Earthshine blog

I have compared two well-exposed flatfields in the B-filter from the dusk session night JD2455856 with a time difference of about 28.5 minutes.

Each flatfield was bias-subtracted with the proper scaled superbias, had a fitted surface subtracted, and was then normalized. A percent difference image was created as the difference compared to the earlier flatfield.
perc = ((late-early)/early)*100

The mean of this image is of course very close to zero (0.0014), but more interesting is the standard deviation (0.53).

The strongest part of the familiar diagonal pattern is still visible in the percent image (see left part of figure). I have used Image J to rotate the percent image and plot the horizontal profile of the yellow box (top right part of the figure) selected to be perpendicular to the diagonal darker areas. This profile is plotted in the bottom right part of the figure.
It can be seen that the diagonal structure is 0.2pp darker than the immediate neighborhood.

The diagonal structure does change within a relatively short time-frame and this will result in an uncertainty in the science frames when they are flatfield corrected. So far it doesn’t seem to be a large change, but more investigations are necessary. The more likely explanation for the change is temperature fluctuations. Perhaps it is possible to investigate if the diagonal structure changes with a period comparable to the change in bias level…

Peter sent me images of the ideal crescent moon for a range of Earth albedos — 0.0, 0.29, 0.30 and 0.31. I made a map of the scattered light to see how much brighter the ES (Earthshine) is than the scattered light from the crescent. Over most of the dark side, ES does indeed dominate over the scattered component — which augers well for measuring its ratio relative to the BS (bright side).

Top left: ideal moon outside the atmosphere with NO earthshine at all (Earth albedo = 0) i.e. we see the BS (brightside) only.
Top right: scattered light from the BS only.
Bottom left: scattered light of the same ideal moon, but with the ES for albedo=0.30 included.
Bottom right: the difference between the top right and bottom left images.

Interestingly, one can subtract the modeled BS to obtain the ES over the entire lunar disc. The isophotes of the remaining light are then nicely round around the full, ES-only illuminated disc, as seen in the lower right panel. This might be a way to check that the solutions are coming out the way they should — i.e. fit the BS and scattered light from it only, remove it — and the rest should be the ES. Errors in modeling the BS will show up as asymmetries in the ES that remains. Perhaps we could think about solving for the ES directly, by fitting a lunar model with a BS only, and recovering the ES on the lunar face directly?