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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 $\mathbf{\bar{\hat{x}}}$?

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    $\begingroup$ There seems to be a typo in the second sentence, $x$-direction should be $y$ direction. $\endgroup$
    – user4552
    Commented Jan 3, 2019 at 17:39

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The Lorentz transformation for a boost in the $y$ direction doesn't change $\hat{\textbf{x}}$.

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  • $\begingroup$ 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? $\endgroup$ Commented Jan 3, 2019 at 17:55
  • $\begingroup$ @Dani. Isn't $\mathbf{\hat{x}} = \mathbf{\bar{\hat{x}}}$? $\endgroup$
    – md2perpe
    Commented Jan 3, 2019 at 19:02
  • $\begingroup$ 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 $E_x$? $\endgroup$ Commented Jan 3, 2019 at 20:41
  • $\begingroup$ which is a unit vector along the boost In the question, you define the boost as being in the y direction, not the x direction. I think it boils down to the question to which basis e.g. E¯x is defined. Is this the same basis as for Ex? The y and t basis vectors are different, but the x and z basis vectors are the same in S and S-bar. $\endgroup$
    – user4552
    Commented Jan 3, 2019 at 21:03
  • $\begingroup$ Can you please state how the basis vectors exaclty transform? $\endgroup$ Commented Jan 4, 2019 at 0:14

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