IPhO 2008 · Theory problem 2

Cherenkov Light and Ring Imaging Counter

Download PDFIPhO 2008

Light propagates in vacuum with the speed \(c\). There is no particle which moves with a speed higher than \(c\). However, it is possible that in a transparent medium a particle moves with a speed \(v\) higher than the speed of the light in the same medium \(\frac{c}{n}\), where \(n\) is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed \(v\) in a transparent medium with refractive index \(n\) such that \(v>\frac{c}{n}\), radiates light, called Cherenkov light, in directions forming with the trajectory an angle

\[ \theta=\arccos\frac{1}{\beta n} \tag{1} \]

where \(\beta=\frac{v}{c}\).

Figure 1

1. To establish this fact, consider a particle moving at constant velocity \(v>\frac{c}{n}\) on a straight line. It passes A at time 0 and B at time \(t_1\). As the problem is symmetric with respect to rotations around AB, it is sufficient to consider light rays in a plane containing AB.

At any point C between A and B, the particle emits a spherical light wave, which propagates with velocity \(\frac{c}{n}\). We define the wave front at a given time \(t\) as the envelope of all these spheres at this time.

1.1. Determine the wave front at time \(t_1\) and draw its intersection with a plane containing the trajectory of the particle.

1.2. Express the angle \(\varphi\) between this intersection and the trajectory of the particle in terms of \(n\) and \(\beta\).

2. Let us consider a beam of particles moving with velocity \(v>\frac{c}{n}\), such that the angle \(\theta\) is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length \(f\) and center C, at point S. SC makes with SI a small angle \(\alpha\) (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror.

Explain why with the help of a sketch illustrating this fact. Give the position of the center O and the radius \(r\) of the ring image.

This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator.

Note: in all questions of the present problem, terms of second order and higher in \(\alpha\) and \(\theta\) will be neglected.

3. A beam of particles of known momentum \(p=10.0\ \mathrm{GeV}/c\) consists of three types of particles: protons, kaons and pions, with rest mass \(M_{\mathrm p}=0.94\ \mathrm{GeV}/c^2\), \(M_\kappa=0.50\ \mathrm{GeV}/c^2\) and \(M_\pi=0.14\ \mathrm{GeV}/c^2\), respectively. Remember that \(pc\) and \(Mc^2\) have the dimension of an energy, and 1 eV is the energy acquired by an electron after being accelerated by a voltage 1 V, and \(1\ \mathrm{GeV}=10^9\ \mathrm{eV}\), \(1\ \mathrm{MeV}=10^6\ \mathrm{eV}\).

The particle beam traverses an air medium (the radiator) under the pressure \(P\). The refraction index of air depends on the air pressure \(P\) according to the relation \(n=1+aP\) where \(a=2.7\times10^{-4}\ \mathrm{atm}^{-1}\)

3.1. Calculate for each of the three particle types the minimal value \(P_{\min}\) of the air pressure such that they emit Cherenkov light.

3.2. Calculate the pressure \(P_{\frac12}\) such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of \(\theta_\kappa\) and \(\theta_\pi\) in this case.

Is it possible to observe the ring image of protons under this pressure?

4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at \(10\ \mathrm{GeV}/c\) having a half width at half height \(\Delta p\). This makes the ring image broaden, correspondingly \(\theta\) distribution has a half width at half height \(\Delta\theta\). The pressure of the radiator is \(P_{\frac12}\) determined in 3.2.

4.1. Calculate \(\frac{\Delta\theta_\kappa}{\Delta p}\) and \(\frac{\Delta\theta_\pi}{\Delta p}\), the values taken by \(\frac{\Delta\theta}{\Delta p}\) in the pions and kaons cases.

4.2. When the separation between the two ring images, \(\theta_\pi-\theta_\kappa\), is greater than 10 times the half-width sum \(\Delta\theta=\Delta\theta_\kappa+\Delta\theta_\pi\), that is \(\theta_\pi-\theta_\kappa>10\Delta\theta\), it is possible to distinguish well the two ring images. Calculate the maximal value of \(\Delta p\) such that the two ring images can still be well distinguished.

