# 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....