PHYS 141: Principles of Mechanics

PHYS 141: PHYS 141: Principles of Principles of MechanicsMechanics

1. Definitiona) A blackbody is a perfect absorber of light at all wavelengths.

b) Wien’s Law: The peak wavelength of emission for a bb decreases as temperature increases :

p = 2.898 x 10-3 (m-K)/T, (I.A.1)

c) The Stefan-Boltzmann Law: The power = (E/s) = P/m2 of a blackbody is given by:

Pbb =AT4, (I.A.2)

where = 5.67 x 10-8 W/(m2K4).

I. Blackbody Radiation A. Properties

c) The Stefan-Boltzmann Law

* Further define an energy Flux, F = P/A:

F = T4 (I.A.3)

Hotter blackbodies emit more total power per area. Properties of blackbody described principally by T.

d) Radiance: radiated power as a function of wavelength and temperature (Max Planck).

(I.A.4)

(I.A.5-6)

5 ( / )

2( , ) .

( 1)bb hc kT

6.63 10 .

1.38 10 / .

Visible IR RadioUV

Light bulb off: No light at any .

Visible IR RadioUV

Light bulb on: T = 3000 K.

Fairly bright inRed & Orange

Dim in VioletAnd Blue

max= 1000 nm

The combined brightness at each

(color)determines what

color a BB appears.=

Visible IR RadioUV

Light bulb on: T = 6000 K.

Fairly bright inRed & Violet

Bright in Y&G

max= 500 nm

Visible IR RadioUV

Light bulb on: T = 30,000 K.

Bright inG,Y,O,R

Very Bright inV,I,B

max= 100 nm

e) The Ultraviolet Catastrophe!

* Classical thermal physics predicted infinite flux and infinite intensity at small wavelengths

* The solution: Planck’s hypothesis

* “Electric resonators” responsible for bb radiation have discrete (quantized) energies:

En = nhf. (I.A.7)

where h is Planck’s constant,

B. The Photoelectric Effect1. Einstein proposed the particle model of light:

photons.

Ephoton = hf = nhc/(I.B.1)

2. Extension of Planck’s work on molecular oscillators

photons.

Ephoton = hf = nhc/.(I.B.1)

2. Extension of Planck’s work on molecular oscillators -e

photons.

2. Extension of Planck’s work on molecular oscillators -e-e-e

photons.

2. Extension of Planck’s work on molecular oscillators -e-e

3. Particle-Theory of Light predictions: RIGHT!

a) As the frequency of light increases, the maximum K of electrons also increases:

Kmax = hf - , (B.2)

Where is the “Work Function” that holds the electrons to the plate.

b) If f < f0 = “cutoff frequency” = /h, no electrons ejected.

c) Maximum electron energy (as measured by the “stopping potential ,” VS) is independent of intensity.

4. Example: Light shines on a substance with = 1 eV. If = 500 nm, are electrons ejected from the surface?

Step 1: Find the energy of the photons. For = 500 nm, we have

E = hc/ = (6.6 x 10-34 J-s)(3.0 x 108 m/s)/(5 x 10-7 m) = 4.0 x 10-19 J;E = (4.0 x 10-19 J)(1 eV/1.6 x 10-19 J) = 2.5 eV.

Step 2: Find Kmax = 2.5eV - 1.0 eV = 1.5 eV > 0. So, yes, electrons are ejected.

A. Compton Effect (1923)1. Scattered light has lower frequency than incident light.

2. Recall that relativistic energy is E = pc for a massless particle. Thus, the relativistic momentum for a photon is

p = E/c = hf/c = h/. (A.1)

II. Light, More Light

Conserve momentum & energy for electron initially at rest:

’ = + (h/mc)(1 - cos). (A.2)(h/mc)(1 - cos). (A.3)

and (h/mc) =C = Compton Wavelength. (A.4)

3. Example: How much energy is lost by a 1 MeV photon that scatters off an electron with an angle = 60o?

Ei= hf = hc/ = 1.6 x 10-13 J*; = 0.0012 nm.

