Principle of relativity and point particle in electromagnetic field Equaling relativistic version of 2nd Newton law to Lorentz force
$$
\frac{d}{dt} (\gamma_u m \mathbf{u}) = q(\mathbf{E} + \mathbf{u} \times \mathbf{B})
$$
we can see  that a point particle (of mass $m$ and charge $q$) in a electromagnetic field ($\mathbf{E},\mathbf{B}$) has acceleration 
$$
\mathbf{a} = 
\frac{q}{\gamma_u m} \left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right]
$$
where I don't use nor four vector formalism, wich I don't know (I speak about ordinary acceleration), neither Gauss units. Let's consider the usual $S'$ system (in motion at speed $v$ toward positive $x$). Lorentz transformations give
$$
\mathbf{u'} = \left( \frac{u_x - v}{1-\frac{u_x v}{c^2}} ,
 \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} ,
 \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} \right)
$$
($\gamma_u$ relative to speed of particle in $S$ should not be confused with $\gamma$ relative to the speed of $S'$) from wich we get the gamma factor of the particle by S' measurements. After many simplifications we get
$$
\gamma_{u'} = \frac{1}{\sqrt{1-\frac{u'^2}{c^2}}} = \gamma \gamma_u (1- \beta_{ux} \beta)
$$
where $\gamma=\frac{1}{\sqrt{1-\beta^2}}$, $\gamma_u=\frac{1}{\sqrt{1-\beta_u^2}}$, $\beta=\frac{v}{c}$, $\beta_{u}=\frac{u}{c}$ and $\beta_{ux}=\frac{u_x}{c}$. But
$$
\mathbf{E}' = (E_x , \gamma (E_y - v B_z) , \gamma (E_z + v B_y)) 
$$
$$
\mathbf{B}' = \left(B_x , \gamma \left(B_y + \frac{v}{c^2} E_z \right) , \gamma \left(B_z - \frac{v}{c^2} E_y \right) \right) \label{trab}
$$
Now, I expect that if $S'$ calculus acceleration using Lorentz force with the speed and fields measured by him, he should find the acceleration of $S$ transformed using Lorentz transformations, wich for example for $x$ component gives $a_x' = \frac{a_x}{\gamma^3 \left( 1 - \beta_{ux} \beta \right)^3 } $. But $\frac{q}{m}=\frac{q'}{m'}$, so $x$ components of $S$ and $S'$ should be related by (I use subscript $x$ to denote $x$ componentof vector in square brackets)
$$
\frac{1}{\gamma_{u'}}
\left[ \mathbf{E'} + \mathbf{u'}\times\mathbf{B'} - (\mathbf{E'} \cdot \mathbf{u'}) \frac{\mathbf{u'}}{c^2} \right]_x = 
 \frac{1}{\gamma^3 \left( 1 - \beta_{ux} \beta \right)^3 } \frac{1}{\gamma_u}
\left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right]
_x 
$$
wich can be arranged in this better way
$$
\gamma^2 {  (1 - \beta_{ux} \beta)^2 }
\
\left[ \mathbf{E'} + \mathbf{u'}\times\mathbf{B'} - (\mathbf{E'} \cdot \mathbf{u'}) \frac{\mathbf{u'}}{c^2} \right]_x = 
\
\left[ \mathbf{E} + \mathbf{u}\times\mathbf{B} - (\mathbf{E} \cdot \mathbf{u}) \frac{\mathbf{u}}{c^2} \right]
_x 
$$
Substituting $\gamma$, $\beta$, $\beta_{ux}$ and primed quantities written above, I expect to find an identity but this doesn't work: what went wrong here?
 A: Lorentz transformations are designed to relate the particle's properties in two different inertial frames. When you try to make a transformation into a non-intertial frame, e.g. when the Lorentz-factor $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ becomes time dependant with $v = v(t)$ then
$$ \Lambda^\mu_{~~~\nu} \frac{d^2}{d\tau^2} x^\nu \ne \frac{d^2}{d\tau^2} \left(\Lambda^\mu_{~~~\nu}~ x^\nu \right) $$
If you transform your fields E and B with $\gamma = \gamma(t)$, you are calculating the acceleration in S and then transform it to S' (left hand side). This means you calculate the instantaneous acceleration measured by an observer in an inertial frame at that single instance of time in which the particle and the observer in S' travel at the same speed. So strictly speaking, that acceleration is not measured relative to the particle but relative to a different observer which is in an inertial frame.
If you use the right hand side, and equate it to the transformed Lorentz-force, you introduce so called fictitious forces, which you should know form newtonian-mechanics, e.g centrifugal forces. This is the closest you can get with special relativity in describing a frame which is moving with the accelerated particle.
A: It isn't true that the final equation doesn't work: it work. The right side is
$$
 E_x + u_y B_z - u_z B_y - \frac{E_x u_x + E_y u_y + E_z u_z}{c^2} u_x
$$
While the left side is
$$
\gamma^2 { (1 - \beta_{ux} \beta)^2 } \left[ 
 E'_x + u'_y B'_z - u'_z B'_y - \frac{E'_x u'_x + E'_y u'_y + E'_z u'_z}{c^2} u'_x \right]
$$
where (I don't write $B'_x=B_x$ because we don't need it now)


*

*$ E'_x = E_x $

*$ E'_y = \gamma (E_y - v B_z) $

*$ E'_z = \gamma (E_z + v B_y) $

*$ B'_y = \gamma \left(B_y + \frac{v}{c^2} E_z \right) $

*$ B'_z = \gamma \left(B_z - \frac{v}{c^2} E_y \right) $

*$ u'_x = \frac{u_x - v}{1-\frac{u_x v}{c^2}} $

*$ u'_y = \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} $

*$ u'_z = \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)} $

*$ \gamma = \frac{1}{\sqrt{1-\beta^2}} $

*$ \beta = \frac{v}{c} $

*$ \beta_{ux} = \frac{u_x}{c} $


Doing substitution we see that both sides are the same.
In the calculus, they can be useful these intermediate steps:
$$
E'_x u'_x + E'_y u'_y + E'_z u'_z = 
\frac{E_x u_x + E_y u_y + E_z u_z + v (B_y u_z - B_z u_y - E_x) }{1-\frac{u_x v}{c^2}} 
$$
$$
u'_y B'_z - u'_z B'_y = 
\frac{u_y \left( B_z - \frac{v E_y}{c^2} \right) - u_z \left( B_y + \frac{v E_z}{c^2} \right)}{1-\frac{u_x v}{c^2}} 
$$
I did by hand this check two times, I'm sure the last equation of my answer works. I checked $y$ component too (and surely $z$ is ok by symmetry).
