# In an electric field transformation, how come $E_\parallel = E'_\parallel$?

Let's assume we have a system $S$ with two infinite sheets oppositely charged with charge density $\pm\sigma$, moving with a velocity $\vec{v_0} = v_0 \hat{y}$, and another system $S'$ moving relatively to $S$ with velocity $\vec{v} = v \hat{x}$.

So in the sheets' self-reference frame their density charge is $\displaystyle{\sigma_0 = \frac{\sigma}{\gamma_0}}$. The sheets' velocity in $S'$ is: $\displaystyle{\vec{v_0'} = \frac{v_0}{\gamma}}$, and their charge density is:
$\displaystyle{\sigma' = \gamma'_0 \sigma_0 = \frac{\gamma'_0}{\gamma_0}\sigma = \sqrt{\frac{1-\beta_0^2}{1-{\beta'}_0^2}}\sigma}$.

Now, $\displaystyle{v'_0 \neq v_0 \implies {\beta'}_0 \neq \beta_0 \implies\sqrt{\frac{1-\beta_0^2}{1-{\beta'}_0^2}} \neq 1 \implies \sigma' \neq \sigma}$.

But in both systems, $\displaystyle{E_{\parallel} = E = \frac{\sigma}{\epsilon_0};\ E'_\parallel = E' = \frac{\sigma'}{\epsilon_0}}$. So how come (according to Joules-Bernoulli equation) that $E_\parallel = E'_\parallel$?

• 1. I am pretty sure that your expression for $v'_0$ is incorrect. Didn't you mean to use the relativistic velocity addition? 2. Why foes $\sigma$ change after all? Remember that $\sigma$ is the two-dimensional charge density. So I am also pretty sure that $\sigma = \sigma'$ and thus $E_{||} = E'_{||}$. – Prof. Legolasov Sep 23 '16 at 6:25
• @SolenodonParadoxus You're right, I meant that the y component of $v_0'$, which is $v'_{0,y} = \frac{v_0}{\gamma(1-\frac{v_xv}{c^2})} =_{v_x=0} \frac{v_0}{\gamma}$. Thus $v'_{0,y} \neq v_{0,y} = v_0$, and hence $\sigma \neq \sigma'$. – galra Sep 23 '16 at 11:15
• The answer is, assuming $E_{\parallel} = \mathbf{E} \cdot \mathbf{B}/B$, that the term $\mathbf{E} \cdot \mathbf{B}$ is a Lorentz invariant. – honeste_vivere Sep 23 '16 at 16:35