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What exactly makes the light from these distant galaxies so “weak” that we need to point the telescope there for months to see it? Like, isn’t there a constant

by alecbz 5y ago

What exactly makes the light from these distant galaxies so “weak” that we need to point the telescope there for months to see it? Like, isn’t there a constant stream of photons? How does just waiting longer for them make them show up if they wouldn’t have otherwise? Are things blocking most of the photons? Things from within the galaxies they’re coming from, or things in between the galaxy and the telescope? edit: Ah, I suppose the further an object is the fewer photons from it are making it to the telescope, and so we need to wait longer for enough photons to arrive to be able to distinguish it from the random background noise on our end?

photoGrant 5y ago
Yes, it's called NOISE!
- alecbz 5y ago
  But what specifically is it? Just random stuff in deep space between galaxies? I’d have thought most of the space between galaxies is mostly truly empty.
  - site-packages1 5y ago
    Isn't the paradox that if the universe is static in size or shrinking, eventually all the light would fill the sky into a blinding display, rather than dark. So it's a function of the universe expanding. https://en.wikipedia.org/wiki/Olbers%27_paradox https://en.wikipedia.org/wiki/Olbers%27_paradox
  - nprecup 5y ago
    The noise is inherent to our sensor technology... Very sensitive telescopes are cryogenically cooled to reduce the temperature of the focal plane, reducing the noise. As in an above comment, the light from distant objects is more spread out and appears weaker - fewer photos are arriving each second and hitting the focal plane than if the telescope were closer. At a certain point, the signal gets buried in the noise. I worked on the BICEP Array telescope at the south Pole - the light we are observing is actually extremely low energy and we can't see it unless our detectors are colder than the light source we are looking at (which is about 2.7K). We cool our detectors to 0.3K!
    - daniel-cussen 5y ago
      Does the natural cold at the South Pole make it much easier to cool the instruments down that low? I know it's around 200K at its lowest, how does that make it easier and cheaper?
      - nprecup 5y ago
        Yes, there are several factors that make that location particularly good, not only temperature. The cold and absence of sunlight for months at a time causes a relative absence of water vapor in the atmosphere during winter, which is extremely important for CMB observations. The high elevation (~10k ft) means there is less atmosphere to look through as well. During summers the sky heats up, contains more water vapor, and reflects so much more light that our telescopes become substantially less effective. Another advantage to the South Pole is the ability to continuously observe the same part of the sky, as the telescope is located on the earths axis there is no rising and setting of sky overhead. There is also a particularly 'dark' patch of sky in the southern hemisphere called the 'southern hole'. This is located up and away from the plane of our galaxy and contains very few objects standing in the 'path' of the CMB which also helps obtain better quality observations of the CMB. Edit - it is as good as it gets on earth and way cheaper than a satellite! Other benefits are rapid upgrades and repairs with newer technology. A satellite, by the time it is deployed, is already pretty old! But the South Pole still isn't cheap!
  - pgorczak 5y ago
    As a sibling already mentioned, thermal noise occurs in the sensors. And there is also a range of cosmic background radiation coming in, including the cosmic microwave background which is just everywhere and is almost as old as the universe.
  - pjc50 5y ago
    The noise is coming from inside the sensor: https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise Cryogenically cooling helps, but since you can't hit absolute zero you can't entirely eliminate it.
    - alecbz 5y ago
      Right, what I was missing was why how far the source of the light is effects how much the noise matters, but the inverse square photon concentration explains that.
    - passivate 5y ago
      There is read noise from downstream electronics converting photons to electrons, but the "shot noise" is a property of the photon stream and will be present regardless of how much we improve the electronics. Cooling the detector won't help there. https://en.wikipedia.org/wiki/Shot_noise#Optics https://en.wikipedia.org/wiki/Shot_noise#Optics
hexatin 5y ago
The strength of light over a distance obeys an inverse square law[1], causing it to effectively lose power over long distances, since the same amount of energy is being spread over larger and larger spheres as it radiates out. For photons, I think this manifests as there being a lower rate of photons occurring in each section of the sphere since they’re “spread out”. [1] https://en.wikipedia.org/wiki/Inverse-square_law https://en.wikipedia.org/wiki/Inverse-square_law
- site-packages1 5y ago
  Doesn't this fail because we're looking at lots of stars spread out evenly, and so the farther you get away, the more stars fill the space?
  - hypertele-Xii 5y ago
    Space is mostly empty.
    - mananaysiempre 5y ago
      As a sibling comment to yours says, Olbers’ paradox is that it doesn’t matter. If we assume that the universe is homogeneous, infinite, and eternal and the density of stars (possibly averaged over a large but constant scale) is constant, then in any direction on the sky there is some (possibly very faraway) star, occupying a finite (possibly minuscule) solid angle in our field of view. The 1/r² falloff (aka conservation of energy) means that the energy received per unit of solid angle is independent of distance from the emitter, equal for example to that on its surface, so every piece of the sky containing a star means we should see stellar-surface amounts of energy shining upon us from every direction. Assuming some sort of absorbing dust would obscure the stars doesn’t help: if the universe truly is eternal, every dust cloud, being unable to store arbitrarily large amounts of energy, will eventually heat up to the point that it radiates as much as it absorbs and so is as hot as the stars which it obscures (this is the insight behind Kirchhoff’s law and the existence of blackbody radiation). What does help is either implementing “no point in space is special” through a device other than simply a constant stellar density (e.g. having stars distributed on a self-similar fractal of Hausdorff dimension < 3, our falloff argument having included what amounts to a definition of Hausdorff dimension) or abandoning “no point in time is special” (giving the universe a finite age or at least having no stars in the far past). Observations show the second possibility (named the “Big Bang” by its critics in what was meant to show its ridiculousness) to be true. (Modern cosmology has much more direct arguments for a finite age of the universe, but they also require more advanced physics and/or observational technology, so the directness is in the eye of the beholder.)

