Gravitational waves detected!

Difficulty level:   

Physicists are very excited, because the first ever direct detection of gravitational waves has just been announced! The signal matches the prediction for two black holes colliding.  This will likely mean a Nobel Prize for someone. This is a tremendous scientific achievement, representing a vast global collaboration between scientists, advanced technology, government funding, and simple good luck.

The signal lasted for just 1/5 of a second, but scientists have extracted an impressive amount of information from it. This video plays the “chirp” which was detected, converting the gravitational wave signal to sound so you can hear it. The video repeats the chirp 8 times, half of those scaled to a higher frequency where human hearing is more sensitive.

The LIGO detectors have two 4km long pipes housing laser light for detecting gravitational waves. This is the Hanford, Washington instrument
The signal was picked up by the two “LIGO” detectors in the United States. These have two 4km long pipes at right angles, housing laser light which measures the miniscule expansion and contraction of space caused by a passing gravitational wave. This is the Hanford, Washington instrument; the other is in Livingston, Louisiana.

But understand that calling it “sound” is metaphorical, for instance when someone gave a demonstration by playing a cello on Australian Broadcasting Corporation (ABC) TV. A gravitational wave is a ripple through the “fabric” of space itself and travels at the speed of light, whereas a sound wave transmits via air molecules bumping together and travels a million times more slowly. It should also be clarified that the gravitational waves would have been emitted for a far longer period than 0.2 seconds, it’s just they were too weak to be detected by us.

Gravitational waves are a consequence of general relativity, and were first predicted by Einstein in 1916. Though not an area of my research so far, I have looked in-depth at the measurement of distances in relativity, which is somewhat related. I look forward to learning and sharing more.

A new 9th planet?

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There’s news today that some scientists predict the existence of a ninth planet. No one has actually found anything, but this is inferred from the orbits of certain icy objects in the outer Solar System. It may have 10 times the mass of the Earth, and take 10,000 or 20,000 years to orbit the Sun, due to its distance far beyond the known planets.Planet NineScientists from the California Institute of Technology (Caltech) claim “Planet Nine” would have an orbit like the above (yellow) to account for the depicted handful of bodies (purple orbits) lying in one direction. They hope to detect it in the next 5 years. It’s not my area, and I have no opinion on this personally, but am happy to wait and see what consensus forms. Still, it’s an opportunity to share some history of planetary discovery.

Artist's conception of the hypothetical Planet Nine
Artist’s conception of the hypothetical Planet Nine

This has happened before. Neptune was discovered because of irregularities in the orbit of Uranus. Pluto was discovered by the same motivation. (Further irregularities in Neptune and Uranus’ orbits had led to a search. However it turned out Neptune’s mass had been overestimated, and besides Pluto was too small to affect these planets much.) Similarly the unexplained rotation of Mercury’s orbit led to speculation of a new innermost planet “Vulcan”, but just like the Star Trek planet it remains fictional. In fact Einstein successfully explained Mercury’s behaviour using an early version of general relativity.

World Science Festival in Brisbane

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World Science Festival BrisbaneThe World Science Festival is coming to Brisbane on 9–13 March, 2016. This might be the first time it’s been held outside of New York City.

There’s a lot of astrophysics and relativity, including some big names, but it’s expensive. Sean Carroll, author of a good relativity textbook , will discuss the accelerating universe with Nobel prizewinner and Aussie Brian Schmidt (who will also be on “Breakfast with the Brians”), and others. Tamara Davis, my Master’s thesis supervisor, will discuss relativity with string theorist and author Brian Greene and others. The drama “Light Falls” written by Brian Greene about Einstein’s discovery of general relativity looks great, but I don’t want to pay $69/$89. Another drama about Einstein’s personal side was written by Alan Alda, who was the main character of M*A*S*H.

Light Falls dramaMadness Redefined” also looks interesting. But I think I’ll just drop in briefly on the Sunday, for the free documentary Science and Islam and maybe the free “Street Science” demonstration. I’ve also booked tickets for a maths documentary on “predictive analytics” about modelling our lives, maybe from internet data.

