This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Phys 571 , Fall 2010 Solution to Assignment #4. 1. The diﬀerence between the two gauges is A −A= Since 2. 3. (a) E = γ (E − V × B /c) − (γ − 1)V V·E V2 V·B V2 × ( f ) = 0, magnetic ﬁeld is the same. − xyB 2 B = γ (B + V × E /c) − (γ − 1)V (b) There are inﬁnitely many solutions to this problem. If in some reference frame K , which moves with velocity V, E B then in any other frame that moves parallel to K along the common direction of the ﬁelds E and B will be parallel (this follows from equations derived in (a)). Therefore, it suﬃcient to ﬁnd a system K that moves perpendicularly to the plane B, E. Now, E × B = 0 since E B. It is convenient to write eqs of (a) for E and B (just replace E ↔ E , B ↔ B , V → −V) and use E × B = 0, V · B = V · E = 0 to obtain a quadratic equation for V/c. Its solution: E2 + B2 − V =E×B c Using the ﬁeld invariants: 1 E 2 = [E 2 − B 2 + 2 1 B 2 = [B 2 − E 2 + 2 (E 2 − B 2 )2 + 4(B · E)2 ] (E 2 − B 2 )2 + 4(B · E)2 2(E × B)2 (E 2 − B 2 )2 + 4(B · E)2 ] (c) Using the ﬁeld invariants we see that if E > B , there exists a frame in which B = 0 √ and E = E 2 − B 2 . Analogously, if E < B , there exists a frame in which E = 0 and √ B = B 2 − E 2. 2 E > B case: V=c E×B , E2 ˆ E =E E2 − B2 In any other frame moving along E with an arbitrary velocity magnetic ﬁeld is absent. E < B case: V=c 4. B×E , B2 ˆ B = B B2 − E2 ...
View
Full
Document
This note was uploaded on 02/24/2011 for the course PHYS 571 taught by Professor Krill during the Fall '10 term at Iowa State.
 Fall '10
 krill

Click to edit the document details