Physics 2D Lecture Slides Jan 15 Vivek Sharma UCSD Physics - PowerPoint PPT Presentation

Physics 2D Lecture Slides Jan 15 Vivek Sharma UCSD Physics

Relativistic Momentum and Revised Newton’s Laws � � Need to generalize the laws of Mechanics & Newton to confirm to Lorentz Transform = and the Special theory of relativity: Example : p mu Watching an Inelastic Collision between two putty balls S P = mv –mv = 0 P = 0 Before V=0 1 2 v v 1 2 After − − − − v v v v 2 v V v = = = = = = − ' ' v 1 0, v 2 , V ' v 1 v v 2 v v V v 2 v − − − 1 1 1 1 + 1 1 2 2 2 c c c 2 c − 2 mv ' = ' + ' = ' = = − p mv m v , p 2 mV ' 2 mv before 1 2 after 2 v + 1 S’ 2 c ' ' p p ≠ before after v 1 ’=0 1 2 V’ 1 2 v 2 ’ Before After

Definition (without proof) of Relativistic Momentum � � With the new definition relativistic � mu = = γ p mu momentum is conserved in all frames − 2 1 ( / ) u c of references : Do the exercise New Concepts Rest mass = mass of object measured In a frame of ref. where object is at rest 1 γ = − 2 1 ( / ) u c u is velocity of the object NOT of a referen ce frame !

Nature of Relativistic Momentum � � mu � m = = γ p mu u − 2 1 ( / ) u c With the new definition of Relativistic momentum Momentum is conserved in all frames of references Good old Newton

Relativistic Force & Acceleration � �   � dp d mu d du d = =   = F use �   dt dt dt dt du − 2 1 ( / ) u c   � mu � = = γ p mu   − − m mu 1 2 u du − 2 1 ( / ) u c   = + × F ( )( ) ( )   3/ 2 2 2 c dt − 2 − 2 1 ( / ) u c 1 ( u c / )     Relativistic 2 − 2 + 2 mc mu mu du   =  F ( )  3/ 2 dt − 2 2 c 1 ( u c / ) Force   And   m du   =  F : Relativistic For ce ( ) Acceleration  3/ 2 dt − 2 1 ( / ) u c   � � d u Since A ccel e r a tion a = , d t � � F 3/ 2   ⇒ − 2 Reason why you cant a = 1 ( / ) u c   m quite get up to the speed � → → Note: As / u c 1, a 0 !!!! of light no matter how Its harder to accelerate when you get hard you try! closer to speed of light

A Linear Particle Accelerator - + F q E= V/d F=eE E d V Charged particle q moves in straight line � � Under force, work is done in a uniform electric field E with speed u on the particle, it gains � � Kinetic energy accelarates under f orce F=qE � � � 3/ 2 3/ 2 New Unit of Energy     2 2 � du F u qE u = = − − a 1 = 1     2 2 dt m c m c     1 eV = 1.6x 10 -19 Joules larger the potential difference V a cross plates, larger the force on particle

A Linear Particle Accelerator � � eE 3/ 2   − 2 a= 1 ( / ) u c   m PEP- PEP -II accelerator schematic and tunnel view II accelerator schematic and tunnel view

Magnetic Confinement & Circular Particle Accelerator � � V Classically B 2 v = F m r 2 v � = qvB m r F B r γ dp d ( mu ) du = = = γ = F m quB dt dt dt 2 du u = (Centripetal accelaration) dt r 2 u γ = ⇒ γ = ⇒ = m quB mu qBr p qB r r

Charged Form of Matter & Anti-Matter in a B Field

Circular Particle Accelerator: LEP @ CERN, Geneve

Magnets Keep Circular Orbit of Particles

Inside A Circular Particle Accelerator @ CERN

Accelerating Electrons Thru RF Cavities

Test of Relativistic Momentum In Circular Accelerator � � mu � = = γ p mu − 2 1 ( / ) u c γ = mu qB r qBr γ = m u

Relativistic Work Done & Change in Energy � � x x � dp � 2 2 ∫ ∫ = = W F dx . . dx dt X 2 , u=u x x 1 1 du � m mu dp dt = ∴ = p , substitute i n W , 3 / 2 dt   2 u 2 u − − 1 1   2 c 2 c   x 1 , u=0 du m dt udt u ∫ ∴ = → W (change in var x u ) 3 / 2   2 u 0 − 1   2 c   u 2 mudu m c ∫ = = − = γ − 2 2 2 W mc mc m c 3 / 2 1/ 2     2 2 u u 0 − − 1 1     2 2 c c     W rk d o one is change in Kinetic energy K γ 2 − 2 K = mc mc or γ = + 2 2 Total Ener gy E= mc K mc