’ = + (h/mec)(1 - cos). = 1.2 x 10-12 m + (2.4 x 10-12 m)(1 - .5)

= 2.4 x 10-12 m = 0.0024 nm.

Since the wavelength doubled, the Ef = hc/’ = 1/2Ei = 0.5 MeV.

*1 MeV = 1.602 x10-13 J.

1. And yet: light also behaves as a wave

a) Young Double Slit experiment: interferenceb) Maxwell’s Equationsb) Connection between frequency and wavelength

f = c.(B.1)c) “Electromagnetic Spectrum”

II. Light, More Light B. Light as a Wave

LowEnergy

HighEnergy

1. de Broglie (1923): If light is observed to have wavelike properties some times (interference) and particle-like properties other times (photoelectric effect), then what about matter?

= h/mv “de Broglie Wavelength” (I.C.1)

2. Example: what is for an electron moving at .01c?

= (6.6 x 10-34 J-s)/(9.11 x 10-31 kg*3.0 x 106 m/s); = 2.4 x 10-10 m = 0.24 nm.

II. Light, More Light C. Waves of Matter?

1. Spectral Analysisa) a) SA: The identification of a

chemical substance by its unique spectral lines.

b) b) Joseph Fraunhofer (1815) : Hundreds of dark lines in the Solar Spectrum

c) c) The Value of Spectral Analysis* Composition, abundance* Temperature* Motion

II. Light, More Light D. Atomic Spectra

2. The visible Hydrogen series (Balmer Series)

1/ = R(1/22 -1/n2), n = 3, 4, 5…(I.D.1)

Where R = 1.097 x 107 m-1 is the Rydberg constant.

Balmer lines 656 nm (B, n = 3).are transitions 486 nm (B, n = 4).to/from n = 2 434 nm (B, n = 5).

1. General aspects of the modela) Hydrogen Atom as simple exampleb) Electron orbits are quantized; not all orbits stable.c) To move from one orbit to the next, an electron needs to absorb or emit an exact E.

d) Larger orbital differences mean more energy required to move

e) Packet of energy: a photon!f) Model later modified from strict

orbits about the nucleus to “energy levels”

II. Light, More Light E. The Bohr Model

2. 2. The mathematics of quantizationa) a) Frequency of radiation

absorption/emission:

Eu - El = hf. (I.E.1)

b) b) Quantum condition: only specific values of electron orbital angular momentum allowed:

L = mevrn = n(h/2), n = 1, 2, 3, …(I.E.2)

c) Electrostatic Force:

F = kq1q2/r2

(I.E.2a)PE = -kq1q2/r. (I.E.2b)with

k = 8.99 x 109 Nm2/C2 is the Coulomb constantand the unit of charge is the Coulomb:

e = 1.602 x 10-19 C.

d) Total energy of hydrogen atom:

PE = -ke2/rn. (Electric Potential energy) (I.E.3)KE = (1/2)mev2. (I.E.4)

Thus, E = PE + KE = -ke2/r +(1/2)mev2. Now assume that the electron orbit is circular:

F = ke2/rn2 = mev2/rn = 2rn(KE), so (I.E.5)

KE = ke2/(2rn), and (I.E.6)E = -keE = -ke22/(2r/(2rnn).). (I.E.7)(I.E.7)

-ern+e

e) Orbital sizes: use the quantum condition (E.1):

rn = n(h/2)/mev, andrn

2 = n2(h/2)2/me2v2.

From (E.5), ke2/rn2 = mev2/rn, so v2 = (ke2/me)/rn.

rn = n2(h/2)2/(ke2me). (I.E.8)Define “the Bohr radius” = r1 = r(n=1) = 0.0529 nm, so

rn = n2r1. (I.E.9)

e) Energy levels of orbits:

En = -{mek2e4/(2(h/2)2}(1/n2), or (I.E.9)

En = (-13.6 eV)/n2 = E1/n2. (I.E.10)

Energy required for ionization:

Ei = E(n = infinity) + (13.6 eV)/n2 = (13.6 eV)/n2.(I.E.11)

F.Quantization and atomic spectra (H)a) Ground state: n = 1, E1 = -13.6 eV.b) First excited state: n = 2, E2 = -3.4 eV.c) Energy required to move from E1 to E2:

E = -3.4 eV - -13.6 eV = 10.2 eV.

