Lorentz transformation and Lorentz force

[sweet springs]

Lorentz transformation of electromagnetic field gives the relation $E'=\gamma(E+v\times B)$.
Lorentz force per unit charge is given as $F=E+v\times B$ without $\gamma$.
Don't we need coefficient $\gamma$ for F?

 

[Ibix]

No. Why would you? There's no change of frame involved in the force expression. F' will include a $\gamma$ somewhere, which you can get either from transforming the force or from transforming E, B, and v.

 

[sweet springs]

Thanks. If I am right the transformation of force from a system S where a particle is moving at v_x to any system S' which is moving relative to system S at a speed -v_x, so the particle is at rest there.

F_x'=F_x,\ F_y'=\gamma F_y,\ F_z'=\gamma F_z

In transforming force from S' ( where only E works ) to S, I am not sure how to deal with gamma in three directions.

P.S. Now I know that E and B also has inhomogeneity in directions and cancel out coefficient gamma. I am happy to confirm that Lorentz force formula holds strictly. Thanks again.

 

[vsv86]

I would urge you to start using four-vector notation. These questions do not arise there. I think what you are missing is that momentum here is 'relativistic', i.e. $\vec{Force}=d\vec{p}/dt$ where $\vec{p}=\gamma {m} \dot{\vec{r}}$

Assuimng that vector potential is static one can write the equation of motion for a charged point particle (directly from least action):

$\frac{dp_\alpha}{d\tau}=-qu^{\mu}F_{\mu\alpha}$

Where $p_\alpha$ is the covariant four-momentum, $\tau$ is proper time, $u^\mu$ is four-velocity and $F_{\mu\alpha}$ is the electromagnetic tensor, and $q$ is charge. Now we go into 3-dimensions and focus on spatial components:

$d\tau \to dt/\gamma$
$p_{\alpha=1,2,3} \to -\vec{p}$
$u^{\mu} \to \gamma \left(c, \dot{\vec{r}}\right)^\mu$
$F_{0i} \to \vec{E}/c$
$F_{ij} \to -\epsilon_{ijk}\left(\vec{B}\right)^k$

You thus get

$\frac{d\vec{p}}{dt}=q\left(\vec{E}+\dot{\vec{r}}\times\vec{B}\right)$

Since, by definition $p_{\alpha}=m u_\alpha=m\gamma\left(c,\dot{\vec{r}}\right)_\alpha=(\dots,\vec{p})_\alpha$, we have $\vec{p}=\gamma {m} \dot{\vec{r}}$

 

[vsv86]

I should probably clarify. What analysis above tells you is that the equation above is valid in any reference frame, and that when moving between reference frames the electromagnetic fields should be changed using standard Lorentz transforms, to get $d\vec{p}/dt$ which by definition is the force on the object.