5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light.

5.1. Find out the minimal kinetic energy \(T_{\min}\) of a particle with a rest mass \(M\) moving in water, such that it emits Cherenkov light. The index of refraction of water is \(n=1.33\).

5.2. The radioactive source used by Cherenkov emits either \(\alpha\) particles (i.e. helium nuclei) having a rest mass \(M_\alpha=3.8\ \mathrm{GeV}/c^2\), or \(\beta\) particles (i.e. electrons) having a rest mass \(M_e=0.51\ \mathrm{MeV}/c^2\). Calculate the numerical values of \(T_{\min}\) for \(\alpha\) particles and \(\beta\) particles.

Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few MeV, find out which particles give rise to the radiation observed by Cherenkov.

6. In the previous sections of the problem, the dependence of the Cherenkov effect on wavelength \(\lambda\) has been ignored. We now take into account the fact that the Cherenkov radiation of a particle has a broad continuous spectrum including the visible range (wavelengths from \(0.4\ \mu\mathrm m\) to \(0.8\ \mu\mathrm m\)). We know also that the index of refraction \(n\) of the radiator decreases linearly by 2% of \(n-1\) when \(\lambda\) increases over this range.

6.1. Consider a beam of pions with definite momentum of \(10.0\ \mathrm{GeV}/c\) moving in air at pressure 6 atm. Find out the angular difference \(\delta\theta\) associated with the two ends of the visible range.

6.2. On this basis, study qualitatively the effect of the dispersion on the ring image of pions with momentum distributed over an interval centered at \(p=10\ \mathrm{GeV}/c\) and having a half width at half height \(\Delta p=0.3\ \mathrm{GeV}/c\).

6.2.1. Calculate the broadening due to dispersion (varying refraction index) and that due to achromaticity of the beam (varying momentum).

6.2.2. Describe how the color of the ring changes when going from its inner to outer edges by checking the appropriate boxes in the Answer Sheet.

Official English text of the problem.

Answers

  • 1.1. The common-tangent wave front is a right circular cone with vertex B and axis AB, opening backward along the trajectory. Its axial section consists of two tangent segments meeting at B. For emission beginning at A, the full outer boundary also includes the exposed rear cap of the first spherical wave.
  • 1.2. \[ \boxed{\varphi=\arcsin\frac{1}{n\beta}},\qquad \theta=\frac{\pi}{2}-\varphi. \]
  • 2. Let F be the midpoint of SC. The center O lies in the focal plane through F, in the plane ISC, on the opposite side of SC from I: \[ \boxed{FO\simeq f\alpha,\qquad CO\parallel IS,\qquad r\simeq f\theta.} \] The reference ray along IS reflects at S along SO.
  • 3.1. \[ \boxed{P_{\min}=\frac{\sqrt{1+(Mc/p)^2}-1}{a}.} \] The threshold pressures are \[ \boxed{P_{\min,\mathrm p}\simeq16.3\ \mathrm{atm},\quad P_{\min,\kappa}\simeq4.63\ \mathrm{atm},\quad P_{\min,\pi}\simeq0.363\ \mathrm{atm}.} \] Nonzero emission requires pressure above the threshold.
  • 3.2. \[ \boxed{P_{\frac12}\simeq6.05\ \mathrm{atm},\quad \theta_\kappa\simeq0.0277\ \mathrm{rad},\quad \theta_\pi\simeq0.0554\ \mathrm{rad}.} \] Protons produce no ring at this pressure.
  • 4.1. \[ \boxed{\frac{\Delta\theta_\kappa}{\Delta p} \simeq9.0\times10^{-3}\frac{\mathrm{rad}}{\mathrm{GeV}/c},\qquad \frac{\Delta\theta_\pi}{\Delta p} \simeq3.5\times10^{-4}\frac{\mathrm{rad}}{\mathrm{GeV}/c}.} \]
  • 4.2. The limiting half-width is \[ \boxed{\Delta p_{\mathrm{lim}}\simeq0.296\ \mathrm{GeV}/c.} \] The stated strict separation criterion requires \(\Delta p<\Delta p_{\mathrm{lim}}\).
  • 5.1. \[ \boxed{T_{\min}=Mc^2\left(\frac{n}{\sqrt{n^2-1}}-1\right) =0.5168\,Mc^2\quad(n=1.33).} \]
  • 5.2. \[ \boxed{T_{\min,\alpha}\simeq2.0\ \mathrm{GeV},\qquad T_{\min,e}\simeq0.26\ \mathrm{MeV}.} \] The observed radiation is produced by the \(\beta\) particles, namely electrons.
  • 6.1. \[ \boxed{\delta\theta=\theta(0.4\ \mu\mathrm m)-\theta(0.8\ \mu\mathrm m) \simeq5.9\times10^{-4}\ \mathrm{rad}=0.59\ \mathrm{mrad}.} \]
  • 6.2.1. Dispersion gives a full visible-spectrum angular span of approximately \(0.59\ \mathrm{mrad}\). Momentum spread gives \[ \boxed{\Delta\theta_p\simeq0.11\ \mathrm{mrad}\ \text{(half-width)},\qquad 2\Delta\theta_p\simeq0.21\ \mathrm{mrad}\ \text{(FWHM)}.} \] Corresponding radial widths are obtained by multiplying these angular widths by \(f\).
  • 6.2.2. The inner edge is red and the outer edge is blue/violet. Dominant wavelengths decrease outward; momentum spread blurs and mixes neighboring colors.

