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Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid?
Are there “gaps” in light, or will it hit everywhere?FOV reduction with FresnelApplication of diffraction problem!Photons arriving from the SunHow “wide” is a beam of light? What is its half-diameter?Shimmering from heated air and the speed of lightJustifying assumptions for deriving Poissonian photon statisticsIs it diffraction I see when I closely look at my finger(s)?At any given point in the universe, are there a huge number of photons crossing paths?Magnitude of the faintest star visible to the human eye?Which Particle would win the race?
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I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^45$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^-43$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^41 $ meters is the length of the white line. For reference, that's over $10^14$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
optics electromagnetic-radiation photons
edited Jun 17 at 3:43
David Z♦
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asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
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|
show 2 more comments
$begingroup$
I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^45$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^-43$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^41 $ meters is the length of the white line. For reference, that's over $10^14$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
optics electromagnetic-radiation photons
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
$endgroup$
28
$begingroup$
FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
$endgroup$
– Jahan Claes
Jun 16 at 2:52
6
$begingroup$
@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
$endgroup$
– Reggie Simmons
Jun 16 at 2:58
25
$begingroup$
I like how this is a pretty good question even if it's based on a simple mistake
$endgroup$
– JollyJoker
Jun 16 at 13:10
2
$begingroup$
That is the relevant fact and the right way to think about your answer, yeah.
$endgroup$
– Jahan Claes
Jun 16 at 15:27
2
$begingroup$
In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02
|
show 2 more comments
$begingroup$
I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^45$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^-43$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^41 $ meters is the length of the white line. For reference, that's over $10^14$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
optics electromagnetic-radiation photons
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
$endgroup$
I recently read an interesting article that states that a human being can perceive a flash of as few as 5 or so photons, and the human eye itself can perceive even a single photon. The brain will filter this out, however.
I wanted to calculate how far away you'd have to be standing from our sun for not a single one of its photons to be hitting your pupil over a given second.
The first thing I did was assume that the sun emits $10^45$ photons per second, because, well, that's the only number I could find through internet research.
The next step is to assume that the average angle between photons emitted from the sun is pretty much the same, and is equal to $3.6 × 10^-43$ degrees.
The next step is to assume that the average human pupil diameter is 0.005 meters, and then draw a triangle like so:
The length of the white line through the center of the triangle equals the distance at which two photons from the sun would be further apart than your pupil is wide, meaning not even one photon should hit your eye.
I broke the triangle into two pieces and solved for the white line by using the law of sines, and my final result is ridiculous.
$3.97887×10^41 $ meters is the length of the white line. For reference, that's over $10^14$ times the diameter of the observable universe.
My conclusion says that no matter how far you get from the sun within our observable universe, not only should some of the photons be hitting your pupil, but it should be more than enough for you to visually perceive.
But if I was right, I'd probably see a lot more stars from very far away every night when I looked up at the sky. Why is my calculation inconsistent with what I see?
optics electromagnetic-radiation photons
optics electromagnetic-radiation photons
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
edited Jun 17 at 3:43
David Z♦
64.3k23 gold badges140 silver badges256 bronze badges
64.3k23 gold badges140 silver badges256 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
asked Jun 16 at 0:50
Reggie SimmonsReggie Simmons
4211 gold badge2 silver badges7 bronze badges
4211 gold badge2 silver badges7 bronze badges
28
$begingroup$
FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
$endgroup$
– Jahan Claes
Jun 16 at 2:52
6
$begingroup$
@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
$endgroup$
– Reggie Simmons
Jun 16 at 2:58
25
$begingroup$
I like how this is a pretty good question even if it's based on a simple mistake
$endgroup$
– JollyJoker
Jun 16 at 13:10
2
$begingroup$
That is the relevant fact and the right way to think about your answer, yeah.
$endgroup$
– Jahan Claes
Jun 16 at 15:27
2
$begingroup$
In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02
|
show 2 more comments
28
$begingroup$
FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
$endgroup$
– Jahan Claes
Jun 16 at 2:52
6
$begingroup$
@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
$endgroup$
– Reggie Simmons
Jun 16 at 2:58
25
$begingroup$
I like how this is a pretty good question even if it's based on a simple mistake
$endgroup$
– JollyJoker
Jun 16 at 13:10
2
$begingroup$
That is the relevant fact and the right way to think about your answer, yeah.
