Contents
- Introduction
- Alpha Radiation
- Modified Bessel Functions of the Second Kind
- Classical Energy Loss
- Oscillator Strengths
- Quantum-Mechanical Energy Loss
- Fields in a Dense Medium
- Cherenkov Radiation
- Stopping of Magnetic Monopoles
- Comments and Numerical Examples
- References
This article fills in the details in Jackson's excellent account of the energy loss of charged particles in matter in his Chapter 13, including omitted derivations and mathematical details. This has always been an important subject since the discovery of radioactivity, and is of great importance in experiments and practical radioactivity. Both particle radiations, classically alpha and beta rays, and electromagnetic radiation, gamma rays, cause ionization in their passage through matter. Electrons are freed from neutral atoms, producing an ion pair of electron and positive ion. Of course, this disturbs chemical bonding as well as produces free charge carriers, which is the source of the biological effects of radiation.
It is interesting to ponder the fact that these radiations traverse matter at all, showing that apparent solidity is not solid at all. To these radiations, matter is mainly empty space in which electrons wander, with the massive, but very small, atomic nuclei presenting very little obstruction. The interactions are mainly with electrons, transferring energy to them, which is necessarily lost by the radiation. In fact, in all our analysis we shall treat the transfer of energy to the electrons, and use the principle of the conservation of energy to find the energy lost by the radiation.
The different radiations behave differently in their passage through matter. Particle radiations can be divided into heavy particles--that is, those massive relative to electrons, say more than about 20me in mass--and light particles, which are usually electrons or positrons, usually called beta rays. Both create many ions in frequent interactions with electrons, perhaps thousands per mm in air. Heavy particles, however, lose little of their energy in a single collision, and are practically undeflected, unless by a rare encounter with a nucleus. Light particles can lose a large fraction of their energy in a single collision with an electron. In fact, they can lose all of it, creating a knock-on electron of about the same energy. They are much more easily deflected, so their paths are not as straight as those of heavy particles. Electromagnetic radiation, considered as particle-like quanta of energies greater than 100 keV, does not fritter away its energy in multiple collisions with electrons, but ends its life in infrequent catastrophes, such as the ejection of an electron with disappearance of the quantum, or electron-positron pair production if is energy is greater than 1.05 MeV. Energy can be partially lost in Compton scattering fender-benders as well. In general, electromagnetic radiation finds matter much more transparent than charged particles do. Incidentally, a fast neutral atom quickly loses its electrons and becomes a positive ion, while neutral particles like uncharged pions act more like electromagnetic radiation. The most important factor in producing ionization and energy loss is the electromagnetic field of the charged particle.
The paths of charged particles can be seen macroscopically by the traces they leave in a cloud chamber, a bubble chamber or a photographic emulsion. What is seen is actually the effects of the ionization. In a cloud chamber, water droplets condense on negative ions. In a bubble chamber, the liquid vaporizes for a similar reason. The energy loss in a photographic emulsion causes the crystallites to become developable. Charged particles can also be seen by the elctromagnetic shock wave of Cherenkov radiation when they travel faster than the phase velocity of light in the medium. All these things can be discussed on the basis of the theory that will be presented here.
In this article, as in Jackson, formulas and electrical quantities are expressed in Gaussian units. Therefore, the electronic charge e = 4.803 x 10-10 esu should be used. This is c/10 times the charge in coulomb.
Alpha radiation consists of fast helium nuclei, with charge z = +2 and mass 4.00 amu, or rest mass 3.724 GeV, emitted from heavy nuclei. Their energies range from a maximum of about 10.53 MeV (Po212) down to about 4.05 MeV (Th232), below which the probability of alpha-decay becomes very small. The half-life of Po212 is only 300 ns, while that of Th232 is 1.39 x 1010 y. The half-life is a very strong function of the disintegration energy. It is clear that such natural alpha particles are nonrelativistic. However, this is not true of artificially produced heavy particles, such as protons or deuterons, nor of cosmic radiation, where fast protons are the usual primary radiation.
Alpha particles make thick, straight tracks in a cloud chamber of a definite length, the range R of the alpha particle. Looking more closely, the tracks are occasionally bent through an angle, usually near the ends, and straggle a little in range. The range in air at 15°C and 760 mmHg of the 10.53 MeV alpha from Po212 is about 11.6 cm, while the alphas from Th232 have a range of 2.49 cm. The energy of an alpha particle can be deduced fairly accurately from its range. The empirical relation R = 0.318E3/2, where R is in cm and E is in MeV, seems to hold fairly well for ranges from 3 to 7 cm.
Aluminium, with a density of 2.70 g/cc, has a stopping power about 1700 times that of air at STP. That is, the ranges in aluminium are 1700 times shorter than in air. The Po212 alpha is stopped by 0.07 mm of aluminium. This illustrates the very small penetrating power of heavy charged particles in solid matter.
The speed of an alpha particle of energy E MeV is v = 6.944 x 108√E cm/s. For particles of rest energy Eo, the speed of a particle of kinetic energy T can be found from γ = 1 + T/Eo and β = √(γ2 - 1)/γ. For a 1 GeV proton, γ = 2.07 and β = 0.878. Recall that 1 amu = 931 MeV rest energy. For a 1 GeV electron, γ = 1962 and β = 1.0000.
Modified Bessel functions of the second kind, Kn(x), appear prominently in the theory, so the properties we shall need are reviewed here. Usually, such functions appear in connection with problems with cylindrical geometry, but here they arrive as Fourier transforms. They are not difficult to work with, but should be well-understood. For integral n, they are infinite at x = 0, and decrease like a negative exponential as x increases. We will use K0(x) and K1(x) here, usually multiplied by x. Then, xK0(x) → 0 and xK1(x) → 1 as x → 0.
