A new 9th planet?

Difficulty level:   

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…