Stellar Structure And Evolution: Part 2: Basic Assumptions and Accuracy of Assumptions

In the previous blog I explained why do we need a theory of Stellar Structure and Evolution when we can get information about stars by just observing them. In this blog I will cover-

1. The basic assumptions of theory of Stellar Structure and Evolution

2. Accuracy of Assumptions of theory of Stellar Structure and Evolution

To understand above mentioned aims more clearly, let's first understand( answer these questions - that I generally don't see many writers stress about:

What are the assumptions for a theory ?

Assumptions are basically the foundation stones for a theory. These are taken as the postulates that are generally assumed to be true throughout the explanation.

 

Now, What is the need for assumptions to a theory?

-First of all, because they are reasonable - they offer observational or at least mathematical verification.

-- Secondly, Making assumptions simplify our work in terms of crucial/critical understanding and mathematics.

- Lastly, although it is harsh but true: We only have limited information about things. Thus, as long as these are not true in general, we choose not to include them (esp while teaching) instead of including it with ad-hoc suppositions that lack any observational verification.

Now that we know what are assumptions and why do we need them for building a theory let's see what are the basic assumptions for the theory of Stellar Structure and Evolution

1. Stars are isolated in space - This is a fairly reasonable assumption for most of the single stars in galaxies as this condition is satisfied to a high degree - compare the distance of sun to its nearest star Proxima Centauri. We are ignoring binary stars and stars in dense clusters.

2.Stars are formed with a homogeneous composition- it is again reasonable as the clouds from which stars are formed are well - mixed.

3.Stars have no magnetic field - This is fully reasonable as for most of the stars magnetism plays a notable role only in phenomena related to surface of stars but in overall life cycle they don't play any significant role.

Stellar evolution is fully determined by internal physical processes which take deep inside the star near it's core.

4. Stars are rotating slowly- This one is a lot harder to justify as most of the stars rotate at a considerable fraction of their critical velocity(*1). Since we do not have a theory that shows how Stellar interior rotates at the birth of the star and making this assumption causes a huge mathematical simplification we are going to hold it.

5. Stars are in mechanical equilibrium(*2) : Majority of stars are in such long lived phases of their evolution that no structural changes are observed for them for most period of times. This implies that there is no noticeable acceleration and all the forces balance each other perfectly.

For an isolated, slowly rotating, homogeneous composition star with no magnetic field these forces are gravity and pressure. Thus, all the stars are in hydrostatic equilibrium.

All of the above mentioned assumptions can be tested just by testing for accuracy of hydrostatic and spherical symmetry assumptions.

Accuracy of Hydrostatic Assumption -

Let's first understand the equation of Hydrostatic support using simplest Newtonian dynamics

Balance between gravity and pressure is called hydrostatic equilibrium.

For a given time t, let's consider a spherical mass shell with infinitesimal thickness δr at a distance r from the centre. 

 

Mass of the element δm at this distance is δm= ρ(r)δs δr

ρ(r) = density at radius r

Outward force = pressure exerted by stellar material on lower surface

P(r)δs

Inward force on mass element= Pressure exerted by stellar material on upper surface and gravitational attraction of all stellar material lying within r

P(r+δr)δs+ GM(r)δm/r² = P(r+δr)δs+ GM(r)ρ(r)δs δr/r²

For hydrostatic equilibrium, inward force= outward force

P(r)δs=P(r+δr)δs+ GM(r)ρ(r)δs δr/r²

so, P(r+δr)-P(r)= GM(r)ρ(r)δr/r²

 For infinitesimal element:

P(r+δr)-P(r)/δr = dP(r)/dr

Thus, dP(r)/dr=-GM(r)ρ(r)/r²

which is the equation of Hydrostatic support.

Accuracy of hydrostatic assumption

To answer how valid is that assumption let's consider a situation where inward force and outward force aren't equal which gives rise to acceleration a.

P(r+δr)δs+GM(r)ρ(r)δs δr/r² -P(r)δs = ρ(r) δs δr a

»dP(r)/dr +GM(r)ρ(r)= ρ(r) a

 

this is the generalized form of equation of hydrostatic support.

