Homework 4: Fermi s Golden Rule and Feynman Diagrams

Transcription

1 Homework 4: Fermi s Golden Rule and Feynman Diagrams 1 Proton-Proton Total Cross Section In this problem, we are asked to find the approximate radius of the proton from the TOTEM data for the total proton cross section presented in the plot From the plot, we see that TOTEM measures the total cross section at 7/8 TeV center-of-mass collision energies to be about 100 millibarns, or about 10 5 cm That is, the radius of the proton r p is approximately the square-root of this cross section, or about r p cm = m Indeed, the radius of the proton as measured by the total cross section of proton-proton scattering is about a femtometer Proton Collision Beam at ATLAS In this problem, we re asked to study the properties of the data collected by the ATLAS experiment at the LHC (a) The total luminosity over the data taking period is about 35 fb 1 (recorded by ATLAS) Data were collected over a period of about five and a half months A very useful bit of trivia is that there are (to extremely good approximation) π 10 7 seconds in a year Therefore, data were collected over about 10 7 s Then, the average luminosity is L = 35 fb s = fb 1 s 1 To convert this to cm s 1, we use the conversion that 1 b = 10 4 cm Then, we find that the luminosity is also L = cm s 1 = cm s 1 (b) The total cross section from TOTEM is about 100 mb and the integrated luminosity from ATLAS is 35 fb 1 Therefore, the total number of events collected is approximately 100 mb 35 fb 1 = ( ) ( ) = events Dividing by 10 7 s of time in the data taking run, there were approximately s = events/sec (c) Now, we re asked to estimate the number of protons in each bunch that is collided at the LHC To do this, we need to determine the volume of a bunch We are told that protons are accelerated in an RF field with a frequency of 400 MHz This corresponds to a wavelength of: λ = c f = m = m 1

2 We expect the length of the bunches of protons to be about half of this wavelength; protons are stuck in the troughs of the wave For order of magnitude estimates, we ll call this length of the bunch 10 cm Using the radius of the bunch as 10 micrometers, the volume of the bunch is approximately Vol = m 3 = m 3 The protons at the LHC are traveling at about the speed of light, so the relative velocity of the two colliding beams of protons is very nearly v A v B c = m/s Also, we assume that the bunches in the colliding beams are identical, so we just call the number of protons in a bunch N p The density of protons in a bunch is then N p /Vol From part (a), the luminosity is L = m 1 s 1 Therefore, using the expression for luminosity derived in class, L = N p Vol Vol v A v B, we can solve for the number of protons N p as a function of everything else: Vol L N p = v A v B Plugging in all of the values, we find N p 10 8 (d) Now, we re asked to study the Stairway to Heaven plot i From the plot, the measured cross section of the process pp Z at 13 TeV collisions is about pb Multiplying by the integrated luminosity of 35 fb 1, we find N pp Z = ( ) events ii The measured cross section of the process pp H is about 70 pb Therefore, the number of events observed is N pp H = ( ) events iii Finally, the measured cross section of the process pp t t is approximately 10 3 pb Then, the number of events observed is approximately N pp t t = (10 6 ) 35 = events

3 e p p 0 e q e + e k k 0 time p p 0 e + e e + q k k 0 time e + 3 e + e e + e Scattering We are asked to draw the Feynman diagrams for e + e e + e scattering, at lowest order There are two Feynman diagrams that can be drawn, which have distinct routing of fermion lines, consistent with maintaining the direction of the arrows on the lines The diagrams are: 4 Non-relativistic limit of Feynman diagrams In this problem, we study the nonrelativistic limit of e µ e µ scattering (a) Let s focus on the electron-photon vertex at the top of the diagram External momentum flows forward in time, and so momentum q and p flow out of the vertex, while momentum p flows in Therefore, enforcing conservation of momentum, p = p + q, or q = p p Equivalently, we could look at the bottom vertex, for which we would find q = k k (b) Now, we re asked to describe the physical situation when the momentum flowing through the photon, q, goes on-shell This occurs when q = (p p ) = p + p p p = (m e p p ) 0, where m e is the mass of the electron That is, the photon goes on-shell when p p = m e 3

4 However, this must mean that p = p! Therefore, if the photon goes on-shell, the electron and muon are not deflected from one another in the scattering: they keep traveling in a straight line (c) Now, we re asked to consider the non-relativistic limit of the momentum electric potential with a non-zero photon mass: Ṽ (q) = e q m γ In the non relativistic limit, the four-momentum p is p (m e, p) Then, the four-momentum q becomes q = p p (m e, p) (m e, p ) = (0, p p ) Then, q p p and so Ṽ (q) = e p p m γ = e q + m γ (d) Finally, we re asked to Fourier transform this non-relativistic momentum potential to a position potential, that we re more familiar with The expression of this Fourier transform is dq V (r) = (π) Ṽ (q)eiq r dq e = e iq r 3 (π) 3 q + m γ In spherical coordinates, the Fourier transform is V (r) = 0 π 1 1 e dq dφ d cos θ 0 1 (π) 3 q e iqr cos θ, q + m γ where we have denoted q q and note that q r = qr cos θ, where r is the radial coordinate Now, we can integrate over φ and replace cos θ u and we have V (r) = e (π) 0 1 dq du 1 q e iqru q + m γ The integral over u can be done (it s just the integral of an exponential), and we find V (r) = e q e iqr e iqr dq = e q dq e iqr 4π 0 q + m γ iqr 4π ir q + m γ 4

