1 / 13

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.

Download Presentation

Why is there Magnetism?

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Why is there Magnetism? Einstein gave the answer: Electricity + motion = Magnetism

  2. 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

  3. 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

  4. 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

  5. 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)

  6. 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

  7. 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

  8. 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

  9. 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/

  10. 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)

  11. 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.

  12. 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

  13. 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.

More Related