3 ms·

Well the photons leaving earth are scattered in almost all directions. So at 1500 light years there might not be any photons arriving from that exact moment and

Well the photons leaving earth are scattered in almost all directions. So at 1500 light years there might not be any photons arriving from that exact moment and place on earth because you might be down to say 1 photon arriving per hour or even less. Then the photons coming back would be all mixed in with other photons from the black hole so even if the same photons did go to where earth will be, the picture would be mangled beyond repair, and the darkness of the data you’re trying to capture is so low that again it might be less than a photon an hour.

baxtr 2y ago
Easy. We just use AI to “augment” any photon that is missing and voila we’ll be able to watch opening day.
- exe34 2y ago
  at this point we can do it without the photon!
  - pavlov 2y ago
    Homeopathic astronomy? Those quack homeopathy treatments operate on the assumption that water has a "memory", and it's somehow enough that it's been in contact with some molecule in the past. The astronomy version would be that space has a memory, and you can observe anything if you know there was a relevant photon out there at some point.
    - dayjaby 2y ago
      Water has memory? We should build a homeopathic computer first of all.
      - lukan 2y ago
        Not homeopathic, but maybe close enough: https://en.m.wikipedia.org/wiki/Phillips_Machine https://en.m.wikipedia.org/wiki/Phillips_Machine
        exe34 2y ago
        Hah that's where GNU Terry got it from then.
- noisy_boy 2y ago
  Enhance! - 2024 AI edition
rachofsunshine 2y ago
We can work out a ballpark value here. Suppose we're trying to see one square meter of Earth's surface on a sunny day, in the sense that we simply want to get our photons close enough to the black hole to be noticeably affected by it. Under direct sunlight typical illumination at noon is on the order of 1 kW/m^2 (this is a slight overestimate, since Japan is in the mid-latitudes and not the tropics). Typical surfaces reflect roughly 10-20% of the light that strikes them; let's say 10% to make the math easy and to compensate for the previous overestimate. So 100 W of mostly visible light is reflected, i.e., 100 joules of energy per second. A single photon of visible light has an energy of about 4 x 10^-19 J, so that's 2.5 x 10^20 photons reflected from our hypothetical square meter per second. The black hole has a radius of around 30 km, or an apparent area in the sky equivalent to a plane of size pi * (30 km)^2 = 2800 km^2. It's actually a bit bigger than that because the "shadow" of a black hole is larger than the event horizon, and that's what we need to hit, not the horizon itself. So let's say 10,000 km^2 to make it nice and round. That's 10,000 km^2 of coverage on a "sphere" of radius 1500 ly and hence of area 4 * pi * (1500 ly)^2 = 2.5 x 10^33 km^2. So the shadow of the black hole occupies about 4 x 10^-30 of the sky. If we spread out 2.5 x 10^20 photons across the sky evenly, we hit the black hole's shadow with about one-billionth of a photon from our square meter of Earth per second, or about one photon every 30 years. You hit it with a photon from somewhere on Earth at most roughly every ten seconds or so (when the geometry is such that Earth is close to "full" from the black hole's perspective). The situation for actually lensing it properly to get it back to Earth is far worse, since not only do you have to hit the "shadow", you have to hit a particular point in the distorted image of the sky that corresponds to Earth, meaning you effectively have to thread this needle twice. Assuming you just want to hit Earth, which is bigger than the black hole's shadow, you're still trying to hit another 10^-26 or so shot. A single photon from Earth should make that shot roughly every 10^19 years, or about a billion times the current age of the Universe. TLDR: ain't gonna work. ----- In reality, none of this matters for a more fundamental reason: diffraction. The fact that light "smears out" as it travels poses a fundamental limit to resolution based on the size of your telescope, and the resolution we're trying to achieve here is orders of magnitude beyond it. For scale, Hubble is pretty close to the diffraction limit for visible light and a telescope of its size, and it already can't image planets within our own solar system at the resolution we're talking about here. So no reasonable telescope could even image Earth from the black hole, much less image a tiny portion of Earth. ...or that would be true, if you limited yourself to physical telescopes. See, the diffraction limit depends on the size of your telescope, or more properly, on the size of its lens. And the effective size of a gravitational lens can be very, very large indeed. It turns out that current technology is sufficient to reach a distance where the distortion of the Sun's own gravity focuses light - the required distance is a few times the distance to the Voyager probes. From that vantage point, you could theoretically use the Sun itself as a giant telescope. And the resolutions achievable are tight enough that, while you wouldn't be able to image individual square meters, you actually could image details of planets with a resolution of tens of km (the rough equivalent of looking at a globe from a distance of a couple of feet away) across galactic distances. [1] The reason this works is that it's not hitting the gravitational shadow of the gravitationally-lensing object itself - it's hitting a focal "ring" with a very large radius around the object, effectively magnifying the imaged object by ~eleven orders of magnitude. At such magnification even a handful of square meters start hitting with meaningful numbers of photons, though diffraction is still (I think) the limiting factor on actually resolving anything. The black hole here is not much more massive than the Sun, so it wouldn't achieve too much more power. But with careful observation, you could use the hole to image Earth to a scale that would let you map out our world and our civilization to reasonable accuracy - if not the details of our corporate endeavors. ----- [1] https://en.wikipedia.org/wiki/Solar_gravitational_lens https://en.wikipedia.org/wiki/Solar_gravitational_lens See also https://en.wikipedia.org/wiki/WHL0137-LS https://en.wikipedia.org/wiki/WHL0137-LS for an example of this kind of lensing on the scale of galaxies, which allows us to see a single star (or possibly a binary) from literally halfway across the Universe.
smallstepforman 2y ago
A pointy haired executive would say “silly engineers, did you forget to add an enhance button, the solution is so simple”.
bjornsing 2y ago
> you might be down to say 1 photon arriving per hour or even less Have you calculated this or are you just speculating? The quantization of light is a real physical limit that may make this impossible. The rest sounds more like engineering problems.
napo 2y ago
Then let's put one of this well-positioned black hole in all photons' trajectories.

