# Lorentz invariants

1. 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

2. Homework Equations

3. The Attempt at a Solution
What does write the explicitly as invariant contractions mean?

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fzero
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3. 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

fzero
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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

fzero
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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

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

fzero
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$$\epsilon^{\mu\nu a b}$$ 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?

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

fzero
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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.

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

fzero