Physics Exam 3, Spring 2005 - Topics in Special Relativity, Exams of Physics

Download Physics Exam 3, Spring 2005 - Topics in Special Relativity and more Exams Physics in PDF only on Docsity! Phys 4620, Spring 2005 Exam #3 1. If we move toward a source of light we still measure the speed of the light as c but we measure a different frequency and wavelength from that measured in the source frame. Now, we have a formula from Phys 2110 for the change in frequency when we move relative to a source1. Why can’t we use that formula for light? The basic Doppler shift formula assumes that we can measure the velocities of source and observer relative to the medium which carries the wave. In relativity there is no medmedium which carries alight wave and the only meaningful velocity is the relative velocity between source and receiver; light travels at the same speed in all inertial reference frames. y x(3/4)c (3/4)c A B 2. As seen in frame S particle A moves at speed 3 4 c in the −x direction and particle B moves at speed 3 4 c in the +x direction. What is the speed of particle B as measured in the frame of particle A? B has velocity v = +3 4 c wrt frame S and S has velocity v = +3 4 c wrt the frame of A. To find the velocity of B as measured in the frame of A, use the “Einstein velocity addition formula”, VBA = vBS + vSA 1 + vBSvSA/c2 = 3 4 c + 3 4 c 1 + 9c 2/16 c2 = 3 2 c 25 16 = 8 · 3 25 c = 24 25 c 3. A pion has a lifetime of 2.6 × 10−8 s (in its rest frame). If a pion has a speed of 4 5 c, how far (on average) do we expect it to travel after being created? Pion has lifetime 2.6 × 10−8 s in its rest frame. If it moves at 4 5 c wrt the lab then in the lab it lives for a time ∆t = ∆t̄√ 1 − v2/c2 = (2.6 × 10−8 s)√ 1 − (16/25)c2 c2 = (2.6 × 10−8 s)√ 9/15 = 5 3 (2.6 × 10−8 s = 4.33 × 10−8 s Then moving at v = 4 5 c it will travel a distance (in the lab) of d = v∆t = 4 5 (3.00 × 108 m s )(4.33 × 10−8 s) = 10.4 m 4.a) What does it mean when we say that a certain quantity transforms as a “Lorentz vector” ? 1It’s f ′ = ( v0 ± vobs v0 ∓ vsrc ) f . 1 When a quantity (with 4 components) tranasforms as a 4–vector we mean that the components in a new reference frame are reltaed tothe components in the old reference frame via aµ = 3∑ ν=0 Λµνa ν where Λµν is an array of numbers dependent on the relative velocity of the two reference frames; for a new frame moving with velocity vx̂ it is Λµν =   γ −γβ 0 0 −γβ γ 0 0 0 0 1 0 0 0 0 1   . b) What does it mean when we say that a certain quantity is “Lorentz invariant”? If a quantity is Lorentz invariant we mean that when its value is calculated in any inertial frame we get the same value. Example of Lorentz invariants are aµaµ, where a µ is a Lorentz 4–vector, and E · B. 5. A proton (Mpc 2 = 938 MeV) has 800 MeV of kinetic energy. What is its speed? (You can answer as a fraction of c.) If a proton has kinetic energy T = 800 MeV, then its total energy is E = mc2 + 800 MeV = 938 MeV + 800 MeV = 1738 MeV Since E = mc2√ 1 − u2/c2 = (938 MeV)√ 1 − u2/c2 = 1738 MeV , then √ 1 − u2/c2 = 938 1738 = 0.540 ⇒ u 2 c2 = 0.709 ⇒ u c = 0.842 ⇒ u = (0.842)c 6. A proton collides with a proton at rest to produce 3 protons and an antiproton (which has the same