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Dimensions of Stars Having determined that stars are at large distances, high temperatures and very luminous, the next question to ask is how large they are. An Example: Dimensions of Stars – The star Betelgeuse (a Orionis) has the following parameters: – Distance 150 pc – Apparent magnitude 0 – Absolute magnitude -6 – Relative luminosity 104 L – Temperature 3000 K (I.e., a class M star). Dimensions of Stars – The luminosity depends on the surface area of the star (R2) and T 4 (from Stefan’s law). – Hence, the ratio of the radii of Betelgeuse and the Sun is: – R / R = (T / T )2 (L / L)1/2 = 370 – This corresponds to almost 2 AU – I.e., larger than the orbit of Mars (1.5 AU)! 106 Dimensions of Stars Supergiants 104 102 Luminosity (L) – Applying the same technique to other stars allows lines of constant radii to be plotted on the HR diagram. Giants 1 Sun 10-2 10-4 40,000 20,000 10,000 5,000 Temperature (K) 2,500 Dimensions of Stars Note: – Most stars on the main sequence have radii similar to the Sun, but that, as we have shown, the luminous red stars have large radii. – White dwarfs have very small radii – comparable to that of the Earth! Luminosity and Spectral Class Can the luminosity be deduced directly from the spectral class? – No - e.g. Betelgeuse and Barnard’s Star are both class M, but the latter is 18 stellar magnitudes fainter Details of the spectrum provides a handle: Luminosity and Spectral Class Absorption line shapes: – Narrow lines Low density, hence large dimensions – Broad lines High density, hence a compact object Rigel, a B8 supergiant Hg The B8 main sequence component of Algol Hd Luminosity and Spectral Class 106 Such studies lead to the idea of 104 II III Luminosity Classes – Labelled I to V in order of decreasing luminosity – Provides another handle on distance “Spectroscopic Parallax” 102 IV Luminosity (L) I V 1 10-2 10-4 40,000 20,000 10,000 5,000 Temperature (K) 2,500 Spectroscopic Parallax Example: – The star a Leonis (Regulus) has an apparent magnitude of 1.4 and its spectrum shows it to be a B7V star. – This gives an absolute magnitude of -0.6. – The distance modulus is therefore 2, giving a distance of 25 pc. Spectroscopic Parallax Note: – The luminosity classes are broad and poorly defined. Hence, a distance determined by this method should be regarded as an estimate – This method has nothing to do with Parallax! Masses of Stars We have now established methods of measuring : – Distance – Brightness – Composition – Temperature – Size Masses of Stars we require the mass to gain a complete picture of the properties of stars. For example: – are giant and dwarf stars proportionatly more massive than the Sun, or are they about the same? – How does mass vary along the main sequence? Masses of Stars It is impractical to measure the mass of an isolated star. Many stars occur in gravitationally bound multiple systems – in particular, binary stars. By observing the orbits of such stars (and knowing the distance), the total mass can be found Masses of Stars Application of Kepler’s Third Law a3 M1 M2 2 P – Mi in Solar Masses, a in AU and P the orbital period in years M1 a M2 Masses of Stars Example: – The orbit of the binary star 70 Ophiuchi (parallax 0.2 arc sec) has been plotted over many years. – The period is 87.7 years – The semi-major axis is 4.5 arc sec. – What is the sum of the masses of the two stars? Masses of Stars – 70 Ophiuchi Distance = 5pc – ~ 106 AU. Hence, the semi-major axis is ~ 22 AU. The period is 87.7 years The total mass is ~ 1.4 M Masses of Stars The ratio of the masses: – The two stars orbit a common centre of mass – The major axes of the two ellipses are in the inverse ratio of the masses. M1 a1 a2 M2 M1/M2 = a2/a1 Masses of Stars Together with the sum of the masses, the individual masses can now be obtained. – Orbital period Total mass – Ratio of major axes Ratio of masses Masses