by alecbz 5y ago

photoGrant 5y ago

Yes, it's called NOISE!

alecbz 5y ago
But what specifically is it? Just random stuff in deep space between galaxies? I’d have thought most of the space between galaxies is mostly truly empty.
- site-packages1 5y ago
  Isn't the paradox that if the universe is static in size or shrinking, eventually all the light would fill the sky into a blinding display, rather than dark. So it's a function of the universe expanding. https://en.wikipedia.org/wiki/Olbers%27_paradox https://en.wikipedia.org/wiki/Olbers%27_paradox
- nprecup 5y ago
  The noise is inherent to our sensor technology... Very sensitive telescopes are cryogenically cooled to reduce the temperature of the focal plane, reducing the noise. As in an above comment, the light from distant objects is more spread out and appears weaker - fewer photos are arriving each second and hitting the focal plane than if the telescope were closer. At a certain point, the signal gets buried in the noise. I worked on the BICEP Array telescope at the south Pole - the light we are observing is actually extremely low energy and we can't see it unless our detectors are colder than the light source we are looking at (which is about 2.7K). We cool our detectors to 0.3K!
  - daniel-cussen 5y ago
    Does the natural cold at the South Pole make it much easier to cool the instruments down that low? I know it's around 200K at its lowest, how does that make it easier and cheaper?
    - nprecup 5y ago
      Yes, there are several factors that make that location particularly good, not only temperature. The cold and absence of sunlight for months at a time causes a relative absence of water vapor in the atmosphere during winter, which is extremely important for CMB observations. The high elevation (~10k ft) means there is less atmosphere to look through as well. During summers the sky heats up, contains more water vapor, and reflects so much more light that our telescopes become substantially less effective. Another advantage to the South Pole is the ability to continuously observe the same part of the sky, as the telescope is located on the earths axis there is no rising and setting of sky overhead. There is also a particularly 'dark' patch of sky in the southern hemisphere called the 'southern hole'. This is located up and away from the plane of our galaxy and contains very few objects standing in the 'path' of the CMB which also helps obtain better quality observations of the CMB. Edit - it is as good as it gets on earth and way cheaper than a satellite! Other benefits are rapid upgrades and repairs with newer technology. A satellite, by the time it is deployed, is already pretty old! But the South Pole still isn't cheap!
- pgorczak 5y ago
  As a sibling already mentioned, thermal noise occurs in the sensors. And there is also a range of cosmic background radiation coming in, including the cosmic microwave background which is just everywhere and is almost as old as the universe.
- pjc50 5y ago
  The noise is coming from inside the sensor: https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise Cryogenically cooling helps, but since you can't hit absolute zero you can't entirely eliminate it.
  - alecbz 5y ago
    Right, what I was missing was why how far the source of the light is effects how much the noise matters, but the inverse square photon concentration explains that.
  - passivate 5y ago
    There is read noise from downstream electronics converting photons to electrons, but the "shot noise" is a property of the photon stream and will be present regardless of how much we improve the electronics. Cooling the detector won't help there. https://en.wikipedia.org/wiki/Shot_noise#Optics https://en.wikipedia.org/wiki/Shot_noise#Optics

hexatin 5y ago

The strength of light over a distance obeys an inverse square law[1], causing it to effectively lose power over long distances, since the same amount of energy is being spread over larger and larger spheres as it radiates out. For photons, I think this manifests as there being a lower rate of photons occurring in each section of the sphere since they’re “spread out”. [1] https://en.wikipedia.org/wiki/Inverse-square_law https://en.wikipedia.org/wiki/Inverse-square_law

site-packages1 5y ago
Doesn't this fail because we're looking at lots of stars spread out evenly, and so the farther you get away, the more stars fill the space?
- hypertele-Xii 5y ago
  Space is mostly empty.
  - mananaysiempre 5y ago
    As a sibling comment to yours says, Olbers’ paradox is that it doesn’t matter. If we assume that the universe is homogeneous, infinite, and eternal and the density of stars (possibly averaged over a large but constant scale) is constant, then in any direction on the sky there is some (possibly very faraway) star, occupying a finite (possibly minuscule) solid angle in our field of view. The 1/r² falloff (aka conservation of energy) means that the energy received per unit of solid angle is independent of distance from the emitter, equal for example to that on its surface, so every piece of the sky containing a star means we should see stellar-surface amounts of energy shining upon us from every direction. Assuming some sort of absorbing dust would obscure the stars doesn’t help: if the universe truly is eternal, every dust cloud, being unable to store arbitrarily large amounts of energy, will eventually heat up to the point that it radiates as much as it absorbs and so is as hot as the stars which it obscures (this is the insight behind Kirchhoff’s law and the existence of blackbody radiation). What does help is either implementing “no point in space is special” through a device other than simply a constant stellar density (e.g. having stars distributed on a self-similar fractal of Hausdorff dimension < 3, our falloff argument having included what amounts to a definition of Hausdorff dimension) or abandoning “no point in time is special” (giving the universe a finite age or at least having no stars in the far past). Observations show the second possibility (named the “Big Bang” by its critics in what was meant to show its ridiculousness) to be true. (Modern cosmology has much more direct arguments for a finite age of the universe, but they also require more advanced physics and/or observational technology, so the directness is in the eye of the beholder.)