Charged Particle (Homework 9-323)

1. Transformation of force.

Consider a charged-particle moving at speed v as seen in the laboratory. In the laboratory frame, this is an uniform cylinder of charge density ρ, with q the charge of each particle. What force does the laboratory observer see on a charge a distance d from the beam axis? You could have noticed that the charge density and current form a 4-vector (called the 4- current), then transform this 4-current to the rest frame, then find the (purely electrostatic) fields. Then transform the fields back to the laboratory frame to find the Lorentz force from the laboratory fields. Whew. You may find it simpler to again use the 4- vector nature of the 4-current to find the charge density in the charge’s rest frame (there’s no current in this frame). This charge density transformation is very simple, it’s just inverting Lorentz contraction along the beam axis direction. Then you can easily find the E-field in the charge’s rest frame (there’s no B-field). Then you can use the result of (a) to transform the force directly.

Here there is a trick: The electric and magnetic fields are called electromagnetic field because the magnetic filed in a certain chosen reference is just electric field.  Consider a current is a wire and a moving charge in the reference of the wire.

 Suppose the current is given by a positive linear distribution of charge moving to the right +λ and a negative linear distribution of charge moving to the left –λ. The two distributions are the same and therefore there is just magnetic field. Consider now the same wire and moving charge but in the reference of the moving charge.

Because one distribution is moving to the right and the other to the left their absolute speeds are different now

$s_±=(v∓u)/(1∓vu/c^2)$

And therefore the contraction factor $γ_±$ of the wire will be different for the positive distribution of charge than tor the negative distribution of charge. Thus, the charge is standing and the magnetic field was replaced by an electric field due to two non-equal + and – distributions of charge.

Now is simple to solve the question: define the 4 current:

$J=(j_x,j_y,j_z,cρ)=(j ⃗ ,cλS)$ where $λ$ is the charge linear density and $λ=ρS$

In the charge rest frame there is no current (as shown in the example before) and the new charge density is given by the Lorentz contraction of the wire along x (the charge volume is contracted therefore the charge distribution is bigger):

$J’=(0,0,0,c(γλS))$

Hence the electric field is the field of an infinite line of charge

$E_y’ (r)=λ’/(2π(y’ )^2 ϵ_0 )=γλ/(2π(y’ )^2 ϵ_0)$ and $E_x’ (r)=0$

From equations 12.67 we have

$E_y=E_y’*γ=(γ^2 λ)/(2πϵ_0 y^2)$ where $λ=ρS$