## Atmosphere of Exoplanets Sunday, Feb 25 2007

From a NASA press release this week

NASA’s Spitzer Space Telescope has captured for the first time enough light from planets outside our solar system, known as exoplanets, to identify molecules in their atmospheres….

Spitzer, a space-based infrared telescope, obtained the detailed data, called spectra, for two different gas exoplanets. Called HD 209458b and HD 189733b, these so-called “hot Jupiters” are, like Jupiter, made of gas, but orbit much closer to their suns.

The data indicate the two planets are drier and cloudier than predicted. Theorists thought hot Jupiters would have lots of water in their atmospheres, but surprisingly none was found around HD 209458b and HD 189733b. According to astronomers, the water might be present but buried under a thick blanket of high, waterless clouds.

Those clouds might be filled with dust. One of the planets, HD 209458b, showed hints of tiny sand grains, called silicates, in its atmosphere. This could mean the planet’s skies are filled with high, dusty clouds unlike anything seen around planets in our own solar system…

And via BA Blog, we are reminded that “Twenty years ago, astronomers witnessed one of the brightest stellar explosions in more than 400 years. The titanic supernova, called SN 1987A, blazed with the power of 100 million suns for several months following its discovery on 23 Feb., 1987.” Of course, the article doesn’t quite spell it out that the neutrino detectors were telling us something interesting three hours before the explosion was seen ! (See this link too – via Backreaction blog .)

‘Rings’ left by the Supernova Explosion

## A Question in Geography Sunday, Feb 18 2007

A theorem in topology says:

For every continuous map $f: S^n \rightarrow R^n$ there exists a pair of anitpodal points $x$ and $- x$ in $S^n$ such that $f(x) = f(-x)$.

Specializing to the case $n = 2$ , one might conclude that at any point of time there are two antipodal points on Earth’s surface (which is homeomorphic to $S^2$) having same, say, pressure $P$ and temperature $T$ (which together constitute $R^2$ of the theorem) .

Question: Isn’t it that the presence of $\textit{two}$ polar caps (The Arctic and The Antarctica) where the day/night variation in temperature is quite low, may then be looked upon as a kind of `consequence’ of this theorem? (ofcourse the theorem doesn’t say where $x$ is located but at almost all other regions on Earth’s surface, the antipodal points are expected to have quite different temperatures due to day/night variation.)

ps1: One might worry about the validity of the assumption of continuity of $P$ and $T$ since there are wild local fluctuations; but I feel once you coarse-grain things out, this is a reasonable assumption.

ps2: For proof of the theorem, see http://www.mi.ras.ru/~scepin/elem-proof-reduct.pdf

## LaTeX at last ! (Illustrated with a GR calculation) Sunday, Feb 18 2007

After waiting for long, Blogphysica gets $\LaTeX$ . (See the announcement at the wordpress.com Blog )

The Basic syntax is $latex <LaTeX Equation>$. You can find a list of $\LaTeX$ symbols here(pdf) . In particular, if you find the formulae are too small try $latex {\displaystyle <LaTeX Equation>}$ . It’s great !

Update(18/02/07) : See this FAQ for some more options.

Now, to illustrate LaTeX, I’ll take up a particular problem. Consider two equal masses falling towards each other, as shown below, starting from rest.

m—>—o—<—m

The question is this – How much power does this system lose as the two masses fall towards each other ?

Let $r$ be the distance from the centre of mass (which I’ve denoted by an ‘o’ above and I choose it to be the origin). Take the common axis to be z-axis.

Hence the position of two masses are respectively $(x,y,z)=(0,0,r)$ and $(0,0,-r)$

To the zeroth approximation, Newtonian mechanics tells you that

${\displaystyle \frac{d^2 r}{dt^2} = - \frac{G_Nm}{(2r)^2}}$

(We neglect the effect of gravitational wave on the masses)

The quadrupole moment $Q_{ij}$ of a mass distribution is defined by

${\displaystyle Q_{ij}(t) = \iiint d\forall\ \ \ \rho(x,t) \left[ x_i x_j - \frac{1}{3} \delta_{ij} r^2 \right]}$

Where the integral is done over the whole source($\rho$ is the mass density of the source). It is basically negative of the traceless part of the moment of inertia.

The Einstein formula for the power emitted by the source (in the form of Gravitational waves) is

${\displaystyle P = \frac{G_N}{5c^5} \sum\ \frac{d^3 Q_{ij}}{dt^3}\ \frac{d^3 Q_{ij}}{dt^3} }$

where the symbol $\sum$ denotes a sum over $i,j=1,2,3$.

Assuming that the masses are small in size, the components of quadrupole moment in this case are

${\displaystyle Q_{zz}= m\left(r^2 - \frac{1}{3}r^2\right)+ m\left((-r)^2 - \frac{1}{3}r^2\right)= \frac{4}{3}mr^2}$
${\displaystyle Q_{xx}=Q_{yy} = m(0^2 -\frac{1}{3}r^2)+ m(0^2 -\frac{1}{3}r^2)= -\frac{2}{3}mr^2 }$
${\displaystyle Q_{xy}=Q_{yx}=Q_{zx} =\ldots = 0}$

Now to calculate the third time derivative, we first use chain rule to get

${\displaystyle \frac{d^3}{dt^3} r^2 = 6 \frac{dr}{dt}\ \frac{d^2 r}{dt^2} + 2r\ \frac{d^3 r}{dt^3} }$

We can now employ Newton’s Law to get

${\displaystyle \frac{d^3}{dt^3} r^2 = - \frac{G_Nm}{2r^2}\ \frac{dr}{dt}}$

${\displaystyle \frac{d^3 Q_{zz}}{dt^3} = - \frac{2G_Nm^2}{3r^2}\ \frac{dr}{dt}}$
${\displaystyle \frac{d^3 Q_{xx}}{dt^3}=\frac{d^3 Q_{yy}}{dt^3} = \frac{G_Nm^2}{3r^2}\ \frac{dr}{dt}}$
${\displaystyle \mbox{Power emitted} = \frac{G_N^3m^2}{15c^5 r^4}\ \left(\frac{dr}{dt}\right)^2 }$