Faraday's Law and Galilean Invariance

Question

In Jackson's text he says that Faraday law is actually: $$ \oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = -k\iint_{\Sigma} \frac{\partial \mathbf B}{\partial t} \cdot \mathrm{d}\mathbf{S} $$ where $k$ is a constant to be determined.(page 210, third ed.).He claims that $k$ is not an independent empirical constant that must be measured from experiment, but is an inherent constant which for each system of units can be determined by Galilean invariance and also Lorentz force law.He writes the Faraday's law in two frames, lab frame and a moving frame with velocity $\mathbf{v}$, and by writing the above law in each of two frames and assuming :

• electric field in one frame is $\mathbf{E}'$ and in the other is $\mathbf{E}$ (so they are different) , but magnetic field is $\mathbf{B}$ in both frames!

• Galilean invariance needs :$$\iint_{\Sigma} \frac{\partial \mathbf B}{\partial t} \cdot \mathrm{d}\mathbf{S} $$ be equal in two frames deduces that :

• $k=1$

and also

• the electric field in the moving reference frame is $$\mathbf{E}' = \mathbf{E} + \mathbf{v} \times\mathbf{B}$$.

I know that this electric field ($\mathbf{E}'$ ,in the moving frame ) is only an approximation and the real $\mathbf{E}'$ that can be obtained using Lorentz transformations. Now the question is that

• how Galilean transformations which are wrong (are approximately correct) give the correct answer for $k$ ?

• Why we should assume that there are two electric fields ,one in the lab frame and one in the other , but just one magnetic field in both frames?

Answer 1

• how Galilean transformations which are wrong (are approximately correct) give the correct answer for k?

The Lorentz prediction and the Galilean prediction must agree in the limit that $v \to 0$ (or in the limit that $c \to \infty$). This is because $v=0$ corresponds to no transformation at all, so they had better both agree there. So if you take the transformation and evaluate it for smaller and smaller $v$, you'll find that $k=1$ still has to be true.

• Why we should assume that there are two electric fields ,one in the lab frame and one in the other , but just one magnetic field in both frames?

That's just the Galilean Transformation of the EM field. To see how it relates to the relativistic case, the Lorentz Transformation of the EM field is:

$$\mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$

$$\mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \mathbf{B} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$

When you take the limit that $c \to \infty$, we know that $\gamma \to 1$, so it just becomes:

$$\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B}$$

$$\mathbf{B}' = \mathbf{B}$$