# Why is electric field $E^i = E_i$ instead of $E^i = - E_i$?

Let us consider the Minkowski spacetim.

Generally, we know that when we lower or raise the index of the convariant or contravairant tensor, we need to use the metric $$\eta^{\mu \nu}=\eta_{\mu \nu}=(+,-,-,-)$$

Clearly, for the electromagnetic tensor $$F_{0i}= - F^{0i}$$ However, if we define the electric field as $$E_i \propto F_{0i}$$ as Wikipdia did

https://en.wikipedia.org/wiki/Electromagnetic_tensor#Relationship_with_the_classical_fields

we also get $$E_i \propto F_{0i} =- F^{0i}$$ But from Wikipedia again, the same page says that $$E^i \propto - F^{0i}$$

Thus, we get

$$E^i = E_i \propto F_{0i} =- F^{0i}$$

My question is that why do $$E^i = E_i$$ instead of $$E^i = - E_i$$? Is there a reason we should not identify

$$E^i = \eta^{ia} E_{a} = - E_i?$$

This also relates to the fact whether we write the energy E as

E $$\propto -(F_{0i} F^{0i}+...) =-(F_{0i} (- F_{0i})+...) = (F_{0i}^2+...)=((E_i)^2+...)$$

E $$=((E_i)^2+...)=((E^i)^2+...)=((E_i)(E^i)+...)$$

Generally, we knew that

E $$\propto((E_i)^2+(B_i)^2)=((E^i)^2+(B^i)^2)$$

Question now: Generally, do we treat $$E^i = E_i$$ or $$E^i = - E_i$$ as a 1-vector, 1-(contra)vector or a 2-tensor?

• Please don't post screenshots. They don't work for blind people, and they don't work for searching. – user4552 May 20 '19 at 14:01
• My post is readable even without the screenshot -- I post screenshots just to save people time to link to Wiki page – annie marie heart May 20 '19 at 15:40
• @annieheart If you just copy the relevant portion, your question would be far easier to read and understand, as well as working for blind people and searching. – Mike May 20 '19 at 18:43

Electric field is not a four vector. They are just three vectors which posses components along three spatial dimension. Their transformation to one form is defined by a $$3\times 3$$ identity matrix(Euclidean metric).$$E^i = \delta^{ia} E_{a} = E_i.$$ They are also components of elctromagnetic field tensor. Here $$E^i$$ depicts the $$i^{th}$$ component of electric field vector is same as the $$-c F^{0i}$$ th term of electromagnetic field tensor.

• +1 thanks so much -- how about the case for magnetic B field? – annie marie heart May 20 '19 at 18:22
• @annieheart: Neither E nor B is a four-vector. – user4552 May 20 '19 at 19:05

Why is $$E^i = E_i$$ instead of $$E^i = - E_i$$?

The fundamental reason would be that the electric and magnetic fields, $$E$$ and $$B$$ do not form four vectors. Rather they are three-dimensional vectors without a fourth component as explained here. The transformation between co- and contra- forms is identity transformation. This is due to different vectorial nature of the respective fields. The electric field is a polar vector (or true vector) because it changes the sign if coordinates are reversed, $$\mathbf{r} \rightarrow \mathbf{-r}$$. In contrast, the magnetic field given by

$$\mathbf{B}(\mathbf{r}) = \frac{1}{c} \int \dfrac{(\mathbf{r}-\mathbf{r'})\times \mathbf{J}(\mathbf{r})}{|\mathbf{r}-\mathbf{r'}|^3} \text dV'$$

remains unchanged against coordinate inversion since both $$(r - r')$$ and current density $$J(r)$$ change sign. The magnetic field is a pseudo-vector (or axial vector).

A detailed calculation of how it turns out to be the same in Minkowski space-time metric regardless of its form is shown here.

The signs of the components of the electromagnetic tensor $$F^{\mu \nu}$$ and $$F_{\mu \nu}$$ depend on the metric convention. However, the mixed tensor $$F^\mu{}_\nu$$ is independent of such a choice. Considering $$c = 1$$, we have

$$F^\mu{}_\nu=\left(\begin{array}{cccc}0&E_x&E_y&E_z\\E_x&0&B_z&-B_y\\E_y&-B_z&0&B_x\\E_z&B_y&-B_x&0\end{array}\right)$$

Here $$\mu,\nu~\in~\{0,1,2,3\}$$ and $$i~\in~\{1,2,3\}$$.

Defining $$\eta_{\mu \nu} = \text{diag}(-1,+1,+1,+1) = \eta^{\mu \nu}$$, we know $$F_{\mu \nu} = \eta_{\mu \rho}F^\rho{}_{\nu}$$ so we obtain $$E_i = -F_{0i}$$

Since, $$F^{\mu \nu} = \eta^{\mu \rho}\eta^{\nu \lambda}F_{\rho \lambda}$$, for $$i \ne 0$$ we obtain $$E^i = F^{0i} = \eta^{00}\eta^{i i}F_{0 i} = -F_{0i} = E_i$$

Now, it is fairly straight-forward to prove that if we use $$\eta_{\mu \nu} = \text{diag}(+1,-1,-1,-1)$$, the result will be $$E^i = -F^{0i} = - \eta^{00}\eta^{i i}F_{0i} = F_{0i} = E_i$$

In both the cases it turns out that $$E_i = E^i$$.

• It seems to me that this misses the point, which is simply that the electric field is not a four-vector, so it doesn't make sense to talk about raising and lowering its indices. – user4552 May 20 '19 at 14:03
• @BenCrowell Thanks and yes, I should have stressed directly on the fact that field not being a four-vector. I just wanted to show via calculation as to how does it not change sign. I added more details. – Abhay Hegde May 20 '19 at 17:01
• +1 thanks so much -- how about the case for magnetic B field? – annie marie heart May 20 '19 at 18:22
• @annieheart As already explained, B is not a four-vector. So it would have the same transformation as electric field. – Abhay Hegde May 21 '19 at 3:03

In Euclidean metric, such as standard 3D, the distinction between co- and contra is not useful and $$E_i =E^i$$. This can be confusing if the Minkowski metric is chosen as a space inversion. However, the alternative gives unintuitive minus signs (my opinion).

E is not a real 3D vector as its sign changes under time reversal. B also is not a true vector as its sign does not change under space inversion. It is called an axial vector. Another example of this is angular momentum.

• +1 thanks so much -- how about the case for magnetic B field? – annie marie heart May 20 '19 at 18:22