Gravitational Radiation 3: GW151226, Encore from LIGO

gw151226-image
Figure 1: A filtered, reconstructed image of the gravitational wave detected by LIGO on December 26, 2015 (UTC).  The signal was generated from the merger of two black holes, of approximately 14.2 and 7.7 solar masses, roughly 1.4 billion light years from  Earth.  Figure taken from Abbott et al, Ref. 3.

LIGO (Laser Interferometer Gravitational-Wave Observatory) consists of two gargantuan Michelson-like interferometers, with arms 4 km long, located 3000 km apart in Livingston LA and Hanford WA.  Its first observing session began on September 12, 2015.  Incredibly, just two days later, it captured the fleeting signal (designated GW150914) produced by the merger of two inspiraling black holes more than 1 billion light years from Earth.  (Ref. 1)  That groundbreaking discovery, announced on February 11, 2016, ended a 60 year race to directly detect gravitational waves, and came 100 years after the phenomenon was first predicted by Albert Einstein.

On Christmas Day (in the USA), LIGO scientists detected a second event, GW151226 (Ref. 3), proving that the earlier discovery was not a fluke, and that black hole mergers are frequent events.  (Ref. 4,5)  Since January, LIGO’s sensitivity has been improved, and it will begin a new observing run in Autumn, 2016.  This time, the two US detectors will be joined by VIRGO, a similar instrument with arms 3 km long, located near Pisa, Italy.  The addition of VIRGO will greatly assist in localizing – by triangulation – the sources of gravitational radiation.  In the near future, when LIGO and VIRGO reach their full design sensitivities, they may be capable of detecting one or more black hole collisions daily!  Indeed, the 21st Century seems destined to become the era of gravitational wave astronomy.

In this brief post, we will use the mathematical treatment outlined in Ref. 2 to analyze the published data given in Ref. 3 for GW151226.  Our educational goal is to add to the collection of homework exercises related to gravitational waves that are suitable for first year physics students.  In this way, we hope to inspire instructors to include this exciting topic in their mechanics syllabi.

 

Analyzing GW151226

In Ref. 2, the chirp mass of the binary black hole system was defined as \large \mathfrak{M}=\frac{(m_{1}m_{2})^{3/5}}{(m_{1}+m_{2})^{1/5}}, where \large m_{1} and \large m_{2} are the masses of the black holes.  This quantity is important because it can be found directly from the time dependence of the gravitational wave’s (GW) frequency (Eqn. 10 of Ref. 2):

\large \mathfrak{M}=\frac{c^{3}}{G}\left ( \frac{5}{96}\: \pi ^{-8/3}f^{-11/3}\: \frac{df}{dt} \right ) .

(1)

Following Ref. 2, define \large A=\frac{c^{5}}{G^{5/3}}\: \frac{5}{96}\: \pi ^{-8/3}=5.45\times 10^{56} (SI units), and integrate Eqn. 1 over the time interval \large \Delta t=t_{2}-t_{1} to obtain

\large \mathfrak{M}^{5/3}\Delta t=A\int_{t_{1}}^{t^{_{2}}}f^{-11/3}df=-\frac{3}{8}Af^{-8/3}\mid _{t_{1}}^{t_{2}} .

(2)

1.  Referring to Fig. 1 above, at 0.80 s before the two black holes coalesce, the frequency of the GW was 39.2 Hz.  Later, 0.40 s before the merger, the frequency was 50.3 Hz. (Ref. 6)  Calculate the chirp mass and express your answer in terms of \large M_{Sun}.  (Ans: \large 9.7\: M_{Sun}.)

2.  Show that the total mass \large M=m_{1}+m_{2} of the binary system before coalescence was at least \large 22\: M_{Sun}.  (Hint: Let \large m_{2}=\alpha m_{1}.  Express \large M\large \mathfrak{M}, and \large M/\mathfrak{M} in terms of \large m_{1} and \large \alpha.  Show that the ratio is a minimum when \large \alpha=1.)

3.  The GW frequency at the moment of coalescence of the two black holes was 420 Hz.  (See Fig. 1.)  Recall from Ref. 2 that the orbital frequency of the binary system is half the frequency of the GW.   Use the total mass given in Ref. 3, \large M=22\: M_{Sun}, to find the distance between the bodies just before they merged.  (Ans: 120 km)

4.  Assume that one black hole was twice as massive as the other, i.e., \large m_{2}=2m_{1}, which is about what the LIGO team concluded.  If the orbits were circular, what were their radii just before coalescence?  What were their speeds? Ignore relativity.   (Ans: 40 and 80 km; \large v_{1}=0.35\, c)

References:

