The flaw in my derivation of relativistic momentum/mass?

Question

I am trying to derive relativistic momentum in a manner similar to that of Max Planck in "The Principle of Relativity and the Fundamental Equations of Mechanics". I begin with a frame of reference $S$ with coordinates $x$, $y$, $z$, and $t$. In that frame of reference a charged particle of charge $\epsilon$ travels along the $x$-axis with a velocity $v$. There is another frame $S'$ with coordinates $x'$, $y'$, $z'$ and $t'$, which is moving at a constant velocity $v$ along the axis, that is, it initially moves together with the charged particle. A purely electric field is present in S' ($\vec{E}=[E_x, E_y ,E_z]$), exerting a force on the charge. Since in the frame $S'$ the charge is not moving, the equations of motion reduce to Newton's classical equations of motion: $$m\dfrac{d^2x'}{dt'^2}=\epsilon E_x'$$ $$m\dfrac{d^2y'}{dt'^2}=\epsilon E_y'$$ $$m\dfrac{d^2z'}{dt'^2}=\epsilon E_z'$$ Transforming these equations of motion into $S$ coordinates, using the transformation of the electromagnetic field (in Gaussian Units) and the Lorentz transformations: $$m\gamma^3 \dfrac{d^2x}{dt^2}=\epsilon E_x$$ $$m\gamma^2 \dfrac{d^2y}{dt^2}=\epsilon \gamma\left(E_y-\frac{v}{c}B_z\right)$$ $$m\gamma^2 \dfrac{d^2z}{dt^2}=\epsilon \gamma\left(E_z+\frac{v}{c}B_y\right)$$ Where $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ And a magnetic field is present in S such that $\vec{B}=[B_x, B_y, B_z]$. If we make it so that the force present of the $S$ frame will be the Lorentz force due to the Electromagnetic Field is equivalent to the electric force in the $S'$ frame, but not necessarily equal. The components of the Lorentz Force are: $$F_x=\epsilon E_x+\frac{\epsilon}{c}(v_yB_z-v_zB_y)$$ $$F_y=\epsilon E_y+\frac{\epsilon}{c}(v_zB_x-v_xB_z)$$ $$F_z=\epsilon E_z+\frac{\epsilon}{c}(v_xB_y-v_yB_x)$$ Since the field in $S'$ is purely electric, electromagnetic field in $S$ is given by $$E_x=E_x'$$ $$E_y=\gamma E_y'$$ $$E_z=\gamma E_z'$$ $$B_x=0$$ $$B_y=-\gamma \frac{v}{c}E_z'$$ $$B_z=\gamma \frac{v}{c}E_y'$$ With these and the fact that in the $S$ frame the charge has no $y$ or $z$ conponents of velocity, we get that the Lorentz Force equations in $S$ are: $$F_x=\epsilon E_x$$ $$F_y=\epsilon \left(E_y-\frac{v}{c}B_z\right)$$ $$F_z=\epsilon \left(E_z+\frac{v}{c}B_y\right)$$ With these, the original equations of motion can be transformerd into $$m\gamma^3 \dfrac{d^2x}{dt^2}=F_x$$ $$m\gamma \dfrac{d^2y}{dt^2}=F_y$$ $$m\gamma \dfrac{d^2z}{dt^2}=F_z$$ From the $y$ and $z$ components I get the usual 'Relativistic Mass' formula $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ But I get a different 'mass' in the $x$-direction, one of $$m=\frac{m_0}{\left(\sqrt{1-\frac{v^2}{c^2}}\right)^3}$$ similar to what Einstein got in his original 1905 paper on Special Relativity using a different definition of force. Where is it that I went wrong?

Answer 1

As the link I provided in the comments explains, the force on an object has different proportionalities to mass and acceleration in the longitudinal and transverse directions. Specifically,

$$\mathbf{F} = \gamma^3 m_0\mathbf{a_{||}}+\gamma m_0\mathbf{a_{\perp}}$$

which is derived from $\mathbf{F} = \frac{d\mathbf{p}}{dt}$ by letting $\mathbf{p}=\gamma m_0\mathbf{v}$ and taking the product rule when differentiating $\gamma\mathbf{v}$.

As is plain to see, you obtained the correct relations, so there is no problem.