3. Hydrogenic atoms: Z = number of protons in nucleus.

En = (Z/n)2E1. (F.1)rn = (n2/Z)r1. (F.2)

3. Photon frequency/wavelength:

f = E/h = [{meke2e4}/{4(h/2)3}](1/nf

2 - 1/ni

f = cR(1/nf2 - 1/ni

2), and (I.F.3)

1/ = R(1/nf2 - 1/ni

2). (I.F.4)

3. Quantization and atomic spectraf) Spectral types explained!

Unique electron orbits => unique energy differences =>

unique patterns

Absorption/Emission spectra understood as upward/ downward transitions

of electrons.

G. The Wave Nature of Matter (revisited)1. de Broglie’s wave description of electrons

can now be understood in terms of standing waves surrounding a nucleus: only waves that close back on themselves can constructively interfere and “survive”

A. The Failure of the Bohr Model

1. Couldn’t be applied to more complicated atoms2. Couldn’t explain “fine structure”3. Couldn’t explain solids, liquids, molecules4. Quasi-classical mechanics

The Bohr Model, while substantial, was incomplete.

III. Quantum Mechanics

1. Define a Wave Function, , that describes a particle.

a) Wave function contains information about a particle’s state: position, speed, momentum & energy, spin, etc.

b) “Matter Wave” or “Matter Field”c) Depends on position and time: (x,y,z,t).d) Probabilities:

“Probability density”

2 ~ Probability of finding an electron at position (x,y,z) and time t.

III. Quantum Mechanics B. Wave Mechanics

2. Calculations: Solution to the “Schroedinger Wave Equation”

a) One dimensional case, Cartesian coordinates. Probability of finding a particle between x and x + dx

b) Normalization: The particle must be SOMEWHERE. Therefore

2( ) .P x dx

2 1.dx

2. Calculations: Solution to the “Schroedinger Wave Equation”

c) Radial Probability distribution. Probability of finding a particle between r and r + dr

d) Normalization: The particle must be SOMEWHERE. Therefore

2 2( ) 4 .P r r dr

4 1.r dr

3. Example: Ground State of Hydrogen

a) Wave function for ground state

b) Step 1: Find the constant by normalization

0/( ) .r ar Ae

02 /2 2

4 1.r ar A e dr

02 /22

4r ar e dr

301 / .A a

(B.4a,b)

3. Example: Ground State of Hydrogen

c) Now ask: What is the probability of finding an electron inside the Bohr Radius, a0? (see pg. 649 and Appendix A the for integral)

0 02 / 2 /2 2 20 3

4( ) 4 0.323

a ar a r aP r a r A e dr r e dr

C. Heisenberg Uncertainty (Indeterminancy) Principle

1. There is an inherent uncertainty in pairs of correlated variables

2. Momentum-Position Uncertainty Principle

xp ≥ (h/2). (III.C.1)

The original argument comes from measurement of an electron. The uncertainty in the electron’s position is roughly the wavelength of observation (), and the photon will give some unknown portion of its momentum to the electron (h/) when it strikes the electron. Trying to narrow the position means using smaller , but that imparts more (unknown) momentum to the electron.

III. Quantum Mechanics C. HUP

3. Energy-Time Uncertainty Principle

Et ≥ (h/2).(III.C.2)

In this UP, the energy of a particle is uncertain (E), over a time scale t ~ (h/2)/E.

Warning: Part of the problem is that we tend to view electrons, etc. exclusively as particles. Indeterminancy is a direct result of electrons, protons, etc. having both wave AND particle properties.

