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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.
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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 • 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 A surprise: Recall if v = c then P = E/c 2E dE/dt – 2c2P • dP/dt = 0 E = Pc but recalling P = Ev/c2 we can say E2 – P 2c2 = m2c4 dE/dt = v • dP/dt dP/dt acts like F = 0 Light has m = 0 (Just like Newtonian physics; Power = Fv)
How P and F change from frame to frame Px/ = g(Px – VE/c2) x/ = g(x – Vt) t/ = g(t – Vx/c2) E/ = g(E – VPx) • (Px , E) form a Lorentz pair exactly like (x, t): Py/ = Py y/ = y Pz/ = Pz z/ = z • The forces, dP/dt, are more complicated: dPx/ dPx/dt – (V/c2) v • dP/dt g(dPx – (V/c2) dE) = = dt/ g(dt – Vdx/c2) 1 – (Vv/c2) dPy/ dPy dPy/dt = = g(dt – Vdx/c2) g(1 – (Vv/c2)) dt/ • Special case: if v = V = (V, 0, 0) then dPx//dt/ = dPx/dt dPy//dt/ = g dPy/dt
Interaction of charge and currentI. 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 vd / • 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 lgl • 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.
Charge moves with speed v Current I in wire creates B Net charge density lnet = 0 No electric force since E = 0 Net force is magnetic, equal to Charge is stationary Current in wire creates B/ Net charge density lnet/ is not 0 ; charge densities of – and + are different due to different speeds and hence different Lorentz contractions No magneticforce since v = 0 Net force is electric, equal to Comparisons Lab Frame Charge Frame 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 non-accelerated 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.