Age of the Earth



Paul AveryPeter HirschfeldPHY3101Oct. 4, 2018Version 1How Does the Sun Shine?Table of contents TOC \o "1-2" 1Age of the Earth PAGEREF _Toc373498350 \h 12Age of the Sun PAGEREF _Toc373498351 \h 22.1Chemical processes PAGEREF _Toc373498352 \h 22.2Shrinking of the sun PAGEREF _Toc373498353 \h 32.3Material falling into the sun PAGEREF _Toc373498354 \h 33Solar Luminosity and Nuclear Physics PAGEREF _Toc373498355 \h 43.1Energies of nuclear reactions PAGEREF _Toc373498356 \h 43.2Basic nuclear reactions in the sun PAGEREF _Toc373498357 \h 53.3Lifetime of the sun from nuclear physics PAGEREF _Toc373498358 \h 64The Fate of the Sun PAGEREF _Toc373498359 \h 7References PAGEREF _Toc373498360 \h 8 Age of the EarthClassical estimates: Greek philosopher Aristotle thought the earth to be eternal, so no need to understand its origin. Roman poet Lucretius believed the earth’s origin was quite recent since no historical records predated the Trojan war.Bible estimate: Talmudic scholars, Martin Luther, etc. made estimates based on Genesis and its chronologies to give a creation date of a few thousand years ago. The most famous exposition of this approach was the 1654 estimate of Bishop Ussher who derived the creation date as 4004 BCE or ~6000 years ago. This estimate was remarkably influential and accepted by many educated Europeans for about two centuries.Geology: Scientists in the 17th century began studying rock strata (layers of rock) which had specific kinds of fossils for each type of strata, implying changes occurring over long time intervals as one rock layer covered another and different forms of animals emerged. As understanding of geologic processes (e.g., deposition, sedimentation, erosion) and their rates improved over the 18th and 19th centuries, it became clear that the earth’s age was 10s of million and even over 100 million years old (late 19th century estimate), even though their assumptions of perfect uniformitarianism (constant rates for geologic processes throughout the earth’s history) were later found to be incorrect.Evolution: Darwin’s theory of evolution (1859) required long periods for life forms to evolve, again requiring enormous spans of time, 10s to 100s of millions of years, though no one knew for sure.Physics: Kelvin in 1862 estimated the age of the earth by considering the time it would take a ball of molten rock to cool sufficiently to today’s temperatures, obtaining 20M – 400M years. (However, he did not know about radioactive elements deep in the earth whose decay products provide large amounts of heating. He also didn’t take into account convection in the earth’s core which moves heat closer to the surface.) By 1897, he had modified his estimate to 20M – 40M years. George Darwin (son of Charles) published a figure of >56M years based on the assumption that the moon had broken away from the earth during rapid rotation and using tidal friction as a mechanism transferring angular momentum from the earth’s rotation to the lunar orbit to estimate how long it would take the earth – moon system to reach its current configuration.Radiometric dating: We now have abundant evidence from radioactive materials that the age of the earth is ~4.55B years old, far older than even the largest estimates at the end of the 19th century.Age of the SunGiven the increasing estimates of the earth’s age from geologic and evolutionary evidence (10s or million years to over 100M years), it became reasonable to ask how the sun could shine for such a long time. Several assumptions could be made about the origins of the sun’s luminosity, which was known reasonably well in the 19th century.The sun’s temperature could be estimated from the wavelength at peak radiance from Wien’s law and the luminosity can be calculated from the Stefan-Boltzmann formula , giving a total solar luminosity of W. But Wien’s law was not formulated until 1896. One could also obtain the approximate luminosity by measuring the flux F (power/m2) at the earth’s surface and extrapolate using , where dsun is the earth-sun distance.Chemical processesCan the solar luminosity be accounted for by chemical processes? Let’s make an estimate. The sun’s mass is kg. If this is mostly hydrogen (close enough), that’s aboutatoms. Chemical reactions release about 1–2 eV per molecule at most. Assuming the high number (2 eV), we obtain as an (over)estimate for the total chemical energy in the sunJThe sun could thus burn at its current luminosity for a total of (1 year = 31,500,000 sec):Given the lifetime of the earth, chemical energy as a source for the sun’s luminosity is orders of magnitude too small.Problem: Go online to find a more typical chemical energy release per kg, and apply it to the sun’s luminosity to calculate the lifetime of the sun.Shrinking of the sunKelvin in the late 19th century was aware that chemical energy could not account for the sun’s luminosity. He proposed a model where the sun slowly shrank in size, converting its gravitational potential energy into radiant energy.The calculation is fairly straightforward. The gravitational potential energy of a constant density sphere of mass M and radius R can be calculated (imagine mass shells being laid down one after another, changing the potential energy) to beIf the radius is changing at a rate then the rate of energy release is(The quantity on the RHS is positive because is negative.) Using kg and m, we find, using the current luminosity, that m/s. This seems small, but the approximate time it would take to move through a solar radius isThis is still not long enough, even considering how unlikely it is that such a mechanism could adjust the shrinkage rate to maintain a constant luminosity for so long. Problem: Assuming a constant solar luminosity at today’s value, calculate the ratio of the sun’s radius 20M years ago to its radius today. Hint: Find the total energy emitted over this time and compare it to the difference in gravitational potential energy.Material falling into the sunAnother one of Kelvin’s ideas was that the sun’s luminosity might be due to the kinetic energy of material falling into the sun. The kinetic energy of a mass falling from infinity to the sun’s surface can be found from conservation of energyIf all this energy is converted to radiation, then the luminosity can be expressed aswhere is the rate of matter (kg/sec) falling into the sun. Using the sun’s luminosity of W, we obtain kg/sec or kg/year, about 0.01 earth masses per year. But in 20M years, the additional material would be have a mass of ~200,000 times that of earth or 0.6 solar mass, which would completely alter the sun’s properties. In 100M years, the sun’s mass would quadruple!Problems: (1) Show that the formula for kinetic energy is correct from energy conservation.