ F_{\mu \nu}= q
\begin{pmatrix}
0 & \frac{1}{2r^2} & 0 & 0 \\
\frac{-1}{2r^2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 & 0 &0
\end{pmatrix}
$$
It seems that we just have $E_1 = \frac{-q}{2r^2}$ (which is correct in its dimension).

(d)About integration, I think it is sounds some how like Stokes' theorem:
$$
\int_V d\omega = \int_{\partial V} \omega
$$
which $\phi$ goes from 0 to $2 \pi$ and $\theta$ goes from 0 to $\pi$

 

[TSny]

Using Field strength tensor we can read electrical and magnetic indices as following:
$$
F_{\mu \nu} =
\begin{pmatrix}
0 & -E_1 & -E_2 & -E_3 \\
E_1 & 0 & B_3 & -B_2 \\
E_2 & -B_3 & 0 & B_1\\
E_3 & B_2 & -B_1 & 0
\end{pmatrix}
$$

The signs can be confusing here. From Carroll's equation (1.94), $F^{0i} = E^i$. Lowering the index $i$ on both sides gives $F_{0i} = E_i$ (without a negative sign on the right side). The usual way to express the components of the electric field vector $\mathbf E$ would be to use upper indices $E^{i}$. See Carroll's equation (1.93), for example. Thus, $\mathbf E = E^r \hat r + E^{\theta} \hat \theta + E^{\phi} \hat \phi$. Here, $E^r = F^{01} = F^{tr} = -F_{tr}$.

and now $F$ is :
$$ F_{\mu \nu}= q
\begin{pmatrix}
0 & \frac{1}{2r^2} & 0 & 0 \\
\frac{-1}{2r^2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 & 0 &0
\end{pmatrix}
$$
It seems that we just have $E_1 = \frac{-q}{2r^2}$ (which is correct in its dimension).

How did you get a factor of $2$ in the denominator? You should end up with a very familiar electric field.

(d)About integration, I think it is sounds some how like Stokes' theorem:
$$
\int_V d\omega = \int_{\partial V} \omega
$$
which $\phi$ goes from 0 to $2 \pi$ and $\theta$ goes from 0 to $\pi$

Yes

 

[Pouramat]

$$
F^{\mu \nu} =
\begin{pmatrix}
0 & E_1 & E_2 & E_3 \\
-E_1 & 0 & B_3 & -B_2 \\
-E_2 & -B_3 & 0 & B_1\\
-E_3 & B_2 & -B_1 & 0
\end{pmatrix}
$$
lowering indices $F_{\mu \nu} = \eta_{\mu \alpha} \eta_{\nu \beta} F^{\alpha \beta}$
in matrix language(in Minkowski 4-dim):
$$
F_{\mu \nu} =
\[\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 &1
\end{pmatrix}
\begin{pmatrix}
0 & E_1 & E_2 & E_3 \\
-E_1 & 0 & B_3 & -B_2 \\
-E_2 & -B_3 & 0 & B_1\\
-E_3 & B_2 & -B_1 & 0
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 &1
\end{pmatrix}
=
\begin{pmatrix}
0 & -E_1 & -E_2 & -E_3 \\
E_1 & 0 & B_3 & -B_2 \\
E_2 & -B_3 & 0 & B_1\\
E_3 & B_2 & -B_1 & 0
\end{pmatrix}\]
$$
$$
F_{\mu \nu} =
\begin{pmatrix}
0 & -E_1 & -E_2 & -E_3 \\
E_1 & 0 & B_3 & -B_2 \\
E_2 & -B_3 & 0 & B_1\\
E_3 & B_2 & -B_1 & 0
\end{pmatrix}
$$
and now $F$ is :
$$ F_{\mu \nu}= q
\begin{pmatrix}
0 & \frac{1}{r^2} & 0 & 0 \\
\frac{-1}{r^2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 & 0 &0
\end{pmatrix}
$$

 

[TSny]

I'm still not following how you are handling the signs. In many applications outside of GR, we use lower indices for the components of vectors. Thus, the cartesian components of the velocity of a particle might be written $V_x$, $V_y$, and $V_z$. However, in tensor analysis, we distinguish between vectors and one-forms (or dual vectors) as in sections 1.4 and 1.5 of Carroll. Components of vectors are written with upper indices, as in Carroll's equation (1.37) while the components of one-forms are written with subscripts, as in Carroll's (1.47). In Carroll, the components of the electric field vector in spherical coordinates would be denoted with upper indices as $E^r$, $E^\theta$, and $E^\phi$.