Tetrad for Schwarzschild metric, in terms of e

Difficulty level:   ★ ★ ★

The following is a natural choice of orthonormal tetrad for an observer moving radially in Schwarzschild spacetime with “energy per unit rest mass” e:

    \begin{align*} e_{\hat 0}^\mu &= \left(e\Schw^{-1},\pm\eroot,0,0\right) \\ e_{\hat 1}^\mu &= \left(\pm\Schw^{-1}\eroot,e,0,0\right) \\ e_{\hat 2}^\mu &= \left(0,0,\frac{1}{r},0\right) \\ e_{\hat 3}^\mu &= \left(0,0,0,\frac{1}{r\sin\theta}\right) \end{align*}

The components are given in Schwarzschild coordinates. (The ± signs are not independent — they must be either both +1 or both -1. Note that e does not distinguish between inward and outward motion. There is additional freedom to define any of these vectors as their negative.)

We normally think of e as invariant, where there is a presumption of freely falling / geodesic motion, but even if not we can regard it as an instantaneous value.

\evec{0} is the 4-velocity computed previously. The other vectors can be obtained from substituting \gamma=e\Schw^{-1/2} and V=-\frac{1}{e}\sqrt{e^2-1+\frac{2M}{r}} into the tetrad here. \gamma is determined from -\fvec u\cdot\fvec u_{\rm obs}=\gamma and the equation for e above, then V follows from inverting \gamma\equiv(1-V^2)^{-1/2}. This orthonormal frame is useful for determining the object’s perspective, e.g. tidal forces, visual appearances, etc.

Tetrad for Schwarzschild metric

Difficulty level:   ★ ★ ★ ★

Suppose an observer u moves radially with speed (3-velocity) V relative to “stationary” Schwarzschild observers, where we define V<0 as inward motion. Then one natural choice of orthonormal tetrad is:

    \begin{align*} (\evec{0})^\alpha &= \left(\gamma\Schw^{-1/2},V\gamma\Schw^{1/2},0,0\right) \\ (\evec{1})^\alpha &= \left(V\gamma\Schw^{-1/2},\gamma\Schw^{1/2},0,0\right) \\ (\evec{2})^\alpha &= \left(0,0,\frac{1}{r},0\right) \\ (\evec{3})^\alpha &= \left(0,0,0,\frac{1}{r\sin\theta}\right) \end{align*}

where the components are given in Schwarzschild coordinates. This may be derived as follows.

The Schwarzschild observer has 4-velocity

    \[\fvec u_{\rm obs}=\left(\Schw^{-1/2},0,0,0\right)\]

because the spatial coordinates are fixed, and the t-component follows from normalisation \fvec u_{\rm obs}\cdot\fvec u_{\rm obs}=-1 (Hartle §9.2).

Now the Lorentz factor for the relative speed satisfies -\fvec u\cdot\fvec u_{\rm obs}=\gamma, and together with normalisation \fvec u\cdot\fvec u=-1 and the assumption that the θ and φ components are zero, this yields \evec{0}\equiv\fvec u given above.

We obtain \evec{1} by orthonormality: \evec{1}\cdot\evec{0}=0 and \evec{1}\cdot\evec{1}=1, and again making the assumption the θ and φ components are zero. Note the negative of the r-component is probably an equally natural choice. Then \evec{2} and \evec{3} follow from simply normalising the coordinate vectors.

Strictly speaking this setup only applies for r>2M, because stationary timelike observers cannot exist inside a black hole event horizon! Yet remarkably the formulae can work out anyway (MTW …) .  An alternate approach is local Lorentz boost described shortly.

