Physics 2D Lecture Slides Oct 8 Vivek Sharma UCSD Physics
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
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 - Parallel Plates + F q E= V/d F= eE E d V Charged particle q moves in straight line Under force, work is done � � on the particle, it gains in a uniform electric field E with speed u � � Kinetic energy accelarates under f orce F=qE New Unit of Energy � � � 3/ 2 3/ 2 ⎛ ⎞ ⎛ ⎞ 2 2 � du F u qE u = = − − a 1 = 1 ⎜ ⎟ ⎜ ⎟ 1 eV = 1.6x10 -19 Joules 2 2 dt m c m c ⎝ ⎠ ⎝ ⎠ 1 MeV = 1.6x10 -13 Joules 1 GeV = 1.6x10-10 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
Accelerating Electrons Thru RF Cavities
Circular Particle Accelerator: LEP @ CERN, Geneve
Magnets Keep Circular Orbit of Particles
Inside A Circular Particle Accelerator @ CERN
Test of Relativistic Momentum In Circular Accelerator � � mu � = = γ p mu − 2 1 ( u / ) c γ = = mu qB r p qB r γ = mu
Relativistic Work Done & Change in Energy � x � x � dp � 2 2 ∫ ∫ = = W F dx . . dx dt x x X 2 , u=u 1 1 du � m mu dp dt = ∴ = p , substitute i n W , 3 / 2 dt ⎡ ⎤ 2 u 2 u − − 1 1 ⎢ ⎥ 2 2 c c ⎣ ⎦ du x 1 , u=0 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 ⎣ ⎦ ⎣ ⎦ Work d one is change in Kinetic energy K γ − 2 2 K = mc mc or γ = + 2 2 Total Energy E= mc K mc
But Professor… Why Can’s ANYTHING go faster than light ? 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 Energy
A Digression: How to Handle Large/Small Numbers • Example: consider very energetic particle with very large Energy E + 2 E mc K K γ = = = + 1 2 2 2 mc mc mc 1/ 2 ⎡ ⎤ u 1 = − • Lets Say γ = 3x10 11 , Now calculate u from � 1 ⎢ ⎥ γ 2 c ⎣ ⎦ • Try this on your el-cheapo calculator, you will get u/c =1, u=c due to limited precision. • In fact u ≅ c but not exactly!, try to get this analytically 1 1 γ = = − β + β In Quizzes, you are − β (1 )(1 ) 2 1 Expected to perform u β ≅ + β = Since = 1, 1 2 Such simple c approximations 1 γ ≈ − β 2 1 1 ⇒ − β = = × − = β 24 1 5 10 , u c γ 2 2 ⇒ = u 0.999 999 999 999 999 999 999 995c !! Such particles are routinely produced in violent cosmic collisions
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 2 E p c m c m u c m c ( 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 ( m c ) ........important relation F or particles with zero rest mass like pho ton (EM waves) E E= pc or p = (light has momentu m!) c − = 2 2 2 2 4 Relativistic Invariance : E p c m c : In all Ref Frames Rest Mass 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 , just like spacetime • 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 • This exchange rate never changes !
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 ⎠
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