Download Why is there Magnetism?

Why is there Magnetism?
Einstein gave the answer:
Electricity + motion = Magnetism
From two points of view: Lorentz transformations
• Lab frame of reference: x, y, z, t
• Moving frame of reference: x/, y/, z/, t/
• Frames coincide at the origin when t = t/ = 0
Describing a moving mass in each frame
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Note that if v = c, then v/ also = c!
The speed of light is the same for all inertial observers!
This was one of the postulates of special relativity
The other: the laws of physics are the same for all inertial observers
Relativistic energy & momentum
• It follows that P = Ev/c2
• In particular, when v = c, P = E/c
• Note that P and E  as v  c
The “Pythagorean theorem” for energy and momentum
Note the negative!
Differentiating both sides with respect to t gives
2E dE/dt – 2c2 P • dP/dt = 0
but recalling P = Ev/c2 we can say
dE/dt = v • dP/dt
A surprise:
Recall if v = c then P = E/c
E = Pc
E2 – P 2c2 = m2c4
dP/dt acts like F
(Just like Newtonian physics; Power = Fv)
=0
 Light has m = 0
How P and F change from frame to frame
• (Px , E) form a Lorentz pair exactly like (x, t):
Px/ = g(Px – VE/c2)
x/ = g(x – Vt)
E/ = g(E – VPx)
t/ = g(t – Vx/c2)
Py/ = Py
y/ = y
Pz/ = Pz
z/ = z
• The forces, dP/dt, are more complicated:
dPx/
dt/
dPy/
dt/
=
=
g(dPx – (V/c2) dE)
g(dt – Vdx/c2)
dPy
g(dt – Vdx/c2)
=
=
dPx/dt – (V/c2) v • dP/dt
1 – (Vv/c2)
dPy/dt
g(1 – (Vv/c2))
• Special case: if v = V = (V, 0, 0) then
dPx//dt/ = dPx/dt
dPy//dt/ = g dPy/dt
Interaction of charge and current
I. from the Lab frame
• The densities of pluses and minuses are equal and opposite;
l+ = l ; l– = – l
• The net charge density is zero; lnet = l+ + l– = 0
• The electric field is thus zero
• The electric force FE = qE = 0
• The magnetic field B is not zero, but by Ampère’s Law
B = 2k/I/R
• The magnetic force on the charge is FB = qvB toward the wire
From the Lab frame, continued…
• Fx = 0
• Fy = qvB = qv (2k/I/R)
• I = (Coulombs/sec) = (Coul/meter)(meters/sec) = (l–)(–vd) = +lvd
Fy = qv • 2k/ • lvd /R
Lab Frame: magnetic force toward wire
II. From the Charge frame
• From the charge frame, the charge is stationary
• There can be no magnetic force on the charge, as its v = 0
• In this frame, the positives are moving to the left with speed v
• In this frame, the negatives are moving to the left with speed v d /
• The negative and positive charges experience different Lorentz contractions
• There is a net charge density l/
• There will be an electric field, E/ and an electric force FE/
From the Charge frame, continued…
• Since lengths  (lengths/g), and l is proportional to 1/length, it follows lgl
• The new positive charge density l+/ = +lg(v)
• The old negative charge density was already contracted due to its velocity vd
• First uncontract by dividing by g(vd), and then contract with g(v/);
l–/ = – (l/g(vd))g(v/) = – l(g(v/)/g(vd))
• Use the formula for v/ from before;
v/ = (v + vd)/(1 + (vvd /c2))
• g(v/) given by the usual formula, and after algebra (left to the interested reader!)
g(v/) = g(v) • g(vd) • (1 + (vvd /c2))
• Then
l–/ = – lg(v) • (1 + (vvd /c2))
• The charge density l/ = l+/ + l–/ = – lg(v)•(vvd /c2)
From the Charge frame, concluded…
• FB/ = 0 since the charge’s v = 0
• FE/ = qE/ = qE/ (–y) (toward the wire)
• By Gauss’ Law, E/ = 2kl//R/, and l/= – lg •(vvd /c2)
• R/ = R since R is perpendicular to the velocities
FE/ = q2kl//R = g • q• v• (2k/c2)•lvd /R
• To put this back into the Lab frame, use the result about Fy/: Fy/ = gFy ;
 FE = FE//g = q•v• (2k/c2)•lvd /R
Charge frame: electric force towards the wire
This is precisely the same as the previous result, provided k/ = k/c2.
Comparisons
Lab Frame
Charge Frame
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Charge moves with speed v
Current I in wire creates B
Net charge density lnet = 0
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Charge is stationary
Current in wire creates B/
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No electric force since E = 0
Net force is magnetic, equal to
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Net charge density lnet / is not 0 ;
charge densities of – and + are
different due to different speeds
and hence different Lorentz
contractions
No magnetic force since v = 0
Net force is electric, equal to
2k/qlvdv /R
2(k/c2)qlvdv /R
These are equal if k/ = k/c2
Conclusion
• An example was given to show that, at least in one case, the effects of
a magnetic field on a moving charge may be completely described as
the effects of an electric field on that same charge in its rest frame
• This is a general truth: magnetic effects are merely electric effects in
an appropriate reference frame, with the rules for transforming
between frames given by relativity
• Historically, the Lorentz transformations were found by Einstein as a
necessary condition that Maxwell’s equations be true for all nonaccelerated observers
• References: E.M. Purcell, Electricity and Magnetism, and T.M.
Helliwell, Introduction to Special Relativity, and of course A. Einstein,
“On the Electrodynamics of Moving Bodies”, 1905.