Momenerg yA2290-25 1MomenergyRelativity and AstrophysicsLecture 25Terry HerterA2290-25 Momenergy 2Outline Momenergy Momentum-energy 4-vector Magnitude & components Invariance Low velocity limit Concept Summary Reading Spacetime Physics: Chapter 7 Homework: (due Wed. 11/04/09) 6-4, 6-7, 7.3, and 7.9Momenerg yA2290-25 2A2290-25 Momenergy 3The Power of It All The conservation law all us to solve for quantities without knowing the details We don’t have to know how the objects deform in the sticking case We don’t need to know about the details for the collisions at all (for the completely inelastic and elastic cases) In cases where this a potential we can include this in the energy conservation equations. For instance, using Newton’s law of gravity we can write a conservation of energy equation which relates velocity to distance from the Earth – we don’t have to solve the details of the acceleration For a vertically falling object we have Where the object starts at rowith zero velocity. Note that we lose information. We don’t know how long it takes to travel over this distance – just the speed at the end.2rGMmF rGMmV rrGMmmvo11212A2290-25 Momenergy 4Momenergy Momenergy Relativity combines momentum and energy into a single concept, momentum-energy (or momenergy) This quantity is conserved in a collision Momenergy proportional to mass Consider different mass pebbles hitting a windshield Momenergy is a directed quantity It matters which direction the pebble come from Momenergy is a 4-vector Expect space and time components due to the unity of spacetime (three spatial parts and one time part_ Space part represent momentum , time part represents energy Points in the direction of a particles spacetime displacement Momenergy is reckoned using proper time for a particle Momenergy is independent of reference frame Looks like Newtonian momentum but modified for Einstein’s relativityntdisplacemefor that eproper timmass ntdisplaceme spacetimemomenergy Momenerg yA2290-25 3A2290-25 Momenergy 5Magnitude of Momenergy Don’t confuse a 4-vector with its magnitude The proper time is the magnitude of the spacetime displacement The fraction is a unit 4-vector pointing in the direction of the worldline of the particle The magnitude of momenergy is its mass.spacetimeenergymomentumMomenergyenergymomentumMomenergyMomenergyenergy(momentum = 0)A2290-25 Momenergy 6Components of Momenergy Let t stand for proper time then the components of momenergyare Let’s look at its magnitude Or more compactly written which is just he equation for a hyperbola in spacetime again. At right is a plot of the momenergy4-vector for a single particle observedin 5 different inertial reference frames.ddtmE ddxmpxddympyddzmpz22222zyxpppEmagnitude 222222ddzdydxdtm2222mpEarrowmomenergyofmagnitude menergyx-momentum222ddm2mMomenerg yA2290-25 4A2290-25 Momenergy 7Momentum: “Space Part” Consider a particle moving along the x-axis with a velocity v in the lab frame The displacement of the particle is x = vt, or for small displacements, dx = v dt. The proper time is: So that the relativistic expressions for energy and momentum are For low velocities the momentum expression becomes very close to the Newtonian value 2/12222/122dtvdtdxdtd mddtmE xxmvddtdtdxmddxmp &/12/12dtvdt 2/121 vA2290-25 Momenergy 8Momentum Units Relating velocities (dimensionless vs. conventional units) For a momentum in dimensionless units For momentum in conventional units Convert from momentum in units of mass to conventional units by multiplying by c, the speed of light.cvvconvmvpNewtonmvpValid for low speedValid at any speedconvNewtonNewtonconvmvmvccppconvconvmvmvcpcpValid for low speedValid at any speedMomenerg yA2290-25 5A2290-25 Momenergy 9Energy: “Time Part” For a particle moving along the x-axis with a velocity v in the lab frame the energy is Which can be compare with the Newtonian expression (using K as the symbol for kinetic energy) How does the relativistic expression for energy compare with the Newtonian for kinetic energy? At low velocities, v = 0, we have Which is called the rest energy of the particle.  Rest energy is simply the mass The relativistic energy does not go to zero like the K! So to define a kinetic energy above and beyond a particles rest energy we havemddtmE 221mvK mErest1mmEEEKrestA2290-25 Momenergy 10Energy Units Note that if we divide the momentum and energy we get the speed of the particle To convert energy in units of mass to energy in conventional units we have The rest energy (and perhaps the most famous equation in physics) is The kinetic energy is At low speeds (v << 1), we havemddtmE mvp&Epv 22mcEcEconvValid at any speed122mccEEKrestconvValid at any speed2mcErestParticle at rest2222121convNewtonconvmvcmvK Valid at low speed2/121 v1211~12 v221vMomenerg yA2290-25 6A2290-25 Momenergy 11Sample Problem 7-2 (pg. 202) Consider a 3 kg mass object which moves 8 meters in the x direction in 10 meters of time. What is its energy and momentum? What is its rest energy? What is its kinetic energy? Compare this to the Newtonian KE. Verify the velocity equals its momentum divided by its energy. The speed is The energy and momentum are The rest energy is The relativistic and Newtonian kinetic energies are The correct relativistic result is quite a bit larger that the Newtonian prediction, and in fact the correct result grows with out limit as v approaches 1 (more on next slide). Finally the velocity can be recovered frommE 8.0m 10m 8txv&Epv /kg 3mErestkg 2restEEK211v&28.0116.0136.0135mvp35kg 3 kg 5358.0kg 3 kg 4&25.0 mvKNewton8.05/428.0kg 35.0  kg 96.0A2290-25 Momenergy 12Energy in the low-velocity limit In terms of momentum the