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## Origin, Evolution and Structure of Stars, Physics tutorial

:IntroductionStars are held together via gravitational attraction applied on each portion of the star through all other parts. Collapse of the star is resisted through internal thermal pressure and gravitational force.

Such two forces play the principal role in finding out the stellar structure as they should be balanced.

:Dynamics of StarsFor our stars that are isolated, static and spherically symmetric. There are four fundamental equations to illustrate their structures. All the physical quantities based on the distance from the centre of the star alone.

1) The equation of hydrostatic equilibrium: at each and every radius, forces due to pressure differences balance gravity.

2) The conservation of mass.

3) Conservation of energy flux equivalent to local rate of energy release.

4) Equation of energy transports the relation between the energy flux and the local gradient of the temperature.

These fundamental equations are supplemented by:

a) The equation of state (that is, pressure of a gas as a function of its density and temperature).

b) Opacity (how opaque the gas is to the radiation field).

c) The core nuclear energy generation rate.

:Hydrostatic EquilibriumThe equation of hydrostatic support the balance between gravity and internal pressure is termed as the hydrostatic equilibrium.

Mass of element ∂m = P(r)∂s∂r

Here, P(r) = density at r

Taking the forces acting in radial direction:

1) Outward force is equivalent pressure applied by stellar material on the lower face: P(r) ∂s

2) Inward force: pressure applied via a stellar material on the upper face and gravitational attraction of all stellar material lying in 'r'

P(r + dr) ∂s + [Gm(r)/r

^{2}] ∂mP(r + dr) ∂s + [Gm(r)/r

^{2}] P(r) ∂s∂rThe two opposing forces are at work in a star such which gravity pulling inward wishes for to make the star contract and pressure pushing outwards wishes to make the star expands.

Pressure and Gravity work on one other:

If there is exact balance between the two, we contain a condition of Hydrostatic Equilibrium. In this case, the star neither expands nor contracts. The outer layers press down on the inner layers. The deeper one goes to a star, the bigger the pressure.

:Equation of Mass ConcentrationThe mass m(r) contained in a star of radius 'r' is determined through the density of the gas and can be represented as follows:

Whenever we consider a thin shell (a star) with inside radius 'r' and outer radius r + ∂r

∂v = 4πr

^{2}∂r∂m = ∂vp(r)

∂m = 4πr

^{2}∂rp(r)∂m(r)/∂r = 4πr

^{2}p(r)The above is the equation of mass concentration.

The Dynamical Timescale:

If we allowed the star to collapse that is, set d = r and replace:

g = Gm/r

^{2}t = 1/√β [(2r

^{3})/(Gm)]^{1/2}Supposing β = 1

Then, t

_{d }= (2r^{3}/Gm)^{1/2}t

_{d }is termed as the dynamical time.Minimum Value for Central Pressure of Star:We have just two of the four equations and no knowledge yet of material composition or physical state of stars however we can represent a minimum central pressure. Why in principle, do you suppose this necessitate being a minimum value? Given what we known, what is this probable to base on?

dP(r)/dr = - GM(r)P(r)/r

^{2}dM(r)/dr = 4πr

^{2}p(r)On dividing the above two equations:

[dP(r)/r]/[dM(r)/dr] ≡ dP/dM = - GM/4πr

^{4}On integrating the above equation, we get:

This gives P

_{m}- P_{o}=_{o}∫^{mo}(GM/4πr^{4}) dM_{o}∫^{mo}(GM/4πr^{4}) dM >_{o}∫^{mo}(GM/4πr_{o}^{4}) dM = GM_{o}^{2}/8πr_{o}^{4}Therefore, we have:

P

_{m}- P_{o}> GM_{o}^{2}/8πr_{o}^{4}We can estimate the pressure at the surface of the star to be zero;

P

_{o}> GM_{o}^{2}/8πr_{o}^{4}For illustration, for the sun:

P

_{c}= 4.5 x 10^{13}Nm^{-2}= 4.5 x 10^{8}Atmospheres:The Virial theoremNow, take the two equations of hydrostatic equilibrium and mass conservation and divide them:

[dP(r)/dr]/[dM(r)/dr] ≡ dP/dM = - GM/4πr

^{4}Now multiply both sides via 4πr

^{2}and we get:4πr

^{8}dP = - (GM/r) dMTaking integration over the whole star:

3

_{Po}∫^{Pc}VdP = -_{o}∫^{Mo }(GM/r) dMHere V = volume contained in radius 'r'

By using the integration by parts, integration of the LHS of equation above leads to:

3[PV]

_{o}^{c}- 3_{Vo}∫^{Vc}PdV = -_{o}∫^{Mo }GM/r dMAt centre, V

_{C}= 0 and on the surface P_{s}= 0Therefore we have:

3

_{o}∫^{V}PdV +_{ o}∫^{Mo}GM/r dM = 0At centre, V

_{c}= 0 and on the surface P_{s}= 0Therefore we get:

3

_{o}∫^{Vo}PdV +_{o}∫^{Mo}GM/r dM = 0Now the right hand term equivalent to total gravitation potential energy of the star or it is the energy liberated in making the star from its components dispersed to the infinity.

