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This article explores the relationship between special relativity and the laws of motion, specifically focusing on momentum, energy, and the famous equation E=mc^2. It also discusses relativistic kinematics and the concept of spacetime vectors.
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Topics • The Law of Motion • Momentum & Energy • E = mc2 • Relativistic Kinematics • Summary
t' t S’ frame P x' Q S frame x O B The Law of Motion According to Principle 1, the laws of physics should hold true in all inertial frames. In particular, this should be true of the 2nd law of motion: However, this law cannot be valid at speeds near that of light
t' t S’ frame P x' Q S frame x O B The Law of Motion However, if Newton’s 2nd law of motion is expressed the way Newton did originally: then the 2nd law holds true in special relativity provided that…
t' t S’ frame P x' Q S frame x O B Momentum & Energy …momentum is defined as follows where u is the speed of the object relative to a frame of reference. WARNING: the g-factor here differs from the g-factor arising from the relative motion of the frames
Momentum & Energy Extra credit: show that implies where due 8-Feb-08
Momentum & Energy Consider the 2nd law of motion in 1-dimension In Newtonian physics the kinetic energyKarises from the work-energy theorem:
Momentum & Energy We shall try the same definition of kinetic energy in special relativity. Consider a particle that starts from rest relative to some inertial frame and is accelerated by a force F
Momentum & Energy Einstein’s interpretation of the result is that E =g mc2 is the total energy of the object E =g mc2=K+ mc2 From this he arrived at the momentous conclusion: the minimum energy of a particle is E = mc2
E = mc2 An Example
How Long will the Sun Shine? Power Output of Sun • 3.826 x 1026 Watts Unit of Power • 1 Watt = 1 Joule/second
How Long will the Sun Shine? Mass of H nucleus (i.e, a proton) 1.007276 amu Mass of He nucleus (i.e., an a-particle) 4.0015 amu 1 Atomic Mass Unit (amu) is 10-3 kg of Carbon-12 / NA = 1.66 x 10-27 kg NA = 6.022 x 1023 is Avogadro’s Number
How Long will the Sun Shine? Mass Deficit (of reaction 4 H -> He) 4 x 1.007276 amu= 4.0291 amu(4H ) - 4.0015 amu(He) = 0.0276 amu That is, 0.0276 / 4.0291 =0.007 of the mass of a proton disappears in this reaction! The mass is converted to energy, in the form of photons and neutrinos
How Long will the Sun Shine? Available Hydrogen Fuel • 0.1 of the Sun’s mass is hot enough to fuse hydrogen to helium • 0.007of that mass is converted to energy fuel = (0.1) x (0.007) x (2 x 1030) kg = 1.4 x 1027 kg that is, about 233times the Earth’s mass
How Long will the Sun Shine? • Mass Destroyed Per Second • m = E / c2 = 4 x 1026 J/s / (3 x 108 m/s)2 =4.3 x 109 kg / s • Mass Destroyed Per Year • m = (3.15x107 s/yr) x (4.3 x 109 kg/s) = 1.4 x 1017 kg / yr
How Long will the Sun Shine? Estimated Lifetime Available Fuel / Rate of Fuel Consumption 1.4 x 1027 kg / (1.4 x 1017 kg/yr) 10 billion years
Spacetime Vectors The values ct, x, y, z can be regarded as the components of a vector in spacetime, a 4-vector relative to some frame of reference. The same vector can be expressed in a different frame of reference using the Lorentz transformation
Spacetime Vectors S frame -> S’ frame Lorentz Transformation
Spacetime Vectors The energy together with the momentum also form a 4-vector which transforms between reference frames in a manner similar to the 4-vector X
Spacetime Vectors S frame -> S’ frame Lorentz Transformation
b = 0.01 b = 0.5 Example - Micrometeorite micrometeorite m = 10-9 kg moves past Earth at b = v/c = 0.01. What are the energy and momentum as viewed by an observer in a starship moving at b = 0.5 relative to the Earth? Starship micrometeorite
b = 0.01 b = 0.5 Example - Micrometeorite Earth frame E =g(u)mc2 = (1.00005) x (10-9kg) c2 J = 1.00005 x 10-9 c2 J px=g (u)mux= (1.00005) x (10-9kg) x (0.01 c) kg m/s = 1.00005 x 10-11 c kg m/s
Example - Micrometeorite Starship frame E’ = g (E – v px)= 1.1547 x [1.00005x10-9c2 - (0.5 c) (1.00005 x 10-11c) ] = 1.14898 x 10-9 c2 J p’x= g (px – vE/c2) = 1.1547 x [1.00005 x 10-11c - (0.5 c) (1.00005x10-9c2)/c2 ] =-56.6 x 10-11 c kg m/s Starship measures energy 15 % greater and momentum -57 times greater
A Word on Units • Energy – electron-Volt (eV) • 1 eV = 1.6 × 10-19 Joules • 1 kW•hr = 3.6 × 106 Joules • Mass – electron-Volt / c2 (eV/c2) • 1 eV/c2 = 1.78 × 10-36 kg • Electron mass = 0.511 MeV/c2 • Proton mass = 938.3 MeV/c2 • Neutron mass = 939.6 MeV/c2 • Top quark mass = 174.0GeV/c2
Spacetime Vectors Like vectors in space, one can add, scale and take the dot product of 4-vectors. However, the dot product of two 4-vectors A and B is defined differently: The dot product is an invariant, that is, it has the same value in all reference frames
Spacetime Vectors Class Problem Consider the momentum 4-vector: Given that and compute the dot product of P with itself The square root of this particular dot product is called the invariant mass
Zero Mass Particles The energy and momentum of a particle are related as follows E2 = (pc)2 + (mc2)2 This suggests that particles with m = 0 are possible. For them, E = pc. Moreover, in vacuum they move at the speed of light. Indeed, such particles exist, for example: the photon, the quantum of light, and the gluon, the quantum of the strong force
But Zero + Zero ≠ Zero! Consider two photons with 4-vectors and The 4-vector of the pair is just So the squared mass of the pair is which is not necessarily zero! Review Example 2-12
Conservation of Energy & Momentum Energy and momentum are defined relative to a frame of reference and, as we have seen, can change as one goes from one frame to another. However, as in Newtonian physics, within each frame of reference, energy and momentum are conserved.
Conservation of Energy & Momentum Space is filled with photons, called the Cosmic Microwave Background (CMB), with a typical wavelength of 2.64 mm in our frame of reference. If a cosmic proton is sufficiently energetic, the reaction proton + photon -> delta can occur in space. The delta is an unstable particle. What proton energy (in our frame of reference) is needed to trigger this reaction? Extra Credit: due 15 Feb
Summary • Newton’s 2nd Law holds in relativity provided one defines momentum as p = g(u) m u • Energy and mass are related by E = mc2. This discovery solved the mystery of sun shine. • Particles with zero mass exist, e.g., the photon in vacuum. • Energy and momentum can be combined into a 4-vector, whose dot products are invariant.