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3:16 PM
Ok, what can I do for you?
 
Hi there :-)
I would like to have your opinion about the following fact.
Is seems to me that the fact that a 2-form in Minkowski spacetime cannot be either self-dual of anti-self-dual implies that the electric and the magnetic field cannot be the same (i.e. equal magnitude, equal direction) in the same point in space
Does this look obvious to you, or do you want me to explain myself?
(Before I go on)
 
I think you'd need to explain that a little bit. Is it something you've concluded from considering duality of the Faraday 2-form and how this interchanges the electric and magnetic fields?
 
Yes, precisely.
I'm of course supposing that the correct way to formalize the electromagnetic field in Minkowski spacetime is through the Faraday 2-form.
"The correct way" meaning the most fundamental one: define the electromagnetic quadripotential as a four-vector field, transform it into a 1-form, then exterior-derive it.
 
Hm, it does not seem obvious to me that this should be true. Is it something you're only considering for vacuum (so that the dual 2-form and the Faraday 2-form both obey the same differential equations)?
 
It doesn't depend upon the dynamical equations. It is a straightforward fact about 2-forms in Minkowski spacetime. I show you the calculations (they really are two equations).
Let us say that we want the electric field $\vec{E}$ to be equal to the magnetic field $\vec{B}$ in some point of space.
Gosh, don't equations work in the chatroom?
Do you know whether there is a way of writing equations in here?
 
3:28 PM
No idea.
 
Mmh. I'll send them to you as a jpeg. Give me one minute.
Of, first of all, we have the Faraday 2-form. In standard Minkowski coordinates:
Then, we are requiring that
And the same goes for the opposite indices
 
And how does duality forbid the electric and magnetic fields from being equal?
 
But, we have
I really hope that the latter are true (the "1" must be replaced with "i")
Then we have
 
Hm, I think there might be a sign error here.
 
But we know that this is not possible, unless the 2-form is the zero form.
There might be, but even if there was, it would only mean that they are anti-self-dual, doesn't it?
So the conclusion is the same
*It is anti-self-dual, it being Faraday's two form
 
3:45 PM
No, I don't think so. Consider F = E_1 dx^0 \wedge dx^1 - B_1 dx^2 \wedge dx^3. Then the dual is E_1 dx^2 \wedge dx^3 + B_1 dx^0 \wedge dx^1. There's a sign change on one of the components, but not both.
(It's precisely for that reason that there are no self-dual or anti-self-dual 2-forms.)
 
I have the signs switched
 
On one but not both.
 
Yes. I'm checking that
You're absolutely sure that your signs are correct (they look to make sense to me, but I'd need to recalculate them to be sure about it myself)
?
 
Yes, quite.
 
Yes, right. It seems that you are correct.
One of the two expressions in the cases need a minus sign, I think
 
3:56 PM
Yeah, your equations relating the components to the dual components imply F = star F for any F.
Sorry, F = star star F.
 
While it should be with a minus sign, right?
 
Yes.
 
F=-star star F
Ok, got it.
At least it makes sense of the fields being equal
 
All right, is there anything else I can do here?
 
Yes. I was asking myself whether mine was an error of calculation or a bad definition. The dual operator is defined in the same way in Minkowski spacetime as in a Riemanniang manifold, isn't it?
I mean, apart from the sign of the determinant
 
4:01 PM
I mean, I believe it is. The sign of the determinant is the main difference from Reimannian to pseudo-Riemannian, as far as I know.
In your question, you yourself have the correct sign change (though there should be factors of 1/2 that aren't there).
 
It should be a standard definition for tensor spaces anyway, doesn't it?
 
Yeah, it all depends on the Levi-Civita really.
 
Not sure about the 1/2
Wouldn't them spoil auto-duality?
Notice that I used two conventions for the indices
 
You have in your question (star omega)_{01} = omega_{23} - omega_{32}
Which if the indices are antisymmetric introduces a factor of 2 that shouldn't be there.
But you also have (star omega)_{23} = omega_{10} - omega_{01}
Again, should be a factor of 1/2 there, and you notice that the 0 and 1 are now switched compared to the first equation.
 
Yes, I'm pretty sure that it is due to the double convention
I used omega_{10}-omega_{01}=(omega)_{10}
 
4:05 PM
Which is - omega_{01} if indices are antisymmetric.
There's the minus sign.
 
And, as for the wedge product, I used (a wedge b)=ab-ba
Without the 1/2
 
I mean, the 1/2 isn't that crucial here.
 
It should be -(omega)_{01}
 
I'm just saying that you had the correct signs at one point and just lost one minus sign somewhere later.
 
With the brackets
Yes, I got it
Thank you so much for your time
 
4:07 PM
You're welcome, let me know if there's anything else I can help with
 
Thank you again
I'd give you another up-vote, if it were possible :-)
 

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