:Mean Temperature of a StarWe have noticed that the pressure 'P' is a significant term in the equation of hydrostatic equilibrium and the viral theorem. We have derived a minimum value for the central pressure (P

_{c}= 4.5 x 10^{8}atmos)What are the physical procedures which give rise to this pressure and which are the most significant?

a) Gas pressure P

_{g}b) Radiation pressure P

_{r}We should illustrate that P

_{r }is negligible in the stellar interior and pressure is dominated through P_{g}To do this we initially require estimating the minimum mean temperature of a star. Consider the Ω term that is the gravitational potential energy:

- Ω =

_{ o}∫^{Mo}(GM/r) dMWe can get a lower bound on the RHS by noting: that at all points within the star r < r

_{s}Therefore, 1/r > 1/r

_{s}=>

_{o}∫^{Mo }(GM/r) dM >_{o}∫^{Mc}GM/r dM = GM_{o}/2r_{o}Now dM = PdV and the Virial theorem can be written as:

- Ω = 3

_{o}∫^{Mc}PdV = 3_{o}∫^{Mo}(P/p) dMNow, pressure is the sum of radiation pressure and gas pressure: P = P

_{g }+ P_{r}Suppose, for now, that stars are comprised of ideal gas having negligible P

_{r}P = nKT = kpT/m

The equation of state of ideal gas:

Here,

n = Number of particles per m

^{3}m = Amount of mass of particles

k = Boltzmann's constant

Therefore we obtain:

- Ω = 3

_{o}∫^{mo}(P/p) dm = 3_{o}∫^{mo}(kT/m) dmAnd we might make use of the inequality derived above to represent:

- Ω = 3

_{o}∫^{mo}(kmT/m) dm > Gm_{o}^{2}/2r_{o}=>

_{o}∫^{mo }TdM > GM_{o}^{2}m/6kr_{o}We can consider the LHS as the sum of the temperature of the entire mass elements dM that make up the star. The mean temperature of the star T‾ is then just the integral divided by the total mass of the star M

_{s}=> m

_{o}. T‾ =_{o}∫^{mo}TdMT‾ > GM

_{o}m/6kr_{o}:Minimum mean temperature of the SunAs an illustration for the Sun we obtain:

T‾ > 4 x 10

^{6}(m/m_{H}) kHere m

_{H}= 1.67 x 10^{-27}KgAs hydrogen is the most plentiful element in stars and for a completely ionized Hydrogen star (m/m

_{H}) = 1/2(Since, these are two particles, n + e for each and every H atom). And for any other element: m/m

_{H}is greater.=> T‾ > 2 x 10

^{6}K:Physical State of Stellar MaterialWe can as well find out the mean density of the Sun by using the given equation:

P

_{av }= 3M_{o}/4πr_{o}^{2 }= 1.4 x 10^{3}Kgm^{-3}The mean density of sun is just a little higher than water and other ordinary liquids. We are familiar such liquids become gaseous at 'T' much lower than 'T

_{o}'. As well the average K.E of the particles at T_{o}is much higher than the ionization potential of Hydrogen. Therefore the gas should be highly ionized, that is, plasma. This can therefore withstand greater compression with no deviation from an ideal gas. It will be noted that an ideal gas demands that the distances between the particles are much greater than their sizes, and nuclear dimension is 10^{-13}m as compared to the atomic dimension of around 10^{-10}m.Now, revisit the issue of radiation as gas pressure we supposed that the radiation pressure was negligible. The pressure applied by photons on the particles in a gas is as follows:

P

_{rad }= aT^{4}/3Now take gas pressure and radiation pressure at a typical point in the sun

P/P

_{o}= (aT^{4}/3)/kTP/m = maT^{3}/3kPAssuming, T ≈ T

_{o}= 2 x 10^{6}K, P - P_{o}≈ 1.4 x 10^{3}Kgm^{-3}And m = (1.67 x 10

^{-27})/2 KgThis gives P

_{o}/P_{a }≈ 10^{-5}Therefore, radiation pressure appears to be negligible at a typical (or average) point in the Sun.

In conclusion, without knowledge of how energy is produced in stars we have been capable to derive a value for the internal temperature of sun and illustrate that it is comprised of near ideal gas plasma having negligible radiation pressure.

:Mass Dependency of Radiation on Gas PressureThough we should later notice that the P

_{r }does become significant in the higher mass stars. To provide a fundamental idea of this dependency we substitute 'P' in the ratio equation above:P

_{r}/P_{g}= (maT^{3})/[3k(3M_{s}/4πr_{s}^{3})] = 4πmar_{s}^{3}T^{3}/gkM_{s}And from the Virial theorem: T ≈ M

_{s}/r_{s}=> P

_{r}/P_{g}α m_{o}^{2}That is, P

_{r}becomes more important in higher mass stars.:Energy Generation in StarsSuppose the origin of the energy that is, the conversion of energy from some form in which it is not instantly available to some form which it can radiate. How much energy does the sun require to produce in order to shine by its measured flux?

l

_{o}= 4 x 10^{25}, ω = 4 x 10^{25}Js^{-1}The Sun has not modified flux in 10

^{9}yr (3 x 10^{6}s)=> Sun has radiated 1.2 x 10

^{10 }JE = mc

^{2}=> m = 10

^{26}Kg = 10^{-4}moThe main sources of this energy are as: (a) Cooling or contraction (b) Chemical (c) Nuclear reaction.

