Electromagnetic tensor - invariant

[parton]

Homework Statement

Hi,

I have to calculate the invariant: $$\tilde{F}^{\mu \nu} \, F_{\mu \nu}$$
where F is the electromagnetic field tensor and $$\tilde{F}$$ the dual one.

Homework Equations

First, the contravariant components of the electromagnetic field tensor are given by:
$$F^{\mu\nu} = \begin{bmatrix} 0 & -E_1 & -E_2 & -E_3 \\ E_1 & 0 & -B_3 & B_2 \\ E_2 & B_3 & 0 & -B_1 \\ E_3 & -B_2 & B_1 & 0 \end{bmatrix}$$

and the covariant by:
$$F_{\mu\nu} = \begin{bmatrix} 0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & -B_3 & B_2 \\ -E_2 & B_3 & 0 & -B_1 \\ -E_3 & -B_2 & B_1 & 0 \end{bmatrix}$$

And last but not least, the contravariant components of the dual elm. field tensor:
$$\tilde{F}^{\mu\nu} = \begin{bmatrix} 0 & -B_1 & -B_2 & -B_3 \\ B_1 & 0 & E_3 & -E_2 \\ B_2 & -E_3 & 0 & E_1 \\ B_3 & E_2 & -E_1 & 0 \end{bmatrix}$$

The Attempt at a Solution

First I note:
$$\tilde{F}^{a0} = B^{a}$$

$$\tilde{F}^{ab} = \epsilon^{abi} E_{i}$$

$$F_{a0} = - E_{a}$$

$$F_{ab} = \epsilon_{abc} B^{c}$$

Now, if I use these relations I obtain the wrong solution:

$$\tilde{F}^{\mu \nu} \, F_{\mu \nu} = 2 \tilde{F}^{a0} F_{a0} + \tilde{F}^{ab} F_{ab} = - 2 B^{a} E_{a} + \epsilon^{abi} E_{i} \epsilon_{abc} B^{c} = 0$$
where I used the relation: $$\epsilon^{abi} \epsilon_{abc} = 2 \delta^{i}_{c}$$

Of course, if i simply insert the components explicitly (the "matrix elements") I get the result $$\tilde{F}^{\mu \nu} \, F_{\mu \nu} = - 4 \vec{B} \cdot \vec{E}$$ and everything is fine.

I used the following convention: $$\epsilon^{0123} = 1$$, latin indices: {1,2,3}, greek indices: {0,1,2,3}

Further I calculated: $$\epsilon_{123} = \eta_{\alpha 1} \, \eta_{\beta 2} \, \eta_{\gamma 3} \, \epsilon^{\alpha \beta \gamma} = - 1$$ where $$\eta = diag(1, -1, -1, -1)$$ is the metric tensor.

I think there has to be a wrong sign, but I don't find it. Does anyone have an idea?

 

[gabbagabbahey]

$$F_{ab} = \epsilon_{abc} B^{c}$$

I think you might want to double check this one

 

[parton]

hmmm, I don't see anything wrong here !?

If I check it I obtain the right elements, for example: $$F_{12} = \epsilon_{123} B^{3} = - B^{3}$$.

 

[gabbagabbahey]

hmmm, I don't see anything wrong here !?

If I check it I obtain the right elements, for example: $$F_{12} = \epsilon_{123} B^{3} = - B^{3}$$.

$$\epsilon_{123}=+1$$

 

[gabbagabbahey]

Further I calculated: $$\epsilon_{123} = \eta_{\alpha 1} \, \eta_{\beta 2} \, \eta_{\gamma 3} \, \epsilon^{\alpha \beta \gamma} = - 1$$

This assumes that $$\epsilon_{abc}$$ transforms as a tensor; but it doesn't. It is a tensor density with weight -1.

 

[parton]

ok, $$\epsilon_{ijk}$$ is a pseudotensor, but where is the difference between covariant $$\epsilon_{ijk}$$ and contravariant $$\epsilon^{ijk}$$ components?

In the case with 4 indices there is a difference, for example: $$\epsilon^{0123} \neq \epsilon_{0123}$$. Why should it be different with 3 components?

 

[gabbagabbahey]

ok, $$\epsilon_{ijk}$$ is a pseudotensor, but where is the difference between covariant $$\epsilon_{ijk}$$ and contravariant $$\epsilon^{ijk}$$ components?

There isn't a difference in the convention that I'm used to seeing.

In the case with 4 indices there is a difference, for example: $$\epsilon^{0123} \neq \epsilon_{0123}$$. Why should it be different with 3 components?

I'm assuming that you are using the same convention (the most common convention) http://planetmath.org/encyclopedia/LeviCivitaPermutationSymbol3.html .

In that case, then $$\epsilon^{0123} = \epsilon_{0123}$$.

If you are using some other convention, whereby $$\epsilon_{123}=-1$$ , then your problem is in the equation $$\epsilon^{abi} \epsilon_{abc} = 2 \delta^{i}_{c}$$ instead.

 

[parton]

i've found something, have a look at this (page 9): http://www.worldscibooks.com/phy_etextbook/6938/6938_chap01.pdf"

Further I found something in another wiki (but it's not in english). There is a remark that in relativity there you have to differ between co- and contravariant indices. It is just convention whether you use $$\epsilon^{0123}$$= +1 or -1.

But all this doesn't explain my problem

 

[gabbagabbahey]

i've found something, have a look at this (page 9): http://www.worldscibooks.com/phy_etextbook/6938/6938_chap01.pdf"

The convention used in the above link is not very common; but if that is the convention you use in your course then you can stick with it.

According to that convention, $$\epsilon_{ijk}=-e^{ijk}$$ and $$e_{123}=1$$. Therefor, $$\epsilon^{abi} \epsilon_{abc} = -2 \delta^{i}_{c}$$ in this convention. In addition, epsilon is a tensor in this convention.

So the question you need to ask yourself is which convention is used in your course text/notes?