Relativistic Application of the Lorentz Force

[VRCAM]

Hi!

I would really appreciate some help on the relativistic Lorentz force as applied to a charged particle; it's my first post on this site btw. Thanks in advance ;)



QUESTIONS

(a) A charged particle with total energy E0 is injected into a region of space containing a uniform electric field at right-angles to its direction of motion. It is deflected by the electric field and after a time is traveling at 45 degrees to its original line of flight. Show that the energy, E1, of the particle at that point is given by (where m is the mass of the particle):
E1^2 = 2*E0^2 - m^2*c^4

(b) The component of the particle’s velocity in the direction of its original line of flight decreases with time, even though no force acts upon the particle in that direction? Explain why this is so. Obtain an expression for this velocity component at the point where the particle is moving at 45 degrees to the original line of flight.



NOTATION

gamma(u) is the Lorentz factor for velocity u

I have taken the particle to be traveling initially in the +z direction at velocity u0, with the electric field acting in the +x direction. At 45 degrees the particle travels at a velocity u.



MY ATTEMPTED SOLUTION FOR (a)

Use the the energy momentum invariant:

E0^2 = {gamma(u0)*u0*mc}^2 + {mc^2}^2

E1^2 = {gamma(u)*u*mc}^2 + {mc^2}^2

E1^2 = {gamma(u)*mc}^2*{u0^2(1+cot^2(45))} + {mc^2}^2
since the z - velocity is unchanged in the IRF??

E1^2 = 2 * {gamma(u)*u0*mc}^2 + {mc^2}^2

The stated answer follows from algebra



MY ATTEMPTED SOLUTION FOR (b)

The momentum 4-vector for the particle at 45 degrees is:

P = (E1, px', py', pz')

P = (E1, gamma(u)*m*u0, 0, gamma(u)*m*u0)

So the z velocity component is gamma(u)*m*u0 where u^2 = 2*u0^2

The z velocity is different despite the absence of a force in the z direction due to the E-field having a z-component in the particle's IRF??



The reason I'm really unsure of my answer is that it is unclear which frames I am supposed to be calculating all the forces in, and I'm confused as to the geometric relationship between the different velocities in them.

Thank you!

VR

 

[Jhero]

Hi VR,

Thank you for your post and for reaching out for help on this topic. The relativistic Lorentz force is a fundamental concept in electromagnetism and plays a crucial role in understanding the motion of charged particles in electric and magnetic fields. Let's dive into your questions and see if we can clarify some of the confusion you have.

(a) Your attempted solution for this part is on the right track. The energy-momentum invariant equation you have used is a good starting point. However, there are a couple of things that need to be addressed. Firstly, the Lorentz factor, gamma(u), is a function of velocity, not a function of time. So it is incorrect to use gamma(u) at one point and gamma(u0) at another point. The Lorentz factor is defined as gamma(u) = 1/sqrt(1 - u^2/c^2). Secondly, the z-velocity component is not unchanged in the IRF (inertial reference frame) of the particle. In fact, the z-velocity component is always changing due to the presence of the electric field. This is what causes the particle to be deflected in the first place. So the correct equation for E1^2 should be:

E1^2 = {gamma(u)*u*mc}^2 + {mc^2}^2

= {gamma(u)*mc}^2*{u0^2(1+cot^2(45))} + {mc^2}^2

= {gamma(u)*mc}^2*{2*u0^2} + {mc^2}^2

= 2*{gamma(u)*u0*mc}^2 + {mc^2}^2

= 2*E0^2 - m^2*c^4

So your final answer is correct, but the steps to get there needed some clarification.

(b) You are correct in saying that the z-velocity component is different at 45 degrees compared to the initial z-velocity component. This is due to the Lorentz transformation that occurs when we change reference frames. The electric field is acting in the +x direction in the laboratory frame, but in the IRF of the particle, it appears to have a z-component as well. This is known as the electric field's "magnetic component" and it is what causes the particle to be