Now consider there is a resultant force on element with the sum being a small fraction of gravitational term(β)

Inward acceleration a = β.g

Spatial displacement from rest after time t = d= 0.5.a.t² =0.5.β.g.t²

If if allow star to collapse or expand, by setting d=r we obtain

t=(2.r³/G.M.β)^1/2

Assuming beta =1 we obtain

t=(2.r³/G.M)^1/2

This is the dynamical time scale of the star.

Of course each mass shell will be accelerated at different rate so this should be taken as an average value for star to collapse at radius R.

Since average density is we can also write this t to be

½.(G.ρ)^1/2

For sun we obtain a value of 1600 sec or about half an hour. Thus, any significant departure from hydrostatic equilibrium should lead very quickly to an observable phenomenon (sudden collapse or explosion of the star ) . But age of sun is already

This is much smaller than the age of sun - 10^17 secs - by 14 orders of magnitude. Thus if this assumption have been wrong we would have noticed a significant collapse or explosion of sun much earlier.

This assumption is very much accurate.

Accuracy of spherical symmetry assumption

Let's see if average rotation rate of stars is causing significant departure from spherical symmetry.

Consider a star of mass M and radius R with an element of mass δm near the surface of the star.

star of mass M and radius R with an element of mass δm near the surface rotating at angular speed w

Gravity supplies the extra centripetal force to make the object move around in a circular path.

Thus, for mass δm, centripetal force = gravitational force

There will be a departure from spherical symmetry if there will be any difference between gravitational and centripetal force i.e.

(δmω²r)/(GMδm/r²) <<1

or ω²<<GM/r³

note RHS of last equation is similar to t

>> ω²<<2/t² (ω=2π/T) where T =rotation period

 for spherical symmetry to be valid T>>t

for example, for sun t~2000 sec and T~21 month

Thus, for majority of stars, departures from spherical symmetry can be ignored.

With this we complete the basic assumptions of theory of Stellar Structure and Evolution and see that these assumptions are qualified to build a theory of Stellar Structure and Evolution.

 

 

*1.Definition of Critical velocity : Stars have a maximum speed at which they can spin. If stars exceed the critical rotation, the outward force caused by their spinning will overcome the inward gravitational force that keeps the star together.

If stars get to that limit, they will begin to fly apart.

*2. Mechanical equilibrium : A state of rest or unaccelerated motion in which sum of all the forces acting on a particle is 0. In case of stars, this state is reached when pressure forces are balanced by gravity. In astronomy, this is called hydrostatic equilibrium.

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Stellar Structure and Evolution Part : 1

Aim of this Blog is to explain why we need a theory of Stellar Structure and Evolution- A war for need of theory over observation.

Before coming to the central question let us look at the answers to some simple looking but very crucial questions:

Q1. What is meant by Stellar Structure and Evolution?

 To understand structure and evolution of stars using laws of physics.

Q2. What are stars ? 

A star is an object that - 

1. Radiates energy from an internal source 

2. It is bound by its own gravity 

3. Star should evolve ? But why ? 

Stars are born within the clouds of dust scattered throughout most galaxies. 

Evolution is the change in properties of star with time. In stars it occurs due to burning of fuel to balance the forces of gas and pressure.  This evolution is highly dependent on mass of stars - On average, Greater the mass shorter it's life

Now coming to the central question -

What is the need for theory for stellar evolution when we can have so much information by just observation of Stars ?

To answer this let's understand what all we can gather by the known observational techniques that we use to study stars -

1. Photometric measurements (Photometry, in astronomy, the measurement of the brightness of stars and other celestial objects) yield the apparent brightness of a star, i.e. the energy flux ( f )  received on Earth, in different wavelength bands.

(I have covered more of the terminologies here ) 

2. Distances to nearby stars can be measured with the help of parallax. As Hipparcos satellite has measured parallaxes with 1 milliarcsec accuracy of more than 105 stars.

                                         

3. Spectroscopy (Spectroscopy is the study of the interaction between matter and electromagnetic radiation as a function of the wavelength or frequency of the radiation.) at sufficiently high resolution gives detailed information about the physical conditions in the atmosphere. With detailed spectral-line analysis using stellar atmosphere models one can determine the photospheric properties of a star ( like effective temperature, surface gravity , rotation velocity etc.)