of Stars Binary systems containing red giants: – show masses of red giants fall into the same range as main sequence stars confirming our conjecture they have low densities) White dwarfs are found to have masses about 1 M – indicating these are very dense objects! Is there a relationship between luminosity and mass? Luminosity, L (Solar Luminosities) Mass-Luminosity Relationship L = M 7/2 Mass, M (Solar Masses) Mass-Luminosity Relationship L = M 7/2 – Approximate relationship – Valid for low mass stars up to about 1 solar mass – L = M 3 gives better fit for high mass stars – Note for Main Sequence stars only Stellar Interiors Basic Ideas – Hydrostatic Equilibrium – Classical and Quantum Gases Electron and photon gases – Ionisation The Saha Equation – Heat Transfer Mechanisms – Fusion Hydrostatic Equilibrium Since main sequence stars are stable, and are neither expanding or contracting, they are clearly in equilibrium – Internal pressure supports against gravity Hydrostatic Equilibrium We wish to examine: – The relationship between mean pressure and potential and kinetic energies The Virial Theorem – The consequences of equilibrium of nonrelativistic and ultra-relativistic gases Hydrostatic Equilibrium Consider a volume element in a spherical system of mass M and radius R R r P.DA g(r)DM (P+DP)DA Mass DM Hydrostatic Equilibrium Net force due to the pressure gradient dP DP DA Dr DA dr – Noting that the mass, DM, of the element is: DM = r(r)Dr DA, and including the gravitational force, g(r) we have the acceleration of the volume element: d 2r 1 dP 2 g r dt rr dr Hydrostatic Equilibrium In equilibrium, net acceleration is zero. – Hence: g r 1 dP rr dr We can write g(r) as Gm (r ) g r 2 r Where m(r) is the mass contained within a radius r Hydrostatic Equilibrium Therefore, if the star is to be in hydrostatic equilibrium, the pressure gradient at any distance r from the centre is: dP Gm r rr 2 dr r Hydrostatic Equilibrium We can use this expression to relate the average pressure to the total gravitational potential – Multiply both sides by 4pr3 – Integrate from 0 to R R 0 2 R Gm r 4 pr rr dr dP 3 4 pr dr 0 dr r = dm Hydrostatic Equilibrium The right hand side is simply the gravitational potential: M E g 0 Gm r dm r The left hand side can be evaluated using integration by parts: df x dg x g x dx dx f x g x f x dx dx Hydrostatic Equilibrium The left-hand side then becomes: P r 4pr =0 3 R 0 R 2 3 P r 4 pr dr 0 = -3<P >V = dV Hydrostatic Equilibrium Hence, the average pressure in a system in hydrostatic equilibrium and with gravitational potential Eg is: 1 Eg P 3V The Virial Theorem Hydrostatic Equilibrium Non-relativistic (p << mc) vs. ultrarelativistic (p >> mc) gases. – Fundamental relationships: P n 3 p v Ep p c m c 2 2 2 2 4 Hydrostatic Equilibrium Non-relativistic gases: – p.v. = mv2 – Hence: P n 3 mv 2 2n 1 mv 2 3 2 Pressure = 2/3 kinetic energy density Hydrostatic Equilibrium Consequences: 2 E KE 1 Eg P 3 V 3V Hence: 2E KE E g 0 Hydrostatic Equilibrium Since: E total E KE E g E total 1 E KE E g 2 i.e., the system is bound with binding energy -Etotal = internal kinetic energy Hydrostatic Equilibrium Implications: – Tightly bound systems are hot – Energy loss from a gas cloud causes contraction and heating 1/2 loss of gravitational energy goes into radiation, 1/2 into raising the temperature Increase in temperature provides pressure to counter gravity – Excess energy production in a star causes expansion and cooling Hydrostatic Equilibrium Ultra-relativistic gases: – p.v. = pc – P >>mc, so EKE = pc – Hence: P n 3 pc Pressure = 1/3 kinetic energy density Hydrostatic Equilibrium Consequences: – EKE = -Eg – System only in equilibrium if Etotal = 0 – Hydrostatic equilibrium of ultra-relativistic particles (e.g., photon gas, high energy electron gas) easily disrupted. Next Lecture: Classical and Quantum Gases The Saha Equation – Derivation – Consequences for ionisation and absorption