# Formulation of the Bianchi identity in EM

I'm trying to understand, as a self learner, the covariant formulation of Electromagnetism. In particular I've been stuck for a while on the Bianchi identity. As I've come to understand, when we introduce the field strength tensor $$F_{\mu\nu}$$ we can write the Maxwell's equation in an explicitly covariant form using observable quantities as the aforementioned tensor (because of its Gauge invariance). Maxwell's equation thus reduce to one single 4-tensor equation (Gauss's + Maxwell-Ampere's laws), plus an identity (Gauss's law for magnetism and Faraday-Maxwell's law): I don't understand how to derive the identity in terms of the field strength tensor. This is my reasoning up until now.

I will use latin indices that range from 1 to 3 and greek ones that range from 0 to 3. We can say that:

$$F_{ij} = B_{ij} = \varepsilon_{ijk} B^k \Longrightarrow \varepsilon^{ijk} F_{ij}= \varepsilon_{ijk}\varepsilon^{ijk} B^k \Longrightarrow B^k = \frac{1}{3!} \varepsilon^{ijk} F_{ij}$$

and

$$F_{0i} = -\frac{E_i}{c} \Longrightarrow E_i = -cF_{0i}$$

Let us consider a flat space-time, then Gauss's law for magnetism can be written as:

$$\nabla \cdot \vec{B} = \partial_k B^k = 0 \Longrightarrow \partial_k \frac{1}{3!} \varepsilon^{ijk} F_{ij} = 0 \Longrightarrow \varepsilon^{ijk} \partial_k F_{ij} = 0$$

and Faraday-Maxwell's law can be written as:

$$\nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = \vec{0} \Longrightarrow \varepsilon^{kij} \partial_i E_j + c \partial_0 B^k = 0 \Longrightarrow \varepsilon^{kij} \partial_i \left( -c F_{0j} \right) + c \partial_0 \frac{1}{3!} \varepsilon^{ijk} F_{ij} = 0\Longrightarrow$$

$$- \varepsilon^{kij} \partial_i F_{0j} + \frac{1}{3!} \varepsilon^{ijk} \partial_0 F_{ij} = 0$$

Now, if I'm not mistaken, $$\varepsilon^{0 \beta \gamma \delta} = \varepsilon^{0ijk} = \varepsilon^{ijk}$$, therefore I can sobsitute the three dimensional Levi-Civita symbol with its four dimensional variant:

$$\varepsilon^{0ijk} \partial_k F_{ij} = 0$$

$$-\varepsilon^{0kij} \partial_i F_{0j} + \frac{1}{3!} \varepsilon^{0ijk} \partial_0 F_{ij} = 0$$

From here I don't know how to continue. The latter of the two equations really seems to involve some kind of Levi-Civita symbol, because of the swapping of the $$0$$ index between the 4-gradient component and the field tensor component, but the $$\frac{1}{3!}$$ is in the way. Moreover, even if I simplified this equation I wouldn't know how to unify the two to obtain the final identity in terms of the field strength tensor:

$$\epsilon^{\alpha \beta \gamma \delta} \partial_{\beta} F_{\gamma \delta} = 0$$

I shall note that in my answer I consider $$c=1$$. If that is confusing, please let me know. The Gauss Law reads $$\partial_i B^{i}=0$$, which implies $$\partial_1 F_{23} + \partial_{2} F_{31} + \partial_{3} F_{12}=0.$$ Faraday's Law tells us that $$\nabla \times \vec{E} + \partial_{t} \vec {B}=0$$. With indices, we can rewrite this as $$\varepsilon_{ijk} \partial_{j} E^{k} + \partial_{0} B^{i}=0.$$ From the above we can read off $$\partial_2 E^3 - \partial_3 E^2 + \partial_0 B^1=0 \iff \partial_2 F_{03}+ \partial_{3} F_{20} + \partial_{0} F_{32}=0;$$ $$\partial_3 E^1 - \partial_1 E^3 + \partial_0 B^2=0 \iff \partial_3 F_{01} + \partial_{1} F_{30} + \partial_{0} F_{13}=0;$$ $$\partial_{1} E^{2} - \partial_{2} E^{1} + \partial_{0} B^3=0 \iff \partial_1 F_{02} +\partial_{2} F_{10} + \partial_{0} F_{21}=0.$$ Based on the above, altogether, for $$F$$, we have: $$\partial_{\alpha} F_{\beta \gamma} + \partial_{\beta} F_{\gamma \alpha} + \partial_{\gamma} F_{\alpha \beta}=0.$$ Multiplying this equation by $$\frac{1}{3}$$, it holds that: $$\frac{1}{3} \left(\partial_{\alpha} F_{\beta \gamma} + \partial_{\beta} F_{\gamma \alpha} + \partial_{\gamma} F_{\alpha \beta} \right)=0.$$ Using the symmetry properties of the Faraday tensor: $$\frac{1}{2 \cdot 3} \left(\partial_{\alpha} (F_{\beta \gamma}- F_{\gamma \beta})+\partial_{\beta}(F_{\gamma \alpha} - F_{\alpha \gamma})+ \partial_{\gamma}(F_{\alpha \beta} - F_{\beta \alpha})\right)=0.$$ Simplifying and regrouping gives: $$\frac{1}{3!}\left(\partial_{\alpha} F_{\beta \gamma} + \partial_{\beta} F_{\gamma \alpha}+ \partial_{\gamma} F_{\alpha \beta} - \partial_{\beta} F_{\alpha \gamma} - \partial_{\alpha} F_{\gamma \beta} - \partial_{\gamma} F_{\beta \alpha} \right)=0.$$ Now contracting the equation with the Levi-Civita tensor gives the result. For the relation with what you have done can be seen here
• Thank you very much for your answer: it was very helpful! No problem in using $c=1$ Sep 22 at 18:19