Course on Inverse Problems Albert Tarantola First Lesson: - - PowerPoint PPT Presentation

Princeton University

Department of Geosciences

Course on Inverse Problems

Albert Tarantola

First Lesson: Introduction to Inverse Problems

The Bayes-Popper approach. Using observations to infer the values of some parameters cor- responds to solving an inverse problem. Practitioners some- times seek the best solution implied by the data, but observa- tions should only be used to falsify possible solutions, not to deduce any particular solution.

Name

• Sq. km
• Sq. miles

Population 636,267 1,505 28,748 17,666 2,381,745 197 465 1,246,700 . . . 245,664 581 11,097 6,821 919,354 76 180 481,226 . . . 15,551,358 22,000 2,590,600 6,730 18,250,000 30,600 35,460 6,920,000 . . . Afghanistan Åland Albania Aleutian Islands Algeria American Samoa Andorra Angola . . . States, Territories, and Principal Islands of the World first digit Benford model actual statistics frequency 1 2 3 4 5 6 7 8 9 400 300 200 100

speed of light in vacuum . . . Newtonian constant of gravitation Planck constant elementary charge . . . c = 299 792 458 m s-1 . . . G = 6.673(10) 10-11 m3 kg-1 s-2 h = 6.626 068 76(52) 10-34 J s = 4.135 667 27(16) 10-15 eV h = 1.054 571 596(82) 10-34 J s = 6.582 118 89(26) 10-16 eV e = 1.602 176 462(63) 10-19 C e/h = 2.417 989 491(95) 1014 A J-1 . . . CODATA recommended values of the fundamental physical constants first digit Benford model actual statistics frequency 1 2 3 4 5 6 7 8 9 80 60 40 20

2 1.5 1 0.5

• 0.5
• 1

100 50 20 10 5 2 1 0.1 0.5 0.2 X = 10x x = log10 X

infinitely resisting wires infinitely conducting wires

W1 W2

R = 1 Ω R = e Ω R = e2 Ω R = 1/e Ω R = 1/e2 Ω C = 1 S C = 1/e S C = 1/e2 S C = e S C = e2 S dl = |dR|/R = |dr| dl = |dC|/C = |dc| dl r = 0 r = 1 r = 2 r = - 2 r = - 1 c = 0 c = - 1 c = - 2 c = 2 c = 1

physical quantities are coordinates over abstract spaces (spaces that represent physical qualities) which is the distance between two electric wires? given two wires W1 and W2, which is the average wire? space of electric wires:

C8 - 4186.0 Hz - 238.89 µs B7 - 3951.1 Hz - 253.10 µs A7# - 3729.3 Hz - 268.15 µs A7 - 3520.0 Hz - 284.09 µs G7# - 3322.4 Hz - 300.98 µs G7 - 3136.0 Hz - 318.88 µs F7# - 2960.0 Hz - 337.84 µs F7 - 2793.8 Hz - 357.93 µs E7 - 2637.0 Hz - 379.22 µs D7# - 2489.0 Hz - 401.76 µs D7 - 2349.3 Hz - 425.65 µs C7# - 2217.5 Hz - 450.97 µs C7 - 2093.0 Hz - 477.78 µs B6 - 1975.5 Hz - 506.19 µs A6# - 1864.7 Hz - 536.29 µs A6 - 1760.0 Hz - 568.18 µs G6# - 1661.2 Hz - 601.97 µs G6 - 1568.0 Hz - 637.76 µs F6# - 1480.0 Hz - 675.69 µs F6 - 1396.9 Hz - 715.86 µs E6 - 1318.5 Hz - 758.43 µs D6# - 1244.5 Hz - 803.53 µs D6 - 1174.7 Hz - 851.31 µs C6# - 1108.7 Hz - 901.93 µs C6 - 1046.5 Hz - 955.56 µs B5 - 987.77 Hz - 1012.3 µs A5# - 932.33 Hz - 1072.5 µs A5 - 880.00 Hz - 1136.4 µs G5# - 830.61 Hz - 1203.9 µs G5 - 783.99 Hz - 1275.5 µs F5# - 739.99 Hz - 1351.4 µs F5 - 689.46 Hz - 1431.7 µs E5 - 659.26 Hz - 1616.9 µs D5# - 622.25 Hz - 1607.1 µs D5 - 587.33 Hz - 1702.6 µs C5# - 554.37 Hz - 1803.9 µs C5 - 523.25 Hz - 1911.1 µs B4 - 493.88 Hz - 2024.8 µs A4# - 466.16 Hz - 2145.2 µs A4 - 440.00 Hz - 2272.7 µs G4# - 415.30 Hz - 2407.9 µs G4 - 392.00 Hz - 2551.0 µs F4# - 369.99 Hz - 2702.7 µs F4 - 349.23 Hz - 2863.5 µs E4 - 329.63 Hz - 3033.7 µs D4# - 311.13 Hz - 3214.1 µs D4 - 293.66 Hz - 3405.2 µs C4# - 277.18 Hz - 3607.7 µs C4 - 261.63 Hz - 3822.3 µs

A musical note N can be characterized, for instance, by the frequency ν or by the period T . Which is the distance between two musical notes N1 and N2 ? D(N1, N2) = | log ν1 ν2

| = | log T1

T2

|

33฀1/3 45

Examples of Jeffreys quantities

frequency ν = 1 τ

−

period τ = 1 ν wevelength λ = 1 κ

−

wavenumber κ = 1 λ temperature T = 1 k β

−

thermodynamic parameter β = 1 k T

ρ = 10 g/cm3 ρ = 20 g/cm3 ρ = 0 r = 0 r = 0.5 r = 1 r = 1.5

r = log10(ρ/ρ0) ( ρ0 = 1 g/cm3 ) ∆ρ/ρ฀constant ⇒ volumetric probability f(ρ)

ρ = 1 g/cm3 ρ = 10 g/cm3 ρ = 3.2 g/cm3 ρ = 32 g/cm3

∆ρ฀constant ⇒ probability density f(ρ) _