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Announcements Proper time. An object is moving past you with a relative velocity u=0.8c (as measured by you). When your watch shows 1 second has elapsed how much time does a clock held by an observer in the objects rest frame show has elapsed? a) 0.6 s b) 1 s c) 1.7 s d) this measurement can’t be made The proper time is the time in the objects frame and is shorter than your time by a Lorentz factor t’=t/g Today’s class • Reading for Friday: 2.4 - 2.6 • HW 4 Due next Wed. noon. Some of the topics on this will be on the exam! • First Midterm is on the 16th. Will cover relativity (Practice questions are on CULearn later today. Solutions will be added soon.) • Extra Office Hours. Duane F619 – Monday 1-4pm – Tuesday 10-12pm (I will also be in the help room 34:30) Conservation of Momentum Problem solved in moving frame Chapter 2: • Momentum • Relativistic forces • Relativistic energy m m m v v m v u’f=0 v u’f≠0 Depending on where you solve this problem you get different answers. This is no good! Relativistic momentum This means that the (classical) momentum ptot=Σ(m∙u) of a system is not invariant under Lorentz transformations. Classical definition: Need to change the definition of the momentum, but we want that: Say we measure the mass 'm' in its rest-frame ('proper mass' or 'rest mass'). Since we measure 'm' it's restframe we agree on the same value for 'm' in all frames. 1. At low velocities the new definition of p must match the classical definition of momentum. Assume we take the derivative with respect to the proper time tproper , which has the same meaning in all frames. 2. We want that the total momentum (Σp) of an isolated system of bodies is conserved in all inertial frames. Relativistic definition: pm dr dt pm dr dt proper This definition fulfills the conservation of momentum in SR! To prove it you can apply the Lorentz transformations. 1 Relativistic momentum Relativistic momentum The time dilation formula tells us that dt = γdtproper for a frame where the particle is moving. We can therefore rewrite the definition of the relativistic momentum as follows: p g m 1 g dr g mu dt 1 u2 c2 An important consequence of the Lorentz-factor γ is, that no object can be accelerated past the speed of light. Example: Classical vs. Relativistic momentum Particle A has half the mass but twice the speed of particle B. If the particles’ momenta are p A and pB, then a) pA > pB b) pA = pB c) pA < pB Newton’s second law is: F = ma This is equivalent to: F = dp/dt only ~1% change! u 0.1c p=m·u classical 0.1 p=γm·u relativistic 0.101 difference [%] 1 0.5c 0.9c 0.99c 0.5 0.9 0.99 0.57 2.06 7.02 14 129 609 Example: Relativistic force A charged particle (charge q) with mass m is at rest at x = 0 in a uniform electric field ℇ . Plot the velocity u of the particle as a function of time t (assume the particle is released at t = 0). Force acting on the particle: F = q·ℇ Relativistic dynamics: F ≡ d(γ·m·u)/dt Therefore: Or: g is bigger for the faster particle. Relativistic “Newton’s 2nd Law” An electron has a mass m ≈ 9·10-31kg. The table below shows the classical and relativistic momentum of the electron at various speeds (units are 2.733·10-22kg·m/s): ~67,000,000 mph q·ℇ = d(γ·m·u)/dt q·ℇ dt = d(γ·m·u) Integrating both sides: q·ℇ dt = d(γ·m·u) yields: q·ℇ · t = γ·m·u p gmu B A Using the definition of the relativistic momentum we obtain a suitable definition for a relativistic force: F dp d g m u , with g dt dt 1 1 u2 c2 Example: Relativistic force (cont.) γ·m·u = q·ℇ·t Now: Solve Dividing by g 1 1 u2 c2 for the velocity u. yields: m·u = q·ℇ·t (1-u2/c2)0.5 Square both sides: m2·u2 = q2·ℇ2·t2(1-u2/c2) Bring u to the left: u2(m2c2 + q2·ℇ2·t2)= q2·ℇ2·t2·c2 Divide by term in bracket and take squareroot: u Classical qℇ ct c u ( qℇ t ) 2 ( mc ) 2 Short time: u ≈ q·ℇ·t/m (const acceleration) Long time: u ≈ c (but just a little smaller) 0 t 2 Ultimate speed demonstration Ultimate speed demonstration Clear “table top” demonstration that the speed of light is unattainable for massive particles. This is a result of Lorentz transformations. WILLIAM BERTOZZI HARVARD Quiz: On the reading Energy 0 0 E= u d) E c) c 0 c u 1. At low velocities, the value E of the new definition must match the classical definition. E b) 2 γmc = K Similar to the definition of the relativistic momentum we want to find a definition for the energy E of an object that fulfills the following: mc2 +0 c u c u 2. The total energy (ΣE) of an isolated system of bodies must be conserved in all inertial frames. E E Which graph best represents the total energy of a particle (particle's mass m>0) as a function of its velocity u, in special relativity? a) Larger potential → Bigger force →larger kinetic energy →faster speed? 0 Kinetic energy Relativistic kinetic energy The relativistic kinetic energy K of a particle with a rest mass m is: The work done by a force F to move a particle from position 1 to 2 along a path s is defined by: 2 K = γmc2 - mc2 = (γ-1)mc2 W12 F ds K 2 K1 1 Note: This is very different from the classical K= ½mv2 . K1,2 being the particle's kinetic energy at positions 1 and 2, respectively (true only for frictionless system!). For slow velocities the relativistic energy equation gives the same value as the classical equation! Remember the binomial approximation for γ: γ ≈ 1+ ½v2/c2 Using relativistic force we can now find the relativistic kinetic energy of the particle. K = (γ-1)mc2 ≈ (1 + ½v2/c2 – 1)mc2 = ½ mv2 3 Total energy Rest energy of a particle We rewrite the equation for the relativistic kinetic energy and define the total energy of a particle as: E = γmc2 = K + mc2 This definition of the relativistic mass-energy E fulfills our condition of conservation of total energy. (Not proven here, but we shall see several examples where this proves to be correct.) Relation between Mass and Energy v m E2 = γmc2 = K + mc2 E1 = γmc2 = K + mc2 In the particle's rest frame, its energy is its rest energy, E0. What is the value of E0? A: B: C: D: E: 0 c2 mc2 (γ-1)mc2 ½ mc2 Note: This suggests a connection between mass and energy! Equivalence of Mass and Energy v -v m E = γmc2 -v m m Conservation of the total energy requires that the final energy Etot,final is the same 2 as the energy Etot, before 2 thetot collision. final Therefore: initial E =γ Mc = 2K + 2mc Total energy: Etot,final = Mc2 ≡ 2K + 2mc2 = Etot,initial Etot = E1+E2 = 2K + 2mc2 We find that the total mass M of the final system is larger than the sum of the masses of the two parts! M>2m. Potential energy inside an object contributes to its mass!!! This argument isn’t a “proof”. But let’s see if it’s correct. Example: Rest energy of an object with 1kg E0 = mc2 = (1 kg)·(3·108 m/s )2 = 9·1016 J 9·1016 J = 2.5·1010 kWh = 2.9 GW · 1 year This is a very large amount of energy! (Equivalent to the yearly output of ~3 very large nuclear reactors.) Enough to power all the homes in Colorado for a year! 4