 

4.Mass of stars -one of its most fundamental properties can 'only' be measured indirectly by using binary stars (spectroscopic binaries)

Spectroscopic binaries

As you might have noticed, all of these( Mass, Temperature , Rotation Velocity, Distance etc.)  are only surface properties. Thus, we need to build a theory of Stellar Evolution to derive internal properties of stars as observational techniques seem to fail in that !?

Well Game isn't over yet ! Observation always provides astronomers  a window to interior of stars like -

1. Neutrinos: which escape from interior of stars without any interaction. But neutrinos interact little with matter regardless of energy. Moreover, beyond certain temperature, there is a decrease in relative flux of neutrinos

2. Oscillations: Yes, I am talking about Seismology here. Stars are musical instruments. You can refer  my blogpost on Helioseismology  to know more. Here is a brief -  The surfaces of stars oscillate with a particular time period and this can give us valuable information about size, age and mass of stars

Why we need a theory for stellar structure and evolution when we can just decode information from observation? 

It is true that Astronomical observations can yield information about important stellar parameters. But these are like snapshots of the life of star as timespan of these observations is much smaller than the age of stars. Thus, any of these observations cannot give us a complete picture of Stellar Evolution. 

Moreover, a theory is also need to explain some of the most important results of Astronomy such as  mass - luminosity relationship and mass - radius relationship that we get from HR Diagram of stars. (I am going to cover HR Diagram in my future Blogs so don't worry about that :) 

Thus overall, we see that the observational techniques we use cannot provide us with 'all' of the necessary information about stars.

Thus a theory is needed to explain Stellar evolution and results of Stellar Observation.

In the next blog I will cover -

The basic assumptions of theory of Stellar Structure and Evolution

Accuracy of Assumptions of theory of Stellar Structure and Evolution

stellar evolution

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How to destroy a Cluster of Stars/ Globular Cluster

 This blog includes - 

1.1 What are Globular clusters ?

1.2 What are Binary Stars ?

1.3 What is escape velocity ?

1.4 How to Destroy a Globular Cluster ?

1.5 Summary of Topic 

In this blog we will destroy a Cluster of Stars called Globular Cluster!

The main focus of this Blog is to make you understand how you can destroy a Globular cluster for which we will run from 1.1 to 1.3 to gather all the material to understand the topic. If you know about all of this already you can directly skip to section 1.4 to know the real tea 🍵

1.1) What are Globular clusters ? 

The name is derived by Latin word - Globulus - which means a small sphere (they are really big though- approximate size of a Globular cluster is - 300 light years) 

Globular clusters are a group of stars all formed at approximately same time and held together by gravity. ( Fig 1) They are called 'museums of stars' as they have held the stars intact since they were formed and thus help astronomers in studying age and properties of various stars( Bonus : As the stars in clusters are all formed at same time, the globular clusters help in comparing the properties of different mass stars formed at approximately same time, which is an excellent opportunity for astronomers ! ) 

fig 1

GLOBULAR CLUSTER

Although they look dense, collisions between stars in globular clusters are very rare as distances between individual stars are greater than approximate distance between stars. Moreover,  velocities of stars at any point inside the cluster are fairly random, in direction as well as size. The stars form what is called a 'collision less gas'.

Soon we will see how to boil away or destroy these museums of stars 😈

1.2) What are Binary Stars ? 

A binary system consists of two stars orbiting each other around a common centre of mass.

BINARY STARS

Given the stars in a binary system are themselves spherical, both of their orbits will be ellipses, just as for planets in the Solar System ( Bonus : for a spherical star system, the gravitational field is as such as if all of the mass is concentrated in center, when this is not the case - like when stars come close to each other tidal forces might deform the shape and resultant orbits will be complex then )

Escape velocity from a binary system : 

Let's consider a binary system of two stars orbiting around a common center of mass as shown below

Two binary stars (in purple) in orbit around a common center of mass

For simplicity, let's consider orbits to be circular and masses of both stars to be equal = M

Radius of circle = R 

In order for one of star to escape from gravitational field of other star, the kinetic energy of star should become equal the gravitational potential of the other star.

