Conservation of momentum is one of the most fundamental and most useful concepts of elementary physis Does it apply in special relativity? Consider the.

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Presentation transcript:

Conservation of momentum is one of the most fundamental and most useful concepts of elementary physis Does it apply in special relativity? Consider the following collision process. All balls have mass m and speed v before and after collision m Before m mm After Surely this will conserve momentum Momentum is conserved But what about other frames?

m m mm Momentum in another frame x’ y’ v In this frame, momentum is not conserved!

The problem with momentum How can we explain this failure of conservation of momentum? Three choices: 1.This process is physically impossible; it cannot occur 2.Conservation of momentum is not a conservation law in S.R. 3.Our formula for momentum is wrong in S.R. Why is t given special privilege over other coordinates? We want something that is like time, but not coordinate dependant in the denominator, like proper time 

Does this fix the problem? (1)

Does this fix the problem? (2) Momentum is conserved

What’s missing here... Hint – we’ve got all three dimensions... But we live in four-dimensional spacetime Is there a conserved fourth component of momentum? These quantities are related by Lorentz boost, just as p x and p y are related by rotations If p x, p y and p z are conserved, then so must be p t Is it p t that is conserved, or cp t ? Actually it doesn’t matter; if p t is conserved, so is any multiple of it We will work with a different multiple The ERIC quantity Eric is conserved

What is Eric, really? m1m1 m1m1 m2m2 m2m2 Consider low velocity limit: Not careful enough... E is really energy

Potential and Kinetic Energy E is the total energy – it includes both potential and kinetic energy Total energy is always conserved Potential energy is the energy something has when it’s not moving Also called rest energy More about this later Kinetic energy is how much extra energy there is due to motion

A sample problem... In 2004, the total world consumption of energy was 15 TW. How much fuel would have to be consumed each hour to power the world if we could extract 100% of its potential energy?

Good vs. Bad Notation Eric’s notation: m, called the mass, is a constant, and independent of velocity It does not increase with velocity It does not approach infinity as u  c The famous formula E = mc 2 is not true except when u = 0. Einstein never wrote this formula The quantity  m is then sometimes called the relativistic mass - but I won’t use this term. Satan’s notation: m  m 0 is renamed the rest mass We then define m =  m 0 as the mass In this notation, mass m goes to infinity as u  c In this notation, E = mc 2 is always true Do not use this notation/nomenclature in this class

Momentum and Energy For small velocities,    1, and the momentum is just the usual formula As u  c, the momentum and energy both go to infinity This tells us we can’t reach u = c. u/cu/c E/mc 2 p/mc

Particle physics units Forces are usually induced by using electric fields Charges are usually the same as the electron charge (-e) or simple multiples of this (+e, -2e, etc.) Electric potentials are measured in volts (V) Easy to work with units of electron charge times volts, or eV’s (called electron volts) Metric multiples are also common Momentum has units of energy/velocity, so it is often expressed in units of eV/c Mass has units of energy/velocity 2, so it is often expressed in units of eV/c 2

A sample problem... A Z-particle at rest (mass m Z = GeV/c 2 ) decays to a B- meson and an anti-B meson (both mass m B = GeV/c 2 ). What is the energy and velocity of the resulting B-mesons? Z BB Since the initial Z is at rest, p Z = 0 and E Z = m Z c 2 = GeV The B-meson and anti-B-meson must have equal & opposite momenta Since they have the same mass, this means they have equal speeds They therefore have equal energies Note: Mass is not conserved!

Two Useful Formulas Written this way, both sides are dimensionless Square and subtract them: Divide them:

Different types of particles Must the combination E 2 – c 2 p 2 always be positive? Maybe m 2 needn’t always be positive 1.If m 2 > 0 we call the particle a massive particle Electrons, protons, rockets, people E > cp, so u < c: Always slower than light 2.If m 2 = 0 we call the particle a massless particle Photons (particles of light) E = cp, so u = c: Always at the speed of light 3.If m 2 < 0 we call the particle a tachyon No known particles E c: Always faster than light

Massless and nearly massless particles First two equations make no sense for massless particles – don’t use them If mass is very small (mc 2 much smaller than E or cp), then often easier to treat as massless Massless versions of these equations: The Large Electron-Positron collider (LEP) accelerates electrons (m e = 0.511MeV/c 2 ) to an energy of E = 100 GeV. How fast are they moving?

Sample problem A Z-particle at rest (m Z = 91.2 GeV/c 2 ) decays to an electron (m e = MeV/c 2 ) with energy 25.0 GeV, a positron (m e ) with energy 20.0 GeV moving at 71° angle compared to the electron, and an unknown X particle. What is the momentum, energy, mass, and velocity of X? Z e-e- e+e+ X Electrons are nearly massless 71  Conservation of energy: Continued...

Sample problem continued What is the momentum, energy, mass, and velocity of X? Conservation of momentum: Velocity formula: Mass formula:

Lorentz boost of four-momentum Suppose I measure the momentum and energy of a particle or collection of particles. What does a moving observer measure? Far easier than the book makes it out Difference in coordinates  (  x,  y,  z,  t) satisfy same equations as (x, y, z, t) These formulas work for either a single particle or for the sum of all particles