Solution

Angles are measured in radians. The mirror geometry is treated to first order in the small angles. The relativistic momentum–speed and Cherenkov relations are retained when evaluating the numerical results.

1.1. Wave-front construction

Write the light speed in the medium as \[ u=\frac{c}{n}. \]

Suppose the particle passes C at time \(t_C\). At time \(t_1\), the wave emitted there has radius \[ R_C=u(t_1-t_C), \] while the particle has moved from C to B through the distance \[ CB=v(t_1-t_C). \] Consequently, \[ \frac{R_C}{CB}=\frac{u}{v}=\frac{1}{n\beta}, \] independent of C.

In any plane containing AB, the spherical waves appear as circles. Their radii are proportional to their centers’ distances from B. They therefore have the same two tangent lines through B. Rotating these tangents about AB produces a right circular cone with its vertex at B, opening backward.

The required sketch is constructed as follows:

  1. Draw AB horizontally, with B to the right and \(AB=vt_1\).
  2. Draw the largest wave circle centered at A, with radius \(ut_1\).
  3. Draw smaller wave circles centered at intermediate points C, with radii \(u(t_1-t_C)\).
  4. Draw the upper and lower common tangents from B to these circles. Mark a tangency point T on the circle centered at A and draw \(AT\perp BT\).

For the finite emission interval A–B, the conical portion extends from B to the tangency circle on the first sphere. If the whole outer boundary of the disturbed region is shown, its rear portion is the exposed spherical cap centered at A.

1.2. Wave-front angle and Cherenkov angle

In the right triangle ATB, \[ AT=\frac{c}{n}t_1,\qquad AB=vt_1. \] The acute angle between the tangent BT and the trajectory line is \(\varphi\), so \[ \sin\varphi=\frac{AT}{AB} =\frac{c}{nv} =\frac{1}{n\beta}. \] Thus \[ \boxed{\varphi=\arcsin\frac{1}{n\beta}.} \]

The light rays are normal to the wave front. Therefore the angle \(\theta\) between a ray and the forward particle direction satisfies \[ \theta+\varphi=\frac{\pi}{2}, \] and hence \[ \boxed{\cos\theta=\sin\varphi=\frac{1}{n\beta}.} \] This establishes equation (1).

2. Formation, center and radius of the ring

Here C denotes the mirror’s center of curvature. Since the mirror is spherical, \[ SC=2f. \] Let F be the midpoint of SC; the paraxial focal plane is perpendicular to SC at F.

First consider a reference light ray traveling along the particle-beam direction IS and striking the mirror at S. The normal to the mirror at S is SC. By the law of reflection, the reflected ray SO makes the same angle \(\alpha\) with SC as SI does, on the opposite side.