Why Can’s Anything go faster than light ? Lets accelerate a particle from rest, particle gains velocity & kinetic energy 2     2 2 mc mc ( )   2 = − ⇒ + =  2 2 K mc K mc  1/ 2 1/ 2     2 2 u u   − − 1 1       2 2 c c         2 u − 2   ⇒ − = + 2 4 2 1 m c K mc     2 c   K K − ⇒ = − + 2 u c 1 ( 1) (Parabolic in Vs u ) 2 2 mc mc 1 2 K ⇒ = 2 Non-relativistic case: K = mu u 2 m

Relativistic Kinetic Energy

When Electron Goes Fast it Gets “Fat” = γ 2 E mc v → γ → ∞ As 1, c ∞ Apparent Mass approaches

Relativistic Kinetic Energy & Newtonian Physics γ − 2 2 Relativistic KE = mc mc 1 −   2 2 u 1 u 2 << ≅ − + When u c , 1- 1 ...smaller terms   2 2 c 2 c   2 1 u 1 ≅ − − = 2 2 2 so K mc [1 ] mc mu (classical form recovered) 2 2 c 2 Total Energy of a Pa r ticle = γ = + 2 2 E mc KE mc For a particle at rest, u = 0 ⇒ 2 Total Energy E= m c

= γ ⇒ = γ 2 2 2 2 4 E mc E m c Relationship between P and E = γ ⇒ = γ 2 2 2 2 2 2 p mu p c m u c ⇒ − = γ − γ = γ − 2 2 2 2 2 4 2 2 2 2 2 2 2 2 E p c m c m u c m ( c u ) 2 2 2 4 m c m c − = − = 2 2 2 2 2 4 = ( c u ) ( c u ) m c − 2 2 2 u c u − 1 2 c = + 2 2 2 2 2 E p c ( mc ) ........important relati on For p articles with zero rest mass like photon (EM waves) E E= pc or p = (light has momentum!) c − = 2 2 2 2 4 Re lativistic Invariance : E p c m c : In all Ref Frames Rest M ass is a "finger print" of the particle

Mass Can “Morph” into Energy & Vice Verca • Unlike in Newtonian mechanics • In relativistic physics : Mass and Energy are the same thing • New word/concept : Mass-Energy • It is the mass-energy that is always conserved in every reaction : Before & After a reaction has happened • Like squeezing a balloon : – If you squeeze mass, it becomes (kinetic) energy & vice verca ! • CONVERSION FACTOR = C 2

Mass is Energy, Energy is Mass : Mass-Energy Conservation Examine Kinetic energy Before and After Inelastic Collision: Conserved? K=0 S K = mu 2 Before V=0 1 2 v v 1 2 After Mass-Energy Conservation: sum of mass-energy of a system of particles before interaction must equal sum of mass-energy after interaction = E E be f ore after 2 2 mc mc 2 m + = ⇒ = > 2 Mc M 2 m 2 2 2 u u u − − − 1 1 1 Kinetic energy is not lost, 2 2 2 c c c its transformed into Kinetic energy has been transformed into mass increase   more mass in final state   2 2 K 2 m c   ∆ = = = − 2 M M - 2 m mc   2 2 c c 2 u −   1 2  c 

Conservation of Mass-Energy: Nuclear Fission M M 1 + M 3 M 2 + Nuclear Fission 2 2 2 M c M c M c = + + ⇒ > + + 2 Mc 1 2 3 M M M M 1 2 3 2 2 2 u u u − − − 1 1 1 2 1 3 2 2 2 c c c < 1 < 1 < 1 Loss of mass shows up as kinetic energy of final state particles Disintegration energy per fission Q=(M – (M 1 +M 2 +M 3 ))c 2 = ∆ Mc 2 → 236 14 3 90 1 U Cs + R b +3 n 92 55 92 0 ∆ × = -28 m=0.177537u=2.947 1 10 kg 165.4 MeV What makes it explosive is 1 mole U = 6.023 x 10 23 Nuclei !!

Relativistic Kinematics of Subatomic Particles Reconstructing Decay of a π Meson