<p> Ph 801 — Exercise 5 1. From the expression of the muon energy loss derive the vertical muon intensity as a function of depth X shown in the plot for a - power law spectrum of muons of the form dF/dE = K E . The relation you should obtain is - a +1 vert Ke - bX - a +1 Fm = ҙ exp[- (a - 1)bX] ҙ (1- e ) a - 1 Comment on the behavior of this function looking at the plot: what value does it approach for large depths?</p><p>Guidelines: Integrate the spectrum between the initial minimum energy that a muon must have to propagate through a thickness of material X and the final energy that is clearly 0.</p><p>This is the depth intensity relation for muons from the vertical direction. Notice: the grey hatched band at large depths is due to the neutrino-induced muons with energies above 2 GeV (upper line: horizontal flux, lower line: vertical flux)</p><p>2. Derive the Compton wavelength shift formula ’e (1-cos) </p><p> with e = h/(mc) Compton wave length of the electron. Consider an </p><p> incoming photon of momentum p = h/c and energy E = h and the </p><p> scattered photon with momentum p’ = h’/c and energy E’ = h’. The </p><p> electron is initially at rest and after the collision recoils at an angle e with momentum p. p’ = h’/c </p><p> p = h/c     </p><p> Suggested solution</p><p>1. The minimum energy E of a muon to penetrate at depth X before it stops can be calculated in the following way:</p><p> dEm 0 dE Em,min 1 1 ж b ц = - (a + bE m) Ю X = = = lnз1 + E m,min ч dx тEm,min dE / dx т0 a + bE b и a ш </p><p> a жa + bE m,min ц жe + E m,min ц bX e + Em,min bX e = Ю bX = logз ч = logз ч Ю e = Ю Em,min = e(e - 1) b и a ш и e ш e For small column depths X<<1/b  1/4e-6 g/cm2 = 2.5 kmwe muons lose energy mostly </p><p> on ionization and E,minbX = aX. The muon energy spectrum underground then reflects the surface muon spectrum.</p><p>0 dFm - a vert - a K - a +1 0 K - a +1 K - a +1 bX - a +1 = KEm Ю Fm = K E dE = [E ] = Em,min = e (e - 1) = dE т 1- a Em,min 1- a 1- a m Em,min</p><p>- a +1 - bX bX - bX - a +1 - a +1 - bX - a +1 - a +1 Ke жe e - e ц Ke ж1 - e ц Ke - a +1 = = = e- (a - 1)bX 1- e- bX з - bX ч з - bX ч ҙ ( ) 1- a и e ш 1- a и e ш 1- a</p><p>This is the flux of muons that survives after propagating a depth X in the vertical direction. The first term is a constant term with depth and reflects the muon energy spectrum at surface. The 3rd term is always larger than 1 and approaches 1 at large depths.</p><p>At large depths the second term determines the depth intensity curve which has the -1 exponential shape exp(-X/X0) with X0 = [b(-1)] . For inclined directions the absorbing ground layer increases like 1/cos = sec ( = zenith angle) for a flat overburden, so that for muons from inclined directions the depth- intensity relation exponential term is modified as e-(-1)bsecX.</p><p>2. Energy conservation:</p><p>2 2 2 2 4 2 hn + mc = hn'+ p c + m c = hn'+Ke + mc Ю</p><p>Ke = h(n - n')</p><p>Momentum conservation</p><p> p = p + p’</p><p>2 2 2 м h n h n ' м ж h n ц ж h n ' ц 2 h n n ' 2 2 = cos q + p cos q + cos q - 2 cos q = p cos q 2 2 n e п з ч з ч n 2 n e 2 п c c п и c ш и c ш c ж h n ц ж h n ' ц h n n ' 2 2 2 2 2 2 2 Ю Ю Ю + - 2 cos q n = p Ю p c = h ( n - n ' ) - 2 h n n ' cos q n н н з ч з ч 2 h n ' ж h n ' ц 2 2 2 и c ш и c ш c п sin q n = p sin q e п sin q = p sin q з ч n e о c оп и c ш </p><p> h 2 ( n - n ' ) 2 + m 2c 4 + 2 mc 2h ( n - n ' ) = p 2c 2 + m 2c 4 = h ( n 2 - n ' 2 ) + m 2c 4 2 2 2 -h n n ' + 2 mc h ( n - n ' ) = - 2 h n n ' cos q n 2 Ю mc (n - n') = hnn'(1- cosqn ) 1 1 h h Ю - = 2 (1- cosqn ) Ю l '- l = (1- cosqn ) = l e (1- cosqn ) n' n' mc mc c l = n</p>