Problem 1:
A fixed dipole moment is pointing in the x-direction while moving in the z-direction with a constant velocity v << c. What are the instantaneous electric and magnetic fields at a point (x, y = 0, z) away from the dipole?
Solution:
 | Concepts, principles, relations
that apply to the problem: The electric field of a charge distribution moving with constant velocity v and v << c is the instantaneous Coulomb field. Here it is the instantaneous dipole field. B(r,t) = (v/c2)´E(r,t), (SI units) |
|
Why do they apply? We have a dipole moving with constant velocity and v << c. |
| How do they apply? E(r,t) = (1/(4pe0)) (1/r'3)[3(p×r')r'/r'2 - p], r' = r - R, R = location of the dipole at time t. p = pi, R = (z0 + vt)k, r = xi + zk, r' = r - R = xi - (-z + z0 + vt)k Let z0 = 0. E(r,t) = (1/(4pe0))(x2 + (z - vt)2)-3/2[3px(xi - (z - vt)k)/(x2 + (z - vt)2)- pi] = (1/(4pe0))(x2 + (z - vt)2)-5/2[2px2i - p(z - vt)2i + 3px(z - vt)k] |
| Details of the calculation: B(r,t) = (v/c2)´E(r,t) = (v/c2)k´E(r,t) = (v/(4pe0c2))(x2 + (z - vt)2)-5/2[2px2j - p(z - vt)2j] |
Problem 2:
Cs55137 is a common laboratory radioactive source of electrons and gamma rays. Eight percent of the time Cs55137 beta decays to the ground state of Ba56137. The net atomic mass difference between the two isotopes is 1.18 MeV/c2. A 180 degree spectrometer is used to measure the beta decay spectrum. The spectrometer has a radius R = 3.8 cm.
(a)
Write down the reaction for the beta decay.
(b) Calculate the maximum momentum of the beta decay electron/positron.
Express the result in MeV/c.
(c) What is the vector direction of the spectrometer magnetic field relative
to the drawing?
(d) What is the magnetic field setting of the spectrometer for the maximum
energy of the electron/positron to arrive at the detector?
Provide a numerical answer with units.
Solution:
| Concepts, principles, relations
that apply to the problem: Relativistic "collisions", energy and momentum conservation, motion of a charged particle in a magnetic field |
| Why do they apply? The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. |
| How do they apply? (a) Cs55137 ® Ba56137 + e- + n (n = anti neutrino) (b) The source is at rest. If we assume the antineutrino is massless then the electron has the maximum energy if the energy of the antineutrino is zero. If the antineutrino has mass, then the electron has the maximum energy if the Ba nucleus and the antineutrino have no relative kinetic energy. Since the mass of the antineutrino is then extremely small, we can neglect it compared to the mass of the Ba nucleus. Let M1 be the mass of the Cs nucleus and M2 be the mass of the Ba nucleus. M1c2 = (M22c4 + p22c2)½ + (me2c4 + pe2c2)½ (energy conservation) p22 = pe2 = p2 (momentum conservation) The mass of the Ba nucleus is much greater than the mass of the electron. If we neglect the recoil of the Ba nucleus (M22c4 >> p2c2) then M1c2 - M2c2 = (me2c4 + p2c2)½ The maximum momentum of the electron is then given by p2c2 = (M1c2 - M2c2)2 - me2c4 = (1.18 MeV)2 – (0.511MeV)2 pc = 1.064 MeV (c) The field points into the page. (d) R = p/(qB) B = 1.064* 106 eV * (1.6 *10-19 J/eV)/[ 1.6 *10-19 C * 0.038 m * 3*108 m/s] = 9.33*10-2 T |
| Details of the calculation: None |
Problem 3:
A point magnetic moment m is at rest in frame K' and in that frame produces a vector potential A' = m'´r'/r'3 (Gaussian units) and no scalar potential (f = 0). Frame K' moves with constant velocity v << c along the x-axis of frame K, so that an observer in K sees the moment moving with velocity v = bcx/x. Show that to first order in b the observer in K detects an electric dipole moment p = b´m as well as an undiminished (to first order) magnetic moment m.
