Special relativity - transformation of electromagnetic fields

[Aleolomorfo]

Homework Statement

In a reference frame $S$ there is a particle with mass $m$ and charge $q$ which is moving with velocity $\vec{u}$ in an electric field $\vec{E}$ and in a magnetic field $\vec{B}$. Knowing the relativisitc laws of motion for a particle in an EM field, find the transformation laws for $\vec{E}$ and $\vec{B}$ to a reference frame $S'$ which is moving with velocity $v$ along the $x$ axis.

Homework Equations

Relativisitc laws of motion:
$$\frac{d\epsilon}{dt}=q\vec{u}\cdot\vec{E}$$
$$\frac{d\vec{p}}{dt}=q(\vec{E}+\vec{u}\times\vec{B})$$

3. The Attempt at a Solution
I have to change the equations of motion in order to obtain the same identical equations with primed quantities. I can change the velocities, the coordinates and the momenta, imposing that the equations of motion must be equal to the above ones but with primed quantities I think I can obtain what the exercise want.
I've split the second equation in the three components, knowing that:
$$\vec{u}\times\vec{B}=\begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\u_x&u_y&u_z\\B_x&B_y&B_z\end{vmatrix}=\hat{x}(u_yB_z-u_zB_y)+\hat{y}(u_zB_x-u_xB_z)+\hat{z}(u_xB_y-u_yB_x)$$
Then I've changed the velocities using $u_x=\frac{u'_x+v}{1+vu'_x}$, $u_y=\frac{u'_y}{\gamma(1+vu_x)}$, $u_z=\frac{u'_z}{\gamma(1+vu'_x)}$; and the for the left hand side I've used this change:$\frac{d}{dt}=\frac{dt'}{dt}\frac{d}{dt'}+\frac{dx'}{dt}\frac{d}{dx'}=\gamma\frac{d}{dt}-\gamma v\frac{d}{dx}$ and also the lorentz tranformations for 4-moment. After doing all this changes the equations are messy and I don't find a way to arrange the terms.

 

[Fred Wright]

I suggest studying http://www.feynmanlectures.caltech.edu/II_26.html for a clear and detailed explanation of the issue from the undisputed master of EM field theory.
Peace,
Fred

 

[TSny]

Relativisitc laws of motion:
$$\frac{d\epsilon}{dt}=q\vec{u}\cdot\vec{E}$$ $$\frac{d\vec{p}}{dt}=q(\vec{E}+\vec{u}\times\vec{B})$$

I can get you started on one approach that will work. Multiply these equations by dt to obtain
$$d\epsilon=q\vec{dx}\cdot\vec{E}$$ $$d\vec{p}=q(\vec{E}dt+\vec{dx}\times\vec{B})$$
Here, $\vec{dx}$ is the displacement of the particle during the time $dt$.

The same equations must hold in the primed frame. So,
$$d\epsilon'=q\vec{dx'}\cdot\vec{E}'$$ $$d\vec{p'}=q(\vec{E}'dt'+\vec{dx'}\times\vec{B}')$$
Use transformation equations to transform $d\epsilon'$ and the components of $d\vec{p'}$ on the left to the unprimed frame and to transform $dt'$ and the components of $\vec{dx'}$ on the right to the unprimed frame.

Then see if you can go from there.

 

[Aleolomorfo]

I can get you started on one approach that will work. Multiply these equations by dt to obtain
$$d\epsilon=q\vec{dx}\cdot\vec{E}$$ $$d\vec{p}=q(\vec{E}dt+\vec{dx}\times\vec{B})$$
Here, $\vec{dx}$ is the displacement of the particle during the time $dt$.

The same equations must hold in the primed frame. So,
$$d\epsilon'=q\vec{dx'}\cdot\vec{E}'$$ $$d\vec{p'}=q(\vec{E}'dt'+\vec{dx'}\times\vec{B}')$$
Use transformation equations to transform $d\epsilon'$ and the components of $d\vec{p'}$ on the left to the unprimed frame and to transform $dt'$ and the components of $\vec{dx'}$ on the right to the unprimed frame.

Then see if you can go from there.

Thank you very much for your help. I've found the solution!