We know that Bessel's equation, y" + (1/x)y' + (1 - n2/x2)y = 0, with n integral, has one solution y = Jn(x) that is finite at the origin and for large x behaves like y = (2/πx)1/2cos(x - nπ/2 - π/4). All other solutions that are linearly independent of this one are singular at the origin. The one that is carefully arranged to behave like y = (2/πx)1/2sin(x - nπ/2 - π/4) at large x is called Weber's function, Nn(x). Then, the complex linear combinations Jn(x) ± Nn(x) will behave like e±(x - nπ/2 - π/4), which is very convenient. These combinations are called the Hankel functions.
If we now let x = ix, Bessel's equation becomes y" + (1/x)y' + (1 + n2/x2)y = 0, whose solutions are called modified Bessel functions. The one regular at the origin, Jn(ix) = (i)nIn(x), behaves like (-i)n (1/2iπx)1/2ex for large x, the positive exponential completely dominating the negative exponential. In(x) is the modified Bessel function of the first kind. We now need a linearly independent second solution that behaves like e-x when x is large to complement this solution. To find one, we must wipe out the positive exponential completely, and we can do this by starting with the Hankel function that behaves like eix and then letting x → ix. To simplify the result, we multiply by (π/2)in+1 and call the function Kn(x), which for large x behaves like √(π/2x) e-x. When you see these various Bessel functions defined in a math text, the motivation for the strange choices of constants is seldom clear. The reason is to give certain simple behaviors for large x, or asymptotically.
For small x, K0(x) → ln (1/x) + ln2 + γ, where γ is Euler's Constant, 0.57721 56649 ... . Euler's Constant is the limit as m → ∞ of 1 + 1/2 + 1/3 + 1/4 + ... + 1/m - ln m. The series is the divergent series used as an example of the fact that a series may not converge even if its terms approach zero. If the signs alternate, its sum is ln 2. Here it appears in getting the ln (1/x) from the series. The product xK0(x) → 0 as x → 0, which can be proved by l'Hôpital's Rule and writing it as ln (1/x)/(1/x). Kn(x) goes as [(n - 1)!/2](2/x)n for small x, so xK1(x) → 1 as x → 0. These are the only two functions we will need, and it is only necessary to remember that they go to 0 and 1, respectively.
At x = 1, we have xK0(x) = 0.4210 and xK1(x) = 0.6019. At x = 2, xK0(x) = 0.2278 and xK1(x) = 0.2797. For larger values of x, both functions decrease exponentially.
If ' stands for differentiation with respect to the argument, then K0' = -K1 and K1' = -K0 - K1/x. From these, it is easy to work out (xK0)' = K0 - xK1 and (xK1)' = -xK0. Each of these relations can be turned into an indefinite integral. Furthermore, (xK1K0)' = -x(K02 + K12) and [x2(K12 - K02)]' = -2xK02. These will help us to integrate squares and products of the K's.
The Fourier cosine transform is F(y) = ∫(0,∞)f(x)cos(xy)dx. If f(x) is an even function of x, it is easy to turn the cosine transform into an exponential transform by expressing the cosine in terms of exponentials, and letting x → -x in the second integral. From Erdélyi, et. al. we find that the cosine transform of (1 + x2)-n-1/2 is (y/2)n(√π/Γ(n+1/2))Kn(y). From this, it is not hard to show that ∫(-∞,+∞)eiωxdx/(1 + x2)1/2 = 2K0(ω), and ∫(-∞,+∞)eiωxdx/(1 + x2)3/2 = 2ωK1(ω), since Γ(1/2) = √π and Γ(3/2) = (1/2)Γ(1/2). We'll see that these are the Fourier transforms of the fields of a moving charge.
Another integral yielding K1(ω) can be found by differentiating the integral for K0 with respect to ω, and using the relation K0'(ω) = -K1(ω). This integral is ∫xeiωx/(1 + x2)1/2 = 2iK1(ω). This integral can also be obtained from a sine transform.
A typical energy loss encounter between a particle of charge +ze moving with velocity v and an electron of charge -e at rest is shown at the right. The impact parameter is b. Any analysis in which we use an impact parameter is a classical one, of course. The relativistically correct electric field components E1 and E2 are shown. The field points away from the present position of the charge, but is not isotropically distributed. The field component E1 alternates in direction as the charge passes by, but E2 is constant in direction.
As a first approximation, we suppose that the charge ze moves by before the electron can move any appreciable distance. The electron receives an impulse ∫(-∞,+∞)(-eE2)dt that is equal to the momentum Δp transferred to it. The heavy particle ze also experiences an equal and opposite impulse that deflects it a little, but only a very little, which can be neglected in calculating the energy transfer. The energy transfer to the electron is ΔE = Δp2/2m, and the energy of the particle ze decreases by the same, relatively very small, amount. It would be much more difficult to estimate directly the force on the particle ze, involving the exact trajectories of both particles, but we use conservation of energy, trusting that it will give the correct answer.
The integral is very easy to perform, and the result is Δp = 2ze2/bv. From this, ΔE = 2z2e4/mv2b2. Let's suppose there are N atoms per cm3, each with Z electrons, which we assume are distributed randomly and evenly. The number of electrons in a thickness dx with impact parameters between b and b + db is then dn = 2πNZb db dx. Each of these electrons will experience the same energy transfer ΔE. The total energy transfer in a distance dx will be -dE/dx = 2πNZ ∫ΔE(b)bdb. With the expression for ΔE just derived, -dE/dx = (4πNZz2e4/mv2)∫db/b = (4πNZz2e4/mv2)ln(bmax/bmin). We already have the form of the expression for the stopping power -dE/dx that we shall see over and over.
We cannot simply let bmax = ∞ and bmin = 0, for then the stopping power would diverge. Getting a useful formula depends on making good choices of these parameters. Since they appear in a logarithmic term, order-of-magnitude estimates may be good enough. The energy transfer increases without limit as b → 0, but we know that there is a certain maximum energy transfer that occurs in a head-on collision. We should choose bmin so that this impact parameter corresponds to the maximum energy transfer.
