Confusion Field Tensor and derivation of Maxwell's equations

Hi as I'm reading http://www.maths.tcd.ie/~cblair/notes/432.pdf at page 13 I see that he states that the covariant and contravariant field tensors are different. But how can that be? Aren't they related by

$$F_{\mu \nu} = \eta_{\nu \nu'} \eta_{\mu \mu '} F^{\mu ' \nu '} ?$$

and is not the product of the metric tensor eta with it self the identity? I view this as a matrix product

$$F = \eta \eta F'$$

and writing out either η or the metric g it seems like their product with eachother are the identity such that

$$F=\eta \eta F' = I F'.$$

Where is my reasoning wrong?
At page 14 he derived two of the maxwell equations from a lagrangian. But what about the other two? The author just states them. Are not these derivable from a lagrangian?

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 This is not the product of two metric tensors. In the definition of the contravariant field tensor appears the product: $$\eta_{\nu\nu'}\eta_{\mu\mu'}$$ while the product of two metric tensors should be (with explicit components): $$\left(\eta\eta\right)_{\nu\mu}=\eta_{\nu\nu'} \eta_{\nu'\mu}=\delta_{\nu\mu}$$ just as any other matrix product.

 Quote by Einj This is not the product of two metric tensors. In the definition of the contravariant field tensor appears the product: $$\eta_{\nu\nu'}\eta_{\mu\mu'}$$ while the product of two metric tensors should be (with explicit components): $$\left(\eta\eta\right)_{\nu\mu}=\eta_{\nu\nu'} \eta_{\nu'\mu}=\delta_{\nu\mu}$$ just as any other matrix product.
So I can't compute it using matrix products? How would you suggest computing it?
Writing out the components one by one using the relation?

Confusion Field Tensor and derivation of Maxwell's equations

If your problem is to compute $$F^{\mu\nu}$$ you can do it by components remembering that it is antisymmetric so you have to compute just 6 of them. Otherwise, if you want to calculate the contravariant tensor instead of the covariant one the problem is a very easy one as you just have to low the indexes and remember that
$$B^i=-B_i \;\;\mbox{ and }\;\; E^i=-E_i$$

I hope to remember well

 Quote by Einj If your problem is to compute $$F^{\mu\nu}$$ you can do it by components remembering that it is antisymmetric so you have to compute just 6 of them. Otherwise, if you want to calculate the contravariant tensor instead of the covariant one the problem is a very easy one as you just have to low the indexes and remember that $$B^i=-B_i \;\;\mbox{ and }\;\; E^i=-E_i$$ I hope to remember well

That does not seem right from

http://www.maths.tcd.ie/~cblair/notes/432.pdf

where only the E's not the B's has changed sign.

 I'm sorry you are right. You have to change the sign only to E's components. B's components don't do that because both the gradient and the A field gain a - sign, so the overall change is a + sign. I'm sorry

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 So I can't compute it using matrix products? How would you suggest computing it?
In order for a tensor expression to be writeable as a matrix product, the contractions must occur between adjacent indices. That is,

Cik = AijBjk

can be written as a matrix product C = AB. In your case,

Fμν = ημμ'ηνν'Fμ'ν'

you have to rearrange the factors:

Fμν = ημμ'Fμ'ν'ην'ν

which is then a matrix product

F = ηF'η

 Quote by Bill_K In order for a tensor expression to be writeable as a matrix product, the contractions must occur between adjacent indices. That is, Cik = AijBjk can be written as a matrix product C = AB. In your case, Fμν = ημμ'ηνν'Fμ'ν' you have to rearrange the factors: Fμν = ημμ'Fμ'ν'ην'ν which is then a matrix product F = ηF'η
Ah, thanks! That was a revelation :)

 Quote by center o bass At page 14 he derived two of the maxwell equations from a lagrangian. But what about the other two? The author just states them. Are not these derivable from a lagrangian?
The other two Maxwell equations, the homogenous equations, come from the Bianchi Identity that the field tensor must obey.

 Quote by jarod765 The other two Maxwell equations, the homogenous equations, come from the Bianchi Identity that the field tensor must obey.
But shouldn't all equations of motions be deriable from the action? Why are the homogeneous equations left out? And how is this identity derived?

Recognitions:
 It is perhaps more simply expressed in vector calculus notation. Once we have postulated the 4-potential $A^\mu = (\phi, \textbf{A})$ with $\textbf{B} = \nabla \times \textbf{A}$ and $\textbf{E} = - \nabla \phi \ - \ \partial\textbf{A}/ \partial t$, then $\nabla . \textbf{B} \ = \ \nabla . \nabla \times \textbf{A} \equiv 0$ and $\nabla \times \textbf{E} \ = \ \nabla \times (- \nabla \phi - \partial\textbf{A}/ \partial t) \ = \ - \nabla \times (\nabla \phi) - \partial(\nabla \times \textbf{A})/ \partial t \ = \ - \partial\textbf{B}/ \partial t$, because $\nabla \times (\nabla F) \equiv \textbf{0}$ for any scalar field F. The only condition required by above is that the partial derivative operators for the four dimensions commute with one another, ie $\partial_\mu\partial_\nu \equiv \partial_\nu\partial_\mu$. The identities $\nabla . \nabla \times \textbf{V} \equiv 0$ and $\nabla \times (\nabla F) \equiv \textbf{0}$ are easily proved by writing out these expressions in full with the individual coordinates of the vectors then simplifying the results. For example, the $x$ component of the latter is $\partial_y \partial_z F - \partial_z \partial_y F \equiv 0$.