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Astronomy

Life Cycles of Stars

Stellar Remnants

                     

After the helium flash, a star like the Sun flickers like the thinning filament of an incandescent bulb. Its radius and luminosity fluctuate over periods of \(10\) million years, as solar winds blow loosely bound gas and dust away from its surface. Nearer to the core, Hydrogen and Helium fusion proceeds in shells until the star's outer envelope is sparsely distributed throughout the solar system.

As the Sun breathes its dying breath and the last flicker of fusion dims, about half of its original mass is pressed into its hot, dense core, now containing a plasma of mostly carbon and oxygen. No more fusion is taking place, but with temperatures greater than \(\SI{10^8}{\kelvin}\) these objects--stellar remnants known as a white dwarf--are common, visible objects in our galaxy.

This quiz is focused on stellar remnants.

White dwarf stars are extremely dense mixtures of carbon nuclei, oxygen nuclei, and electrons. Suppose the mass of a white dwarf is \(0.5 M_\text{Sun}.\) From an HR diagram, we can calculate that the radius of such a star is similar to Earth's radius, \(R_\text{Earth}=\SI{6370}{\kilo\meter}.\)

Steel is the densest, relatively common material humans use on Earth. The density of solid steel is \(\SI[per-mode=symbol]{8.05e3}{\kilo\gram\per\meter\cubed}\) . About how many times denser than steel is the average density of a white dwarf?

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Details and Assumptions:

  • Assume the white dwarf is spherical, with volume \(V=\frac43 \pi R_\text{Earth}^3.\)
  • The mass of the Sun is \(\SI{1.99e30}{\kilo\gram}.\)

A white dwarf is truly an alien environment. The density of matter is so high that the buffer around nuclei that is usually reserved for electrons is penetrated. Principles of physics that we would not normally worry about when dealing with ordinary matter start to emerge. In fact, within a white dwarf, quantum mechanics begins to rub up against Einstein's principle of relativity.


Quantum effects usually show their face in electrons first, since electrons are fundamental particles (i.e. not made of smaller particles). We will first work out in a white dwarf how close the electrons are. This is related to their number density—the number of electrons per volume. The higher the number density, the closer the electrons are on average.

How many electrons, on average, are in a \(\SI{1}{\meter\cubed}\) volume of matter in a white dwarf with total mass \(0.5M_\text{Sun}?\)

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Details and Assumptions:

  • Assume for simplicity that the white dwarf consists entirely of carbon nuclei, each with mass \(\SI{1.99e-26}{\kilo\gram}.\)
  • A carbon nucleus contains \(6\) positively charged protons. Assume that for each proton, the star has one negatively charged electron. Thus, the white dwarf has zero net charge (i.e. equal numbers of positive and negative charges).
  • The mass of an electron is \(\SI{9.11e-31}{\kilo\gram},\) which is insignificant compared to the mass of the nucleus.
  • \(M_\text{Sun}=\SI{1.99e30}{\kilo\gram}.\)

If at a moment in time, all the electrons within a cubic meter are evenly spaced on a grid, how far is any given electron from its nearest neighbors?

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Details and Assumptions:

  • While it would be nearly impossible to find the electrons in such an ordered configuration, visualizing the electrons on a grid allows us to estimate their average separation by utilizing the formula for volume of a cube with length \(n\) \((\)i.e. \(V=n^3).\)
  • Express your answer in angstroms \(\big(\SI{1}{\angstrom}=\SI{10^{-10}}{\meter}\big).\)

So far in this course, we have not appealed to quantum mechanics, except as an explanation of atomic spectra. In stellar remnants, the counter-intuitive properties of quantum particles are inescapable. In fact, stellar remnants stand as a pragmatic test of basic quantum principles.

An important condition on dynamics of matter particles is Heisenberg's uncertainty principle, which imposes a condition on the simultaneous information we can know about a particle. For example, if a particle is confined to a volume \((\Delta x)^3,\) then its momentum can only be measured roughly to within a range of \[\Delta p \sim \frac{h}{\Delta x},\] where \(h=\SI{6.63e-34}{\joule\second}\) is Planck's constant, the same mysterious quantity that appeared in the photon's energy.

Using the previous estimate for \(\Delta x\), estimate the momentum uncertainty \(\Delta p\) of an electron in a white dwarf.

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Note: Because momentum is related to an electron's speed \(v\), the uncertainty principle imposes a limit on how accurately we can know \(v\).

