- #1

- 4,807

- 32

[tex]\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}\frac{m\vec{u}}{\sqrt{1-(u/c)^2}} = \frac{m}{\sqrt{1-(u/c)^2}}\frac{d\vec{u}}{dt}+\frac{1}{c^2}\frac{m\vec{u}}{(1-(u/c)^2)^{3/2}}\frac{du}{dt} = \frac{m}{\sqrt{1-(u/c)^2}}\vec{a} + \frac{a}{c^2}\frac{m\vec{u}}{(1-(u/c)^2)^{3/2}} [/tex]

So this means that the acceleration of the particle does no take place in the same diretion as the force. Instead it is in some weird direction dictated by the actual speed of the particle.

Is this right?