Now, let space tmie metric is $$\eta_{\mu\nu}=\text{diag}(+,-,-,-)$$ now $$x_{\mu}=(x^0,-\mathbf{x})$$ and $$x^{\mu}=(x^0,\mathbf{x})$$

and $$x^{\mu}=\eta^{\mu\nu}x_{\nu}$$

also $$\partial_\mu=(\partial^0,\nabla)$$ and finally $$\epsilon^{\mu\nu\rho\lambda}$$ is totally anti symmetric tensor with $$\epsilon^{\mu\nu\rho\lambda}=-\epsilon_{\mu\nu\rho\lambda}$$ now by using this rules i am trying to describe electromagnetic thoery covariantly.

also notice that $$x_i=-\mathbf{x}_{i}$$

we have $$F^{i,0}=E^{i}$$ now the thing i got confused is, $$E^{i}=E_{i}$$ and $$B^{i}=B_{i}$$

also $$\epsilon_{ijk}\partial_{j}A^{k}=B_i$$

now these last two formulas are confusing because, when i have $B$ and $E$ the usual raising and lowering does not work they do not change sign . also How does raising and lowering works for $$\epsilon_{ijk}$$ I know what happens for $$\epsilon^{\mu\nu\rho\lambda}$$ but not $$\epsilon_{ijk}$$ is $\epsilon_{ijk}$ totaly antwi symmetric tensor or what?

  • $\begingroup$ Yes the Levi-Civita tensor, regardless of it's rank, is completely anti-symmetric. $\endgroup$ Commented Jun 19, 2019 at 14:30
  • $\begingroup$ I think he meant to ask whether $\epsilon_{ijk} = \pm \epsilon^{ijk}$. Can you clarify, op, what is the specific doubt you are having? $\endgroup$
    – MannyC
    Commented Jun 19, 2019 at 14:39
  • $\begingroup$ Well, if that's the case then $\epsilon _{ijk} = \epsilon ^{ijk}$ always. Since covariant and contraviant vectors are the same in the Euclidean geometry. The same would hold for any number of raising or lowering. $\endgroup$ Commented Jun 19, 2019 at 14:49
  • $\begingroup$ @thunderbolt but we definitely have $A^i=-A_i$ $\endgroup$
    – physshyp
    Commented Jun 19, 2019 at 14:56
  • 1
    $\begingroup$ I'm not quite clear on what the question is, but you're definitely wandering off track when you write things like $E^i=E_i$. The electric and magnetic fields are not vectors. Together, they're part of the electromagnetic field tensor, which is rank 2. Why does the title talk about "3-pseudovectors?" AFAIK the standard definition of a pseudovector is that it doesn't change under parity. Is that what you mean, or do you mean "something that's not really a vector?" $\endgroup$
    – user4552
    Commented Jun 19, 2019 at 16:30

1 Answer 1


The case for antisymmetric tensors is not different from other generic tensors. Normally one has $$ T_{\mu_1\cdots \mu_k} = \eta_{\mu_1\nu_1}\cdots \eta_{\mu_k\nu_k} T^{\nu_1\cdots \nu_k}\,. $$ If you have a totally antisymmetric tensor this simplifies to $$ \epsilon_{\mu\nu\rho\lambda} = \det(\eta) \,\epsilon^{\mu\nu\rho\lambda}\,. $$ And since $\eta = \mathrm{diag}(1,-1,-1,-1)$ you get the minus, the same would be true for the mostly plus metric (in any even dimension). For the electric field one has to be precise with the definition $$ E^i \equiv F^{i0}\,,\qquad E_i \equiv F_{i0}\,. $$ And you can see that for lowering the $i$ index you pay the price of a minus sign. The same would be true for a mostly plus metric but the sign comes from the $0$.

Finally the $\epsilon_{ijk}$. If you define $$ \epsilon_{ijk} \equiv \epsilon_{0ijk}\,,\qquad \epsilon^{ijk} \equiv \epsilon^{0ijk} = \eta^{00}\eta^{i l} \eta^{jm}\eta^{kn} \epsilon_{lmn}\,, \tag{1}\label{1} $$ then the upper and the lower index version differ in sign. If you instead define $$ \epsilon^{ijk} = \eta^{i l} \eta^{jm}\eta^{kn} \epsilon_{lmn}\,, \tag{2}\label{2} $$ then the relative sign depends on whether you use mostly plus $(+)$ or mostly minus $(-)$ signature. This is not an ambiguity. If you use either definition consistently, when you decompose any four-vector expression into a time and a three-dimensional part you'll get the same result.

For example, $B$ can be defined as $$ \tilde{F}^{\mu\nu} \equiv -\tfrac12 \epsilon^{\mu\nu\rho\lambda} F_{\rho\lambda}\,,\qquad B^i \equiv \tilde{F}^{i0}\,,\qquad B_i \equiv \tilde{F}_{i0}\,. $$ And it also changes sign as any other (presudo)vector. Its decomposition into time and 3d part is $$ B^i = \tfrac12 \epsilon^{0ijk} F_{jk} = s \tfrac12 \epsilon^{ijk} F_{jk}\,,\;\;\qquad B_i = \tfrac12 \epsilon_{0ijk} F^{jk} = \tfrac12 \epsilon_{ijk} F^{jk} \,, $$ where $s=1$ for the definition \eqref{1} and $\eta^{00}$ for the definition \eqref{2}. You can see that, not matter what, $B^i = - B_i$ (in any signature).

Note that if you are in Euclidean space to start with (i.e. you are not in an Euclidean subspace of Minkowski), then the natural metric is the all-plus $\delta_{ij}$, obviously. Therefore the Levi-Civita tensor with upper and lower indices will have the same sign $\epsilon_{i_1\ldots i_n} = \epsilon^{i_1\ldots i_n}$.

  • $\begingroup$ great answer in my calculations for $E$ and $B$ i did not changed sign while raising and lowering. on the other hand I also did the same mistake with $\epsilon^{ijk}$ thus at the end my result were correct(agreed with litreture). thanks a lot i will go through my calculations again and fix those mistakes. $\endgroup$
    – physshyp
    Commented Jun 24, 2019 at 13:10
  • $\begingroup$ Maybe you just defined that $E_i = E^i$. That's still correct as long as you are consistent (e.g. you have to say $E_i = - F_{i0}$). $\endgroup$
    – MannyC
    Commented Jun 24, 2019 at 14:27
  • $\begingroup$ well thats exactly what i am doing $\endgroup$
    – physshyp
    Commented Jun 24, 2019 at 15:13
  • $\begingroup$ if i define $E_i=E^i$ wouldn't I violate raising and lowering rule, in + --- metric. $\endgroup$
    – physshyp
    Commented Jul 14, 2019 at 10:53
  • 1
    $\begingroup$ $E$ is not a four-vector, so you have a bit of freedom on how to relate it to four-vector quantities. You just have to be consistent. $\endgroup$
    – MannyC
    Commented Jul 14, 2019 at 14:03

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