A maximum energy transfer collision, b = 0, is shown in the figure. In the approximate CM system, the electron simply bounces off elastically, its velocity reversed. Returning to the Lab system by adding the velocity v, and using the correct relativistic formula for adding parallel velocities, we find that the electron now has a velocity v' = 2v/(1 + β2). Now, 1/γ'2 = 1 - β'2 = (1 - β2)/(1 + β2). The kinetic energy of the electron is T = (γ' - 1)mc2 = 2mβ2c2γ2 = 2γ2mv2, which is the maximum energy transfer.
Setting ΔE equal to this value, we find bmin = ze2/γmv2. A more exact derivation would show that replacement of b2 by b2 + bmin2 would have the same effect, with the lower limit of integration b = 0.
As b becomes larger and larger, the encounter becomes softer and longer. If the time of the collision is on the order of the frequencies of motion of the electron, we know that the orbits are only perturbed adiabatically, and no change takes place. If we set the period of the motions 1/ω equal to the time of the collision b/γv (the time during which the fields are appreciable), we find that this happens when b/γv = 1/ω or bmax = γv/ω. This is a very rough estimate, and we soon shall do very much better. However, this gives the stopping power -dE/dx = (4πNZz2e4/mv2)ln[(γv/ω)/(ze2/γmv2)] = (4πNZz2e4/mv2)ln(γ2mv3/ze2ω).
The argument of the logarithm is usually denoted B. The formula obtained by Bohr in 1915 is almost exactly the same, except that B is multiplied by a factor 1.123 that has negligible effect, and the meaning of the frequency ω is a bit clearer. For relatively slow, heavy particles that can be treated classically with fair accuracy, it agrees well with experiment. For lighter particles, and for fast particles, it overestimates the energy loss.
Before taking this matter up, we shall handle the energy transfer to a harmonically bound charge in a better manner, and also obtain some useful results for the fields. Let's assume, as in Lorentz's electron theory, that an electron in the atom has an equation of motion x" + Γx' + ωo2x = -(e/m)E(t). The easiest way to solve this equation is to use Fourier transforms, which will reduce it to an algebraic equation that is easily solved. Following Jackson, we'll use the transforms in the form x(t) = (2π)-1/2∫(-∞,+∞)x(ω)e-iωtdω. The inverse transform is the same, except that the sign of the exponent is changed. We transform both the position and the field.
The result is x(ω) = (-e/m)E(ω)(ωo2 - iωΓ - ω2)-1, which should be familiar. If Γ is small, the resonance is sharp. What we want is the energy transfer, which should be given by ΔE = -e∫(-∞,+∞)v·Edt, where v = -iωx, and E is the field at the electron. We now write this expression using the Fourier transforms.
ΔE = -(e/2π)∫dω∫dω'∫e-i(ω + ω')dt (-iω)x(ω)·E(ω'). All limits are from -∞ to +∞. The integral over dt gives δ(ω + ω'), and this delta function can be used to integrate over dω'. The result is ΔE = -e∫dω(-iω)x(ω)·E(-ω). Since E is real, E(-ω) = E*(ω) (the reality condition), so ΔE = -e∫dω(-iω)x·E*(ω). Now split this integral into two integrals, one from -∞ to 0, and the other from 0 to ∞. Let ω = -ω in the first integral, and use the reality conditions again on both x and E. The result can be written most simply as ΔE = 2e Re{∫(0,∞)iωx(ω)·E*(ω)dω}. This is our equation for the energy transfer in terms of the Fourier transforms.
We already have the transform x(ω). Now we must find the transforms of the fields E1(t) and E2(t) at the position of the electron. These fields were given as a function of time in the first diagram above. All we have to do is evaluate the integrals E(ω) = (2π)-1/2∫E(t)eiωtdt. For E2, we have E2(ω) = zebγ(2π)-1/2∫(b2 + γ2v2t2)-3/2eiωtdt. If u = γvt/b, this integral is easily expressed as (ze/bv)(2π)-1/2∫(1 + u2)-3/2exp[i(ωb/γv)u]du. We have already shown the value of this integral in our discussion of the Bessel functions. Therefore, E2(ω) = (ze/bv)√(2/π)[ξK1(ξ)], where ξ = ωb/γv.
The transform of E1, with the same substitution, can be expressed as E1(ω) = -(ze/γbv)(2π)-1/2∫ (1 + u2)-3/2exp[i(ωb/γv)u]udu. This can be integrated by parts, taking w = exp[i(ωb/γv)u], dv = -udu(1 + u2)-3/2. Then dw = i(ωb/γv)ex[[i(ωb/γv)u]du and v = (1 + u2)-1/2. The integrated part wv vanishes at both limits, so E1(ω) = -i(zeω/γ2v2)(2π)-1/2∫ (1 + u2)-1/2exp[i(ωb/γv)u]du. This integral was also given above, so the final result is E1(ω) = -i(ze/bvγ)√(2/π)[ξK0(ξ)], where ξ has the same meaning as in the preceding paragraph. We now have the Fourier transforms of the fields.
Now we go back and substitute x(ω) in the integral for ΔE. The result is ΔE = -(2e2/m) Re{∫ |E(ω)|2(ωo2 - ω2 - iωΓ)-1. The real part of everything inside the integral except the square of the field is -ω2Γ/[(ωo2 - ω2)2 + ω2Γ2]. This function is strongly peaked at ω = ωo. We make the approximation that ω+ωo = 2ωo, and put x = (ω - ωo)/Γ. Then, ΔE = (2e2/m)|E(ωo)|2∫(-∞,+∞)dx/(1 + 4x2). The limits of the integral have been extended to infinity with little error. Its value is π/2, so finally ΔE = (πe2/m)|E(ωo)|2. Since we already have the Fourier transforms of the field, we can find ΔE for any value of ξ = ωb/γv; that is, as a function of the impact parameter b. This has been quite a long pull, but now we do not need to estimate bmax because we have an exact expression for ΔE(b).