baxtr 2y ago

Easy. We just use AI to “augment” any photon that is missing and voila we’ll be able to watch opening day.

exe34 2y ago
at this point we can do it without the photon!
- pavlov 2y ago
  Homeopathic astronomy? Those quack homeopathy treatments operate on the assumption that water has a "memory", and it's somehow enough that it's been in contact with some molecule in the past. The astronomy version would be that space has a memory, and you can observe anything if you know there was a relevant photon out there at some point.
  - dayjaby 2y ago
    Water has memory? We should build a homeopathic computer first of all.
    - lukan 2y ago
      Not homeopathic, but maybe close enough: https://en.m.wikipedia.org/wiki/Phillips_Machine https://en.m.wikipedia.org/wiki/Phillips_Machine
      - exe34 2y ago
        Hah that's where GNU Terry got it from then.
noisy_boy 2y ago
Enhance! - 2024 AI edition

rachofsunshine 2y ago

We can work out a ballpark value here. Suppose we're trying to see one square meter of Earth's surface on a sunny day, in the sense that we simply want to get our photons close enough to the black hole to be noticeably affected by it. Under direct sunlight typical illumination at noon is on the order of 1 kW/m^2 (this is a slight overestimate, since Japan is in the mid-latitudes and not the tropics). Typical surfaces reflect roughly 10-20% of the light that strikes them; let's say 10% to make the math easy and to compensate for the previous overestimate. So 100 W of mostly visible light is reflected, i.e., 100 joules of energy per second. A single photon of visible light has an energy of about 4 x 10^-19 J, so that's 2.5 x 10^20 photons reflected from our hypothetical square meter per second. The black hole has a radius of around 30 km, or an apparent area in the sky equivalent to a plane of size pi * (30 km)^2 = 2800 km^2. It's actually a bit bigger than that because the "shadow" of a black hole is larger than the event horizon, and that's what we need to hit, not the horizon itself. So let's say 10,000 km^2 to make it nice and round. That's 10,000 km^2 of coverage on a "sphere" of radius 1500 ly and hence of area 4 * pi * (1500 ly)^2 = 2.5 x 10^33 km^2. So the shadow of the black hole occupies about 4 x 10^-30 of the sky. If we spread out 2.5 x 10^20 photons across the sky evenly, we hit the black hole's shadow with about one-billionth of a photon from our square meter of Earth per second, or about one photon every 30 years. You hit it with a photon from somewhere on Earth at most roughly every ten seconds or so (when the geometry is such that Earth is close to "full" from the black hole's perspective). The situation for actually lensing it properly to get it back to Earth is far worse, since not only do you have to hit the "shadow", you have to hit a particular point in the distorted image of the sky that corresponds to Earth, meaning you effectively have to thread this needle twice. Assuming you just want to hit Earth, which is bigger than the black hole's shadow, you're still trying to hit another 10^-26 or so shot. A single photon from Earth should make that shot roughly every 10^19 years, or about a billion times the current age of the Universe. TLDR: ain't gonna work. ----- In reality, none of this matters for a more fundamental reason: diffraction. The fact that light "smears out" as it travels poses a fundamental limit to resolution based on the size of your telescope, and the resolution we're trying to achieve here is orders of magnitude beyond it. For scale, Hubble is pretty close to the diffraction limit for visible light and a telescope of its size, and it already can't image planets within our own solar system at the resolution we're talking about here. So no reasonable telescope could even image Earth from the black hole, much less image a tiny portion of Earth. ...or that would be true, if you limited yourself to physical telescopes. See, the diffraction limit depends on the size of your telescope, or more properly, on the size of its lens. And the effective size of a gravitational lens can be very, very large indeed. It turns out that current technology is sufficient to reach a distance where the distortion of the Sun's own gravity focuses light - the required distance is a few times the distance to the Voyager probes. From that vantage point, you could theoretically use the Sun itself as a giant telescope. And the resolutions achievable are tight enough that, while you wouldn't be able to image individual square meters, you actually could image details of planets with a resolution of tens of km (the rough equivalent of looking at a globe from a distance of a couple of feet away) across galactic distances. [1] The reason this works is that it's not hitting the gravitational shadow of the gravitationally-lensing object itself - it's hitting a focal "ring" with a very large radius around the object, effectively magnifying the imaged object by ~eleven orders of magnitude. At such magnification even a handful of square meters start hitting with meaningful numbers of photons, though diffraction is still (I think) the limiting factor on actually resolving anything. The black hole here is not much more massive than the Sun, so it wouldn't achieve too much more power. But with careful observation, you could use the hole to image Earth to a scale that would let you map out our world and our civilization to reasonable accuracy - if not the details of our corporate endeavors. ----- [1] https://en.wikipedia.org/wiki/Solar_gravitational_lens https://en.wikipedia.org/wiki/Solar_gravitational_lens See also https://en.wikipedia.org/wiki/WHL0137-LS https://en.wikipedia.org/wiki/WHL0137-LS for an example of this kind of lensing on the scale of galaxies, which allows us to see a single star (or possibly a binary) from literally halfway across the Universe.

smallstepforman 2y ago

A pointy haired executive would say “silly engineers, did you forget to add an enhance button, the solution is so simple”.

bjornsing 2y ago

> you might be down to say 1 photon arriving per hour or even less Have you calculated this or are you just speculating? The quantization of light is a real physical limit that may make this impossible. The rest sounds more like engineering problems.

napo 2y ago

Then let's put one of this well-positioned black hole in all photons' trajectories.