4. Example: What is the maximum precision for measuring the position of an electron moving with a speed equal to v = (3.00 ± 0.01) x 106 m/s?

xp ≥ (h/2) =>

xmin = (6.6 x 10-34 J-s/2)/[(9.1 x 10-31 kg)(104 m/s)],

= 1.2 x 10-8 m,= 12 nm, or about 100x the size of an

5. Example: What is the lifetime of a particle with a spread in energy equal to 1kev?

Et ≥ (h/2) =>

t = (6.6 x 10-34 J-s/2)/[(1000eV)(1.6 x 10-19 J/eV)],

= 6.6 x10-19 s.

1. Principal quantum number: n (e.g., En = E1/n2).a) Related to energy of electronb) n = 1, 2, 3, …

2. Orbital Quantum number: l a) related to orbital angular momentum vector of

electron

L = {l(l +1)}1/2 , where = h/(2), and (D.1)

l = 0, 1, 2, …, n-1.(D.2)

III. Quantum Mechanics D. Quantum Numbers

3. Magnetic quantum number: ml.a) Related to direction of electron’s angular

momentum- l, - l + 1, - l + 2, …, 0, … l - 2, l - 1, l. (D.3)

b) Angular momentum along the “z-axis:”

Lz = ml (D.4)h

- ml = -1.

ml = 0.

ml = 1.

Total number of states for a given value ofl depends on projection Lz.

3. Magnetic quantum number: ml.c) Energy levels are split in the presence of magnetic field.

n = 2l = 1

ml = 1.ml = 0.ml = -1.

n = 1l = 0 ml = 0.

4. Spin quantum number: ms.a) originally thought of as intrinsic spin of electronb) ms = ±1/2, but s = ½ always (for an electron)c) “fine structure”

n = 2l = 0

ml = 0.

n = 1l = 0 ml = 0.

Electron spin vector is given by (D.5).( 1) .s s S

5. Total Angular Momentum: Vector sum of J = L + S:

For l ≠ 0 j = l ± ½ (parallel vs. antiparallel) (D.6)For l = 0 j = ½ (D.7)

(D.8)Total number of angular momentum states depends on projection Jz:

-j, -j+1, …, j-1, j => 2j + 1 states.

Nomenclature: n(l)j, (D.9)

Where l = 0, 1, 2, 3, 4 => S, P, D, F, G

( 1) .J j j

6. Selection Rules and Probability Distributionsa) “Allowed” transitions: l = ±1.

l and ml change the shape of the probablitiy

distribution function.

1. The Pauli Exclusion Principlea) Electrons in the same atom cannot have the

same quantum state--each electron has a unique set of quantum numbers.

b) Usually electrons are found in the lowest possible state.

c) Similar exclusion principle for any particle that has half integer values of spin (e.g., protons & neutrons): Fermions.

d) Particles with integer values of spin do not obey EP (e.g., photons): Bosons.

III. Quantum Mechanics E. Other Atoms

e) ExamplesHe (Z = 2). 2e: (n, l, ml, ms ) = (1, 0, 0, 1/2).

(n, l, ml, ms ) = (1, 0, 0, -1/2).

Li (Z = 3). 3e: (n, l, ml, ms ) = (1, 0, 0, 1/2).(n, l, ml, ms ) = (1, 0, 0, -1/2).(n, l, ml, ms ) = (2, 0, 0, 1/2).

Be (Z = 4). 4e: (n, l, ml, ms ) = (1, 0, 0, 1/2).(n, l, ml, ms ) = (1, 0, 0, -1/2).(n, l, ml, ms ) = (2, 0, 0, 1/2).

(n, l, ml, ms ) = (2, 0, 0, -1/2).

III. Quantum Mechanics E. Other Atoms

1. Radioactive Decay Equation

N/N0 = e-t, (A.1)

where is the decay rate.

2. Half-life: The time for half the sample to decay from parent isotope to daughter isotope = “t1/2.”

IV. Nuclear Physics A. Radioactive Decay

1/2/0/ 2 .t tN N

1/2t 0.693 / .

Rewriting the RDE, we have

N/N0 = exp(-0.693t/t1/2). (A.4)

3. Define activity as the number of decays/second,

N/t = -N = (0.693/t1/2)N = (N/t)0e-t.