Indeed, |E(ωo)|2 = (2/π)(ze/bv)2ξ2[K1(ξ)2 + K0(ξ)2/γ2], where ξ = ωob/γv. The stopping power is given by the same integral we used long ago, -dE/dx = 2πNZ ∫ΔE(b)bdb. However, we now introduce a new refinement. So far we have discussed only one resonant frequency ωo that corresponds to one electron. We now consider a more correct model of the atom.
Spectroscopy showed that atoms did not contain harmonically bound electrons providing a set of fundamental frequencies and their harmonics. Instead, the spectrum consisted of many discrete lines of apparently unrelated frequencies, and of a wide range of intensities. This was clarified by Bohr's recognition that the observed frequencies ν were related to differences in energy levels through hν = W(a) - W(b), where a is the initial state and b the final state of the transition. It was found that these frequencies could be thought of as corresponding to an oscillating electron with natural frequency ν, provided that the contributions of the oscillators to, for example, the emission of radiation, were those of the classical oscillator times a constant less than unity called the oscillator strength, f(a,b).
The intensity of a spectral line is proportional to the absolute value squared of a matrix element of the dipole moment, -e(a|x|b), where x is a vector and a,b are the initial and final states. A state is characterized by its total angular momentum j, and there are 2j+1 states of the same energy that correspond to the same values of the other quantum numbers. The rate of radiation in erg/s is (64π4ce2σ4/3) Σ|(a|x|b)|2, where σ is the wavenumber, 1/λ, in cm-1. The sum is over all the final states and an average over all the initial states making up the line; it is called the strength S(a,b) of the line.
The oscillator strength of the line is defined as f(a,b) = (8π2m/3e2h)ν(a,b)S(a,b)/(2j+1), where ν(a,b) = [W(a) - W(b)]/h, so that ν(b,a) = -ν(a,b). The reason for the choice of the constant factor will be clear shortly. The momentum and position operators satisfy the commutation relation pr - rp = -3ih', where h' = h/2π, and p, r are vectors, and their product is the scalar product. The Hamiltonian for one electron is H = p2/2m + V(r), so the commutation relation gives us p = (-im/h')(rH - Hr). Taking matrix elements between states a and b, and summing over third states c in evaluating the products rH and Hr, while remembering that H is diagonal, we get (a|p|b) = -2πimν(a,b)(a|x|b). Then, the diagonal matrix components of pr - rp are 4πmΣ(b)ν(a,b)|(a|x|b)|2, and also 3h', since the commutator is a pure number and (a|a) = 1. If we sum over final states by multiplying by 2j+1, we have 4πm(3e2)h/8π2m)Σf(a,b) = 3h/2π This simplifies, miraculously, to Σf(a,b) = 1. On the other hand, if we did not know the constant factor in the definition of f(a,b), we could determine it to make this true (which, of course, was how it was originally done).
The condition Σf(a,b) = 1 is called the Thomas-Kuhn Sum Rule (1925), and means that the effects of all the spectral frequencies characteristic of the atom do in fact add up to one electron overall. Jackson says it is "obvious," but I think it is a rather clever and non-obvious result that allows us to use the classical electron theory with quantum-mechanical atoms. If the atom contains Z electrons in all, then Σf(a,b) = Z, which is probably obvious.
In the preceding section we found the energy transfer ΔE for a particular frequency ωo. Now we consider the atom in the light of oscillator strengths, and conclude that the total energy transfer is Σf(a,b)ΔE(a,b), where we sum over the transitions a,b. Now we have -dE/dx = 2πNΣf(j)∫ΔE(j,b)bdb, where we have replaced a,b with j to avoid confusion with the impact parameter b.
Putting things together, we now have ΔE(j,b) = (πe2/m)|E(b,ωj)|2 = (2e2/m)(ze2/bv)ξ2[K12 + (1 - β2)K0(ξ)], where ξ = ωjb/γv, and 1/γ2 has been replaced by 1 - β2. When this is inserted into the integral, and the variable of integration changed to ξ, one power of ξ disappears and the integrals can be done as shown in the section on Bessel functions. This is quite straightforward, and the result can be written down at once.
-dE/dx = (4πNz2e4/mv2)Σ{ξK1(ξ)K0(ξ) - (β2/2)ξ2[K12(ξ) - K02(ξ)]}, where ξ = ωjbmin/γv. We see our familiar expression for the stopping power reappearing once more. If ξ << dx =" (4πNz
Quantum considerations may affect our results both at large and small impact parameters. We have defined bmax as corresponding to ξ = 1, or bmax = γv/ω. The energy loss at this impact parameter is ΔE = (2z2e4ω2 / γ2mv4)[K12(1) + (1 - β2)K02(1)]. We can express this in a more transparent form by using vo = e2/h' = 2.187 x 108 cm/s, the velocity in the first Bohr orbit, and I = me4/2h'2, the ionization energy of hydrogen, 13.6 eV. The result is ΔE = (z/γ)2(vo/v)4 [(h'ω)2/I][.5395 - 0.17724β2]. Since usually v >> vo, and h'ω <>
We know that energy transfers with an atom or molecule are probable only when the energy transfer approximates the energy difference between two stationary states, so an arbitrarily small amount of energy cannot be absorbed. In spite of this, it is found that the classical result is correct on the average. Most encounters result in no energy transfer, but occasionally one occurs with a large energy transfer, so the average over collisions is the small energy calculated classically. This is encouraged by the use of the oscillator strengths defined quantum-mechanically, which we have explained above. It should be clear that the energy transfer is not necessarily anything like h'ω, but can be much larger at small impact parameters. Therefore, our energy loss formula needs no significant correction at large impact parameters.