So, I think part (c) of problem 9 is asking for these upper-index quantities. And these are given by Carroll's equation (1.94): $E^i = F^{0i}$. So, for example, part of the answer to part (c) would be to give the radial component of the electric field: $E^r = F^{01} = F^{tr}$.

Recall that you found $F = - \large \frac{q}{r^2} \normalsize dt \wedge dr .\,\,\,\,$ So you know that $F_{tr} = - \large \frac{q}{r^2}$.

Therefore, to find the radial component of the electric field vector, you just need to raise the indices $t$ and $r$ in order to get $F^{tr}$.

 

[Pouramat]

I'm still not following how you are handling the signs. In many applications outside of GR, we use lower indices for the components of vectors. Thus, the cartesian components of the velocity of a particle might be written $V_x$, $V_y$, and $V_z$. However, in tensor analysis, we distinguish between vectors and one-forms (or dual vectors) as in sections 1.4 and 1.5 of Carroll. Components of vectors are written with upper indices, as in Carroll's equation (1.37) while the components of one-forms are written with subscripts, as in Carroll's (1.47). In Carroll, the components of the electric field vector in spherical coordinates would be denoted with upper indices as $E^r$, $E^\theta$, and $E^\phi$.

So, I think part (c) of problem 9 is asking for these upper-index quantities. And these are given by Carroll's equation (1.94): $E^i = F^{0i}$. So, for example, part of the answer to part (c) would be to give the radial component of the electric field: $E^r = F^{01} = F^{tr}$.

Recall that you found $F = - \large \frac{q}{r^2} \normalsize dt \wedge dr .\,\,\,\,$ So you know that $F_{tr} = - \large \frac{q}{r^2}$.

Therefore, to find the radial component of the electric field vector, you just need to raise the indices $t$ and $r$ in order to get $F^{tr}$.

I think you are right. Thanks for being such a star.

 

[vanhees71]

The signs can be confusing here. From Carroll's equation (1.94), $F^{0i} = E^i$. Lowering the index $i$ on both sides gives $F_{0i} = E_i$ (without a negative sign on the right side). The usual way to express the components of the electric field vector $\mathbf E$ would be to use upper indices $E^{i}$. See Carroll's equation (1.93), for example. Thus, $\mathbf E = E^r \hat r + E^{\theta} \hat \theta + E^{\phi} \hat \phi$. Here, $E^r = F^{01} = F^{tr} = -F_{tr}$.
How did you get a factor of $2$ in the denominator? You should end up with a very familiar electric field.

Yes

The confusing thing is that $\vec{E}$ and $\vec{B}$ are only vector components under rotations but not under the full Lorentz group. That's why one usually uses the formalism used above, i.e., the $(1/2,1/2)$-representation of the Lorentz group, i.e., the antisymmetric Faraday tensor with the corresponding components.

An alternative way is to use the Riemann-Silberstein formalism, combining the electric and magnetic field components into a complex-valued three-vector $\vec{F}=\vec{E}+\mathrm{i} \vec{B}$. This vector maps the Lorentz transformations uniquely to $\mathrm{SO}(3,\mathbb{C})$ matrices, i.e., matrices $\hat{O} \in \mathbb{C}^{3 \times 3}$ with $\hat{O} \hat{O}^{\text{T}}=\hat{1}$ and $\mathrm{det} \hat{O}=1$. It's a good exercise to show that a rotation free Lorentz boost is represented by a transformation with a purely imaginary angle. The usual rotations, of course, are represented by the usual $\mathrm{SO}(3)$ group, which naturally is a subgroup of $\mathrm{SO}(3,\mathbb{C})$.