4. Decay channels:

-particle emission: He nucleus (2p, 2n) => Z - 2, A - 4-particle emission: e- + => Z+1, A-particle emission: high energy photon => Z, A

Example: 238U as a half-life of 4.5 billion years. If an initial sample contains 1018 atoms, what is the initial activity?

= 0.693/t1/2 ~ 5 x 10-18 s-1.

The initial activity for the sample is

(N/t)0 = N = (2 x 10-18 s-1)(1018 nuclei) = 5 decays/second.

What is the activity after 13.5 billion years?

(N/t) = (N/t)0 e-t = 0.6 decays/second ~ 1/8(N/t)0

Example: 238U as a half-life of 4.5 billion years. If an initial sample contains 1018 atoms, what is the initial activity?

After 13.5 billion years, how much 238U is left?

N/N0 = exp(-0.693t/t1/2) = exp(-0.693(3)) = 1/8.

3 half-lives => (1/2)(1/2)(1/2).

1. Nomenclature

Z ≡ proton (element) numberN ≡ neutron numberA ≡ atomic mass number, e.g., 14

7N or 136C

2. Spina) Each nucleon (p,n) is a fermionb) Nuclear spin is quanitized (I, Iz): resonances & NMRc) Each nucleus may be a fermion or a boson depending on the number of

nucleons. For even and equal numbers of p & n, nucleons are paired=> spin 0

42He => spin 0 (boson)

32He => spin ½ (fermion)

IV. Nuclear Physics B. Nuclear Structure

3. Nuclear stabilitya) Nuclei have “shell structure” analogous to atomic shellsb) Z,N = 2, 8, 20, 28, 50, 82, 126: stablec) PEP applies to n,p =>forces nucleons into energy levels:

Ground State configurations

42He p n s = 0

32He p n s = ½

IV. Nuclear Physics B. Nuclear Structure

Even Z, N minimizetotal energy and increase

stability of nucleus.

1. Reactionsa + X => Y + b. (IV.C.1)X(a,b)Y. (IV.C.2)

2. Reaction energy Q = (Ma + MX - Mb -MY)c2. (IV.C.3)

= KE = KEb + KEY - KEa - KEX.(IV.C.4)

Q > 0: Exothermic (energy released).Q < 0: Endothermic (energy required).

IV. Nuclear Physics C. Energy

1. Reactionsa + X => Y + b. (IV.C.1)X(a,b)Y. (IV.C.2)

2. Reaction energy Q = (Ma + MX - Mb -MY)c2. (IV.C.3)

= KE = KEb + KEY - KEa - KEX.(IV.C.4)

Q > 0: Exothermic (energy released).Q < 0: Endothermic (energy required).

3. Binding Energya) Unified mass

1u = 931.5 MeV/c2.b) Definition: The energy required to hold a nucleus together.

M(proton) = 1.00728 u M(42He) = 4.00153 u

M(neutron) = 1.00866 u

BE = (2Mp + 2Mn)c2 – MHec2

= (4.03188 uc2 - 4.00153 uc2 )(931.5 MeV/uc2) = 28.3 MeV.

4. Reaction Example: n + 147N → 2

1H + X. What is X?

In this example, a neutron strikes a 147N nucleus

producing deuterium (21H) and another element. To

find X, conserve nucleon number and charge:

14 (nitrogen) + 1 (neutron) = A + 2 (deuterium).+7e (in nitrogen nucleus) = qX + 1e (deuterium).

Thus, A = 13 and qX = 6, and X = 136C.

5. Example: What is the minimum proton kinetic energy required for the reaction (1

1H + 136C → 13

7N + n) to take place?

Step one: compute rest energy on both sides:

M(136C) = 13.003355 u M(13

7N) = 13.005738 u

M(11H) = 1.007825 u M(n) = 1.008665 u

(Mafter - Mbefore)c2 = (0.003223 uc2)(931.5 MeV/uc2) = 3.00 MeV.

Thus, the proton would have to have kinetic energy of at least ~ 3.00 MeV.