Its intersection O with the focal plane consequently satisfies \[ FO=f\tan\alpha\simeq f\alpha, \qquad SO\simeq f. \] Because \(CF=SF=f\), the same construction gives \[ CO\parallel IS. \]

Now fix an azimuth around the particle trajectory. All Cherenkov rays emitted at that azimuth have the same direction, even when emitted at different points of the trajectory. They form a parallel bundle, which the mirror focuses to one point in its focal plane.

Changing the azimuth changes the ray direction around a cone of half-angle \(\theta\). A paraxial mirror converts an angular displacement into a focal-plane displacement equal to \(f\) times that angle. Thus the image points have coordinates relative to O of the form \[ X=f\theta\cos\psi,\qquad Y=f\theta\sin\psi, \] where \(\psi\) is the azimuth. Therefore \[ X^2+Y^2=f^2\theta^2, \] which is a circle: \[ \boxed{r=f\theta.} \]

For the sketch, draw S at the right, C to its left, and F halfway between them. Put I above SC. The reference ray reflects at S toward O below SC. Draw the focal plane through F and mark the two ring intersections in the plane ISC at signed distances \[ -f\alpha+f\theta,\qquad -f\alpha-f\theta \] from F, taking the side toward I as positive. These points are separated by \(2f\theta\) and have midpoint O. The corresponding points from all azimuths form the ring.

3.1. Threshold pressures

Relativistic energy and momentum obey \[ E^2=p^2c^2+M^2c^4,\qquad \beta=\frac{pc}{E}. \] It is convenient to define \[ q=\frac{1}{\beta} =\sqrt{1+\left(\frac{Mc}{p}\right)^2}. \] The Cherenkov condition is \[ n\beta>1. \] At the threshold, \[ n_{\min}=q. \] Using \(n=1+aP\), \[ \boxed{P_{\min} =\frac{q-1}{a} =\frac{\sqrt{1+(Mc/p)^2}-1}{a}.} \]

With the supplied masses and \(p=10.0\ \mathrm{GeV}/c\):

Particle \(\beta\) \(P_{\min}\)
Proton \(0.995611\) \(16.3\ \mathrm{atm}\)
Kaon \(0.998752\) \(4.63\ \mathrm{atm}\)
Pion \(0.999902\) \(0.363\ \mathrm{atm}\)

At equality the Cherenkov angle is zero; a nonzero ring requires pressure above the threshold.

3.2. Pressure for a kaon ring of half the pion radius

Since \(r=f\theta\), the required radius ratio gives \[ \theta_\pi=2\theta_\kappa \] to the stated optical accuracy.

For each species, \[ \cos\theta_i=\frac{q_i}{n}, \qquad q_i=\sqrt{1+\left(\frac{M_i c}{p}\right)^2}. \] Using the double-angle identity, \[ \frac{q_\pi}{n} =\cos(2\theta_\kappa) =2\left(\frac{q_\kappa}{n}\right)^2-1. \] Thus \[ n^2+q_\pi n-2q_\kappa^2=0. \] The positive root is \[ n_*=\frac{\sqrt{q_\pi^2+8q_\kappa^2}-q_\pi}{2}. \]

Numerically, \[ q_\kappa=\sqrt{1+0.050^2},\qquad q_\pi=\sqrt{1+0.014^2}, \] giving \[ n_*\simeq1.00163306. \] Therefore \[ \boxed{P_{\frac12}=\frac{n_*-1}{a}\simeq6.05\ \mathrm{atm}.} \]

The corresponding angles are \[ \theta_\kappa=\arccos\frac{q_\kappa}{n_*} \simeq0.027685\ \mathrm{rad}, \] \[ \theta_\pi=\arccos\frac{q_\pi}{n_*} \simeq0.055371\ \mathrm{rad}. \] Hence \[ \boxed{\theta_\kappa\simeq27.7\ \mathrm{mrad},\qquad \theta_\pi\simeq55.4\ \mathrm{mrad}.} \]

The proton threshold is \(16.3\ \mathrm{atm}\), substantially above this pressure. Thus \[ \boxed{\text{No proton ring is produced.}} \]