Cooling and Contraction:

These are closely correlated. Therefore we take them altogether. Assume that the irradiative energy of sun is due to the sun being much hotter if it was formed, and has since been cooling down. We can test how conceivable this is. Or is the sun slowly contracting by subsequent discharge of gravitational potential energy that is transformed to radiation?

For an ideal gas, the thermal energy of a particle (here n

_{s}= number of degrees of freedom that is three (3)) is.Total thermal energy per unit volume = 3KnT/2

Now, according to the Virial theorem:

3

_{o}∫^{Vo}PdV + Ω = 0Supposing that the stellar material is an ideal gas (having negligible P

_{r})=> P = nKT

3

_{o}∫^{Vo}nKTdV + Ω = 0Now let us define U = integral over volume of the thermal energy per unit volume.

Thermal energy per unit volume = 3knT/2

=> 2u + Ω = 0

The negative gravitational energy of a star is equivalent to two times its thermal energy. This implies that the time for which the current thermal energy of the sun can supply its radiation and the time for which the past discharge of gravitational potential energy could have supplied its current rate of radiation vary via merely a factor of two. Negative gravitational potential energy of a star is represented by:

Ω > G M

_{s}^{2}/2r_{s}Ω ≈ G M

_{s}^{2}/2r_{s}Net release of gravitational potential energy would have been adequate to give radiant energy at a rate given by the luminosity of the stars, for a time represented by:

t

_{s}≈ GM_{o}^{2}/L_{o}P_{o}Chemical reaction:

We can promptly rule these out as possible energy sources for the sun. We computed above that we require determining a procedure which can generate at least 10

^{-4}of the rest mass energy of the sun. The chemical reactions such as the combustion of fossil fuels release ≈ 5 x 10^{-10}of the rest mass energy of the fuel.Nuclear reaction:

The only identified method of producing adequately huge amounts of energy to fuel the stars is via nuclear reactions. There are two kinds of nuclear reaction, fission reaction and fusion reaction. The fission reaction, like those which take place in nuclear reactors or atomic weapons can discharge ≈ 5 x 10

^{-4}of rest mass energy via fission of heavy nuclei (that is, Uranium or Plutonium).:Equation of Energy ProductionThe third equation of stellar structure sets up the relationship between the energy discharge and the rate of energy transport.

Suppose a spherically symmetric star as illustrated above in which the time variations are insignificant.

L(r) = Rate of energy flow across the sphere of radius 'r'

L(r + ∂r) = Rate of energy flow across sphere of radius r + ∂r

As the spherical shell is thin:

∂V(r) = 4πr

^{2}∂rAnd ∂m(r) = 4πr

^{2}p(r) ∂rWe can state:

ε = Energy discharge per unit mass per unit volume (WKg

^{-1})Therefore, the energy discharge in the spherical shell is represented as:

4πr

^{2}p(r)∂rεThe conservation of energy leads us to:

L(r + ∂r) = L(r) + 4πr

^{2}p(r) ∂rε=> [L(r + ∂r)/∂r] - L(r) = 4πr

^{2}p(r)∂rεAnd for ∂r → 0

∂L(r)/∂r = 4πr

^{2}p(r)εIt is the equation of energy production in the star. We now encompass three of the equations of stellar structure; though there are five unknown and these are: P(r), M(r), L(r), p(r), ε(r). In order find out them; we require considering energy transport in stars.

:Methods of Energy TransportThere are three modes energy can be transported in stars:

1) Convection: This is the energy transport via mass motions of stellar gas.

2) Conduction: This is energy exchanged throughout the collisions of stellar gas particles.

3) Radiation: Energy transport via the emission and absorption of the photons.

Conduction and radiation are alike methods as they both involve transfer of energy via direction interaction, either between particles or between the photons and particles.

Energy earned through a typical particle E = 3KT/2 is comparable to energy taken through a typical photon. This is represented as: E = hc/λ

However the number density of particles is much bigger than that of photons that would imply conduction is more significant than radiation in the stellar energy transport.

Convection is basically the mass motion of gas elements. This merely takes place if the temperature gradient exceeds certain critical value.

Now, consider these two suppositions:

a) The element increases adiabatically.

b) The element increases at a speed much less than the speed of sound.

For adiabatic method:

PV

^{γ}= constantHere, γ = C

_{p}/C_{v}is specific heat at constant pressure, divided by the specific heat at constant volume.Given that 'V' is inversely proportional to 'P', we can represent:

P/p = constant

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