Conditions are quite different at small impact parameters. We have used a bmin that gave a ΔE equal to the maximum energy transfer. In quantum mechanics, we cannot specify an impact parameter that is less than the distance given by the uncertainty relation, Δx = h'/p. This is certainly a rough and unsubstantiated estimate, but it turns out to give the correct answer. We have a classical bmin,c = ze2/γmv2, and a quantum bmin,q = h'/γmv, and must use whichever one is the larger. The ratio η = bmin,c/bmin,q = ze2/h'v can tell us which to use. For η > 1, the classical value should be used; for η <>
To treat this problem quantum-mechanically, we consider the collision in the CM system. That's the reason we used the electron mass m in the preceding paragraph. Partial-wave analysis can be used to solve the problem in the CM system, and then the energy transfer can be found by transforming back into the lab system. Bethe did this in 1930, obtaining the most-used stopping power formula, -dE/dx = (4πNZz2e4/mv2) [ln(2γ2mv2/h'Ω) - β2]. If we simply take the ratio bmax/bmin,q as the argument of the logarithm, we find that it is γ2mv2/h'Ω, the same except for the factor 2.
For electrons, the CM system is different from that for a massive incident particle. Then, bmin = (h'/mc)√[2/(γ - 1)], and the argument of the logarithm becomes (γ - 1)√[(γ + 1)/2](mc2/h'Ω) instead.
For either Bohr's or Bethe's formula, the stopping power decreases rapidly as v increases, then passes through a minimum and afterwards increases slowly for ultrarelativistic particles, due to the γ2 in the argument of the logarithm. One factor of γ comes from the increased maximum energy transfer in a hard collision, while the other comes from the greater effective range of the compressed fields of a fast particle. However, it is found that the energy loss of very fast particles is less than that predicted by the stopping-power formulas. The reason is in the shielding due to the dielectric constant of the medium, which reduces the fields and so the energy loss. This is called the density effect, because it is largest in dense media.
Now we make another digression, but one that will not only allow us to explain the density effect in reducing the stopping power, but also to consider some other phenomena associated with the passage of fast charged particles through matter, such as Cherenkov radiation. In another article, Relativistic Electrodynamics, we used Maxwell's equations to find the field of a moving charged particle in a fully covariant form. In that case, the medium was vacuum. Now we wish to find the electromagnetic fields of a charged particle in a polarizable medium. Our result will be relativistically correct, of course, since it will be based on Maxwell's equations, but not manifestly covariant.
Here is what we are going to do. First, Maxwell's equations will be written with D, and Fourier transformed from functions of x,t to functions of k,ω, eliminating the space and time derivatives. Then D will be set equal to ε(ω)E. Allowing ε to be a function of ω allows a more realistic characterization of the medium. This will be enough for our purposes; we are not attempting to characterize the medium accurately, and shall be satisfied with the general behavior. This is not the same thing as Fourier transforming ε(t)E(t), incidentally. We shall also set H = B, since they are the same in a nonpermeable medium.
Now we will introduce the scalar potential φ and the vector potential A, from which the fields can be found by differentiation. As is well-known, this identically satisfies the homogeneous Maxwell equations, and uncouples the others, so that we will have equations for φ and A individually. We do this in the frequency domain, although it can also be done in the time domain, of course. The equations satisfied in the frequency domain by φ and A are algebraic equations equivalent to the inhomogeneous wave equations of the time domain. The potentials must be related by the Lorentz condition for this to occur. This is a Lorentz-invariant condition that is equivalent to requiring the fields to be transverse.
The sources proper for a moving point charge will then be introduced into the equations for the Fourier transforms of the potentials, which will give us the Fourier transforms of the field. We already know how to find the stopping power from the field transforms, so all we have to do is carry out the algebra as we have done above with the simpler fields of the vacuum. Now let's do the work.
The figure at the right shows how the Fourier transformed inhomogeneous wave equations are derived. This is a straightforward procedure, and everything needed is in the diagram. In the time domain, the moving point charge is represented by ρ(x,t) = zeδ(x - vt) and J = vρ(x,t). Of course, x, v and J are vectors. In the frequency domain, ρ(k,ω) = (ze/2π)δ(ω - k·v), and A(k,ω) = vρ(k,ω). To find these transforms, simply use the delta function to do the time integration, and remember that the transform of the delta function is unity. We can use these source transforms to find the transforms of the potentials. The transforms of the fields can be expressed in terms of φ(k,ω) alone.
Now we can invert the space part of the transform to get back into ordinary space, while leaving us in the time domain for the frequency part. We assume the particle is moving on the x-axis, and we want the field at (0,b,0), where the electron is. The electric field is then E(x,ω) = (2π)-3/2∫d3kE (k,ω)eibk2. This integral can be evaluated, but the algebra is tedious. However, it is important to see how it is done.
Writing it out in detail, E1(ω) = (2ize/ε)(2π)-3/2∫d3k exp(ibk2)[(ωεv/c2) - k1] [k-2 - ε(ω/c)2]-1δ(ω - vk1). The delta function can be used to integrate over k1, remembering to divide by v by the rule for a delta function of a function. The wave equation denominator becomes (ω/v)2 + k22 + k32 - ε(ω/c) 2. If we introduce λ2 = (ω/v)2 - ε(ω/c)2 = (ω/v)2(1 - β2ε). If βε <> 1, then λ will be pure imaginary, with important effects on the fields. When ε = 1, λ = ω/γv = 1/bmax. Think of λ as the reciprocal of the range of the fields in the polarizable medium.
It is now possible to integrate over k3. The integral is of the form ∫dx/(a2 + x2), and has the value π/(λ2 + k32)-1/2. Finally, the integral over k2 is the one that gives us K0. The final expression is E1(ω) = (-izεω/v2) (2π)-1/2[1/ε - β2]K0(λb). In all this, remember that ε is a function of ω. When ε → 1, we find the same field as before.
