# Lorentz transformation of magnetic field

I am studying relativistic electrodynamics and I am stuck at a certain point regarding the transformation rules of the electric and magnetic fields. The stationary inertial frame will be called $$S$$ and the moving frame at velocity $$\mathbf{v} = v\mathbf{\hat{y}}$$ will be called $$\bar{S}$$, $$\mathbf{\hat{y}}$$ denotes the unit vector in frame $$S$$ along the $$y$$-direction.

My question is very general but I will explain it using an example. Say e.g. we have $$\mathbf{E} = E_0 \mathbf{\hat{z}}$$ and $$\mathbf{B} = B_0\mathbf{\hat{x}}$$. The general transformation rule is as follows (according to Wikipedia):

$$\mathbf{\bar{B}}_\perp = \gamma\Big(\mathbf{B}_\perp - \frac{\mathbf{v} \times \mathbf{E}}{c^2}\Big),$$

where $$\perp$$ denotes with respect to $$\mathbf{v}$$ and $$\gamma=(1-v^2/c^2)^{-1/2}$$ Now suppose I want to find $$\mathbf{\bar{B}}_\perp$$, then I am confused about the fact that the following two expressions seem to contradict.

(1) We note that $$\mathbf{v} \times \mathbf{E} = E_0 v\mathbf{\hat{x}}$$ and $$\mathbf{B}_\perp = B_0 \mathbf{\hat{x}}$$ therefore one finds:

$$\mathbf{\bar{B}}_\perp = \gamma \mathbf{\hat{x}}\Big(B_0 -\frac{vE_0}{c^2}\Big).$$

(2) Method 2 is based on what is stated in Griffiths. He writes the previous expressions component-wise:

$$\bar{B}_x = \gamma\Big(B_x - \frac{vE_z}{c^2}\Big).$$

But if I want to express this in the basis of the underlying vector space of $$\bar{S}$$, this is a different basis than the basis in case of $$S$$, so I should write:

$$\mathbf{\bar{B}}_\perp = \gamma \mathbf{\bar{\hat{x}}}\Big(B_0 - \frac{vE_0}{c^2}\Big).$$

I used $$\bar{B}_z = 0$$.

My question is: Why can we express the magnetic field in both bases using the same component $$\bar{B}_x$$ and how are these expressions consistent when one has $$\mathbf{\hat{x}}$$ in it and the other $$\mathbf{\bar{\hat{x}}}$$?

• There seems to be a typo in the second sentence, $x$-direction should be $y$ direction. – Ben Crowell Jan 3 at 17:39

The Lorentz transformation for a boost in the $$y$$ direction doesn't change $$\hat{\textbf{x}}$$.
• But if we take for instance $\mathbf{E} = E_0 \mathbf{\hat{x}}$ and $\mathbf{B} = \mathbf{0}$ in frame $S$ and boost with $\mathbf{v} = v\mathbf{\hat{x}}$ to frame $\bar{S}$ then we find using $\mathbf{\bar{E}}_\parallel = \mathbf{E}_\parallel$ that $\mathbf{\bar{E}} = E_0 \mathbf{\hat{x}}$. But one could also write $\bar{E}_x = E_0$ and then $\mathbf{\bar{E}} = E_0 \mathbf{\bar{\hat{x}}}$ which is a unit vector along the boost. My question is: how does this then make sense? – Dani Jan 3 at 17:55
• @Dani. Isn't $\mathbf{\hat{x}} = \mathbf{\bar{\hat{x}}}$? – md2perpe Jan 3 at 19:02
• I think it boils down to the question to which basis e.g. $\bar{E}_x$ is defined. Is this the same basis as for $E_x$? – Dani Jan 3 at 20:41