(1/2)*m*v^2 = (G*M*M)/(2*R)  (Kinetic energy = potential energy)

So, v= sqrt(GM/R)

Therefore, energy of 1 star when it has this speed

is 1/2 * m *v^2 = (G*M*M)/(2*R)   ------------ (1) 

We need to learn only this much about binary stars to understand today's topic :)

1.3) What is escape velocity ? 

The minimum velocity which is needed by a non- propelled body to escape from gravitational field of another object is called it's escape velocity. 

Formula - sqrt( 2*GM/ R ) where,  ----------(2)

G - Gravitational Constant = 6.674 x10−11 m3⋅kg−1⋅s−2

M = Mass of object which is to be escaped 

R = Radius of object which is to be escaped

Now that we are equipped with all the necessary tools to destroy a Globular cluster let's dive into it ! 

1.4) How to destroy a Globular cluster?

Let's understand what is meant by destruction of a Globular cluster. It means kicking out every star from the cluster so that it no longer remains a Globular cluster. 

Here's is the recipe, let's do it together 

Step 1. Let's calculate how much energy is needed to kick out stars from a Globular cluster?

 To do this let's take a sample cluster. 

* Mass of cluster Mcl      = Mass of 10^(6) stars, where avg. mass of 1 star = 1 Mo ( 1 solar mass ) 

* Radius of Cluster Rcl   = 50 parsecs ( pc),where 1 parsec = 3.086e+13 km 

* Escape velocity for a single star =  v_escape = sqrt( 2*G*Mcl/ Rcl ) ~ 13 km/s  ----from(2)

* Escape energy for a single star, 

   E_escape = 0.5*mass of 1 star * (v_escape)^2

                   = 2*(10^38) Joules

* Escape or destruction energy for whole cluster! 

   E_escapecluster = 

   Escape energy for a single star × total number of stars =2 × 10^38 × 10^6 

                                                                                          = 2 × 10^44 Joules

To destroy whole cluster, we need to provide 2 × 10^44 Joules energy to it. But how can we do it ?

Answer: Step 2: By forming binary pairs of stars!

Formation of a binary pair releases energy, as the two stars become bound. The energy required to form a binary pair is the same that is required to split it again. 

Thus, From (1) we know that energy released when a binary pair is formed is - (G*M*M)/(2*R)

 Suppose the stars each have one solar mass and are separated by 

the radius of a white dwarf, about 5 × 10^6 m.  Formation of such a system releases energy equivalent to 3 × 10^43 Joules.( using (1) )

Wow! only 10% of the binding energy of the cluster is provided by only one binary system. The formation of a handful of such systems could easily provide enough energy to expand the cluster or even disrupt it!!!

The energy released in this way goes to the third star by Newton's third law ( Newton's Third Law: every action has equal and opposite reaction)  and the third star now has so much energy that it simply shoots straight out of the cluster. ( This is not out of some magic, when two stars form a binary pair energy is released in a same way it is released when a particle is falling onto the star. If third star were not present, they couldn't form a binary pair ) - see the last section of this Blog ( * )to understand more accurately how the energy is actually transferred :)

Finally! We have successfully destroyed the Globular Cluster

Summary of Topic 

Globular clusters are a group of stars all formed at approximately same time and held together by gravity.

Destruction of cluster is possible by formation of a handful of binary pairs. The energy released in this way goes to the third star by Newton's third law and the third star now has so much energy that it simply shoots straight out of the cluster.

Thus one by one, shooting the stars out we can completely destroy the Globular Cluster !

___________________________________________________

                                     ( *)

(Let's take 3 stars in a Globular cluster with masses say 1, 2 and 3 solar mass. We arrange them in mercury - sun distance and give them some small initial velocity so that no star has energy to escape the three - body system. After some time, two stars become bound and the third one is expelled. 

Now how does that happen? 