4.1. Angular broadening from momentum spread

At fixed pressure, \(n\) is constant. Differentiate \[ \cos\theta=\frac{q}{n}, \qquad q=\sqrt{1+\frac{M^2c^2}{p^2}}. \] Since \[ \frac{dq}{dp}=-\frac{M^2c^2}{p^3q}, \] we obtain \[ \frac{d\theta}{dp} =\frac{M^2c^2}{np^3q\sin\theta}. \] Using \(q=1/\beta\) and \(q/n=\cos\theta\), this can also be written \[ \boxed{\frac{d\theta}{dp} =\frac{1-\beta^2}{p}\cot\theta.} \]

For a narrow momentum distribution, its transformation is locally linear, so \[ \Delta\theta\simeq \left|\frac{d\theta}{dp}\right|\Delta p. \] The derivative is positive: increasing momentum increases the Cherenkov angle.

Evaluating at \(p=10.0\ \mathrm{GeV}/c\) and the pressure from 3.2 gives \[ \left.\frac{d\theta}{dp}\right|_\kappa \simeq9.005\times10^{-3} \frac{\mathrm{rad}}{\mathrm{GeV}/c}, \] \[ \left.\frac{d\theta}{dp}\right|_\pi \simeq3.535\times10^{-4} \frac{\mathrm{rad}}{\mathrm{GeV}/c}. \] Therefore \[ \boxed{\frac{\Delta\theta_\kappa}{\Delta p} \simeq9.0\times10^{-3} \frac{\mathrm{rad}}{\mathrm{GeV}/c},} \] \[ \boxed{\frac{\Delta\theta_\pi}{\Delta p} \simeq3.5\times10^{-4} \frac{\mathrm{rad}}{\mathrm{GeV}/c}.} \]

4.2. Largest permitted momentum half-width

The required separation condition is \[ \theta_\pi-\theta_\kappa> 10\left(\Delta\theta_\kappa+\Delta\theta_\pi\right). \] Substituting the linear broadening relations gives \[ \Delta p< \frac{\theta_\pi-\theta_\kappa} {10\left[ (d\theta/dp)_\kappa+(d\theta/dp)_\pi \right]}. \]

Using the unrounded values, \[ \Delta p_{\mathrm{lim}} \simeq \frac{0.027685} {10(0.0090052+0.00035355)} \ \mathrm{GeV}/c. \] Thus \[ \boxed{\Delta p_{\mathrm{lim}}\simeq0.296\ \mathrm{GeV}/c.} \]

This is approximately \(3.0\%\) of the central momentum. Since the problem states a strict inequality, the well-distinguished regime is \[ \boxed{\Delta p<0.296\ \mathrm{GeV}/c.} \]

5.1. Kinetic-energy threshold in water

The relativistic kinetic energy is \[ T=(\gamma-1)Mc^2,\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}. \] At the Cherenkov threshold, \[ \beta_{\min}=\frac{1}{n}. \] Therefore \[ \gamma_{\min} =\frac{1}{\sqrt{1-1/n^2}} =\frac{n}{\sqrt{n^2-1}}, \] and \[ \boxed{ T_{\min} =Mc^2\left(\frac{n}{\sqrt{n^2-1}}-1\right). } \]

For water, \[ n=1.33,\qquad \gamma_{\min}\simeq1.51676, \] so \[ \boxed{T_{\min}\simeq0.5168\,Mc^2.} \] Nonzero radiation requires kinetic energy above this threshold.

5.2. Alpha particles and electrons

For an alpha particle, \[ T_{\min,\alpha} =0.51676(3.8\ \mathrm{GeV}) \simeq1.96\ \mathrm{GeV} \simeq2.0\ \mathrm{GeV}. \]

For an electron, \[ T_{\min,e} =0.51676(0.51\ \mathrm{MeV}) \simeq0.264\ \mathrm{MeV} \simeq0.26\ \mathrm{MeV}. \]

An alpha particle with kinetic energy of only a few MeV is far below its threshold of approximately \(1960\ \mathrm{MeV}\). Electrons with energies of a few MeV can readily exceed their threshold. Consequently, \[ \boxed{\text{The observed Cherenkov radiation is produced by electrons.}} \]