5 Now we re cooking The remaining integral can be done with a contour in the complex plane using Cauchy s theorem If we extend the contour along the real axis (the region of integration) and into the upper half of the complex plane, then the contour can safely be pushed to q +i because the integrand is exponentially suppressed there There s a single pole in the upper half of the complex plane, when q = im γ, because we can factorize q + m γ = (q + im γ )(q im γ ) So, we just need to find the residue of the integrand at q = im γ This is: Res q imγ q e iqr q = Res q + m q imγ γ (q + im γ )(q im γ ) eiqr = πi im γ e mγr = iπe mγr im γ Putting it all together, the potential is then V (r) = e e mγr 4π r This is exactly the potential energy we would calculate in classical electromagnetism for two charges e separated by distance r! (e) If m γ 0, then the effective range of the potential is set by the argument of the exponential When the separation r = m γ 1, the strength of the potential has decreased by a factor of e Therefore the range is r = m γ 1 This is also how Yukawa predicted the existence of the pion! If the mass of photon m γ 0, then the potential becomes V (r) = e 1 4π r This is exactly the potential energy we would calculate in classical electromagnetism for two charges e separated by distance r! 5 Three-body Phase Space In this question, we explore three-body phase space (a) We re first asked to study the variables x i = Q p i Q, for i = 1,, 3 Let s look at the sum of these variables: x 1 + x + x 3 = Q p 1 Q + Q p Q + Q p 3 Q = Q (p 1 + p + p 3 ) Q By momentum conservation, p 1 + p + p 3 = Q, and therefore, Note that this is independent of frame x 1 + x + x 3 = 5

6 (b) Now, we go to the center of mass frame In this frame, Q = (E CM, 0, 0, 0) and so Therefore, Then, the energy E i is x i = Q p i Q Q p i = E CM E i = E CME i E CM E i = x ie CM = E i E CM To find the magnitude of the three-momentum p i, we just use the energymomentum relation of special relativity: p i = E i m i = x i E CM 4 m i (c) Now, we integrate over p 3 using the momentum conserving δ-function We then have d 3 p 1 d 3 p dπ 3 = π δ (E (π) 3 (π) 3 CM E 1 E E 3 ), E 1 E E 3 with p 3 = p 1 p and E 3 = ( p 3 +m 3) 1/ = ( p 1 + p +m 3) 1/ = ( p 1 + p + p 1 p cos θ 1 +m 3) 1/ On the far right of this expression, I have expanded the square of the vector sum p 1 + p in terms of dot products Note that the angle between momenta 1 and appears in this expression We can identify one of the polar angles in spherical coordinates with this angle Writing the remaining integrals in spherical coordinates, we have dπ 3 = 1 (π) 5 d p 1 d cos ψ dφ 1 d p d cos θ 1 dφ p 1 E 1 p E 1 E 3 δ (E CM E 1 E E 3 ) (d) Now, we want to integrate over cos θ 1 using the remaining δ-function This requires changing variables in the delta function, which requires calculating a Jacobian Recall the identity δ(y f(x)) = 1 f (x) δ(x f 1 (y)), 6

7 where f(x) is some function of x, f (x) is its derivative, and f 1 (y) is its inverse function Applying this identity to the remaining energy δ-function, and using the expression for E 3 from earlier, we have δ(e CM E 1 E E 3 ) = δ ( E CM E 1 E ( p 1 + p + p 1 p cos θ 1 + m 3) 1/) Taking the derivative of the argument of the δ-function with respect to cos θ 1, we find d ( p 1 + p + p 1 p cos θ 1 + m d cos θ 3) 1/ = p 1 p 1 E 3 Also, we can re-arrange the argument of the δ-function to solve for cos θ This is cos θ = (E CM E 1 E ) p 1 p m 3 p 1 p Therefore, we can re-write the δ-function as δ(e CM E 1 E E 3 ) = E 3 p 1 p δ ( cos θ 1 (E CM E 1 E ) p 1 p m 3 p 1 p With this, we can then do the integral over cos θ 1 We then have dπ 3 = 1 (π) 5 = 1 (π) 5 = 1 8(π) 5 d p 1 d cos ψ dφ 1 d p d cos θ 1 dφ p 1 E 1 p E 1 E 3 δ (E CM E 1 E E 3 ) p 1 p 1 E 3 d p 1 d cos ψ dφ 1 d p d cos θ 1 dφ E 1 E E 3 p 1 p ( δ cos θ 1 (E CM E 1 E ) p 1 p m 3 p 1 p d p 1 d cos ψ dφ 1 d p dφ p 1 E 1 p E (e) The last three angles are just the Euler angles that rotate the total three-body system We can easily integrate over these angles to find dπ 3 = 1 p 1 p d p 8(π) 5 1 d cos ψ dφ 1 d p dφ = 1 d p E 1 E 3π 3 1 d p p 1 p E 1 E (f) Finally, we want to change integration variables to x 1 and x To do this, we recall that x p 1 = 1 ECM m 1, 4 7 ) )

8 and so the differential object d p 1 = x 1 E CM 4 p 1 dx 1 Also, recall that and so the integrand becomes: E 1 = x 1E CM, d p 1 p 1 E 1 ECM p 1 E CM = dx 1 x 1 = dx 1 4 p 1 x 1 E CM A similar expression exists for x Putting this altogether, we then find dπ 3 = E CM dx 18π 3 1 dx Whew! 8