Order of Magnitude Astrophysics - a.k.a. Astronomy 111. Radiation Basics

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1 1 Order of Magnitude Astrophysics - a.k.a. Astronomy 111 Radiation Basics To relate the information received from electromagnetic waves to the properties of the emitting system, it is necessary to first understand the processes of electromagnetic radiation production, and the nature of the spectrum emitted by the different systems. Radiation from an Accelerated Charge Radiation is any means by which energy can be transported from the emitting object to arbitrarily large distances. The electric field of a stationary point charge, as well as that of a charge moving with uniform velocity, falls as r 2. The energy flux S E 2 r 4, thus approaches zero as r. The situation, however, changes dramatically when a charged particle undergoes an acceleration. The field of an accelerated charge picks up a component which falls only as r 1, usually called a radiation field. A field with E r 1 has an energy flux S r 2 ; and because the surface area of a sphere increases as r 2, the amount of energy flowing through a sphere of any radii is equal, and nonzero. This fact allows the accelerated charge to transfer energy to large distances. Why Does an Accelerated Charge Radiate? Let us consider an electric field in the instantaneous rest frame of the charge. It has to be constructed from q, a, c, and r and hence must have the general form. E = C(θ) qa ( q ) ( c n r = C(θ) a ), (1) m r 2 c n r m 2

2 2 where C is a dimensionless constant, depending only on the angle θ between r and a, and n and m need to be determined. Because v = 0 in the instantaneous rest frame, the field cannot depend on velocity. From dimensional analysis, it immediately follows that n = 2 and m = 1. Hence E = C(θ) qa c 2 r. (2) Thus dimensional analysis and the fact that E must be linear in q and a imply the r 1 dependence for the radiation term. We can determine the factor C(θ) by studying a special case. Consider a charge that was at rest, at the origin, from t = to t = 0 and undergoes a constant acceleration a along the x axis for a short time t. For t > t, it moves with a constant velocity v = a t along the x axis. Let us study the electric field produced by this charge at some time t t. Because t is arbitrarily small, we have a t c and we use the non-relativistic approximation throughout. The news that the charge was accelerated at t = 0 could have travelled only to a distance r = ct in time t. Thus, at r > ct, the electric field should be that due to a charge located at the origin: E = q ˆr (for r > ct). (3) r2 At r ct, the field is that due to a charge moving with velocity v along the x axis. This will be a Coulomb field radially directed from the instantaneous position of the charge E = q r 2 ˆr (for r < ct). (4) Around r = ct there exists a small shell of thickness c t in which neither solution holds true. From Figure (1a) it is clear that the electric field in the transition region should interpolate between the two Coulomb fields. Let E and E be the magnitudes of the electric fields parallel and perpendicular to the direction ˆr, respectively. From the geometry, we have E E = v t c t. (5)

3 3 Fig. 1. (a) The electric field produced by a charged particle that was accelerated for a small time interval t. For t > t, the particle is moving with a uniform velocity v along the x axis. At r > ct, the field is that of a charge at rest in the origin. At r < c(t t) the field is directed towards the instantaneous position of the particle. The radiation field connects these two Coulomb fields in a small region of thickness c t. (b) Pillbox construction to relate the normal component of the electric field around the radiation zone.

4 4 But v = a t and t = (r/c), giving E E = (a t)(r/c) c t = a ( r c 2 ). (6) We can determine the value of E by applying Gauss theorem to a small pillbox, as shown in Figure (1b). This gives E = E r = (q/r 2 ); thus we find that ( r ) q E = a c 2 r = q ( a ). (7) 2 c 2 r This is the radiation field located at r = ct, which is propagating with a velocity c. Comparison with equation [2] shows that C(θ) = sin θ. Power Radiated by an Accelerated Charge In classical electromagnetic theory, radiation is emitted by any charged particle in accelerated motion. The total amount of energy radiated per second in all directions by a particle with charge q moving with acceleration a is given by de dt = 2 q 2 3 c 3 a2 (8) provided the acceleration is measured in the frame in which the particle is instantaneously at rest. This is called the Larmor s formula and can be used to understand a host of classical electromagnetic phenomena. Because d = (qx) is the dipole moment related to an isolated charge located at position x, this formula shows that the total power radiated is proportional to the square of d. In periodic (bounded) motion, if d varies at frequency ω (so that d = ω 2 d), then the energy radiated is given by de dt = 2 d 2 3 c 3 ω4. (9) Different physical phenomena are essentially characterized by different sources of acceleration for the charged particle.

5 5 Thermal Bremsstrahlung Consider a scattering between an electron of mass m e and a proton, with an impact parameter b and relative velocity v in a hydrogen plasma. The acceleration of the electron is a q 2 /(m e b 2 ) and lasts for a time b/v. Such an encounter will result in the radiation of energy E (q 2 a 2 /c 3 )(b/v) q 6 /(c 3 m 2 eb 3 v) q 6 n i /(c 3 m 2 ev), as b n 1/3 i on the average. The total energy radiated per unit volume will be n i E. Because each collision lasts for a time v/b, there will be little radiation at frequencies greater than v/b. For ω < v/b, we may take the energy emitted per unit frequency interval to be nearly constant. Further, in the case of a plasma in thermal equilibrium, v (k B T/m e ) 1/2. Putting all these together, we get ( ) ( ) ( ) de q 6 1/2 me j ω n dωdtdv m 2 ec 3 e n i n 2 T 1/2 for ω k B T. (10) k B T This process is called thermal bremsstrahlung. The bremsstrahlung spectrum is flat for 0 < ω (k B T/ ) and will fall rapidly for ω (k B T/ ), where the upper limit comes from the fact that an electron with a typical energy of k B T cannot emit photons with energies higher than k B T. The total energy radiated, over all frequencies, from such a plasma can be found by integration of this expression over ω in the range (0, k B T/ ). This gives ( ) de = dtdv kb T/ 0 ( ) ( ) ( ) de q 6 1/2 me k B T dω n dωdtdv m 2 ec 3 2 e n i n 2 T 1/2. (11)