6.1. Angular difference across the visible spectrum

At \(P=6\ \mathrm{atm}\), \[ n=1+aP =1+(2.7\times10^{-4})(6) =1.00162. \] The magnitude of the refractive-index decrease across the visible range is \[ \delta n=0.02(n-1) =3.24\times10^{-5}. \]

For \(10.0\ \mathrm{GeV}/c\) pions, the representative Cherenkov angle is \[ \theta_0 =\arccos\left(\frac{\sqrt{1+(0.14/10.0)^2}}{1.00162}\right) \simeq0.05513\ \mathrm{rad}. \]

At fixed particle speed, differentiating \[ \cos\theta=\frac{1}{\beta n} \] with respect to \(n\) gives \[ -\sin\theta\,\frac{d\theta}{dn} =-\frac{1}{\beta n^2}. \] Hence \[ \boxed{\frac{d\theta}{dn}=\frac{\cot\theta}{n}.} \]

To first order in the index change, \[ \delta\theta \simeq\frac{\cot\theta_0}{n}\,\delta n \simeq5.86\times10^{-4}\ \mathrm{rad}. \] Thus \[ \boxed{\delta\theta\simeq5.9\times10^{-4}\ \mathrm{rad} =0.59\ \mathrm{mrad}.} \]

The shorter wavelength has the larger index and therefore the larger angle: \[ \theta(0.4\ \mu\mathrm m)>\theta(0.8\ \mu\mathrm m). \]

6.2.1. Dispersion broadening and momentum broadening

The dispersion broadening is the full angular span between the two visible endpoints: \[ \boxed{w_{\mathrm{disp}}=\delta\theta\simeq0.59\ \mathrm{mrad}.} \] The corresponding radial span is \[ \boxed{w_{r,\mathrm{disp}}=f\delta\theta \simeq5.9\times10^{-4}f.} \]

For the momentum spread, use the derivative from 4.1, now evaluated at \(6\ \mathrm{atm}\): \[ \left.\frac{d\theta}{dp}\right|_{\pi,\,6\,\mathrm{atm}} =\frac{1-\beta_\pi^2}{p}\cot\theta_0 \simeq3.55\times10^{-4} \frac{\mathrm{rad}}{\mathrm{GeV}/c}. \] With \(\Delta p=0.3\ \mathrm{GeV}/c\), \[ \Delta\theta_p \simeq(3.55\times10^{-4})(0.3) =1.07\times10^{-4}\ \mathrm{rad}. \] Therefore \[ \boxed{\Delta\theta_p\simeq0.11\ \mathrm{mrad}} \] is the half-width at half height, while \[ \boxed{w_{p,\mathrm{FWHM}}=2\Delta\theta_p \simeq0.21\ \mathrm{mrad}} \] is the full width at half maximum.

The corresponding radial half-width and FWHM are \[ \Delta r_p\simeq1.1\times10^{-4}f, \qquad w_{r,p,\mathrm{FWHM}}\simeq2.1\times10^{-4}f. \]

The visible-spectrum span from dispersion is about \(2.8\) times the momentum-induced FWHM. Dispersion therefore supplies the larger broadening scale. A unique combined FWHM cannot be calculated without specifying the momentum profile and the spectral weighting.

6.2.2. Color ordering across the ring

Because \[ \frac{dn}{d\lambda}<0, \qquad \frac{d\theta}{dn}=\frac{\cot\theta}{n}>0, \] we have \[ \frac{d\theta}{d\lambda}<0. \] Since \(r=f\theta\), \[ \boxed{\frac{dr}{d\lambda}<0.} \]

Longer wavelengths therefore produce smaller radii. The correct color choice is:

  • Inner edge: red, corresponding to the longer wavelengths.
  • Moving outward: progressively shorter dominant wavelengths, with neighboring colors overlapping.
  • Outer edge: blue/violet, corresponding to the shortest wavelengths.

The momentum spread blurs this ordering, but its characteristic width is smaller than the chromatic span, so the red-inner, violet-outer pattern remains.

The solution was checked against the official solution.