Hartle … Also check no “twisting” etc…

Radial motion in the Schwarzschild metric, relative to stationary observers

Difficulty level:   ★ ★ ★

Last time we derived the 4-velocity u of a small test body moving radially in the Schwarzschild geometry, in terms of e, the “energy per unit rest mass”. Another parametrisation is in terms of the 3-speed V relative to stationary observers. This turns out to be, in Schwarzschild coordinate expression,


To derive this, first consider the 4-velocity of stationary observers:


We know the “moving” body has 4-velocity u of form u^\mu=(u^t,u^r,0,0) since the motion is radial. The Lorentz factor \gamma\equiv(1-V^2)^{-1/2} for the relative speed is

    \[\gamma=-\fvec u\cdot\fvec u_{\rm{Schw}}\]

Evaluating and rearranging yields u^t=\gamma\Schw^{-1/2}. Normalisation \fvec u\cdot\fvec u=-1 leads to u^r=\pm V\gamma\Schw^{1/2}, after some algebra including use of the identity \gamma^2-1=V^2\gamma^2. We allow V<0 also, and define this as inward motion. Carefully considering the sign, this results in the top equation. (An alternate derivation is to perform a local Lorentz boost. Later articles will discuss this… The Special Relativity formulae cannot be applied directly to Schwarzschild coordinates.)

Some special cases are noteworthy. For V=0, γ=1, and u reduces to uSchw. This corresponds to e=-\fvec\xi\cdot\fveclabel{u}{Schw}=\Schw^{1/2}. Also we can relate the parametrisation by V (and γ) to the parametrisation by e via

    \[\gamma=e\Schw^{-1/2}\qquad V=-\frac{1}{e}\sqrt{e^2-1+\frac{2M}{r}}\]

where the leftmost equation follows from the definition e\equiv-\fvec\xi\cdot\fvec u, and subsequently the rightmost equation from γ=γ(V). For raindrops with e=1, the relative speed reduces to V=-\Schwroot.

We would expect the construction to fail for r\le 2M, as stationary timelike observers cannot exist there, and so the relative speed to them would become meaningless. But curiously, it can actually work for a faster-than-light V>1 “Lorentz” boost, as even the authorities MTW (§31.2, explicit acknowledgement) and Taylor & Wheeler (§B.4, implicitly vrel>1 for r<2M) attest. Sometime, I will investigate this further…

Radial motion in the Schwarzschild metric, in terms of e

Difficulty level:   ★ ★ ★

A nice way to parametrise the 4-velocity u of a small test body moving radially within Schwarzschild spacetime is by the “energy per unit rest mass” e:


For the “±” term, choose the sign based on whether the motion is inwards or outwards. All components are given in Schwarzschild coordinates (t,r,\theta,\phi). The result was derived as follows. In geometric units, the metric is:


By definition e\equiv-\fvec\xi\cdot\fvec u, where \fvec\xi\equiv\partial_t is the Killing vector corresponding to the independence of the metric from t, and has components \xi^\mu=(1,0,0,0) (Hartle §9.3). For geodesic (freefalling) motion e is invariant, however even for accelerated motion e is well-defined instantaneously and makes a useful parametrisation.

We want to find u^\mu=(u^t,u^r,u^\theta,u^\phi) say. Rearranging the defining equation for e gives u^t=e\Schw^{-1}. Radial motion means u^\theta=u^\phi=0, so the normalised condition \fvec u\cdot\fvec u=-1 yields the remaining component \abs{u^r}. The resulting formula is valid for all 0<r\ne 2M, and for e=1 the 4-velocity describes “raindrops” as expected.

Relative speed

Difficulty level:   ★ ★

Suppose two observers at the same place and time (that is, “event”) move with 4-velocities u and v respectively, then they measure their relative speed as follows. The Lorentz factor is simply

    \[\gamma=-\fvec u\cdot\fvec v\]

(The dot is not the Euclidean dot product, but uses the metric: g_{\alpha\beta}u^\alpha v^\beta where the indices \alpha and \beta are summed over by the Einstein summation convention.) The proof is based on the axiom that some local inertial frame exists, although interestingly one does not need to explicitly construct it.

The relative 3-speed V, may then be recovered via:


See for instance Carroll (end of §2.5) who terms it “ordinary three-velocity”. Other sources express the first formula more indirectly, in terms of the energy and momentum measured by an observer \fvec u: E=-\fvec u\cdot\fvec p where \fvec p=m\fvec v is the 4-momentum of another observer/object, and combine this with E=m\gamma (MTW Exercise 2.5 in §2.8 term it “ordinary velocity”, or Hartle §5.6, and Example 9.1 in §9.3).