Physics 2170 – Spring 20091 Special relativity Homework solutions are on CULearn Homework set 3 is on the website.

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Physics 2170 – Spring Special relativity Homework solutions are on CULearn Homework set 3 is on the website and is due Wed at 12:50pm. Remember, problem solving sessions today 3-5 and tomorrow 3-4,5-6 in physics help room (G2B90) Announcements: Today we will continue with relativistic energy and the relationship between mass, momentum, and energy.

Physics 2170 – Spring u c u c u c u c E E E E Which of the graphs below is a possible representation of the total energy of a particle versus its speed A. B. C. D. E. None of them Set frequency to DA Clicker question 1 As the velocity u increases towards c, the energy increases. But for u→c,  →∞ so it is impossible for a particle to reach c as it would need infinite energy.

Physics 2170 – Spring At low speeds, this becomes the familiar Total energy: Relativistic momentum and energy At rest,  u =1 so the rest energy is Kinetic energy is always the total energy minus the rest energy: Relativistic momentum: With these definitions for momentum and energy, conservation of momentum and conservation of (total) energy continue to work in isolated relativistic systems This tells us that mass and energy are equivalent. Mass is a type of energy like motion is a type of energy (kinetic energy).

Physics 2170 – Spring Rest Energy is real: Nuclear fission Initial mass of a neutron and a 235 U nucleus. Final products have less mass, but much more kinetic energy. Conversion of mass to kinetic energy. Oh yes, and more neutrons, so the reaction can run wild (chain reaction!).

Physics 2170 – Spring Q. If we could convert mass entirely to energy, how much mass would be required to run a 30W light bulb for a year (a year is approximately s)? One year is s so running a 30 W light bulb for a year requires E = PΔt = 30 W s = J. To get that energy from rest mass we use E=mc 2 and solve for mass. A. 10 kg B. 3 g C. 3 mg D. 10 μg E. 10 ng Set frequency to DA Clicker question 2

Physics 2170 – Spring A relativistic collision problem Object A has mass 9m 0 and speed v A =0.8c (  A =5/3). Object B has mass 12m 0 and speed v B =−0.6c (  B =5/4). The objects collide and stick together (completely inelastic collision) Classically, what is the total initial momentum? From Physics 1110 we know collisions conserve momentum and that inelastic collisions do not conserve kinetic energy But this is an isolated system, so total energy must be conserved. Let’s start with momentum conservation. What is the total relativistic momentum? So it does not end at rest as predicted classically!

Physics 2170 – Spring A relativistic collision problem Object A has mass 9m 0 and speed v A =0.8c (  A =5/3). Object B has mass 12m 0 and speed v B =−0.6c (  B =5/4). Momentum conservation gives us: Remember that m f may not be m A +m B as it would be classically. Now let’s look at the total energy. The initial energy is So conservation of energy gives us: Dividing these two equations: or so Furthermore: conservation of energy equation for m f : so we can solve the

Physics 2170 – Spring Classically, total momentum is 0 but in reality it is A relativistic collision problem Object A has mass 9m 0 and speed v A =0.8c (  A =5/3). Object B has mass 12m 0 and speed v B =−0.6c (  B =5/4). Classically, but in reality, so 8.85m 0 of mass is gained. Classically, final kinetic energy is 0 and the initial kinetic energy is So the change in KE is But in reality, the initial and final kinetic energies are: The “lost” kinetic energy appears as gained mass in the total energy

Physics 2170 – Spring Really, mass gets created! CERN in Geneva, Switzerland Before the LHC (Large Hadron Collider) CERN operated LEP, the Large Electron-Positron collider in the same underground tunnel. Electron and positrons have a mass of 9x kg. They were accelerated to very high energies so when they annihilate, they create a Z 0 particle with a mass of 1.6x kg.

Physics 2170 – Spring Total energy: Relativistic momentum and energy Relativistic momentum: With some algebra we can eliminate the velocity variable from these two relations and obtain the triangle relation: A proton has a mass of 938 MeV/c 2. What is this in kg? A new unit of energy is the electron-volt (eV). It’s the energy obtained by an electron moving through 1 V. It is not an SI unit but is very common. ΔE = qΔV = 1 eV = C 1 V = J Also use eV/c or MeV/c units for momentum Since mc 2 is a unit of energy, dividing energy by c 2 gives a unit of mass. Also, dividing energy by c gives a unit of momentum.