1.  B. P. Abbott et al, “Observation of Gravitational Waves from a Binary Black Hole Merger,”, Phys. Rev. Lett. 116, 061102 (2016)

2.  this blog:  “Gravitational Radiation 2: The Chirp Heard Round the World”

3.  B. P. Abbott et al, “Observation of Gravitational Waves from a 22-Solar-Mass Binary Black Hole Coalescence,”, Phys. Rev. Lett. 116, 241103 (2016)

4.  A. Cho, “LIGO Detects Another Black Hole Crash,” Science 352, 1374, 17 June 2016

5.  D. Castelvecchi, “LIGO Sees a Second Black Hole Crash,” Nature 534, 448, 23 June 2016

6.  We thank Jonah Kanner at the LIGO Open Science Center for providing these numbers, which were calculated using the tutorial on the Center’s webpage:  https://losc.ligo.org.  The numbers obtained from the tutorial differ slightly from those reported in Ref. 3 because the analysis is not identical to the one used for the paper.  (In Ref. 3, \large \mathfrak{M}=8.9\pm 0.3\: M_{Sun}.)  The LIGO Open Science Center is a service of LIGO Laboratory and the LIGO Scientific Collaboration.    LIGO is funded by the U.S. National Science Foundation.

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Proxima Centauri b: An Earth-like Neighbor in our Galactic backyard

proxima-centauri-b (II)
Figure 1:  Artist’s conception of sunset on Proxima Centauri b.  The binary star Alpha Centauri is also visible.  Inset:  Proxima Centauri is the Sun’s nearest neighbor, 1.294 pc (4.224 ly) away.  Image credit:  Pale Red Dot/ESO

In August (2016), astronomers at the European Southern Observatory in Chile announced the discovery of an Earth-sized exoplanet orbiting the red dwarf Proxima Centauri, the nearest star to our Sun (Ref. 1, 2).  The planet, called Proxima Centauri b, was detected by the radial velocity method:  the orbiting planet causes the star to execute a much smaller orbit of its own (as required by momentum conservation), and light from the star is Doppler-shifted due to its radial motion relative to Earth.  The discovery of Proxima Centauri b is exciting because its orbit lies within the “Goldilocks,” or temperate, zone of the star, where the planet’s surface temperature would allow water to exist in liquid form – a likely prerequisite for life.  Red dwarfs are the most common stars in our galaxy, so if this particular red dwarf harbors a life-supporting planet, it is likely that the Milky Way, which contains over 100 billion stars, is brimming  with life.

In this post, we will use the data reported in Reference 1, plus the strategy presented in Section 8.12 of PPE, to calculate the mass and orbital radius of the exoplanet Proxima Centauri b.  Our goal is to help students share the excitement of this new discovery through their understanding of introductory mechanics.

 

A.  Calculating the exoplanet’s mass and orbital radius

In the following, we will assume \large m\ll M, where \large m and \large M are the masses of the planet and star, respectively.  We will also assume that their orbits are circular, which is consistent with observations, and begin by treating the simplest case where the orbits are viewed edge-on from Earth.  (Later, we’ll relax this restriction.)  Radial velocity measurements of Proxima Centauri collected over the past 16 years are compiled in Figure 2.  Careful analysis indicates that the period \large T of the star’s orbit is 11.186 ± .002 d, and its orbital speed \large v is 1.38 ± 0.02 m/s.

figure-2
Figure 2:  Radial velocity measurements of Proxima Centauri, compiled over 16 years of observations.  Note the sinusoidal shape of the best-fit curve.

Kepler’s 3rd law states that

\large \frac{T^{2}}{r^{3}}=\frac{4\pi ^{2}}{GM},

where \large r is the orbital radius of the planet.  Observations of the star’s luminosity and color indicate that its mass \large M=0.120\pm .015\, M_{Sun}.

 

 

 

1.  Find the planetary orbit radius \large r (in AU).  Hint: use  \large T=11.186/365.25=.03063\, \mathrm{yr} and \large r^{3}\propto MT^{2} to compare this orbit to Earth’s orbit about the Sun.  (Ans: \large 0.048 \: \mathrm{AU} = 7.24\times 10^{9}\: \mathrm{m}.

