Lecture 9 - Relativistic Dynamics Read chapter 1 (French) So far we have considered measurements of: position (Ξx) time (Ξt) velocity - kinematics Now we go on to consider: momentum energy mass forces - dynamics Classical Dynamics Newtonβs Laws: πΉ= π π, ππ‘ π = ππ£ 1. If we assume mass is independent of velocity: β΄ πΉ = π × πππππππππ‘πππ 2. Momentum conserved, energy conserved 2. βunder constant F, v increases indefinitely. Does this happen? Do an experiment Accelerate π β under constant πΈ, measure π£ versus t Accelerate π β under constant πΈ, measure π£ versus t v π£2 c π2 t Kinetic energy, K v approaches c, but does not reach it If m is constant, then π£ β π (const.) β π β ππππ π‘ even though πΉ > 0 This is inconsistent with 1. πΉ= π π ππ‘ Relativistic Dynamics It was found that inertial mass π increases and is given by: π= π0 π£2 1β 2 π ½ = πΎπ0 3. π0 = rest mass, π£ = vel. of particle relative to observer We need to use π = πΈππ to conserve momentum. (see French, ch. 6, p 169) Thus, πΉ = π π, ππ‘ π = ππ£, π = πΎπ0 π = πΎπ0 π£ Relativistic Dynamics Let us consider this expression for inertial mass more closely: π = πΎπ0 = Use the binomial expansion: π = π0 π0 π£2 1β 2 π ½ 1 π£2 3 π£4 1+ + +β― 2 π2 8 π4 1 π0 π£ 2 3 π£2 Ξπ = π β π0 = 1+ +β― 2 π2 4 π2 2 1 3 π£ Ξππ 2 = π0 π£ 2 1 + +β― 2 4 π2 The first term corresponds to classical KE Relativistic Dynamics 2 1 3 π£ Ξππ 2 = π0 π£ 2 1 + +β― 2 4 π2 Particle moving under action of force πΉ KE acquired. KE = Work done = β« πΉ β π π y πΉ π£=0 π£=π x Use this relation in SR: πΉ= π π π= (πΎπ0 π£) ππ‘ ππ‘ Relativistic Dynamics πΉ= π (πΎπ0 π£) ππ‘ πΉ = π0 πΎ ππ£ ππΎ + π0 π£ ππ‘ ππ‘ ππΎ ππΎ ππ£ = ππ‘ ππ£ ππ‘ β½ π£2 ππΎ π = 1β 2 ππ£ ππ£ π ππΎ ππ‘ = β3 2 2 π£ π£ = 2 1β 2 π π πΎ 3 π£ ππ£ π 2 ππ‘ 1. πΎ 3π£ = 2 π Relativistic Force ππΎ ππ‘ = πΎ 3 π£ ππ£ π 2 ππ‘ 1. ππ£ π£2 2 πΉ = π0 πΎ 1+ 2πΎ ππ‘ π πΎ2 = 1 π£2 1β 2 π πΉ = π0 πΎ β π£2 1 = 1 β π2 πΎ2 ππ£ 1 1 + 1 β 2 πΎ2 ππ‘ πΎ πΉ = π0 πΎ 3 ππ£ ππ‘ π 2 ππΎ πΉ = π0 (π’π πππ 1. ) π£ ππ‘ Relativistic Kinetic Energy π₯ πΎπΈ = π‘ πΉππ₯ = 0 = π0 π πΉπ£ππ‘ 0 πΎ 2 ππΎ 1 = π0 π 2 (πΎ β 1) Where the upper limit of πΎ ππ 1 ½ π2 1β 2 π Kinetic energy Force πΉ = π0 πΎ 3 Rest mass ππ£ ππ‘ πΎ. πΈ. = π0 π 2 (πΎ β 1) Acceleration π 2 ππΎ πΉ = π0 π£ ππ‘ β πΉπ£ππ‘ = π0 π 2 dΞ³ Relativistic Kinetic Energy KE in a given reference frame can be defined as the difference between the energy when an object is in motion and when it is at rest. πΎ = ππ 2 β π0 π 2 ππ ππ 2 = πΎ + π0 π 2 Call ππ 2 the total energy E. Then πΈ = ππ 2 = πΎ + π0 π 2 Total energy = Kinetic energy + rest energy β’ πΈ = ππ 2 β’ Energy is conserved β’ Energy has mass β’ Mass is conserved (but not necessarily rest mass) β’ Mass and energy are different physical quantities A Useful Relation π = πΎπ0 π£ πΈ = πΎπ0 π 2 β΄ πΈ 2 β π 2 π 2 = πΎ 2 π 0 2 π 4 β πΎ 2 π0 2 π 2 π£ 2 πΈ 2 β π 2 π 2 = π0 2 π 4 πΎ 2 π£2 1β 2 π πΈ 2 β π 2 π 2 = π0 2 π 4 But, π0 and π are invariant under a Lorentz Transform: β΄ πΈ 2 β π2 π 2 is invariant under LT πΈ 2 β π2 π 2 is the same for all observers! Summary π = ππ£ πΈ = ππ 2 = πΎ + π0 π 2 π = πΎπ0 πΈ 2 β π 2 π 2 = π0 2 π 4 Photons π£ = π for all observers Since π = πΎπ0 , π0 = 1 β π£2 π2 ½ π β΄ π0 = 0 β πΈ = ππ since πΈ 2 β π2 π 2 = (0)2 π 4 πΈ βπ£ βπ= = , π π light photon has momentum Example: laser cooling for Bose-Einstein condensation of atoms
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