Solution:
| Concepts, principles, relations
that apply to the problem: Lorentz transformation of 4-vector potential A'm |
| Why do they apply? We transform the 4-vector potential from K' to K and check if it transforms into a vector potential due to a magnetic dipole m and a scalar potential due to an electric dipole p to first order. |
| How do they apply? In the rest frame K' of the magnetic dipole we have A'm = (0,Ax,Ay,Az). We use Gaussian units. The dipole is at rest at the origin in K'. K moves with velocity -bcx/x with respect to K'. In K we have Am = (gbA'x, gA'x, A'y, A'z). Since v << c we have g @ 1. We therefore write Am = (bA'x, A'x, A'y, A'z). xm = (ct', x'+bct', y', z'), i.e. t = t' x = x' + bct', y = y', z = z'. A(r = r'+bctx/x) = A'(r') = A(r - bctx/x), the observer in K detects an undiminished (to first order) magnetic moment m = m' moving with velocity bcx/x. f(r = r'+bctx/x) = bA'x(r') = b×A' = b×(m'´r'/r'3) =(r'/r'3)×(b´m') = (r'/r'3)×p, with p = b´m' = b´m. f(r) = ((r-bctx/x)/|r-bctx/x|3)×p. f(r) is the scalar potential of a dipole p moving with velocity bcx/x. Details of the calculation: None |
Problem 4:
An excited nucleus of 57Fe formed by the radioactive decay of 57Co emits a gamma ray of 1.44 x 104 eV. In the process, there is conservation of energy and m0c2 = gma0c2 + hn, where m0c2 is the initial mass of the nucleus and ma0c2 is its mass after the emission of the gamma ray. There is also conservation of momentum, hn/c = gma0u, where u is the recoil velocity of the iron nucleus. The energy released by the reaction is Er = (m0 - ma0)c2.
(a) Show that
hn
= Er(m0
+ ma0)/(2m0)
= (1 - Er/(2m0c2))Er.
Thus hn
< Er:
part of Er
goes to the photon, and the other part supplies kinetic energy to the recoiling
nucleus.
(b) Set m0 = 57*1.7*10-27 kg, and show that Er/(2m0c2))
~ 1.3*10-7.
Thus the fraction of the available energy Er
that appears as recoil is small.
(c) Mössbauer discovered in 1958 that, with solid iron, a significant fraction
of the atoms recoil as if they were locked rigidly to the rest of the solid.
This is the Mössbauer
effect. If the sample has a mass of 1 gram, by what fraction is the gamma
ray energy shifted in the recoil process?
(d) A sample of normal 57Fe
absorbs gamma rays of 14.4 keV by the inverse recoilless process much more
strongly than it absorbs gamma rays of any nearby energy. The excited
nuclei thus formed reemit 14.4 keV radiation in random directions some time
later. This is resonant scattering. If a sample of activated 57Fe
moves in the direction of a sample of normal 57Fe,
what must be the value of the velocity v that will shift the frequency of the
gamma rays, as seen by the normal nuclei, by 3 parts in 1013?
This is one line width.
(e) A Doppler shift in the gamma ray results in a much lower absorption by a
nucleus if the shift is of the order of one line width or more. What
happens to the counting rate of a gamma-ray detector placed behind the sample of
normal 57Fe
when the source of activated "Fe moves
(i) toward the normal 57Fe,
(ii) away from it?
(f) If a 14.4 keV gamma ray travels 22.5 meters vertically upward, by what
fraction will its energy decrease?
[Gravitation redshift, a thought experiment:
Suppose
particle of rest mass m is dropped from the top of a tower and falls freely with
acceleration g. It reaches the ground with a velocity v = (2gh)1/2,
so its total energy E, as measured by an observer at the foot of the tower is E
= mc2 + (1/2)mv2 + O(v4) = mc2 + mgh
+ O(v4).
