# Electromagnetic dual tensor. General Relativity

1. Feb 15, 2012

### TheSource007

1. The problem statement, all variables and given/known data
I attach a word document with the equations because I dont know how to write them on the post.
My question reads: Show that Maxwell's equations Eq (1) is equivalent to Eq (3).

2. Relevant equations
The first term of Eq 1 reads: F sub alpha beta comma gamma. That means partial of F sub alpha beta with respect to x super gamma.
The book says that Eq 1 and Eq 2 are equivalents. I am just showing Eq 2 because I know that the book's notation is confusing.
Eq 2 means the gradient of F, which is the EM tensor.
Eq (4) is the components electromagnetic tensor.
And I have calculated the components of the dual tensor *F which are Eq (5)

3. The attempt at a solution
I have written Eq (3) in matrix notation, but I dont think it resembles Eq (1) or Eq (2). Also, I know that Eq 2 is a 3rd rank tensor due to the gradient.
I have no idea how to show the equivalency between 1 and 3.
The book Im using is Mister, Thorne, Wheeler Gravitation, but this is a undergraduate level course in general relativity, and I am very new at this tensor stuff, so please be gently.

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Last edited: Feb 15, 2012
2. Feb 15, 2012

### jambaugh

Here, I type-set them for you...
$$F_{\alpha\beta,\gamma} + F_{\beta\gamma,\alpha} + F_{\gamma \alpha,\beta} = 0$$
$$\mathbf{dF} = 0$$
$$\mathbf{\nabla}\bullet {}^\star \mathbf{F}=0$$
$$[F_{\alpha\beta}] = \left[ \begin{array}{c c c c}0 & -E_x & -E_y & -E_z\\ E_x & 0 & B_z &-B_y\\ E_y & -B_z & 0 & B_x\\ E_z & B_y & -B_x 0\end{array}\right]$$
$$[{}^\star F_{\alpha\beta}] = \left[ \begin{array}{c c c c}0 & B_x & B_y & B_z\\ -B_x & 0 & E_z &-E_y\\ -B_y & -E_z & 0 & E_x\\ -B_z & E_y & -E_x 0\end{array}\right]$$

3. Feb 15, 2012

### jambaugh

Your post didn't complete the sentence so I'm not clear on what is your question. But if you want to show equivalence of the first three equations, you have to expand the notation.

$$\mathbf{d} \omega = \mathbf{dx}^\mu \wedge \partial_\mu \omega$$

To translate the matrix format to differential forms, you contract the matrix with the bivector basis:
$$\mathbf{F} =\frac{1}{2} F_{\mu\nu} \mathbf{dx}^\mu\wedge \mathbf{dx}^\nu$$
The factor of 1/2 is not vital since these are homogenous equations but are there to account for the double counting of e.g. $\mathbf{dx}^1\wedge \mathbf{dx}^2=- \mathbf{dx}^2 \wedge \mathbf{dx}^1$

(to properly sum you should be summing, not over each index, but over the set of unique anti-symmetric pairs.... or throw in the factor of 1/2 to get the same thing.)

You now need to parse through the definitions and see what results.

4. Feb 15, 2012

### jambaugh

A second observation. Note the first is a rank 3 tensor equation (three free indices).
The second is a rank-3 tensor equation (the outer differential of a bi-vector is a tri-vector), and the last is a vector equation (the divergence of a 2-tensor is a 1 tensor).

Given each case will be a totally anti-symmetrized tensor, one can count the dimensions of these tensor spaces. Given totally antisymmetized indices in 4 dimensions you have 4 choose r ways to select distinct indices so the dimensions are:
rank 0 (scalar) dim=1
rank 1 (vector) dim=4
rank 2 (bi-vector) dim=6
rank 3 ("tri-vector") dim = 4
rank 4 (The Levi-Civita tensor) dim = 1

The hodge dual swaps these equal dimension tensor spaces.
So you should find each of your equations is a set of 4 equations, either 4-vector or "tri-vector"=rank 3 antisymmetric tensor.

5. Feb 15, 2012

### TheSource007

I updated my post.
Eq 2 means the gradient of the tensor F.
Does your equation dw means the same? Because it seems to me as the differential, and w being a scalar....
As I said (in my update) I am very new at this.
Thank you.

Last edited: Feb 15, 2012