1. Definition: the splitting of one nucleus into multiple nuclei after particle bombardment.

2. Example: 23592U

a) n + 23592U → 236

92U* → 14156Ba + 92

36Kr + 3n. b) n + 235

92U → 8838Sr + 136

54Xe + 12n.

3. Energy released in single fission event:

Efission = Mc2 (D.1)

IV. Nuclear Physics D. Fission

4. Example: What is the energy released by the reaction n + 235

92U → 8838Sr + 136

54Xe + 12n?

Before: Mi = MU + Mn = 235.0439 u + 1.008665 u = 236.053 u.

After: Mf = MSr + MXe + M12n = 235.917 u.

Efission = Mc2 = (0.135 uc2)(931.5 MeV/uc2) = 126 MeV.

5. Chain Reactionsa) Sustained nuclear

reaction requires chain reaction

b) Moderator to slow neutrons

c) Enrichment d) Critical mass ~

kilograms

1. Definition: the fusing of two or more nuclei into new nucleus.

2. Example: Proton-Proton Chain in the Sun.

IV. Nuclear Physics E. Fusion

First Step in “Proton-Proton Chain”

11H + 1

1H → 21H + e+ + e + 0.42 MeV.

Likelihood of interaction: cross-section of interaction between

two protons (t ~ 109 years.)

Second Step in “Proton-Proton Chain”

21H + 1

1H → 32He + + 5.49 MeV.

t ~ 1 second.

Third Step in “Proton-Proton Chain”

32He + 3

2He → 42He + 1

1H + 12.86 MeV

t ~ 106 years.

Total products: 42He + 2e + 2 + 2e+

+26.7 MeV

“Bottlenecks”

Deuterium bottleneck: H to He Fusion

High Mass Bottleneck: no stable nuclei with 5 or 8 nucleons

(protons or neutrons)

e.g, 42He + 4

2He → 84Be is

endothermic & the product has a half-life of 10-

Fusion reactions tend to favor steps built on 4He, rather than

“proton capture,”(but process is SLOW.)

2. Example: Proton-Proton Chain in the Sun.a) “Deuterium Bottleneck”b) Energy produced in one chain: ~ 27 MeV.c) 2 gamma rays + positrons + 2 neutrinos d) Requires high temperature (fast moving

particles): Tcore ~ 15 million K

Use of plasma with similar parameters as solar interior plasma for energy production on Earth is completely impractical - as even modest 1 GW fusion power plant would require about 170 billion tons of plasma occupying almost one cubic mile.

3. Solar life-time estimate: How long can the Sun shine?

Lsun = 4 x 1026 W = 4 x 1026 J/s = 2.5 x 1039 MeV/s.The mass of the Sun’s core is ~5 x 1029 kg, or ~ 2 x1056 protons.

In the p-p chain, 4p => He + 27 MeV. The number of reactions per second is then ~ 2.5 x 1039 MeV/s/(27 MeV) ~ 1038.

Every second, 4 x 1038 H become He (~billion metric tons). How long before there is no more hydrogen?

2 x 1056 p/(4 x 1038 p/s) = 5 x 1017 s ~ 10 billion years.

4. Other Fusion reactions

a) C-N-O cycleb) H-fusion in higher mass

stars and later evolutionof the Sun.

c) 12C is a nuclear catalyst.d) Yield is ~ 27 MeV.

e) Example: How much energy is

released in the reaction13

6C(11H,)14

M = M(13C) + M(1H) - M(14N),= 0.008106 u

E = (0.008106uc2)(931.5 MeV/uc2),= 7.55 MeV = 1.2 x 10-12 J.

> hc/E = (6.6 x 10-34 Js)(3 x 108 m/s)/(1.2 x 10-12 J) = 0.000165 nm.

4. Other Fusion reactions

b) “Triple-alpha”: He-Fusion Ignition T ~ 108 K Occurs in the Sun when

H core is depleted.

4He + 4He => 8Be + 4He => 12C + γ + 7.367 MeV.

