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Let there be two point charges, positive or negative, having velocities in directions perpendicular to each other. I need to evaluate the total interactive force on those (Lorentz force).

I do it by assuming magnetic force is an effect of special relativity on the electrostatic force. In doing so I am finding a value half of the Lorentz force predicted in classical electrodynamics. I have proceeded with many different possible approaches, but the result is always the same. Why is it so?

Here is my reasoning: $$v_\text{rel}= \frac{\vec{V}-\vec{U}}{1-\vec{V}\cdot\vec{U}/c^2}$$ Since the charges are moving perpendicular to each other in the static (lab) frame, $V\cdot U=0$. The Lorentz factor between two frames, (1) the rest frame of the "field charge" $Q$ and (2) the rest frame of the "test charge" $q$ is given by $$\gamma = \frac{1}{\sqrt{1-(u^2+v^2)/c^2}}$$ Now I evaluate the proper force (considering purely electrostatic force) exerted by $Q$ on $q$, in the rest frame of $q$. $$\begin{align} E_\perp q &= \gamma E_{0\perp} q & E_\parallel q &= E_{0\parallel} q \end{align}$$ where $E_0$ is the field at $q$ in the rest frame of $Q$.

After finding these two forces, I resolved them in the X and Y directions, where X is the direction in which $Q$ moves and Y is the direction in which $q$ moves, both in the lab frame. $$\begin{align} \sin\theta &= \frac{u}{\sqrt{u^2+v^2}} & \cos\theta &= \frac{v}{\sqrt{u^2+v^2}} \end{align}$$ Then I get $$\begin{align} F_x &= \frac{1}{2}\frac{v^2}{c^2} E_0 q & F_y &= \biggl(1+\frac{1}{2}\frac{v^2}{c^2}\biggr)E_0 \end{align}$$ Now these forces will produce proper acceleration in their respective directions in the rest frame of $q$. $$\begin{align} \alpha_x &= \frac{F_x}{m_0} & \alpha_y &= \frac{F_y}{m_0} \end{align}$$ where $m_0$ is the rest mass of the particle with charge $q$. After that I proceed to transfer these forces to the lab frame. Acceleration of $q$ with respect to the lab is given by $$\begin{align} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} &= \frac{1}{\gamma^2}\alpha_x & \frac{\mathrm{d}^2y}{\mathrm{d}t^2} &= \frac{1}{\gamma^3}\alpha_y \end{align}$$ Obviously here $\gamma$ is given by $$\gamma = \frac{1}{\sqrt{1-u^2/c^2}}$$ so $$\begin{align} F_{\text{lab},x} &= \gamma m_0 \frac{\mathrm{d}^2x}{\mathrm{d}t^2} \\ &= \frac{1}{\gamma} m_0 \frac{F_x}{m_0} \\ &=\sqrt{1-u^2/c^2}\frac{1}{2}\frac{v^2}{c^2} E_0 q \\ &\approx \frac{1}{2}\frac{v^2}{c^2} E_0 q \end{align}$$ This is just half of the usual result in the X direction that we would classically find in the lab frame.

Also, $$\begin{align} F_{\text{lab},y} &= \gamma^3 \frac{1}{\gamma^3} m_0 \alpha_y \\ &= F_y \\ &= \gamma_v E_0 q \end{align}$$ where $\gamma_v= 1/\sqrt{1-v^2/c^2}$.

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closed as unclear what you're asking by ACuriousMind, Sebastian Riese, John Duffield, user10851, user36790 Feb 23 '16 at 7:11

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    $\begingroup$ maybe you could post your equations here... otherwise its not easy to know what you did wrong. $\endgroup$ – AccidentalFourierTransform Feb 20 '16 at 11:31
  • $\begingroup$ equations contain lot of symbols and signs. i dont know how those can be written here. $\endgroup$ – Kausik Bhaduri Feb 20 '16 at 14:34
  • $\begingroup$ i am describing the steps i followed. with respect to the frame of moving test charge, ie rest frame of TC, i found the electrostatic forces exerted upon the TC. here i consider the relative velocity of principal charge wrt TC frame while computing γ factor . next F parallel= F parallel as it would have been had it been in rest wrt to TC, in the direction parallel to V relat. F perpend= γ times F perpend, as it would have been had it been in rest wrt to TC, in the direction normal to V relat. next i resolved those forces in X direction and Y direction. $\endgroup$ – Kausik Bhaduri Feb 20 '16 at 14:57
  • $\begingroup$ • The velocity of test charge wrt to lab= U • The velocity of field charge wrt lab= V • Hence the velocity of test charge q wrt field charge Q=(V-U)/ 1- V.U /c^2 = V-U , since v and u are perpendicular . • Hence γrel= 1/(1-(v^2+u^2)/c^2 $\endgroup$ – Kausik Bhaduri Feb 22 '16 at 6:10
  • $\begingroup$ • The proper force in perpendicular direction on q = γrel Fperpend • The proper force in parallel direction on q= Fparallel • F=vectorQq/4πε0y^2 • Resolving F in X and Y direction • Fx=1/2*(v^2/c^2)*Qq/4πε0y^2 $\endgroup$ – Kausik Bhaduri Feb 22 '16 at 6:23