mass, but opposite charge from the proton): p + p −→ p + p + p + p̄ Find the smallest possible (threshold) kinetic energy for the incident proton for this reaction to take place. Use Mp̄c 2 = Mpc 2 = 938 MeV and express the answer in MeV. (Hint: Use invariance and conservation and use the fact that at threshold in the CM frame the final particles are at rest.) 2 z x y va 9. a) There are two charged plates at z = 0 and z = a each carrying charge densities +σ and −σ respectively. What is the electric field between the plates? Between big parallel plates with charge densities ±σ the E field is E = σ 0 ẑ, in the lab frame. (This is old stuff; follows from Gauss law arguments.) b) If a charge q moves in the +x̂ direction with speed v, what are the electric and magnetic fields in the frame of this charge? Use the equations for transformation of the fields to get E and B in the moving frame: Ēz = γ(Ez + vby) = γEz = σ/0√ 1 − v2/c2 B̄y = γ(By + v c2 E + z) = v/c2√ 1 − v2/c2 σ 0 = µ0vσ√ 1 − v2/c2 where we’ve used 0c 2 = 1 µ0 . 10. The Maxwell equations can be written in the form ∂F µν ∂xν = µ0J µ ∂G µν ∂xν = 0 Show how the first of these, with the choice µ = 1 (x) gives a certain component of one of the Maxwell equations. With µ = 1, we have (use F 11 = 0) ∂F 1ν ∂xν = ∂F 10 ∂x0 + ∂F 12 ∂x2 + ∂F 13 ∂x3 = ∂(−Ex/c) ∂(ct) + ∂Bz ∂y + ∂(−By) ∂z = − 1 c2 ∂Ex ∂t + ( ∂Bz ∂y − ∂By ∂z ) = − 1 c2 ∂Ex ∂t + (∇× B)x = µ0J x This is the x–component of the Maxwell eqn we’ve been calling Ampere’s law with maxwell’s correction. 11. Evaluate F µνFµν. It will make things easier to use the antisymmetry of F µν and to break up the values of F µν into F 0i and F ij. Recall that when you lower a “0” index you get a minus sign. F µνFµν is a sum on both µ and ν, but we can just sum on unique pairs of the indices, with µ < ν. From antisymmetry the term for ν < µ is the same from two sign switches, (−1)(−1). Breaking up F µν term into F0i and F ij, the explicit sum is F µνFµν = 2[F 01F01 + F 02F02 + F 03F03 + F 12F12 + F 23F23 + F 13F13] 5 Note, F01 = −F 01 from lowering just one zero index, but F12 = F 12. Get: F µνFµν = 2[−E2x/c2 − E2y/c2 −E2z /c2 + B2x + B2y + B2z ] = 2 c2 [−E2 + c2B2] which can also be expressed as F µνFµν = − 2 c2 [E2 − c2B2] This is quantity is invariant, as it must be since F µνFµν is a Lorentz scalar. 6 Useful Equations ∫ b a (∇T ) · dl = T (b) − T (a) ∫ V (∇ · v)dτ = ∮ S v · da ∫ S (∇× v) · da = ∮ P v · dl Spherical: dl = dlr r̂ + dlθ θ̂ + dlφ φ̂ dτ = r 2 sin θ dr dθ dφ Gradient: ∇T = ∂T ∂r r̂ + 1 r ∂T ∂θ θ̂ + 1 r sin θ ∂T ∂φ φ̂ (4) Divergence: ∇ · v = 1 r2 ∂ ∂r (r2vr) + 1 r sin θ ∂ ∂θ (sin θvθ) + 1 r sin θ ∂vφ ∂φ (5) Curl: ∇×v = 1 r sin θ [ ∂ ∂θ (sin θ vφ) − ∂vθ ∂φ ] r̂+ 1 r [ 1 sin θ ∂vr ∂φ − ∂ ∂r (rvφ) ] θ̂+ 1 r [ ∂ ∂r (rvθ) − ∂vr ∂θ ] φ̂ (6) Laplacian: ∇2T = 1 r2 ∂ ∂r ( r2 ∂T ∂r ) + 1 r2 sin θ ∂ ∂θ ( sin θ ∂T ∂θ ) + 1 r2 sin2 θ ∂2T ∂φ2 (7) Cylindrical: dl = dls ŝ + dlφ φ̂ + dlz ẑ dτ = s ds dφ dz Gradient: ∇T = ∂T ∂s ŝ + 1 s ∂T ∂φ φ̂ + ∂T ∂z ẑ (8) Divergence: ∇ · v = 1 s ∂ ∂s (svs) + 1 s ∂vφ ∂φ + ∂vz ∂z (9) Curl: ∇× v = ( 1 s ∂vz ∂φ − ∂vφ ∂z ) ŝ + ( ∂vs ∂z − ∂vz ∂s ) φ̂ + 1 s [ ∂ ∂s (svφ) − ∂vs ∂φ ] ẑ (10) Laplacian: ∇2T = 1 s ∂ ∂s ( s ∂T ∂s ) + 1 s2 ∂2T ∂φ2 + ∂2T ∂z2 (11) 7