The y-component is E2(k,ω) = (-2izek2/ε)δ(ω - k·v)/[k2 - (ω/c)2ε]. The integrals over k1 and k3 are done exactly as for E1. If we let x = k2/λ, then E2(b,ω) = (-izeλ/vε)√(2π) ∫xeibλx/(1 + x2)1/2. The integral is the one given in the last paragraph of the section on Bessel functions, and has the value 2iK1(bλ). The final result is E2(b,ω) = (ze/v)√(2/π)(λ/ε)K1(λb).
Our earlier formula for the energy loss to a single electron oscillator can be generalized using oscillator strengths, which will be more accurate than applying oscillator strengths to an approximation at the end. The generalized formula is ΔE(b) = 2eΣf(j)Re∫(0,∞) iωxj(ω)·E(ω). Instead of using x(ω) from the equation of motion, we can find the equivalent using ε(ω). Now ε(ω)E = E + 4πNp, where p is the dipole moment induced in the atom or molecule by the electric field. This moment is -eΣf(j)xj = (ε - 1)E/4πN, which we can put into the integral for the energy loss. The result is ΔE(b) = (1/2πN) Re∫(0,∞)-iωε(ω) |E(ω)|2 dω. The contribution due to the 1 in ε - 1 vanished because it is pure imaginary. Energy transfer depends on there being an imaginary part of ε, so the integral has a real part.
The stopping power is, as before -dE/dx = 2πN∫(a,∞)ΔE(b)bdb, where a is a cutoff distance like bmin. Combining the energy loss and this equation, we find Fermi's result: (-dE/dx)(b > a) = (ze/v)2(2/π) Re∫(0,∞) iωλ* a K1(λ*a)K0(λa) (1/ε - β2)dω. In order to evaluate this formula, the behavior of ε(ω) must be known. With reasonable assumptions, it gives the same results as we have already obtained, but it also includes the density effect that was not present in the previous formulas.
We won't go into the rather complicated details of evaluating the density effect in general, but only state the results for the extreme relativistic case, where it is most important. The relativistic limit of Fermi's equation is -dE/dx = (zeωp/c)2ln(1.123c/aωp), where ωp is the plasma frequency, [4πNZe2/m]1/2. Note that there is no power of γ in the argument of the logarithm, so the energy loss does not increase as rapidly with energy. Also, the energy loss does not depend on the details of atomic structure, only on the total number of electrons present. The stopping power without the density effect can be obtained from the usual classical expression (Bohr's) by putting v = c. The result is -dE/dx = (zeωp/c)2[ln(1.123γc/aΩ) - 1/2]. The difference, for very relativistic particles is Δ(dE/dx) = (zeωp/c)2[ln(γωp/Ω) - 1/2], which increases as ln γ.
Cherenkov (usually spelled Cerenkov in English, but I agree with Jackson in preserving the sound in the spelling) radiation is named after P. A. Cherenkov, who reported it in 1934. It is radiation emitted by a charge moving faster than the local speed of light c/n in a medium. The explanation of it was given by I. Frank and I. Tamm in 1937, who calculated its spectrum. It is a weak radiation, but of considerable interest. We can give a good account of it from the fields we found in the density effect.
Cherenkov radiation is easily observed in the cooling ponds of nuclear reactors. Spent fuel rods are surrounded by a bluish-white glow in the water, giving visual evidence of their strong activity. These days it is difficult to get near enough to a reactor to see this, but I can assure you it is impressive and eerie. The glow comes from fast electrons and positrons created by energetic gamma rays absorbed in the water. Homer Simpson's uranium rods exhibit a greenish glow, but nuclear materials do not glow at all normally, and the reputation for luminosity is unfounded.
Cherenkov radiation also occurs in nature: it comes from the sky. Cosmic rays of TeV energies (1012 eV) produce showers of fast particles in the atmosphere, and beneath these showers there is a nanosecond burst of Cherenkov radiation a hundred yards or so in diameter. The chief problem is separating the Cherenkov radiation from the background. TeV gamma rays are not influenced by electric fields, so they preserve their direction of approach. Cherenkov radiation from gamma showers can locate the direction to within a degree or so. One source was found to be the Crab Nebula pulsar, a neutron star with a rotational period of 33 ms that emits energetic gamma rays. Cherenkov counters are useful in nuclear and particle physics.
We are now interested in the fields at large distances, for bλ large instead of small, as in the density effect. The fields we need are easily obtained from our previous expressions by using the asymptotic forms of the Bessel functions. They are: E1(b,ω) = (izeω/c2)(1 - 1/β2ε)e-λb/√(λb), E2(b,ω) = (ze/εv)√(λ/b)e-λb), and B3(b,λ) = βεE2. The energy loss to distances greater than a can be calculated by Poynting's Theorem as well as by our previous method. The formula that results is -dE/dx = -ca Re∫(0,∞)B3*E1 dω. If we insert the Fourier transforms for the fields, the energy loss turns out to be -dE/dx = (ze/c)2 Re[-i√(λ*/λ)]∫ω (1 - 1/β2ε)e-(λ + λ*)adω.
Now, λ = (ω/v)√(1 - β2ε), so if β2ε > 1, λ is pure imaginary, say λ = -iκ. In this case, λ*/λ = -1 and λ + λ* = 0, so we have -dE/dx = (ze/c)2∫ω(1 - 1/β2ε)dω, where the integral is over the frequencies for which β2ε is greater than unity. This expression no longer contains the distance a, so the energy is lost to infinite distance; that is, it is radiated. The integrand is the spectrum of the Cherenkov radiation, Frank and Tamm's result. The shaded area in the diagram at the left illustrates the spectrum in the case of a resonance in the ultraviolet, the usual case in most transparent media. It is easy to see why Cherenkov radiation in water is blue.
