# Lorentz Transformation of the 3-force

Recall from class that we get
\begin{align}
F^\mu \eta_{\mu\nu}U^\nu = F^0U^0-F^iU^i &= 0\\
\gamma F^0-\gamma u^i\frac{\text{d}p^i}{\text{d}\tau}&=0\\
F^0=u^i\frac{\text{d}p^i}{\text{d}\tau} = \gamma_u u^i \frac{\text{d}p^i}{\text{d}t} &= \gamma_u(\mathbf{f\cdot u})
\end{align}

Here $\gamma_u = 1/\sqrt{1-u^2}$ where $u$ is the velocity of the particle in the $O$ frame, and $\mathbf{f}$ is the 3-force. So to transform the force into another frame $O'$ moving with velocity $v$ in the $x$-direction, we do the Lorentz transformation on the 4-force $F^\mu$.

\begin{align}
F'^\mu =\sum_{\nu} {\Lambda^\mu}_\nu F^\nu\implies F'^{\,x} &= {\Lambda^x}_0 F^0 + {\Lambda^x}_x F^x\\
\frac{\text{d}p'^{x}}{\text{d}\tau} &=-\gamma_v\,v(\gamma _u\;\mathbf{f\cdot u})+ \gamma_v\frac{\text{d}p^{x}}{\text{d}\tau}\\\\
\gamma_{u'} \frac{\text{d}p'^{x}}{\text{d}t'} &= \gamma_v\left(\frac{\text{d}p^{x}}{\text{d}\tau} - \gamma_u v \; \mathbf{f\cdot u}\right)\\\\
&=\gamma_v\left(\gamma_u\frac{\text{d}p^{x}}{\text{d}t} - \gamma_u v \; \mathbf{f\cdot u}\right)\\\\
\frac{\text{d}p'^{x}}{\text{d}t'}&=\frac{\gamma_v\gamma_u}{\gamma_{u'}}\left(\frac{\text{d}p^{x}}{\text{d}\tau} - v \; \mathbf{f\cdot u}\right)\\
\end{align}

The prefactor can be evaluated explicitly with some work:

\begin{align}
\frac{\gamma_v\gamma_u}{\gamma_{u'}} = \sqrt{\frac{1-u'^2}{(1-v^2)(1-u^2)}} = \sqrt{\frac{1-\left(\dfrac{u-v}{1-uv}\right)^2}{(1-v^2)(1-u^2)}} = \frac{1}{1-uv}\sqrt{\frac{1+(uv)^2-u^2-v^2}{1-v^2-u^2+(vu)^2}} = \frac{1}{1-uv}
\end{align}

Hence, we get

$$f'_x = \frac{f_x-v(\mathbf{f\cdot u})}{1-uv}$$

For the $y$ and $z$ components,
\begin{align}
F'^{\,y} &= {\Lambda^y}_0 F^0 + {\Lambda^y}_y F^y\\\\
\frac{\text{d}p'^{\,y}}{\text{d}\tau} &=\frac{\text{d}p^{\,y}}{\text{d}\tau}\\\\
\gamma_{u'}\frac{\text{d}p'^{\,y}}{\text{d}t'} &=\gamma_{u}\frac{\text{d}p^{\,y}}{\text{d}t}\implies f_y'= \frac{f_y}{\gamma_v(1-uv)}
\end{align}