The momentum uncertainty you just estimated imposes an uncertainty on the speeds of electrons in the white dwarf. Because momentum \(p,\) speed \(v,\) and electron mass \(m_e=\SI{9.11e-31}{\kilo\gram}\) are related by \(p=m_e v,\) the speed of an electron can only be known up to \[\Delta v \sim \frac{\Delta p}{m_e} \approx \SI[per-mode=symbol]{10^8}{\meter\per\second}.\] In order to satisfy Heisenberg's uncertainty principle, electrons' speeds in a white dwarf are much larger than we would expect at a temperature of \(\SI{10^8}{\kelvin}.\) What effect does this have on the pressure?

You just found the source of degeneracy pressure in stellar remnants. This pressure stabilizes the collapse of stellar cores in stars like the Sun. After the collapse, the Sun's core will remain as a white dwarf, radiating leftover thermal energy for many billions of years until it cools to the background temperature of the universe.

However, more massive stars have a different fate. Electrons are only allowed to move as fast as the speed of light: \(c=\SI[per-mode=symbol]{3e8}{\meter\per\second}.\) If the density rises enough, the separation between electrons \(\Delta x\) can become so small that the uncertainty condition on \(\Delta p\) cannot be met. All of the available quantum mechanical electron states fill up, and the electron speeds approach the speed of light before the pressure can stabilize the collapse.

This catastrophic scenario is predicted for stellar remnants with mass \(M\gt 1.4M_\text{Sun}.\) Gravity overcomes electron degeneracy pressure, and electrons combine with protons to form neutrons, separated only by the separation of particles in an atomic nucleus. The resulting object is called a neutron star.

A neutron star may have a mass of \(2\) solar masses and a radius of just \(\SI{10}{\kilo\meter}\). A teaspoon measure has a volume of \(\SI{5}{\milli\liter}\), or \(\SI{5e-6}{\meter\cubed}\). What would be the mass of a teaspoonful of neutron star?

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Details and Assumptions:

  • Assume the neutron star is spherical. The volume of a sphere with radius \(R\) is \(\frac43 \pi R^3.\)
  • Assume that the neutron star has a uniform density.

When electron degeneracy pressure cannot support the collapsing core of a massive star, the matter is crunched into a neutron star, losing a whole lot of potential energy in the process.

How does the gravitational energy lost compare to the largest thermonuclear weapon ever detonated, Tsar Bomba?

Details and Hint:

  • Tsar Bomba's explosion released about \(\SI{2e17}{\joule}\) of energy \((\SI{50}{megatons}),\) the most humans have ever released in a localized event.
  • Use a mass of \(\SI{4e30}{\kilo\gram}\) (about 2 solar masses). The initial radius of the core when it is supported by electron degeneracy pressure is \(\SI{6500}{\kilo\meter}\) and it collapses to a radius of \(\SI{10}{\kilo\meter}.\)
  • The gravitational potential energy of a mass \(m\) a distance of \(R\) from the center of a mass \(M\) is \(U_g=-\frac{GmM}R.\)
  • Consider the energy released when a kilogram of matter is dropped to the surface of a neutron star with this mass and radius. If you built the neutron star up like this, how would the energy change with each kilogram you added?

An enormous amount of energy is released during the collapse in the previous question. Did we just discover a supernova?

The luminosity of a supernova decreases from its peak on the scale of months. Assume that this amount of energy is released as entirely visible radiation over the period of \(1\) month. How many times more luminous than the Sun would this collapsing star appear?

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Details and Assumptions:

  • The average luminosity is the total energy released \(\Delta E\) divided by the time: \(\frac{\Delta E}{\Delta t}.\)
  • The luminosity of the Sun is \(\SI{3.8e26}{\watt}.\)

What shines with more light than a trillion Suns? Answer: A supernova!

When the core of a highly massive star collapses at the end of the star's lifetime it forms a neutron star, and along the way, we get a Type II supernova. This is a little different from the Type Ia supernova we used as a standard candle in the distances chapter, but similarly dramatic.

Although Type II supernovae are tremendously bright, only a fraction of their energy is emitted as visible radiation. They also emit other electromagnetic waves, neutrinos, and crucially, heavy elements. Helium is formed in the normal lifetime of stars. Other lighter elements like oxygen and nitrogen, all the way up to iron, can be formed in the red giant stage. But elements above atomic number 26 can only be formed in supernovae. If you have ever held a piece of copper, silver, lead, or any of the other heavier elements, you have held the remnants of a supernova explosion.

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