The answer is in gravitational energy.When two stars become more tightly bound, they release energy, just as a particle falling onto a star releases energy . This energy goes into the motion of the third star by Newton’s third law which says that every action has equal and opposite reaction. By this law, the force exerted on the pair by the third star is exerted back on the third star by the pair: which results in binding the pair more tightly and expelling the third. If the third star were not present, the first two could not form a tight binary pair: they would fall towards one another and then recede to the same distance. )

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Dark Matter Slowing Down Milky Way

Topics of this Blog include- 

1.1- A brief about Milky Way Spiral Galaxy and Dark Matter 

1.2- Details about Topic -Dark Matter Slowing down Milky Way

1.3 - Conclusion and Learnings from this discovery 

1.4 - Summary of Topic

1.1) A brief about Milky Way galaxy and Dark Matter

       Milky Way Galaxy  

Milky way ( or Akashganga ) is our home galaxy. Approximately two-thirds of all spiral galaxies contain a bar, so does our galaxy - It is a barred spiral galaxy which means that it has a central bar with spiral body structure ( Fig. 1and 2) 

Fig 1

Fig 2

                                                                      

This bar-shaped core region is surrounded by disk of gas, dust and stars

       Dark matter 

- To know more about dark matter you can refer to this link : Dark matter 

Dark matter constitutes 27 % of content of whole universe. It is made up of 'something' that cannot be detected by electromagnetic force ( basically by touch or sight ). It's gravitational effects are responsible for flattening of velocity vs radius curves in the outer regions of galaxies.  According to theories It forms a spherical halo around the galaxy. 

1.2) Details about Topic - How dark matter is slowing down Milky Way

"Astrophysicists have long suspected that the spinning bar at the center of our galaxy is slowing down, but we have found the first evidence of this happening," study co-author Ralph Schoenrich, in a statement.

Galaxy models have long predicted that galactic bars slow down by losing angular momentum to their postulated dark haloes.

Hercules stream is a group of stars currently rotating away from center of Milky Way. This stream is in resonance with the central bar - i.e. it is gravitationally trapped by the central bar and thus, revolves around at the same rate as bar's spin ( like Jupiter's Trojan ) 

If the bar's rotation slows, the stars in the stream move outward to have orbit in sync with rotation of bar. Researchers investigated the chemical composition of stars in Hercules stream and found them to be rich in heavier elements which is possible only when they have been formed close to the center of Galaxy.

Thus they have been sweeping out of the center of galaxy which implies that galactic rotation has been slowing down and has slowed down by 24%.

 The one possible and seemingly correct explanation is that dark matter has been slowing down the spin by dynamical friction. Loss of the bar's angular momentum has been attributed to dark matter.

This possible explanation, although seeming the most correct till now has run into rivals.  Schoenrich said "Our finding also poses a major problem for alternative gravity theories — as they lack dark matter in the halo, they predict no, or significantly too little slowing of the bar ". There are also alternating theories of gravity which propose tweaks in Einstein's relativity and disregard dark matter possibility. But Einstein's theory has passed all of the tests till now ( Our GPS works on basis of that, the famous black hole picture that we have obtained is also consistent with his theory ) 

1.3) Conclusion and Learnings from this discovery 

• This model supports the presence of Dark Matter, while physical detection is yet to be done.

•It again put a shade on alternating theory of gravity and dark matter. ( Like MOND )

• We have got new constraints on galactic history and the unique opportunity to differentiate between different dark matter models.

• Due to position of  Sun’s position far from the center we currently see only the outer region of the resonance of spin of bar and Hercules stream . By performing the analysis at a spatial coordinate closer to the Lagrange points, we could probe deeper into the inner region of the resonance, where we may find traces of events that happened in the early epoch of bar formation and also determine the size of the initial core of the resonance which stems from the formation of the bar. However author mentions that this will be possible in the future with extended data covering the full range of resonance and a proper chemo-dynamical model predicting the age-dependent effects. 

1.4) Summary of Topic 

Analysing data from Gaia, a European Space Agency Mission to map position of stars in milky way, researchers at University College London (UCL) and the University of Oxford have shown that the spin of the Milky Way's galactic bar has slowed by about 24 % or a quarter since its formation. Most plausible explanation is Dark matter acts as a counterweight slowing the spin. 

Also Check :

Milky way : https://en.m.wikipedia.org/wiki/Milky_Way

Dark matter: https://en.wikipedia.org/wiki/Dark_matter

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