# Proof that $\partial_\mu G^{\mu\nu}=0$? [closed]

Given the electromagnetic field tensor
$$F^{\mu\nu}=\left(\begin{matrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{matrix}\right)$$ and the dual electromagnetic tensor $G^{\mu\nu}=\frac{1}{2} \epsilon^{\mu\nu\lambda\rho} F_{\lambda\rho}$, prove that $\partial_\mu G^{\mu\nu}=0$.

Attempt:

I ended up with the matrix representation of $G$ being
$$G^{\mu\nu}=\left(\begin{matrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z & -E_y \\ B_y & -E_z & 0 & E_x \\ B_z & E_y & -E_x & 0 \\ \end{matrix}\right)$$

When finding the partials of these components I got, for example,
$$\partial_0 G^{0i}=\left(\frac{\partial(-B_x)}{\partial t},\frac{\partial(-B_y)}{\partial t},\frac{\partial(-B_z)}{\partial t}\right)=-\frac{\partial\vec{B}}{\partial t}$$

$$\partial_1 G^{12}+\partial_3 G^{32}=\frac{\partial E_z}{\partial x}-\frac{\partial E_x}{\partial z}=\left(\nabla\times\vec{E}\right)_y$$

As you can see, I keep getting values in the form of the Maxwell-Faraday equation. This would be fine if it were electrostatics, since we would know that the values are equal and zero, but how can we get $\partial_{\mu}G^{\mu\nu}=0$ from these components given that this isn't electrostatics?

## closed as off-topic by Kyle Kanos, John Rennie, stafusa, Jon Custer, sammy gerbilDec 6 '17 at 19:08

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• The equation $\partial_{\mu}G^{\mu\nu}=0$ tells you that for example $\partial_{0}G^{02}+\partial_{1}G^{12}+\partial_{3}G^{32}=-\frac{\partial B_{y}}{\partial t}+\left(\nabla\times\vec{E}\right)_{y}=0$, which is perfectly general and accounts not only to electrostatics. – eranreches Dec 3 '17 at 20:37
• I'm not sure how you can expect $\partial_\mu G^{\mu \nu}=0$ to give anything other than a partial differential equation in the force fields, or how you can expect said PDE to be anything other than a subset of Maxwell's equations. – Emilio Pisanty Dec 3 '17 at 20:52
• Moreover, the Faraday induction law isn't really a part of electrostatics (that would require $\nabla\times\vec E=0$ because $\partial\vec B /\partial t=0$ in quasistatic situations), so I don't understand why you think there's a contradiction there. – Emilio Pisanty Dec 3 '17 at 20:53
• Look at Belzebu's answer. You need to prove it equals zero, not assume it. Think about the anti-symmetry of the epsilon tensor and the symmetry of partial derivatives. – kηives Dec 4 '17 at 2:53

You can also notice $F_{\lambda \rho}= \partial_{\lambda} A_\rho-\partial_{\rho} A_{\lambda}$, so that $\partial_\mu G^{\mu \nu}$ is made up of two pieces, each containing a symmetric combination of derivatives ($\partial_\mu \partial_\lambda$ and $\partial_\mu \partial_\rho$ ). This vanishes when contracted with the antisymmetric tensor $\epsilon ^{\mu \nu \lambda \rho}$.
• By symmetric combination of derivatives, do you mean that $\partial_\mu\partial_\lambda=\partial_\lambda\partial_\mu$, or is it a different kind of symmetry? – T. Walker Dec 4 '17 at 22:34
Einstein summation is great, but when you're starting out it can obscure the mathematical operations you actually need to do. I think your confusion stems from a misinterpretation of what $\partial_\mu G^{\mu \nu}$ means. This equation stands for four separate equations, one for each possible value of $\nu$: \begin{align} \nu &= 0: & \partial_0 G^{00} + \partial_1 G^{10} + \partial_2 G^{20} + \partial_3 G^{30} &= 0 \\ \nu &= 1: & \partial_0 G^{01} + \partial_1 G^{11} + \partial_2 G^{21} + \partial_3 G^{31} &= 0 \\ \nu &= 2: & \partial_0 G^{02} + \partial_1 G^{12} + \partial_2 G^{22} + \partial_3 G^{32} &= 0 \\ \nu &= 3: & \partial_0 G^{03} + \partial_1 G^{13} + \partial_2 G^{23} + \partial_3 G^{33} &= 0 \end{align} Be sure to study the pattern of the indices carefully in these equations, and make sure you understand why $\partial_\mu G^{\mu \nu}$ stands for them.
So, for example, the $\nu = 1$ equation reduces to $$0 = \partial_0 G^{01} + \partial_2 G^{21} + \partial_3 G^{31} = -\frac{\partial B_x}{\partial t} + \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} = \frac{\partial B_x}{\partial t} + [\vec{\nabla} \times \vec{E}]_x.$$ The last of these quantities vanishes even for dynamical fields, not just for static fields, since it's the $x$-component of Faraday's Law $\partial \vec{B}/\partial t + \vec{\nabla} \times \vec{E} = 0$. The $\nu = 2$ and $\nu = 3$ components work much the same way; the interpretation of the $\nu = 0$ equation is a little different, but I'll leave you to figure out what that is.
• Michael, that makes much more sense than what I was previously thinking. I'm kind of teaching myself this, which is how I failed to understand the notation properly. So for $\nu=0$, am I correct in assuming that this goes along with the "no magnetic monopole" law? – T. Walker Dec 3 '17 at 21:13
• @T.Walker: Yes, that's the law corresponding to the $\nu = 0$ component. – Michael Seifert Dec 3 '17 at 21:53