Answer to the Question 02/03
REFERENCE FRAMES
The 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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