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Life Cycles of Stars

Black Holes


In the last quiz, we looked at white dwarfs—high-density objects supported by electron degeneracy pressure. At a high enough mass, this pressure can be overcome, causing the stellar remnant to collapse into an even denser neutron star, supported by neutron degeneracy pressure.

Can we eventually overcome neutron degeneracy pressure as well? It turns out that we can. If a stellar remnant has a mass of more than about \(3\) solar masses (corresponding to a main-sequence star of \(30\) solar masses or more as a result of mass shedding), its gravity will be so strong that it will collapse down to a singularity—an object believed to have no volume at all.

Such an object is a black hole. In this quiz, we first will understand why these tiny dense objects were first believed to be black. We will see that, in truth, they are not, and along the way, we will figure out what observational signatures they leave for astronomers to detect.

A black hole is a stellar remnant so massive that degeneracy pressure cannot support the surface, and consequently, it shrinks down to nearly nothing. To start, we will work out the gravitational potential energy of an object on its surface.

Let's begin somewhere more familiar. On the surface of Earth, things have gravitational potential energy that we can calculate: \[U_g=\frac{-GM_\text{Earth}m_\text{body}}{R_\text{Earth}},\] where \(m_\text{body}\) is a mass.

If the radius of Earth were smaller, its density would be higher because its mass would be compressed into a smaller volume. How would the potential energy of your body be different on a denser Earth?

Gravitational potential energy is conventionally negative to remind us that a body's gravity may trap nearby objects. However, if you can gain enough speed, you can leave the influence of a star or planet (without external propulsion). This is called gravitational escape.

If we were to launch ourselves from Earth's surface with speed \(v\), our total energy would be the sum of our kinetic energy \(\frac12 m_\text{body} v^2\) plus our (negative) gravitational potential energy.

As in the Star Formation quiz, if \(E_\text{tot} \gt 0\), we can escape Earth's gravity (ignoring the effects of air resistance). If \(E_\text{tot} \lt 0\), we fall back to Earth. Thus, we define the escape speed \(v_\text{esc}\) as the smallest possible speed a body would need to leave Earth's gravitational influence without external propulsion.

By setting total energy equal to zero, derive an expression for \(v_\text{esc}.\) Which of the parameters in the choices does not appear in your expression?

The expression for escape speed we just derived applies to stars and stellar remnants, too. Our theoretical understanding of black holes is that they take up no volume—all the mass is concentrated at a single point called a singularity. (In the chapter Cosmology, we will explore a bit more of what this means.)

If you imagine the radius of a degenerate star collapsing from \(\SI{10}{\kilo\meter}\) to \(\SI{1}{\kilo\meter}\) to \(\SI{0.001}{\kilo\meter}\) and even smaller, what happens the escape speed as the radius approaches \(0?\)

Assume for concreteness that the mass of the degenerate star is \(3M_\text{Sun}\) and that it does not change during the collapse.

Since escape speed does not depend on an object's mass, not even light traveling at a finite speed of \(c\) can escape the singularity at the center of a sufficiently dense object. In fact, there is a region with radius \(R\) around the singularity where \(v_\text{esc} \gt c\). Everything within this region is invisible to us since light emitted or reflected within this region cannot escape. This argument led astrophysicists to believe black holes would be entirely invisible.

Set escape speed equal to the speed of light \(c\) in the expression we derived to find the radius \(R\) of the region near a singularity from which no light can escape.

The distance you just calculated is called the Schwarzschild radius. Any object that strays within this radius of a black hole will never be seen or heard from again. The Schwarzschild radius is the location of the black hole's event horizon—the point of no return for ordinary matter.

What is the Schwarzschild radius of a black hole with mass \(5M_\text{Sun}?\)

Details and Assumptions:

  • Express your answer in \(\si{\kilo\meter}.\)
  • \(M_\text{Sun}=\SI{1.99e30}{\kilo\gram}.\)
  • \(c=\SI[per-mode=symbol]{3.0e8}{\meter\per\second}.\)
  • \(G=\SI[per-mode=symbol]{6.67e-11}{\meter\cubed\per\kilo\gram\per\second\squared}.\)

What lies beneath the event horizon is a speculative subject, but astronomers attempting to spot a black hole would be looking for observable signals from above the event horizon. Astronomers would be trying to detect radiation emitted by objects as they fall toward the Schwarzschild radius.

Consider a hydrogen atom with mass \(\SI{1.67e-27}{\kilo\gram}.\) How much potential energy does it lose as it falls to the Schwarzschild radius of the black hole we considered in the previous question?

Details and Assumptions:

  • The atom is so far away when it starts falling toward the black hole that its initial potential energy is effectively \(0.\)
  • Express the lost energy as a positive number \((\)i.e. the absolute value of \(\Delta U_g).\)
  • The mass of the black hole is \(5M_\text{Sun}\approx \SI{10^{31}}{\kilo\gram}.\)

As the atom falls toward the black hole, according to the virial theorem, half of its energy is radiated away, while the other half is retained by the atom as kinetic energy.

If we imagine the atom is part of a cloud of gas with temperature \(T\), find the temperature of the particles that have fallen to this radius, and hence the wavelength of radiation they give off as they reach the event horizon.

What type of electromagnetic radiation is emitted?

Details and Assumptions:

  • The average energy of a gas is related to temperature by \(E = \frac{3 k_B T}2.\)
  • Wien's law states \(\lambda_\text{max} T = \SI{0.0029}{\meter\cdot\kelvin}.\)
  • Use the above diagram to identify the type of radiation.

The highest-energy radiation is emitted close to the event horizon, but objects falling toward a black hole emit continuously along their trajectory. The material can form an accretion disk, a huge disk of glowing matter orbiting the black hole. Friction and collisions between particles in this disk cause the material to fall inward and consequently radiate energy.

The total amount of energy emitted depends on the rate at which mass crosses the event horizon. Suppose our \(5M_\text{Sun}\) black hole is swallowed one Earth-mass each year. Estimate its luminosity.

Details and Assumptions:

  • By the virial theorem, assume half of the potential energy lost during its fall is radiated.
  • \(M_\text{Earth} = \SI{5.97e24}{\kilo\gram}.\)
  • Express your answer as a multiple of the Sun's luminosity \(L_\text{Sun}=\SI{3.83e26}{\watt}.\)
  • Although accretion disks can be millions of times larger than Earth, assume the mass entering the disk initially has zero potential energy.

While this is a huge power output, we have seen even bigger ones out there. Some of the brightest objects in the universe are known as quasars and are thought to be formed of enormous super-massive black holes surrounded by accretion disks, swallowing mass equivalent to hundreds of Sun-like stars every year. The brightest quasars are thousands of times more luminous than our entire galaxy.

Residing a mere \(26000\) light-years from Earth is the nearest super-massive black hole. Deep in our galaxy's past, our local black hole may have been swallowing matter in the way we've seen in distant quasars, but for now it's just lurking in the center of our galaxy. Astronomers now believe that every large galaxy has at its center a giant black hole of this kind.

As one of the universe's most extreme and exotic objects, black holes are subjects of intense interest and research. To study them in more detail requires use of Einstein's theory of general relativity to map their warping of space and time.


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