5 ms·

Isn't this just an issue of comparing a countable and uncountable infinity? The number of points on the unit sphere is uncountable, but the number of stars is c

by pontus 6y ago

Isn't this just an issue of comparing a countable and uncountable infinity? The number of points on the unit sphere is uncountable, but the number of stars is countable. As such there are in some sense more points on the sphere than there are stars, even though there are an infinite number of each. Take this together with the fact that intensity falls off as the square of the distance and it seems like the sky should be dark.

oconnor663 6y ago
I don't think this is an issue. A star isn't a single point, and each one maps to a small-but-not-infinitessimal area on the unit sphere.
cambalache 6y ago
No. At no distance "R" from the earth does a start becomes a point. A point does not have a surface area, a star does (even if it is very far away).
pontus 6y ago
I guess the fall-off is even worse. At the beginning the intensity falls off as 1/r^2, but eventually the intensity becomes so small that you're talking about individual photons. At some point the intensity will then fall from a single photon to zero. So, after some critical distance the intensity will actually drop to zero. More formally, each star emits some amount of power in each frequency band: P(f) so that \int_0^\infty P(f) df = P_total. For each frequency then, we have a total of P(f)/(hf) photon emitted per second. The total number of photons emitted per second by the star is then \int_0^\infty df P(f)/hf which is a finite number. The total number of photons received per unit area a distance r away from the star would then be \frac{1}{4\pi r^2} \int_0^\infty df P(f)/hf If your detector has an area A (e.g. your retina or some other device), you'd expect to see \frac{A}{4\pi r^2} \int_0^\infty df P(f)/hf photons per second from the star. As r gets really large, you'd see this drop arbitrarily low. Conversely, the amount of time you'd need to wait to see a single photon from that star then grows, making the star dark.
- pdonis 6y ago
  > after some critical distance the intensity will actually drop to zero. No, it won't. If you're going to use a quantum model of light (which you have to to use the concept of "photon"), then you have to use the quantum interpretation of "intensity". The quantum interpretation of "intensity" is the probability of detecting a photon; and this is a continuous quantity which can get smaller and smaller indefinitely without ever dropping to zero.
  - pontus 6y ago
    The probability can then get arbitrarily small, meaning that the expected amount of time needed before the probability of having observed a photon would get progressively larger. My argument above is semi-classical, but it shouldn't change with a full quantum mechanical approach.
    - pdonis 6y ago
      > The probability can then get arbitrarily small, meaning that the expected amount of time needed before the probability of having observed a photon would get progressively larger. Yes, but the probability is never zero, and the expected time is never infinite. So saying "the intensity drops to zero" is never correct.
      - pontus 6y ago
        The point is not that the probability needs to hit zero, it's that it's not correct to say that you receive a quarter of the power as you move twice as far from the source. It's still true that the expected number of photons per second drops by a factor of 4, but it can drop so far as to render the source dark for an appreciable amount of time. The paradox claims that the sky should appear bright, which I take to mean that a detector should be receiving light from each point in the sky at each moment in time. It does not say that the detector will receive light from each part of the sky at some point, but that you may need to wait a million years before a particular point flickers and that, even then, there's nothing that guarantees that all points will flicker at the same time.
colanderman 6y ago
Fall-off of light intensity does not factor in: while the amount of light reaching an observer from any given star does indeed fall off with the square of the distance, so does the apparent size of that star; its apparent surface brightness thus does not change with distance. (Think of day-to-day experience: people who walk away from you do not darken!) I agree that countability vs. uncountability seems like it should come into play.
- pontus 6y ago
  Great point, thanks! Makes perfect sense. Turns out though that the intensity actually falls off faster than 1/r^2 toward the end due to quantization effects. Feels silly to include quantization, but I guess when were talking about stars that may be arbitrarily far away this would need to be part of the story.
  - pdonis 6y ago
    > Turns out though that the intensity actually falls off faster than 1/r^2 toward the end due to quantization effects. No, it doesn't. See my other post in response to you upthread.
4ad 6y ago
No, the rational numbers are dense.

by pontus 6y ago

oconnor663 6y ago

I don't think this is an issue. A star isn't a single point, and each one maps to a small-but-not-infinitessimal area on the unit sphere.

cambalache 6y ago

No. At no distance "R" from the earth does a start becomes a point. A point does not have a surface area, a star does (even if it is very far away).

pontus 6y ago

I guess the fall-off is even worse. At the beginning the intensity falls off as 1/r^2, but eventually the intensity becomes so small that you're talking about individual photons. At some point the intensity will then fall from a single photon to zero. So, after some critical distance the intensity will actually drop to zero. More formally, each star emits some amount of power in each frequency band: P(f) so that \int_0^\infty P(f) df = P_total. For each frequency then, we have a total of P(f)/(hf) photon emitted per second. The total number of photons emitted per second by the star is then \int_0^\infty df P(f)/hf which is a finite number. The total number of photons received per unit area a distance r away from the star would then be \frac{1}{4\pi r^2} \int_0^\infty df P(f)/hf If your detector has an area A (e.g. your retina or some other device), you'd expect to see \frac{A}{4\pi r^2} \int_0^\infty df P(f)/hf photons per second from the star. As r gets really large, you'd see this drop arbitrarily low. Conversely, the amount of time you'd need to wait to see a single photon from that star then grows, making the star dark.

pdonis 6y ago
> after some critical distance the intensity will actually drop to zero. No, it won't. If you're going to use a quantum model of light (which you have to to use the concept of "photon"), then you have to use the quantum interpretation of "intensity". The quantum interpretation of "intensity" is the probability of detecting a photon; and this is a continuous quantity which can get smaller and smaller indefinitely without ever dropping to zero.
- pontus 6y ago
  The probability can then get arbitrarily small, meaning that the expected amount of time needed before the probability of having observed a photon would get progressively larger. My argument above is semi-classical, but it shouldn't change with a full quantum mechanical approach.
  - pdonis 6y ago
    > The probability can then get arbitrarily small, meaning that the expected amount of time needed before the probability of having observed a photon would get progressively larger. Yes, but the probability is never zero, and the expected time is never infinite. So saying "the intensity drops to zero" is never correct.
    - pontus 6y ago
      The point is not that the probability needs to hit zero, it's that it's not correct to say that you receive a quarter of the power as you move twice as far from the source. It's still true that the expected number of photons per second drops by a factor of 4, but it can drop so far as to render the source dark for an appreciable amount of time. The paradox claims that the sky should appear bright, which I take to mean that a detector should be receiving light from each point in the sky at each moment in time. It does not say that the detector will receive light from each part of the sky at some point, but that you may need to wait a million years before a particular point flickers and that, even then, there's nothing that guarantees that all points will flicker at the same time.

colanderman 6y ago

Fall-off of light intensity does not factor in: while the amount of light reaching an observer from any given star does indeed fall off with the square of the distance, so does the apparent size of that star; its apparent surface brightness thus does not change with distance. (Think of day-to-day experience: people who walk away from you do not darken!) I agree that countability vs. uncountability seems like it should come into play.

pontus 6y ago
Great point, thanks! Makes perfect sense. Turns out though that the intensity actually falls off faster than 1/r^2 toward the end due to quantization effects. Feels silly to include quantization, but I guess when were talking about stars that may be arbitrarily far away this would need to be part of the story.
- pdonis 6y ago
  > Turns out though that the intensity actually falls off faster than 1/r^2 toward the end due to quantization effects. No, it doesn't. See my other post in response to you upthread.

4ad 6y ago

No, the rational numbers are dense.