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

+ 1 like - 0 dislike
127 views

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

enter image description here

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

enter image description here 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?

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user annie marie heart
asked May 20, 2019 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
Please don't post screenshots. They don't work for blind people, and they don't work for searching.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user user4552
My post is readable even without the screenshot -- I post screenshots just to save people time to link to Wiki page

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user annie marie heart
@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.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user Mike

3 Answers

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user walber97
answered May 20, 2019 by walber97 (20 points) [ no revision ]
+1 thanks so much -- how about the case for magnetic B field?

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user annie marie heart
@annieheart: Neither E nor B is a four-vector.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user user4552
+ 2 like - 0 dislike

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 $E_i = E^i$.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user Abhay Hegde
answered May 20, 2019 by Abhay Hegde (20 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user user4552
@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.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user Abhay Hegde
+1 thanks so much -- how about the case for magnetic B field?

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user annie marie heart
@annieheart As already explained, B is not a four-vector. So it would have the same transformation as electric field.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user Abhay Hegde
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user my2cts
answered May 20, 2019 by my2cts (10 points) [ no revision ]
+1 thanks so much -- how about the case for magnetic B field?

This post imported from StackExchange Physics at 2020-11-09 19:26 (UTC), posted by SE-user annie marie heart

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