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The following question was posed at the end of Maury Goodman's June 2012 long-baseline neutrino newsletter.

During the Venus transit of the sun, were more solar neutrinos absorbed in Venus, or focused toward us by gravity?

As I don't know off the top of my head I thought I'd pass it along. Maybe latter I'll have time to look up the relevant figures for order of magnitude estimates.

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this is an extremely good question. But now that you mention this, do we have any measurements for this for solar-moon eclipses? i would assume that the statistics and the error bars are far better! –  lurscher Jul 10 '12 at 19:51
    
The atmospheric neutrino community has been able to image the moon (I think I've seen this plot from SNO and from IceCube. Can't recall if I've seen it from Super-K), but what they see is the moon intercepting the cosmic rays, not the moon intercepting neutrinos which would be hard. –  dmckee Jul 10 '12 at 21:06
    
On imaging the sun with neutrinos: blogs.discovermagazine.com/cosmicvariance/2006/10/13/sun-shots. Note that the image took Super-K 500 days to accumulate. Trying to image the transit of Venus or a lunar eclipse this way would be hopeless. –  dmckee Jul 10 '12 at 21:11
    
i understand that matter-neutrino absortion is extremely tiny, but i was thinking in the same venue as your question, the gravity of the moon should be focusing more neutrinos than usual into the detectors. As far as we know the equivalence principle should beat the neutrino small matter coupling hands down, but i don't think this has ever been confirmed –  lurscher Jul 10 '12 at 21:11
    
in this case, deflection of neutrinos by moon gravity would be confirmed by a lot less bins than those required for creating that beautiful Super-K image, so the statistics should be somewhat better, what you think? –  lurscher Jul 10 '12 at 21:14
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1 Answer 1

An answer requires two order of magnitude estimates.

Absorption by the planet

The fraction $f_a$ lost to absorption is roughly $$ f_a = -\sigma N $$ where $N$ is the mean number of nucleons along the neutrinos path and $\sigma$ is the neutrino nucleon cross-section at the appropriate energy (a few MeV). We can expand the number to $$ f_a = -\sigma \frac{M_\text{venus}}{\pi R_\text{venus}^2} N_A $$

At solar energies the neutrino cross-section for neutrino--nucleon interactions is of order $10^{-45}\text{ m}^2$, the mass of he planet is around $5 \times 10^{24}\text{ kg}$ and it's radius is about 6000 km.

So $$ f_a = -\left( 10^{-45}\frac{\text{m}^2}{\text{nucleon}} \right) \frac{5 \times 10^{27}\text{ g}}{10^{14}\text{m}^2} \left(6 \times 10^{23} \frac{\text{nucleons}}{\text{g}} \right)$$ punching the number we get $$ f_a = -3 \times 10^{-8} $$

Gravitation focusing

Wikipedia tells me I can use $$ \theta_E = \sqrt{\frac{4GM_\text{venus}}{c^2} \frac{d_s - d_l}{d_s d_l}}$$ for the Einstein angle. Lets compute that first and then we'll try to figure the amplification.

The first term under the radical is twice the Schwarzschild radius for the planet or about $2.5\text{ cm}$.

The distance to the source $d_s$ is $1\text{ AU} = 1.5 \times 10^{11}\text{ m}$ and $d_l$ is about a quarter of that. So $$ \theta_E = \sqrt{(2.5 \times 10^{-2}\text{ m}) \frac{3}{1.5 \times 10^{11}\text{ m}}} = \sqrt{5 \times 10^{-13}} \approx 10^{-6} $$ (that is in radians of course).

The wikipedia then goes on to define a amplification in terms of the ratio $u$ of the angular separation (between the source and the lens) and the Einstein angle. Getting that exactly is tricky. I seem to recall that at closest approach Venus appears several times it's own diameter from the center of the sun (and we can treat the neutrinos as coming only from the very center as the fusion rate is strongly dependent on the temperature and pressure), and that Venus was very much smaller than the angular size of the sun.

Taking the angular size of the sun to be 0.01 radians, and the angular size of Venus to therefore be about 0.00005 radians I shall arbitrarily set $$ u = \frac{\theta_{ls}}{\theta_E} = \frac{0.0002}{10^{-6}} = 200 , $$ so that $$A = \frac{u^2 + 2}{u\sqrt{u^2 + 4}} = \frac{40002}{200\sqrt{40004}} = 1 + 10^{-9}$$

So the fraction gain is on order of $f_g = +10^{-9}$.

Caveats

Like I said in the gravitation section, the neutrinos come mostly from the very center and the planet passed a bit to one side, so the level of absorption may have been much lower than calculated and if the planet came close than I guesstimated the level of focusing could have been higher.

This makes me reluctant to declare absorption the winner despite the number.

Take it for what you will.

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