2.  The orbital radius of the star \large R can be found from its period and radial velocity \large v\large vT=2\pi R.  Using  \ 1\: \mathrm{d}=8.64\times 10^{4}\: \boldsymbol{\mathrm{s}}, find \large R.  (Ans: \large 2.12\times 10^{5}\: \mathrm{m})

3.  The planet and star co-orbit their center of mass.  Find the mass of the exoplanet.  (Ans:   \large m=MR/r=2.93\times 10^{-5}\, M=6.99\times 10^{24}\, \mathrm{kg}=1.18\, m_{Earth})

If the orbit is not viewed edge-on, but is inclined to the line of sight (LOS), the measured radial speed of the star is less than the actual orbital speed \large v used in the above analysis:  \large v_{meas}=vsin(i), where i, the angle of inclination, is unmeasurable.  See Figure 3.  In this case, \large 2\pi R=Tv=Tv_{meas}/sin(i)).

figure-3
Figure 3:  In the general case, the orbits of the star and planet are not viewed edge-on (i = 90 deg) from Earth.  The measured speed is less than the actual speed by a factor of sin(i).

4.  Prove that, in the general case, the radial velocity measurements are only sufficient to determine \large msin(i).  Is the exoplanet mass you found in Question 3 the maximum, or minimum, mass of the planet?

5.  Because red dwarfs are much cooler than the Sun, their temperate zones are much closer to them than the Earth is to the Sun.  In addition, red dwarfs are less massive than the Sun.  Explain why both of these factors favor the detection of small habitable exoplanets.  (See Ref. 2 for a brief discussion.)

6.  Proxima Centauri subtends an angle of 1.02 ± .08 mas (milliarcseconds) when viewed from Earth.  (See PPE, Section 1.7)  Find its radius \large R_{s}.  (Ans: \large 1.0\times 10^{8}\: \mathrm{m}=0.14\, R_{\mathrm{Sun}})

If Proxima centauri b transits, or passes in front of, its star, as viewed from Earth, it will intercept some of the starlight and allow its radius to be determined.  Additionally, if the planet has an atmosphere, it will absorb certain wavelengths of light preferentially, enabling astronomers to detect the presence of atmospheric gases such as water, oxygen, carbon dioxide, and methane.  Methane (CH4), in particular, is often associated with the presence of life.

The probability of transit is found from the geometric argument outlined in Figure 4.  In the figure, the horizontal dotted line is the LOS from Earth, and the vertical line represents the plane perpendicular to the LOS passing through the star’s center.  For an orbit to be transiting, its normal (shown as a red line for each of the two orbits depicted) must lie within \large \pm \theta _{max}=\pm R_{s}/r of the vertical plane, where \large R_{s} is the star’s radius.  But the normal to the orbit does not need to lie in the plane of the figure: any transiting orbit rotated by any angle (0 – 2π) about the LOS would remain a transiting orbit.  The total solid angle available for transiting orbits is therefore \large 2\pi \cdot 2\theta _{max}=4\pi R_{s}/r, whereas the total solid angle associated with all possible orientations is \large 4\pi.  Therefore, the probability of a transiting orbit is \large R_{s}/r.  (Check: if \large r=R_{s}, then the probability equals 1.)  For a fuller discussion, see  http://kepler.nasa.gov/Science/about/characteristicsOfTransits/ .

figure-4
Figure 4:  In a transiting orbit, the planet passes in front of the star, as seen from Earth.  For this to happen, the radius of the planet’s orbit r must be less than the stellar radius Rs.

7.  How probable is it that Proxima Centauri b transits its star?  (Ans: 1.5%)

B.  How Warm is Proxima Centauri b?

The following questions are only suitable for students who have studied blackbody radiation in a course of thermal physics.

8.  The luminosity of a star (its total radiated power) is proportional to its surface area and the fourth power of its surface temperature: \large L_{star}\propto R_{star}^{2}T_{star}^{4}.  Proxima Centauri’s surface temperature is about 3050 K, whereas the Sun’s is 5780 K.  Show that \large L_{Proxima}\simeq 0.0015\, L_{Sun}.  This indicates that the temperate zone lies much closer than 1 AU to the star.

9.  The fraction of the star’s radiation that is intercepted by a planet is equal to the solid angle subtended by the planet divided by 4π:  \large \pi R_{p}^{2}/4\pi r^{2}.  The absorbed radiation warms the planet, which reaches a steady state temperature \large T_{p} when the blackbody radiation emitted by the planet equals the radiation absorbed from the star:
\large 4\pi R_{p}^{2}\sigma _{B}T_{p}^{4}=L_{Proxima}\pi R_{p}^{2}/4\pi r^{2}, where  \large \sigma _{B} is called the Stefan-Boltzmann constant.  (Its value is not needed to answer this question.)  Use this information to compare the surface temperature of Proxima Centauri b to that of the Earth, assuming both planets are perfect black bodies (or have the same albedo).  (Ans:   \large T_{p}/T_{Earth}=0.90)

References

1.    G.  Anglada-Escudé et al, Nature 536, 437 (25 August, 2016)

2.   A. P. Hatzes, Nature 536, 408 (25 August, 2016)