Lorentz Invariants and Field Strength Tensor Fuv

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Homework Statement



The Field strength tensor Fuv encodes the electric and magnetic fields via:
Ei=-cF0i, Bi=-1/2 eijkFjk, i=1,2,3 Show that E^2-c^2B^2 and cE.B are invariant under lorentze transformations, by writing them explicitly as invariant contractions using the tensors Fuv and euvab

Homework Equations





The Attempt at a Solution


What does write the explicitly as invariant contractions mean?
 
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jc09 said:

The Attempt at a Solution


What does write the explicitly as invariant contractions mean?

A Lorentz invariant quantity can't have any free index. A quantity like A_{\mu\nu\rho}v^\mu}B^{\nu\rho} would be invariant, but A_{\mu\nu\rho}B^{\nu\rho} would not.
 
ok so for the first one E^2-c^2B^2 how would I start the problem. I know the equation for E and B are given in terms of the metric tensor so I would write them out but what would be the next step
 
Well what sort of contractions can you form from F^{\mu\nu} and \epsilon_{\mu\nu ab}? Note that the quantities you're supposed to obtain are quadratic in the fields.
 
I know I can contract FuvUb=Va but not sure how to proceed at all
 
jc09 said:
I know I can contract FuvUb=Va but not sure how to proceed at all

The expression you've written has free indices \mu, \nu and b on the LHS and an index a on the RHS, so it is incorrect. Also U and V are undefined and have nothing to do with the question. You do not need any other tensors other than F^{\mu\nu} and \epsilon_{\mu\nu ab}. Try to form contractions that have no free indices. For example, \epsilon_{\mu\nu ab} \epsilon^{\mu\nu ab} is a Lorentz invariant, but it's not one of the ones you're looking for.
 
so does euvxy=euvexy make more sense? are these correct? I know these are still not what I used in the question but is the idea right here. Then FuvFuv
 
jc09 said:
so does euvxy=euvexy make more sense? are these correct? I know these are still not what I used in the question but is the idea right here. Then FuvFuv

sorry that isn't right I think maybe this is euveuvexyexy
 
\epsilon^{\mu\nu a b} is the Levi-Civita symbol, you might want to brush up on what it means here: http://en.wikipedia.org/wiki/Levi-Civita_symbol You can't write \epsilon^{\mu\nu a b} =\epsilon_{\mu\nu} \epsilon_{ a b} for any definition of \epsilon_{\mu\nu}, the RHS does not have all of the symmetries that the LHS does.

I wrote the expression with 2 \epsilon's to illustrate a Lorentz-invariant combination, don't get hung up on trying to rewrite that. You mentioned F^{\mu\nu}F_{\mu\nu}. Why don't you try writing that in terms of E and B?
 
  • #10
hi so if I write FuvFuv in terms of E and B I get FuvFuv=FolFol+FijFij=cE.B is this correct?

Then to get the other one I can write 1/2euvxyFuvFxy in terms of E and B
 
  • #11
jc09 said:
hi so if I write FuvFuv in terms of E and B I get FuvFuv=FolFol+FijFij=cE.B is this correct?

No, that's not right. Remember that Ei=-cF0i, Bi=-1/2 eijkFjk, i=1,2,3.

Then to get the other one I can write 1/2euvxyFuvFxy in terms of E and B

That's one of the invariants that you need. You should see what expression that gives in terms of E and B.
 
  • #12
fzero said:
No, that's not right. Remember that Ei=-cF0i, Bi=-1/2 eijkFjk, i=1,2,3.



That's one of the invariants that you need. You should see what expression that gives in terms of E and B.

So for the first bit if I write FuvFuv=F0iF0i+FjkFjk is that better. sorry I'm having quite a lot of trouble with these
 
  • #13
jc09 said:
So for the first bit if I write FuvFuv=F0iF0i+FjkFjk is that better. sorry I'm having quite a lot of trouble with these

That part was fine, the part that was wrong was what you wrote for the E and B expression. You should use the relationships between E, B and the components of Fuv to explicitly compute that, don't just try to guess.
 
  • #14
so for F0iF0i does that equal -2/c E.B then to finish I add on the second part of it?
 
  • #15
ah wait sorry I see the error of my ways I had written down the second matrix wrong so I had B's where there were meant to be E's etc. Think I have the first one now.
 
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