Suppose an observer has some magical method of converting all this energy into a
photon of the same energy. Upon its arrival at the top of the tower with
energy E the photon is again
magically changed into a particle of rest mass
m =
E/c. Energy conservation requires
that m = m. Therefore
E/E = mc2/(mc2
+ mgh + O(v4)) = 1 - gh/c2 + O(v4).]
(g) A normal 57Fe absorber located at this height must move in
what direction and at what speed in order for resonant scattering to occur?
Solution:
| Concepts, principles, relations
that apply to the problem: Relativistic "collisions", energy and momentum conservation, the Doppler shift |
| Why do they apply? The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. The Doppler shift can compensate for the gravitational redshift to produce optimum absorption. |
| How do they apply? (a) energy conservation: m0c2 = gma0c2 + hn, momentum conservation: hn/c = gma0u Define: Er = (m0 - ma0)c2. Show: hn = Er(m0 - ma0)/(2m0) m0c2 - hn = gma0c2, m02c4 + h2n2 - 2m0c2hn = g2ma02c4, -2m0hn = g2ma02c2 - m02c2 - h2n2/c2 = g2ma02c2 - m02c2 - g2ma02u2. = g2ma02(c2 - u2) - m02c2 = ma02c2 - m02c2. 2m0hn = (ma0 - m0)(ma0 + m0)c2 = Er(ma0 + m0). hn = Er(ma0 + m0)/(2m0) = Er(ma0 - m0 + m0 + m0)/(2m0) = Er(1 - Er/(2m0c2)). (b) Set m0 = 57*1.7*10-27 kg, hn = 1.44 x 104 eV. Er - Er2/(2m0c2)) - hn = 0. Er2 - 2m0c2Er + 2m0c2hn = 0. Er = m0c2 - ((m0c2)2 - 2m0c2hn)1/2. m0c2 = 8.72*10-9 J, hn = 2.3*10-15 J, Er = 2.3 *10-15 J. Er/(2m0c2) ~ 1.3*10-7. |
| Details of the calculation: (c) If m0 = 10-3 kg, then m0c2 = 9*1013 J, Er/(2m0c2) ~ 1.27*10-29. The fraction that appears as recoil energy is now negligible. The gamma ray energy is now hn = Er. It has increased by Er2/(2m0c2)) = 1.3*10-7Er. Dn/n = 1.3*10-7. (d) We want Dn/n = 3*10-13. Dn/n = (n' - n)/n = [(1 + v/c)/(1 - v/c)]1/2 -1 if the samples approach each other. We can solve for v/c ~ 3*10-13. v ~ 10-4 m/s. (e) The counting rate of a gamma-ray detector increases for both directions, because the sample 57Fe absorbs less. (f) A 14.4 keV gamma emitted from 57Fe ray travels vertically upward in a uniform gravitational field. We have Dn/n = gDh/c2 from E2/E1 = 1 - g(h2-h1)/c2. Dn/n = -(9.81 m/s2)(22.5 m)/(3*108 m/s)2 = -2.45*10-15. Such a shift can also be produced if the absorber moves away from the source. We would need Dn/n = [(1 - v/c)/(1 + v/c)]1/2 -1 = -2.45*10-15, v/c = -2.45*10-15. (g) The absorber has to move towards the source with v/c = 2.45*10-15 in order to increase, in its own frame, the frequency for optimum absorption, because the frequency was decreased by the gravitational redshift. If the absorber is at rest we still have absorption, since Dn/n|g << 3*10-13. |
Problem 5:
A 30 GeV proton passes 10-7cm away from a
hydrogen atom.
(a) Estimate the peak magnitude of the electric field and the duration of the
electric field pulse to which the atom is subjected.
(b) Do the same for a 30 GeV electron passing at the same distance.
You may use mpc2 = 1GeV and
mec2 = 0.5
MeV.