The electric field is transverse to the distance from a point on the path to the observer. This distance makes an angle θ with the path, and tan θ = -E1/E2. Using the expression for λ, it is not hard to show that tan θ = √(β2ε - 1), or cos θ = 1/β√ε = c/nv, where n is the index of refraction. The wavefronts are cones whose normals make this angle with the path of the particle. The radiation is polarized in the plane defined by the path and the observation point. For water, n = 4/3, so Cherenkov radiation is emitted by particles moving faster than 0.75c, or 2.25 x 1010 cm/s. For electrons, the kinetic energy must be greater than 261 keV for the emission of Cherenkov radiation. Of course, electrons of this energy will not get far in water. Fast particles emit the radiation more and more perpendicularly to their paths. Geometrically, it's like the shock wave from a supersonic projectile.
A Cherenkov telescope is shown in the figure. Three photomultipliers and three plane mirrors are arranged at 120° around the axis. The baffle stops radiation that is not emitted around the angle θ. When the three photomultiplers have an equal output, the axis of the telescope is directed toward the source. Therefore, the telescope is sensitive to both particle energy and direction.
P. A. M. Dirac showed in 1931 that the existence of a magnetic monopole would imply the quantization of electric charge. A magnetic monopole is a free magnetic charge, that probably would be of two kinds, N and S, and is the source of a magnetic field H just as electric charge is a source of E. Such monopoles have not been found (alas) but speculation about them is interesting.
If there were only magnetic charge, Maxwell's equations would appear as at the left. We can get these equations from the duality transformation E' = -E, B' = B, and similarly for D and H, and changing electric charge to magnetic. The magnetic field H produced by a magnetic charge g is the same as the electric field produced by an electrical charge q. Incidentally, we shall consider H = B and E = D for simplicity, so our monopole exists in a vacuum. The H of a point monopole is the same as the -E of a point charge, even relativistically, so -E1 becomes H1, and -E2 becomes H2. B3 is then βE2, so that βg replaces ze in the expression for the field. When a monopole zips by an electron, the electron will be knocked to the side, not toward the path of the monopole, as for an electric charge.
Dirac showed that consistency required (ge/h'c) = n/2, where n could be ±1, ±2, ... . For a given g, this means that electric charge is quantized in units of e. Taking the known value for e, we can solve the equation for the pole strength g, g = h'c/2e, where we have taken n = 1 to get the value analogous to the electronic charge. From this expression, we find that g = 3.2913 x 10-8 emu. This is quite large compared to e = 4.803 x 1010 esu, and shows that elementary monopoles will interact strongly.
We can use all our equations for the stopping of charged particles if we make the replacement ze = βg. There is a small approximation in that we have no B1 field, but it contributes very little. The charge equivalent to a monopole is then z = βg/e = 1/2α, where α is the fine-structure constant e2/h'c. Numerically, for β = 1, this is z = 68.5! A fast elementary monopole would be stopped like a heavy nucleus. The Bohr or Bethe formulas (let's choose the Bethe formula) would give us -dE/dx = (4πNZe2/m)(ge/c)2[ln(2γ2mv2/h'Ω) - β2). The main thing to notice is that the β in &geta;g has wiped out the velocity from the coefficient and replaced it by c. The only velocity dependence is now in the logarithm, so the variation is small. Energy loss for a monopole increases slowly with γ, and does not show the rapid rise in energy loss at small velocities, which is exactly cancelled by the decrease in the electric field at small velocities. This is certainly one way to distinguish monopoles from charges.
The formulas for Cherenkov radiation cannot be used, however, because they depend on the dielectric constant ε which becomes the permeability in the case of monopoles, which is unity. In a permeable material, there would, of course, be rather strong radiation. The dielectric constant does affect the electric field E3, however. This would require working through the theory again with this assumption, which I have not done. However, it surely has been done, but I do not know the results at present.
Some general characteristics can, however, be predicted. There will be the same Cherenkov cone, but the radiation will be polarized with the magnetic field in the plane containing the path of the monopole and the observer, and the electric field perpendicular to that. This difference in polarization should be easy to detect, and will be a distinct signature of the monopole. Because of the large size of g, the radiation should not be very weak, but quite prominent.
Air is an important stopping medium. The number density N/V at 15°C and 760 mmHg is pNA/RT = 2.5482 x 1019 per cc. N2, with Z = 14, makes up a fraction 0.78, O2, with Z = 16, makes up a fraction 0.21, and A, with Z = 18, makes up 0.01. Therefore, the value of NZ for air is 3.792 x 1020 per cc. The plasma frequency for air is ωp = 1.0985 x 1015 s-1. Hence the factor ωpze/v = 1.0552 x 106 cm/s / v. If we use B = 2γ2mv2/h'Ω for the argument of the logarithm, and estimate h'Ω = 35 eV, γ = 1, we find B = 1.143 x 105E, where E is in MeV. For the fast alphas from Po212, energy 8.78 MeV, and v = 2.05 x 109 cm/s. Hence, ln B = 13.82. Putting this all together, we find that -dE/dx = 3.66 x 10-6 erg/cm, or 2.29 MeV/cm, which is certainly of the correct order of magnitude.
The value of 35 eV that we used for h'Ω is only a very rough guess. It is the average energy loss per ion pair created in air. Not all interactions create an ion pair, of course; most transfer much less energy. The ionization potential of N2 is 15.576 eV, and that of O2 is 12.2 eV. Both molecules dissociate more easily than they ionize, with dissociation energies of 7.373 eV and 5.080 eV, respectively. High-energy particles hardly notice the molecular associations of the atoms.
Electrons have a further energy loss mechanism, the emission of radiation when they are scattered by nuclei. This is called Bremsstrahlung, brake radiation, and is an important energy loss for fast electrons. The formulas for bremsstrahlung loss look very much like the energy-loss formulas in this article, except that they involve the classical electron radius ro, and the maximum impact parameter is determined by shielding of the nuclear charge by the atomic electrons. In fact, it is used in electron synchrotron X-ray sources. Jackson treats bremsstrahlung in his Chapter 15. Cherenkov radiation, incidentally, has negligible stopping effect.
