# How should MTW's derivation of the Maxwell-Faraday formula be interpreted?

In the following derivation, I am not sure exactly how the components of the final vector equation are established. I suspect this is a situation analogous to the vector addition of infinitesimal rotations. The discussion is in reference to section 3.4 beginning on page 80 of Gravitation by Misner, Thorne and Wheeler.

The electromagnetic field tensor is determined by writing the Lorentz force law as a derivative of 3-momentum with respect to proper time, and writing the derivative of energy with respect to proper time as the time component of the particle's momentum 4-vector. That is, the electromagnetic field tensor is defined without appeal to Maxwell's equations.

The objective is to derive the Maxwell-Farady formula, beginning with the invariant vanishing of the divergence of the magnetic field, and the transformation laws for the electric and magnetic fields. We consider an infinitesimal Lorentz boost $$\vec{\beta}$$ in the positive $$X$$ direction. The following form of the magnetic part of the electromagnetic field transformation is established by applying the Lorentz transformation to the electromagnetic field tensor and equating terms

$$\begin{bmatrix}\overline{\mathfrak{B}}_{\parallel}\\ \overline{\mathfrak{B}}_{\perp} \end{bmatrix}=\begin{bmatrix}\mathfrak{B}_{\parallel}\\ \gamma\left(\mathfrak{B}_{\bot}-\overset{\rightharpoonup}{\beta}\times\mathfrak{E}_{\bot}\right) \end{bmatrix}=\begin{bmatrix}B_{x}\\ \gamma\left(B_{y}+\beta E_{z}\right)\\ \gamma\left(B_{z}-\beta E_{y}\right) \end{bmatrix}. \tag1$$

Since our boost is infinitesimal, we may write $$\gamma=1$$. The transformation of the partial derivatives with respect to space coordinates for an infinitesimal boost in the $$X$$ direction are

$$\frac{\partial}{\partial\overline{x}}=\frac{\partial}{\partial x}+\beta\frac{\partial}{\partial t};\frac{\partial}{\partial\overline{y}}=\frac{\partial}{\partial y};\frac{\partial}{\partial\overline{z}}=\frac{\partial}{\partial z}.$$

In terms of the barred system, the divergence law of the magnetic field takes the form

$$\overline{\nabla}\cdot\overline{\mathfrak{B}}=0=\frac{\partial B_{\overline{x}}}{\partial\overline{x}}+\frac{\partial B_{\overline{y}}}{\partial\overline{y}}+\frac{\partial B_{\overline{z}}}{\partial\overline{z}}.$$

We rewrite this by replacing the barred terms with their unbarred equivalents:

$$0=\frac{\partial B_{x}}{\partial x}+\beta\frac{\partial B_{x}}{\partial t}+\frac{\partial B_{y}+\beta E_{z}}{\partial y}+\frac{\partial B_{z}-\beta E_{y}}{\partial z}.$$

We group factors of $$\beta$$

$$0=\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z}+\beta\left(\frac{\partial B_{x}}{\partial t}+\frac{\partial E_{z}}{\partial y}-\frac{\partial E_{y}}{\partial z}\right),$$

then apply the magnetic field divergence law in its unbarred form

$$\nabla\cdot\mathfrak{B}=0=\frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z},$$

to arrive at the condition

$$\frac{\partial B_{x}}{\partial t}+\frac{\partial E_{z}}{\partial y}-\frac{\partial E_{y}}{\partial z}=0.$$

Had the velocity of the transformation been directed in the $$y-$$ or $$z-$$ directions, a similar equation would have been obtained for $$\partial B_{y}/\partial t$$ or $$\partial B_{z}/\partial t$$. In the language of three-dimensional vectors these three equations reduce to the one equation $$\frac{\partial\mathfrak{B}}{\partial t}+\nabla\times\mathfrak{E}=\mathfrak{0}. \tag2$$

If the boosts under consideration were relativistic in magnitude, I don't believe we could consider the results as constituting independent vector components. My uncertainty in how to interpret the quoted statement is that it implies that we are obtaining our results one at a time, but combining them as if they all exist simultaneously. Would someone please help me understand this?

$$\let\b=\beta$$
There are two possible replies to your doubt. The first is that there is no need to consider high-$$\b$$ boosts. For the sake of the argument calculation is only required to first order in $$\b$$. To this order boosts in different directions do have additive effects.
Another way to see same thing is to modify eq. (1) (I've edited your post in order to have relevant equations tagged). Assume $$\vec\b$$ has an arbitrary direction, i.e. components along all three axes. I leave you to write the second passage to take this into account. If you develop the ensuing calculation you will directly arrive at eq. (2).