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Radiation from an accelerated charge Erad P P’ • q is initially at rest at P • q is accelerated and moves from P to P’ • After some time has passed, the electric field at point is changed by an amount Erad 1 29:130 S 2012 Notes for 4.12.13 y M’ d M N’ P2 P1 O • +q at rest at O; E lies along line OL L • Q given uniform accel. a along Oy for a short time N interval . At the end of this short interval q will be at P1 and will be moving with velocity v = a , where v << c. • After an interval t >> , q will reach P2, where P1P2 = v t. • The “kink” in the electric field line OL, due to the accel. of q, moves outward along OL with velocity c. At the end of the interval + t, it will have reached N. • After passing P1, q moves with uniform velocity v and a point on the electric field line which leaves q at P1 has moved uniformly outward and the line is now in the position P2M, almost parallel to OL; also OP1 and P1P2 are << compared to the radii of the two spheres, which are separated by an amount d = c . 2 analysis c M’ M Er Et N’ c E N • The outer spherical shell moves outward at c. In front of this shell, the E field is stationary, while behind it, there is a moving field. • A disturbed field moves outward to change the field due to the acceleration. The disturbed field has radial and tangential components– Er and Et. Et MN Er MM MM c P2 , M MN OP2 sin vt sin , since t N’ O 3 Et vt sin q , but Er , 2 Er c 4 o r vt sin q Et , 2 c 4 o r but t r c , and v a, qa sin Et 2 4 o c r • • This is the radiated electric field due to the accelerated charge. Note that if a = 0, Et = 0 an accelerated charge radiates electromagnetic waves. is the angle between the velocity of the charge and the observer; there is no radiation in the forward direction ( = 0). 4 The magnetic field can be obtained simply by realizing that at large distances the radiation fields must conform to plane waves, so that B = E c ,so with Erad qa sin 1 qa sin , Brad . 2 2 4 o c r c 4 o c r The Poynting vector is then: S= Erad Brad o qa sin 1 3 2. 4 o c r 2 The angular distributionof the radiation is sin 2 . 5 Angular distribution of radiated power 6 Total Radiated power The total power radiated by the accelerated charge is given by the integral of S over a sphere centered at the instantaneous position of the charge. 2 qa 1 P S da 3 4 o c 2 0 sin 2 2 2 r sin d 2 r 2 qa 2 3 qa 2 4 3 0 sin d 3 . 4 o c 4 o c 3 1 Writing o , this can be expressed as 2 oc 2 o q 2 a 2 P= . 3 4 c This is the famous Larmor power formula. 7