Charged particle moving relativistically through E field

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The discussion revolves around the behavior of a charged particle moving at 0.8c through a uniform electric field. Participants express confusion regarding the decrease in the particle's x-velocity and momentum when it enters the electric field, despite initial assumptions that no magnetic field would affect it. Clarification is provided that while the electric field appears stationary in the particle's frame, Lorentz contraction leads to a magnetic force that influences the particle's motion. The relationship between energy and momentum is highlighted, suggesting that as the particle gains velocity in the y-direction, its x-velocity must decrease to conserve momentum. This understanding aligns with the principles of special relativity and the Lorentz force equation.
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Right now I'm taking an introductory E&M class that uses Purcell's book "Electricity and Magnetism." The chapter we're covering focuses on deriving the Lorentz force law for moving charges using SR arguments. This is confusing because we've never covered SR before, so I have a lot of difficulty going through the appropriate transformations and understanding what's going on. One problem that is troubling me a lot goes like this:

Consider the trajectory of a charged particle moving with a speed 0.8c in the x direction when it enters a large region in which there is a uniform electric field in the y direction. Show that the x velocity of the particle must actually decrease. What about the x component of momentum?

This is totally counterintuitive to me, and if someone could explain it for me I would really appreciate it.
 
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In the coordinate system where the particle is moving at .8c, is the electric field stationary? If so, there should be no magnetic field, and thus no reason for the momentum of the particle in the x direction to change.


If the electric field was moving, it'd be a different story.
 
I'm not sure what you mean by the field being stationary. If you're asking if the electric field varies with time, then the answer is no. The observer seeing the electron moving past at 0.8c measures no magnetic field. In the reference frame of the particle, however, there is a magnetic force caused by the lorentz contraction of the field lines. In fact, the Lorentz force equation is the definition for the magnetic field. From my book: "It will turn out that a field B with [the properties of the Lorentz force equation] must exist if the forces between electric charges obey the postulates of special relativity."
 
I think I just figured it out. The quantity c^2p^2 - E^2 is not changed by the Lorentz transformation, so it seems like a good equation to start from. Since the electric field does work on the particle the energy term gets bigger, which means the momentum term must get smaller. Because the particle gains y-velocity, the x-velocity must decrease if the momentum is going to decrease. Does this sound reasonable?
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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