Field strength tensor / matrix

1. Jan 27, 2014

Abigale

In my note,
we have written the field strength tensor as:

$F^{\mu\nu} =\partial ^\mu A^\nu -\partial ^\nu A^\mu = \begin{pmatrix} 0&E_x &E_y&E_z \\ -E_x&0 &B_z &-B_y \\ -E_y&-B_z &0 &B_x \\ -E_z&B_y &-B_x&0 \end{pmatrix}$

But if I look into another book or wiki it is written as:

$F^{\mu\nu} =\partial ^\mu A^\nu -\partial ^\nu A^\mu = \begin{pmatrix} 0&-E_x &-E_y&-E_z \\ E_x&0 &-B_z &B_y \\ E_y&B_z &0 &-B_x \\ E_z&-B_y &B_x&0 \end{pmatrix}$

Why is it possible to write the field strength tensor in both notations?
And are both notations really equal?

THX
Abby

2. Jan 27, 2014

ChrisVer

It depends on how you define your metric and thus the $∂^{μ}$

3. Jan 27, 2014

Abigale

I think it comes from that $$F^{\alpha \beta}$$ is an antysymmetric tensor. So I can use $$F^{\beta \alpha} =-F^{ \alpha \beta}$$ but I am not sure.
Need help^^

Both definitions of $$\partial ^\mu$$ are in both cases equal.

4. Jan 27, 2014

ChrisVer

Well then try for example to write the first line:
$F^{0i}=∂^{0}A^{i}-∂^{i}A^{0}$
Now I am not sure about minus/plus conventions I would write it:
$F^{0i}=\frac{∂A^{i}}{∂t}-∇_{i}Φ= Ε^{i}$
So that's what I'd use.....
it has to do I guess with how you define covariant and contravariant vectors.

5. Jan 28, 2014

samalkhaiat

Because you can use either one to write the correct form of Maxwell equations:
$$\partial_{ \mu } F_{ 1 }^{ \mu \nu } = - \partial_{ \mu } F_{ 2 }^{ \mu \nu } = - e J^{ \nu }$$

How can they be equal? $F_{ 1 }^{ \mu \nu } = - F_{ 2 }^{ \mu \nu }$.

6. Jan 28, 2014

ChrisVer

well the problem is that both notations are practically equivalent- you will just have to define differently the current 4vector....
In the Lagrangian what appears is the $F_{μν}F^{μν}$, and whether you have them with a minus each, they will lead in the same equations of motion....