Monday, Feb. 16, 2015PHYS 3313-001, Spring 2014 Dr. Jaehoon Yu 1 PHYS 3313 – Section 001 Lecture #8 Monday, Feb. 16, 2015 Dr. Jaehoon Yu Relativistic Energy.

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Monday, Feb. 16, 2015PHYS , Spring 2014 Dr. Jaehoon Yu 1 PHYS 3313 – Section 001 Lecture #8 Monday, Feb. 16, 2015 Dr. Jaehoon Yu Relativistic Energy Relationship Between Relativistic Quantities Binding Energy Quantization Discovery of the X-ray and the Electron

Monday, Feb. 16, 2015PHYS , Spring 2014 Dr. Jaehoon Yu 2 Announcements Reading assignments: CH3.9 Homework #2 –CH3 end of the chapter problems: 2, 19, 27, 36, 41, 47 and 57 –Due Wednesday, Feb. 25 Quiz #2 Monday, Feb. 23 –Beginning of the class –Covers CH1.1 – what we finish this this Wednesday, Feb. 18

How do we keep momentum conserved in a relativistic case? Redefine the classical momentum in the form: This Γ (u) is different than the  factor since it uses the particle’s speed u  What? How does this make sense?  Well the particle itself is moving at a relativistic speed, thus that must impact the measurements by the observer in the rest frame!! Now, the agreed form of the momentum in all frames is (  is the proper time): Resulting in the new relativistic definition of the momentum: When u  0, this formula becomes that of the classical. What can the speed u be to maintain the accuracy of the classical momentum to 1%? Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Relativistic Energy Due to the new idea of relativistic mass, we must now redefine the concepts of work and energy. –Modify Newton’s second law to include our new definition of linear momentum, and force becomes: The work W done by a force F to move a particle from rest to a certain kinetic energy is Resulting relativistic kinetic energy becomes Why doesn’t this look anything like the classical KE? Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Only is right! and are wrong! Big note on Relativistic KE Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Total Energy and Rest Energy Rewriting the relativistic kinetic energy: The term mc 2 is called the rest energy and is denoted by E 0. The sum of the kinetic energy and rest energy is interpreted as the total energy of the particle. (note that u is the speed of the particle) Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Relativistic and Classical Kinetic Energies Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

We square this formula, multiply by c 2, and rearrange the terms. Relationship of Energy and Momentum Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu Rewrite

Recall that a photon has “zero” rest mass and the equation from the last slide reduces to: E = pc and we may conclude that: Thus the speed, u, of a massless particle must be c since, as 0, and it follows that: u = c. Massless Particles have a speed equal to the speed of light c Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Units of Work, Energy and Mass The work done in accelerating a charge through a potential difference V is W = qV. –For a proton, with the charge e = × 10 −19 C being accelerated across a potential difference of 1 V, the work done is 1 eV = × 10 −19 J W = (1.602 × 10 −19 )(1 V) = × 10 −19 J eV is also used as a unit of energy. Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Other Units 1) Rest energy of a particle: Example: Rest energy, E 0, of proton 2) Atomic mass unit (amu): Example: carbon-12 What is 1u in eV? Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Binding Energy The potential energy associated with the force keeping a system together  EB.EB. The difference between the rest energy of the individual particles and the rest energy of the combined bound system. Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu

Examples 2.13 and 2.15 Ex. 2.13: A proton with 2-GeV kinetic energy hits another proton with 2 GeV KE in a head on collision. (proton rest mass = 938MeV/c 2 ) Compute v, , p, K and E for each of the initial protons What happens to the kinetic energy? Ex. 2.15: What is the minimum kinetic energy the protons must have in the head-on collision in the reaction p+p    +d, in order to produce the positively charged pion (139.6MeV/c 2 ) and a deuteron.(1875.6MeV/c 2 ). Monday, Feb. 16, PHYS , Spring 2014 Dr. Jaehoon Yu