5
$\begingroup$

@SteveLinton's nicely written and sourced answer about using strong gravitational lensing by the Sun or even Jupiter as a kind of telescope to resolve the surfaces of exoplanets is really interesting, and the link cited there is indeed quite readable.

For the bare minimum resolution I considered of 1E-10 to make a exo-Jupiter 10 pixels wide at 7 light years, or even the 1E-12 he mentioned to do roughly the same at 1000 light years, what would a telescope like that be like?

Is it a maneuverable JWST that is so far from the Sun that it just scans back and forth building up an image of whatever is behind the Sun (or Jupiter) or a rigid 10x10 array of JWSTs?

Would each telescope just be a light collector for that pixel, or would this be more like a "light field camera" where the angular information from the telescope's focal plane at each telescope position could be used to improve the spatial resolution of the whole thing by "computational de-blurring"?

Would each telescope need a coronograph to block the light from the Sun (or Jupiter) while collecting the light from the far dimmer Einstein ring-like structure surrounding it?

The reason I ask that is that gravitational lenses are barely "lenses" from the point of view of telescopy. The deflection of a ray by a "normal" or thin lens increases linearly proportional to distance from the axis, whereas for a point gravitational source the deflection is inversely proportional. In order to image an extended source like a planet or star's disk rather than just just collect the light of the Einstein ring from an unresolved star, you have to play games with geometry.

For more on light fields, see the question and great answer to Is a “Light Field” useful in mathematics, or just in marketing?

I'm not interested in just opinions, but instead answers that are built from reasonable, cited sources, or unsourced but derived from solid physics and math principles.

enter image description here

Source

A remote light source passing behind a gravitational lens. There is a large point mass in the center acting as a lens. The aqua circle is how we would see the light source if there was no lens, while the white spots/circle is the light source as seen through the lens. If the light source is collinear with the earth and lens, the image is an "Einstein ring". When the source is off this line we see a double image. As it moves far away, one of the images gets fainter while the other one is almost not affected by the lens any more (thus coinciding with cyan circle).


enter image description here

Source

Stanford plenoptic camera array used to research light fields

$\endgroup$
  • 1
    $\begingroup$ JWST is Jamess Webb Space Telescope, right? $\endgroup$ – J. Doe Nov 5 '18 at 13:01
  • $\begingroup$ @J.Doe yes indeed; I use that as an analogy for a modern, large space telescope with substantial station-keeping maneuverability, excellent attitude control, and built with much of it's optics to work at very low temperature due to long-term protection from solar radiation. $\endgroup$ – uhoh Nov 5 '18 at 13:05
3
$\begingroup$

Here are some papers that describe what a gravitational lens telescope would look like.

"FOCAL Mission Concept" http://kiss.caltech.edu/workshops/ism/presentations/FOCAL%20Mission%20Concept%20JOHNSON.pdf

"Mission to the Gravitational Focus of the Sun: A Critical Analysis" https://arxiv.org/abs/1604.06351

Note that Jupiter is a much more difficult option for a gravitational lens -- the sun's lens starts at 550AU away, while Jupiter's starts at 6100AU.

I will update this answer with more information from the links as I have time later today.

$\endgroup$
  • 1
    $\begingroup$ Thanks, the first link is quite thin, really just "we are going to think about this". The second is the exact same "readable" ArXiv paper I mention in the first paragraph that (at)SteveLinton uses in his answer. Hopefully you can add some more here, but you are at least on the right track. Welcome to Space! $\endgroup$ – uhoh Nov 5 '18 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.