Solution:
| Concepts, principles, relations
that apply to the problem: The electromagnetic fields of a moving point charge, the Lorentz transformation of the electromagnetic fields |
| Why do they apply? In the rest frame of the proton the electric field is the Coulomb field and the magnetic field is zero. We can transform those field to a frame in which the proton is moving with velocity vi. |
| How do they apply? (a) gmc2 = 30 GeV, g » 30. The electric field in the proton frame is the Coulomb field, E = kqe/r2(r/r). In the frame of the hydrogen atom we have E'|| = E||, E'^ = gE^. In the proton frame the maximum field the hydrogen atom is subjected to is Emax = E^max = kqe/b2. Therefore in the frame of the hydrogen atom E'max = E'^max = gkqe/b2. E'^max = [30*9*109*1.6*10-19/10-18](N/C) = 4.3*1010 N/C. To estimate the duration of the electric field pulse we estimate the time interval during which E > Emax/2. In the proton frame E = kqe/(b2 + (vt)2), so when vt = b the E = Emax/2. Dt = 2b/v is the duration of the pulse in the proton frame. The duration of the pulse in the frame of the hydrogen atom is a proper time interval. Dt' = Dt = 2b/(gv). Dt' = 2*10-9/(30c) = 2.22*10-19 s |
| Details of the calculation: For a 30 GeV electron g » 6*104. Emax = 8.6*1013 N/C, Dt' = 1.1*10-22 s |
Problem 6:
A line of charge with charge density
l C/m is fixed at rest along the x’
axis of a reference frame S’. A test charge q is at rest in S’ at (0,0,z’ =
d). S’ is in constant motion with velocity v = vi with respect to
a reference frame S.
(a) Calculate the electric field of the line of charge in the rest frame S’ and
the force on q.
(b) Calculate the electric and magnetic fields of the line of charge measured
by an observer at rest in S.
(c) Calculate the force measured by the observer in S on the test charge q.
Solution:
| Concepts, principles, relations
that apply to the problem: Lorentz transformation of the electromagnetic fields |
| Why do they apply? We are asked to transform the electromagnetic fields from the frame S' to a frame moving with uniform velocity v = -vi with respect to S'. |
| How do they apply? (a) In frame S’: E’ = l/(2pr)(r/r), F’ = qE’ = kql/(2pd). (b) S moves with velocity –vi with respect to S’. Transformation of the fields: E'|| = E||, B'|| = B||, E'^ = g(E + v´B)^, B'^ = g(B - (v/c2)´E)^. (SI units) Here || and ^ refer to the direction of the relative velocity. We therefore have in S: E|| = B|| = 0, E = gl/(2pr)(r/r), B = gvl/(2c2pr)(f/f), where (f/f)is a unit vector encircling the x axis according to the right-hand rule. At (0,0,z’ = d) it points into the –j direction. (c) Force on the test charge in S: F = kgql/(2pd) - kgqv2l/(2c2pd) = kg-1ql/(2pd) = g-1F’. |
| Details of the calculation: Explain the difference: F = dp/dt. The momentum is a 4-vector. The components perpendicular to the relative velocity are the same in two frames moving with respect to each other. We have Dp = Dp’, dt = dt’ = dt/g. |
Problem 7:
A long superconducting solenoid with radius R is at rest in frame K. It
has its axis along the z-axis and a current flowing on the surface produces a
uniform field inside.
(B = Bk, r < R; B = 0, r > R.)
An observer moves with uniform velocity v = vi (v << c)
along the x-axis. Write down the electric and magnetic fields in the rest
frame K' of the observer for r < R and r > R.
Solution:
| Concepts, principles, relations
that apply to the problem: Lorentz transformation of electric and magnetic fields |
| Why do they apply? We are asked to transform the electromagnetic fields from the laboratory frame K to a frame moving with uniform velocity v = vi with respect to K. |
|
How do they apply? In the laboratory frame K we have E = 0, B = Bk, r < R; B = 0, r > R. In K' we observe E'|| = E||, B'|| = B||, E'^ = g(E + v´B)^, B'^ = g(B - (v/c2)´E)^. Since v << c we have g @ 1. Therefore E'|| = 0, B'|| = 0, E'^ = v´B, B'^ = B. E' = -vBj, B' = Bk, r < R; E' = 0, B' = 0, r > R. |
| Details of the calculation: None |