Side process: 12C + 4He → 16O+ γ

Saturn Orbit ~ 3 billion km

5. Massive Stars & Nucleosynthesis: 25 MSUN Star

H fusion

He fusion C fusionO fusionNe fusionSi fusionFe

Earth Diameter ~ 12, 000 km

5. Massive Stars & Nucleosynthesis: 25 MSUN Star

BBN spallation

Stellar fusion

Neutron capture, decay products

Built from alpha capture

Built from identical nucleus fusion + n, p release

1. General Relativitya) Spacetime is curved by the mass-energy in it.

“Matter tells space how to curve, Space tells Matter how to move.”

2. The Cosmological Principleb) At the largest scales, the

Universe is both isotropic and homogeneous.

V. Cosmology A. The Pillars of Modern Cosmology

The Shape of Space: Given GR and the CP, mass-energy density of the Universe allows only three possible “geometries”

1.“Spherical” (positive curvature)

2.“Hyperbolic” (negative curvature)

3.“Flat”—Euclidean (zero curvature)

1. General Relativitya) Spacetime is curved by the mass-energy in it.

“Matter tells space how to curve, Space tells Matter how to move.”

2. The Cosmological Principleb) At the largest scales, the

Universe is both isotropic and homogeneous.

1. Observational Data and Cosmological Theorya) Galaxy Recessionb) Elemental Abundancesc) Cosmic Microwave Background

V. Cosmology B. Observations

2. Galaxy Recession: Why are Galaxies Moving Away from Us?

a) Vesto Slipher (1910’s)

b) Hubble & Humason (1920’s)

c) Doppler Shift

vz isticnonrelativ

Similar shift in frequency to smallerfrequencies for recession and larger

frequencies for approach.

3. The Hubble Law

vCosmo = H0d, (B.4)

where v is in km/s and d is in Mega-light years or Mega-parsecs.

Current WMAP measurements give:

0 22 1H (km/s)/Mly.

4. Age of the Universe

We can estimate the ageof the Universe using theHubble constant:

[H0] = s-1, so

t(Universe) ~ 1/H0 = 13.7 ± 0.1 Gyr.

1. Standard Model (prior to ~1980)a) Combines CP, GRb) Hubble Constant => Approximate Agec) Temperature decreases with timed) Expansion => mass-energy & Dark Energy

t < 10-43 s: “after Big Bang”: General Relativity breaks down.t = 10-43 s: strong-electro-weak force + gravity, T = 1032 K.t = 10-35 s: electro-weak force + strong force + gravity, T = 1027 K.t = 10-12 s: four forces separate, T = 1015 K.

V. Cosmology C. The Big Bang

1. Standard Model (prior to ~1980)

t = 10-6 s: Quark confinement & production of p & n, T = 1013 K.t = 1 s: Universe becomes transparent to neutrinos, T = 1010 K.t = 3 min: Cosmic nucleosynthesis everywhere, T = 109 K.

E ≥ ħ/t.

Side note on particles:

Leptons (F)Hadrons: Baryons (F) & Mesons (B)Field Particles (B)Higgs (B)

+2/3 +2/3

Also: particles and anti-particles

1. Standard Model (prior to ~1980)

t = 10-6 s: Quark confinement & production of p & n, T = 1013 K.t = 1 s: Universe becomes transparent to neutrinos, T = 1010 K.t = 3 min: Cosmic nucleosynthesis everywhere, T = 109 K.t = 4 x 105 yrs: Universe becomes transparent to photons, T = 3000 Kt = 4 x 108 yrs: First galaxies, T = 30 K.

E ≥ ħ/t.

V. Cosmology D. Universal Composition

VI. The END

A. The Ultimate Fate of the Universe1. Before 1999

a) The Big Good Bye (?)b) The Big Crunch (100 billion years)c) The Big Freeze (+1014 years)

2. Post 1999a) The Big Rip?

(20 Billion years)

VI. The END

A. The Ultimate Fate of the Universe1. Before 1999

a) The Big Good Bye (?)b) The Big Crunch (100 billion years)c) The Big Freeze (+1014 years)

2. Post 1999a) The Big Rip?

(20 Billion years)

PHYS 141: Principles of Mechanics

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Transcript of PHYS 141: Principles of Mechanics

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