(2) Show that the infalling rate is as shown here.Solar Luminosity and Nuclear PhysicsIncreasingly better understanding of slow geologic and evolutionary processes at the end of the 19th century presented scientists with an age for the earth (100M years or larger) that was far longer than could be accommodated by solar luminosity models based on known physics of the time. However, experiments on radioactivity in the 1890s and afterwards gave the first hints of what we now know to be nuclear processes.Energies of nuclear reactionsWe can estimate the energies of nuclear processes by applying the uncertainty principle to the motion of nucleons within the nucleus. An average nucleus has a radius of ~4 fm, so by the uncertainty principle in each coordinate and the kinetic energy of a typical nucleon is therefore approximately(where we used ). Other evidence of these high energies comes from radioactive decays of elements like Co60, radon, etc. which also involve energies of the order of a few MeV (see REF _Ref373349118 \h Figure 1).Figure SEQ Figure \* ARABIC 1: Beta and gamma nuclear transitions in Co60 to Ni60The overall evidence is that nuclear reactions involve energies ~1 million times larger than chemical energies. We discuss in the next subsection how such reactions are responsible for the sun’s luminosity.Basic nuclear reactions in the sunThe basic nuclear reactions we need to know take place in the sun’s core (~0.24 solar radius) where the temperature is about 15,000,000 K and the density is about 150 times that of water ( REF _Ref373351657 \h Figure 2)Figure SEQ Figure \* ARABIC 2: Cross-section of the sun, showing the core to approximate scale90% of the sun’s energy comes from the “proton-proton” (p-p) cycle, which is comprised of the following nuclear reactions.(We show the entire atom in these reactions even though only the nucleus takes part.) Energy is released for each of these steps, which we describe asTwo protons fuse to create deuterium, releasing a positron (, an antiparticle of the electron that annihilates when it comes in contact with it) and a neutrino (, a neutral partner of the electron that barely interacts with matter). This is a very slow step.Deuterium fuses with a proton to form helium-3 plus a photonTwo helium-3 nuclei fuse to form helium-4 and 2 protons.Note that each of the first 2 reactions must take place twice for every occurrence of the last reaction (this is like chemistry!). The net reaction can thus be written as ( REF _Ref373349155 \h Figure 3)Figure 3: The p-p cycle in the center of the sun, showing the relative rates of the three reactions in terms of the length of time an average nucleus of that type has to wait before interacting. The first step limits the rate of the overall reaction.The energy Q released in this overall process is calculated from the difference in masses between the initial and final state, using the mass-energy relation and adding the energy of annihilation as each positron annihilates an electron ( MeV per annihilation). This gives Q = 26.73 MeV for the net reaction. Since an average of 0.5 MeV is taken away by the neutrinos which escape from the sun without interacting, we have 26.2 MeV of available energy per p-p cycle or 6.55 MeV per proton.Lifetime of the sun from nuclear physicsWe can use this basic energy release per proton to calculate the total number of hydrogen atoms fusing into helium per second. The actual solar luminosity is W, so the number of hydrogen atoms fused per second isThe total mass per second of hydrogen converted to helium is thusor about 600 millions tons of hydrogen per second. It’s also interesting to calculate the total mass per second converted completely to energy. This can be done directly from the luminosity and the mass-energy relation yielding kg/sec or about 4.3 million tons/sec!The sun’s lifetime can be approximately calculated, using the above hydrogen fusion rate, asA more realistic estimate uses the fact that the sun at its origin was 75% hydrogen and that only a fraction of the total mass can be converted through fusion before other stellar evolution processes interfere. These factors lead to a lifetime closer to 10B years, still an enormous time that easily accounts for the fact that the sun has been shining since the earth formed. The sun is approximately halfway through its full lifecycle.The Fate of the SunOver 4.55B years the sun has converted about 4% of its total mass into helium, all of it in the core. Here is what we can expect.In ~5B more years the core will have depleted its store of hydrogenThe core, consisting mainly of helium, will not be able to sustain a sufficient fusion rate to support the layers above it. The core will shrink somewhat, raising its temperature, and hydrogen will begin fusing in a shell just outside the core.Hydrogen burning at such a large radius will cause tremendous expansion of the layers above it. The sun becomes a red giant, with its outer layers reaching past the orbits of Mercury and perhaps Venus and earth’s surface utterly destroyed.The helium core will continue to shrink until helium begins fusing to form carbon. This is the phase of “helium burning” through the triple-alpha process ( REF _Ref373394478 \h Figure 4)The core will continue to burn helium, building up a large amount carbon and oxygen.The sun’s mass is too low to generate the temperatures and pressures necessary to burn carbon, so the core will continue to shrink and blow off the sun’s outer layers until it forms a hot, dense “white dwarf” about the size of the earth but a density 105–106 times that of water. The star is now supported by “degeneracy pressure” from electrons with energies from 0 to the Fermi energy EF.The white dwarf will continue to cool until the end of the universe, the fate of ~99% of all stars.Figure SEQ Figure \* ARABIC 4: Fusing of helium to form carbon through the “triple-alpha” processReferences ................
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