J. D. Jackson, Classical Electrodynamics, 2nd ed. (New York: John Wiley & Sons, 1975). Chapter 13.
I. Kaplan, Nuclear Physics (Cambridge, MA: Addison-Wesley, 1955). Chapter 13.
E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra (Cambridge: Cambridge University Press, 1935). p. 108f. Oscillator strengths and the Thomas-Kuhn sum rule.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions (Washington, DC: USGPO, 1964). pp. 374-379 and pp. 416-429 (numerical tables).
A. Erdélyi, et. al., Tables of Integral Transforms, Vol. I (New York: McGraw-Hill, 1954). p. 11, 1.3(7).
For example, a 2P3/2 state has s = 1/2, l = 1 and j = 3/2 in the usual notation. In labelling states, S,P,D and F mean l = 0,1,2 and 3, the superscript is 2S + 1, and the subscript is J. The g-factor for this state is g = 1 + [(3/2)(5/2) + (1/2)(3/2) - (1)(2)]/2(3/2)(5/2) = 4/3. In a magnetic field B, each of the 4 states now has a different energy (4Z/3)m, which makes a symmetrical set of equally-spaced states. Similarly, a 2S1/2 state has g = 2 since l = 0, and splits into two states of energies 2Zm = +Z and -Z. Each spectral line in the observed Zeeman pattern of the line that is a transition between the P state and the S state starts on one of the P levels and ends on one of the S levels, obeying the selection rule Δm = ±1 for s-polarized lines, and 0 for p-polarized lines. The selection rules are given by quantum mechanics, and are conditions that the transition matrix elements not vanish from symmetry. The consequent pattern is composed of six lines instead of three, as shown in the diagram. Relative intensities of the lines are shown above the arrows representing the transitions.
Putting this all together, the spin-orbit interaction energy Γ = (a/2)[j(j+1) - l(l+1) - s(s+1)], where the constant a = Rα2Z4/n3l(l+1/2)(l+1) cm-1, in terms of the Rydberg constant R (109737 cm-1) and the fine-structure constant α (1/137). For Z = 1, the numerator is 5.85 cm-1. For the level n = 2, l = 1 we have the states 2P3/2 and 2P1/2. From our formula, Γ = 0.122[j(j+1) - 11/4], which gives +0.122 for j = 3/2 and -0.244 for j = 1/2. The "doublet" is shown at the right. The numbers to the left of the states give the degeneracy. The level of smallest j lies lowest, and the "center of gravity" is at the level without spin-orbit coupling, since 4(0.122) + 2(-0.244) = 0. This is called a "normal doublet." The total splitting is 0.366 cm-1, a very small amount. Careful experiments give 0.364 cm-1, so the agreement is excellent.
The scattering cross section is dσ/dΩ = ro2 |ε·ε'|2. Let's suppose that the wave vector of the unpolarized incident wave is directed along the +z-axis, so that the electric fields lie in the xy-plane, and their polarizations are defined by the unit vectors in the coordinate directions, with half the power in each. Polarization directions can be chosen for radiation scattered into θ,φ with one in the xy-plane, and the other in the plane of the scattered wave vector and the incident wave vector. These will be, respectively, ε1 = -εx sinφ + εy cosφ, and ε2 = cosθ(εx cosφ + εy sinφ) + εz cosφ. It is now easy to pick out the contributions from the two incident polarizations. The result is cos2θcos2φ + sin2φ and cos2θsin2φ + cos2φ. On adding, φ drops out, and we have just 1 + cos2θ, which must be divided by two, since half the power is in each incident polarization.
A diagram of such a collision in the laboratory system is shown at the right. The electron of rest mass m is initially at rest, while the incident X-ray photon approaches from the left. The scattered photon leaves at an angle φ, the recoil electron at an angle θ. For each particle, the energy and momentum are shown in the diagram. These quantities are the relativistic values, with β = v/c and γ = (1 - β2)2, where v is the recoil velocity of the electron. Conservation of energy gives h(f - f') = mc2(γ - 1), where f is the frequency of the X-ray before collision, and f' the frequency after. Conservation of momentum is expressed by two equations. One is (f - f'cosφ) = (γmc2β/h) cos θ = K cosθ, and the other is f'sinφ = K sinθ. Squaring the two equations and adding, we get f2 -2ff'cosφ + f'2 = K2, which eliminates the angle of recoil θ. Subtracting the result of squaring the equation resulting from the conservation of energy gives 2ff'(1 - cosφ) = K2 - [mc2(γ - 1)/h]2. The right-hand side reduces to 2(mc2/h)2(γ - 1) = 2(mc2/h)(f - f'), using γ2(β2 - 1) = -1, and the conservation of energy again. The result is f - f' = ff'(h/mc2)(1 - cosφ). Since f = c/λ, (f - f')/ff' = (1/c)(λ' - λ), so finally we get λ' - λ = (h/mc)(1 - cosφ), a quite elegant result.
The magnetic field also holds the electron in a circular orbit, and equating the magnetic force to the centripetal acceleration times the mass gives γmv2/r = eBv/c, or γmv = reB/c. Note that γ does not change in circular motion, since v is constant, so we can do this. This is a second expression for the momentum, coming from Newton's Law. If we equate our two expressions for the momentum γmv, we find eΦ/2πrc = reB/c, or B = Φ/2πr2. This is the condition that the magnetic field at the orbit, B, will be just right to keep the accelerating electrons in an orbit of fixed radius. Since the average value of the magnetic field inside the orbit is Φ/πr2 = , we have = 2B, the famous 1:2 condition. The average field inside the orbit must be twice the value of the field at the orbit. This can be arranged by properly shaping the pole faces, as shown in the diagram. It is a good exercise to verfy that the sense of revolution in the orbit is consistent with the magnetic field, and that the induced electric field is in the correct direction to accelerate the electrons when the flux increases. Don't forget that the electronic charge is negative.