# How do we get $\nabla \cdot \mathbf{B} = 0$ from $\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B}$? [closed]

I am currently studying the textbook Physics of Photonic Devices, second edition, by Shun Lien Chuang. Section 2.1.1 Maxwell's Equations in MKS Units says the following:

The well-known Maxwell's equations in MKS (meter, kilogram, and second) units are written as $$\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B} \ \ \ \ \text{Faraday's law} \tag{2.1.1}$$ $$\nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial{\mathbf{D}}}{\partial{t}} \ \ \ \ \text{Ampére's law} \tag{2.1.2}$$ $$\nabla \cdot \mathbf{D} = \rho \ \ \ \ \text{Gauss's law} \tag{2.1.3}$$ $$\nabla \cdot \mathbf{B} = 0 \ \ \ \ \text{Gauss's law} \tag{2.1.4}$$ where $$\mathbf{E}$$ is the electric field (V/m), $$\mathbf{H}$$ is the magnetic field (A/m), $$\mathbf{D}$$ is the electric displacement flux density (C/m$$^2$$), and $$\mathbf{B}$$ is the magnetic flux density (Vs/m$$^2$$ or Webers/m$$^2$$). The two source terms, the charge density $$\rho$$(C/m$$^3$$) and the current density $$\mathbf{J}$$(A/m$$^2$$), are related by the continuity equation $$\nabla \cdot \mathbf{J} + \dfrac{\partial}{\partial{t}}\rho = 0 \tag{2.1.5}$$ where no net generation or recombination of electrons is assumed. In the study of electromagnetic, one usually assumes that the source terms $$\rho$$ and $$\mathbf{J}$$ are given quantities. It is noted that (2.1.4) is derivable from (2.1.1) by taking the divergence of (2.1.1) and noting that $$\nabla \cdot (\nabla \times \mathbf{E}) = 0$$ for any vector $$\mathbf{E}$$.

So we are told that Gauss's law $$\nabla \cdot \mathbf{B} = 0$$ is derivable from Faraday's law $$\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B}$$ by taking the divergence and noting that $$\nabla \cdot ( \nabla \times \mathbf{E}) = 0$$ for any vector $$\mathbf{E}$$. I did not understand why/how $$\nabla \cdot ( \nabla \times \mathbf{E}) = 0$$, and nor did I understand how we then get $$\nabla \cdot \mathbf{B} = 0$$ from $$\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B}$$. So, since this seemed like a mathematical issue, I asked here. However, based on the comments, I am told that this actually isn't valid. So is the author just wrong here?

From the above point, the author then continues as follows:

Similarly, (2.1.3) is derivable from (2.1.2) using (2.1.5). Thus, we have only two independent vector equations (2.1.1) and (2.1.2), or six scalar equations as each vector has three components. However, there are $$\mathbf{E}$$, $$\mathbf{H}$$, $$\mathbf{D}$$, and $$\mathbf{B}$$, 12 scalar unknown components. Thus, we need six more scalar equations. These are the so-called constitutive relations that describe the properties of a medium. In isotropic media, they are given by $$\mathbf{D} = \epsilon \mathbf{E} \ \ \ \ \ \ \ \ \ \ \mathbf{B} = \mu \mathbf{H} \tag{2.1.6}$$ In anisotropic media, they may be given by $$\mathbf{D} = \epsilon \cdot \mathbf{E} \ \ \ \ \ \ \ \ \ \ \mathbf{B} = \mu \cdot \mathbf{H} \tag{2.1.7}$$ where $$\epsilon$$ is the permittivity tensor and $$\mu$$ is the permeability tensor: $$\epsilon = \begin{bmatrix} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix} \ \ \ \ \ \ \ \ \ \ \mu = \begin{bmatrix} \mu_{xx} & \mu_{xy} & \mu_{xz} \\ \mu_{yx} & \mu_{yy} & \mu_{yz} \\ \mu_{zx} & \mu_{zy} & \mu_{zz} \end{bmatrix} \tag{2.1.8}$$

For electromagnetic fields at optical frequencies, $$\rho = 0$$ and $$\mathbf{J} = 0$$.

Is all of this correct?

• It looks wrong to me also.
– Dale
Dec 9, 2021 at 7:46
• Perhaps there's an assumption that you are solving the wave equation, in which case you guess harmonic solutions for some of these quantities, like $\vec{B}=\vec{B}_0e^{i\omega t}$. In that case, since the divergence of a curl is zero (show this using brute force calculation if you'd like), then that would imply that the divergence of $\vec{B}$ must be zero. There are likely some hidden assumptions in here. Dec 9, 2021 at 7:49
• @march Interesting. Would you post an answer to elaborate on this and show what this would look like with the wave equation assumption? Dec 9, 2021 at 7:55
• While $\nabla\cdot(\nabla\times\mathbf{E})=\epsilon_{ijk}\partial_i\partial_jE_k$ just follows from symmetries, this technique only shows $\partial_t\nabla\cdot\mathbf{B}=0$.
– J.G.
Dec 9, 2021 at 8:09
• The divergence of the curl is zero is a well known theorem. See this link for a proof.
– ludz
Dec 9, 2021 at 8:18

Taking the divergence of both sides gets

$$\nabla \cdot (\nabla \times \vec{E}) = \nabla \cdot \frac{\partial \vec{B}}{\partial t}$$

The left hand side is zero. This is a vector calculus identitiy that you can check by writing out the derivatives. On the right hand side you can rearrange the space and time derivative to get

$$\frac{\partial (\nabla \cdot \vec{B})}{\partial t} = 0$$

If the result is integrated with respect to time and the constant of integration is taken to be 0, this is identical to gauss' law of divergence $$\nabla \cdot \vec{B} = 0$$

In essence, Faraday's law states that there is no change in the divergence of $$\vec{B}$$. Gauss' law puts a stronger constraint that states that the divergence of $$\vec{B}$$ is always 0. So while they are not exactly independent, I wouldn't go so far to say that all the information in Gauss' law is encapsulated in Faraday's law.

• It would be worth making explicit that you can derive $\nabla\cdot\mathbf B = \mu_0\rho_m$, but that equating the magnetic charge density to zero cannot be derived as that's an external constraint. Dec 9, 2021 at 8:20
• While this sounds legit, the integration you are talking about is an integration with respect to time. That means that you perform the integral at every point in space - you can choose a spatially varying constant of integration, i.e. a completely arbitrary field. The time-derivative of any constant field is 0, the field need not be homogeneous in space. Dec 9, 2021 at 8:22
• Thanks for the answer. Where is this "Gauss' law of divergence" that says that $\nabla \cdot \vec{B} = 0$? I can't see it here en.wikipedia.org/wiki/Gauss%27s_law#Differential_form or en.wikipedia.org/wiki/Divergence_theorem Dec 9, 2021 at 8:39
• @The Pointer I made a mistake in calling it Gauss' law of divergence, that is a different theorem. I meant to say Gauss' law of magnetism which is the aforementioned $\nabla \cdot \vec{B} = 0$ Dec 9, 2021 at 8:44
• @CyrusTirband Ahh, ok, got it en.wikipedia.org/wiki/… Dec 9, 2021 at 8:45