Field Theory exam II â Solutions

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(a) By applying a Lorentz-boost from the rest frame of the particle to the laboratory frame, prove that the electromagnetic fields are given by ⃗E = γq (⃗x − t⃗v) r ′3 ⃗B = γq ⃗v × ⃗x cr ′3 r ′2 = γ 2 (x 1 − vt) 2 + x 2 2 + x 2 3 (b) Compute the corresponding Poynting vector ⃗ S in terms of ⃗v, ⃗x, t, and r ′ . Also compute | ⃗ S|. Required knowledge • If K and K ′ are two inertial coordinate frames, and K ′ moves with velocity βc in in x-direction with respect to system K, and the coordinate axes are not rotated against each other, then the relativistic transformation is the Lorentz-boost given by x ′ 0 = γ(x 0 − βx 1 ) x ′ 1 = γ(x 1 − βx 0 ) (48) x ′ 2 = x 2 x ′ 3 = x 3 • The electromagnetic fields transform under the above Lorentz-boost like E 1 = E 1 ′ B 1 = B 1 ′ E 2 = γ (E 2 ′ + βB 3) ′ B 2 = γ (B 2 ′ − βE 3) ′ (49) E 3 = γ (E ′ 3 − βB ′ 2) B 3 = γ (B ′ 3 + βE ′ 2) • The electric field of a charge q at rest at the origin is given by ⃗E = q ⃗x (50) |⃗x| 3 • The Poynting vector in vacuum (in Gauss units) is given by Solution ⃗S = c 4π ⃗ E × ⃗ B (51) (a) In the following, the quantities with a prime are defined with respect to the rest frame of the particle, unprimed ones with respect to the laboratory frame. Since the particle moves with constant velocity, it’s rest frame is an inertial frame. In the rest frame of the particle, there is no magnetic field and the electric field is given by ⃗E ′ = q r ′3 ⃗x′ r ′ ≡ |⃗x ′ | (52) 6
Using the transformation rule (49) and the Lorentz-boost (48), we get the electric field in the laboratory frame ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ E 1 ′ ⃗E = ⎝γE 2 ′ ⎠ = q x ′ 1 ⎝γx ′ ⎠ r ′3 2 = q γ (x 1 − βx 0 ) ⎝ γx r ′3 2 ⎠ (53) γx 3 γE ′ 3 = qγ r ′3 ( ⃗x − x 0 ⃗ β ) γx ′ 3 = qγ (⃗x − t⃗v) (54) r ′3 For the magnetic field, we get ⎛ ⎞ 0 ⃗B = ⎝−βγE 3 ′ ⎠ = β ⃗ × E ⃗ = qγ βγE 2 ′ r ⃗ ) ′3 β × (⃗x − x 0β ⃗ (55) = qγ r ⃗ ′3 β × ⃗x = qγ ⃗v × ⃗x cr ′3 (56) Using (48), we finally find r ′2 = x ′2 1 + x ′2 2 + x ′2 3 = γ 2 (x 1 − vt) 2 + x 2 2 + x 2 3 (57) (b) Using (51), (55), and (54) we obtain ⃗S = c E 4π ⃗ × B ⃗ = 1 = 1 4π E 4π ⃗ ( × ⃗v × E ⃗ ) = 1 ( ( ) ) ⃗E 2 ⃗v − ⃗E · ⃗v ⃗E 4π (58) ( qγ r ′3 ) 2 (⃗y 2 ⃗v − (⃗y · ⃗v) ⃗y ) (59) where ⃗y ≡ ⃗x − t⃗v. Further, Problem 4 | S| ⃗ = 1 ( qγ 4π = 1 4π ) 2 √ r ′3 ⃗y 4 ⃗v 2 − 2⃗y 2 (⃗y · ⃗v) 2 + ⃗y 2 (⃗y · ⃗v) 2 (60) ( qγ ) 2 √ |⃗y| r ′3 ⃗y 2 ⃗v 2 − (⃗y · ⃗v) 2 (61) A particle with charge q, mass m, and velocity ⃗v crosses a region with a homogeneous magnetic field ⃗ B. Assume that the field is weak such that the particle moves approximately on a straight line with constant velocity. (a) Show that the instantaneous power radiated by the particle in terms of the force F ⃗ acting on the particle is given by P = 2q2 γ 2 ( ( ) ) 2 ⃗F 2 3m 2 c 3 − ⃗F · β ⃗ (b) From this, compute the instantaneous power radiated by the particle while inside the field region. (c) Compute the total energy W radiated by the particle in terms of the proper time τ F the particle spent inside the field region. 7
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Field Theory exam II â Solutions