Answer to the Question 02/03
REFERENCE FRAMESThe question was:
A positively charged particle starts moving with velocity V0 parallel to a uniformly positively charged infinite plate. The plate generates field E perpendicular to it. The motion of the particle can be described by a parabola; after a while it will acquire velocity V1 in the vertical direction and will continue to move with velocity V0 in the horizontal direction, as depicted in Fig. (a)
Now let us look at this motion in the reference frame moving with velocity V0 to the right. As depicted in Fig. (b), the particle moves with increasing velocity in the vertical direction, but does not move in the horizontal direction. However, in this reference frame, the infinite plate moves to the left with constant velocity and generates a magnetic field of size V0E/c in the direction perpendicular to the plane of the picture. As a result, there is a force acting on the particle in the horizontal direction, and consequently it must accelerate!
Find the error in the above reasoning, and explain the apparent paradox.
(5/03) The problem has been solved (4/2/2003) by Srikant Marakani a graduate student the University of Chicago (e-mail srikant@midway.uchicago.edu), (10/2/03) by Dimitrios Vardis (e-mail vardis@ote.net), (8/3/03) by Alexander Shpunt (e-mail AlexanderS@radvision.com), (12/3/03) by Yoga Divayana an undergraduate Student of School of Electrical and Electronics Engineering Nanyang Technological University (Singapore) (e-mail divayana@pmail.ntu.edu.sg), (30/5/03) by Ivan Sirakov from Laboratoire de Rheologie des Matieres Plastiques at CNRS, St. Etiene, France (e-mail Yvan.Sirakov@univ-st-etienne.fr) - see his very detailed solution in the following PDF file, (24/6/2003) jointly by Roberto D'Agosta (e-mail dagosta@fis.uniroma3.it) and Martino De Prato (e-mail deprato@fis.uniroma3.it), and (26/06/03) by Carlos Soria from Sevilla (Spain) (e-mail cshoyo@us.es).
The answer:
The answer to the problem is that the velocity V0 is not constant in the original reference frame since its relativistic momentum mV0/sqrt(1-V2/c2), where V is the total velocity, would then be increasing. Hence, V0 must be decreasing which would give the same direction for the acceleration as the force due to the magnetic field in the second frame. One can directly verify this statement quantitatively.
Dimitrios Vardis sent us the following detailed description of the process as is seen from each of the reference frames:
Let K be the inertial frame relative to which the plate is at rest and K' the inertial frame relative to which the plate is moving with velocity - Vo. The relativistic equation of motion dp/dt = F is invariant, i.e. it has the same form in all inertial frames of reference. It is more convenient in our case to write this equation in the form.
dp/dt = F or d(mu)/dt = F or mdu/dt + dm/dt u = F or ma = F - dm/dt u
but mc2= E or c2dm/dt = dE/dt = F.u so the last of the above equations takes the form ma = F- (F u/c2)u.
In the K frame the projection of the above equation along the direction of the relative motion of the two frames (x-axis) is
max = Fx - (qEV1/c2)ux
or
max = - (qEV1/c2)Vo at the moment the velocity along x-axis is Vo)
In the K' frame relative to which the charged plate is moving it generates an electrical field perpendicular to the plate and a magnetic field parallel to the plate and at right angle to the page. Taking the y-axis perpendicular to the plate, the electric and magnetic field vectors in the K' frame are
E'y = gy
B'y=-uEg/c2 (g = Lorentz factor)
The equation of motion in K' is max = -qVoEg/2 (along x-axis at the moment the velocity of charged particle relative to K is Vo) etc. Therefore there is an acceleration of the particle along the x-axis in both frames as requires the principle of relativity...
Alexander Shpunt performed a detailed calculation of the actual motion described in the following postrscript file. In his derivation you can actually see that horizontal velocity is decreasing due to increasing mass of the particle.
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