$endgroup$
– Jahan Claes
Jun 16 at 15:27
2
$begingroup$
In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02
28
28
$begingroup$
FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
$endgroup$
– Jahan Claes
Jun 16 at 2:52
$begingroup$
FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
$endgroup$
– Jahan Claes
Jun 16 at 2:52
6
6
$begingroup$
@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
$endgroup$
– Reggie Simmons
Jun 16 at 2:58
$begingroup$
@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
$endgroup$
– Reggie Simmons
Jun 16 at 2:58
25
25
$begingroup$
I like how this is a pretty good question even if it's based on a simple mistake
$endgroup$
– JollyJoker
Jun 16 at 13:10
$begingroup$
I like how this is a pretty good question even if it's based on a simple mistake
$endgroup$
– JollyJoker
Jun 16 at 13:10
2
2
$begingroup$
That is the relevant fact and the right way to think about your answer, yeah.
$endgroup$
– Jahan Claes
Jun 16 at 15:27
$begingroup$
That is the relevant fact and the right way to think about your answer, yeah.
$endgroup$
– Jahan Claes
Jun 16 at 15:27
2
2
$begingroup$
In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02
$begingroup$
In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02
|
show 2 more comments
5 Answers
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$begingroup$
The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = fracN4pi r^2.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_texteye = fracN4pi r^2A_texteye.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = fracN4pi r^2A_texteye lt 1$$
Solving for $r$ gives
$$r > sqrtfracNA_texteye4pi$$
Plugging in your numbers gives
$$r > sqrtfrac(10^45)pi(0.005,textrmm/2)^24pi = 4cdot10^19 ,textrmm approx 4000,textrmlight-years$$
This distance is still well within our own galaxy.
edited Jun 17 at 9:33
Loong
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answered Jun 16 at 1:20
Mark HMark H
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There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.
Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).
Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.
But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).
The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.
Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
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2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
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show 4 more comments
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If the sun emits $N=10^45 $ photons per second, an observer at distance $R$ will receive $N/4pi R^2$ photons per second per unit area. If human eye has radius about $r=10^-3 m$, it will receive 1 photon per second if
$$fracNpi r^24pi R^2 =1, $$
$$R sim sqrtN r sim 10^20 msim 1 kpc. $$
This is less than the size of the universe.
answered Jun 16 at 1:26
atarasenkoatarasenko
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Besides the other errors pointed out in other posts, including the assumption of a 2d universe, the assumption that we can perceive single photons, the assumption that the photon remains of the same wavelength by the time it reaches us, there is still one more rather big factor, namely the assumption that space is completely empty, otherwise we have to take attenuation into account.
This even has a name, interstellar extinction, referring to the extinction of photons via absorption or scattering by interstellar dust or gas along their paths.
[Wikipedia] The amount of extinction can be significantly higher than this in specific directions. For example, some regions of the Galactic Center are awash with obvious intervening dark dust from our spiral arm (and perhaps others) and themselves in a bulge of dense matter, causing as much as more than $30$ magnitudes of extinction in the optical, meaning that less than $1$ optical photon in $10^12$ passes through.
Note that the Milky Way is not flat, and we sit roughly in the middle of one of its spiral arms, so any light has to pass through a significant amount of interstellar matter in our own galaxy before it can reach us. See here for a simple diagram. Beyond that, there is also quite some dust at the edge of our galaxy, and intergalactic dust between galaxies.
In fact, the observed low stellar EBL (extragalactic background light) is mainly due to intergalactic opacity, having a much greater effect than obscuration by galaxies. And both of these, together with the comparatively minor correction for finite age and expansion of the universe, appear to account for the observed EBL.
Of course, we also have atmospheric extinction, and you can also calculate roughly how much gets lost by the time it reaches sea level.
answered Jun 18 at 9:57
user21820user21820
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The observable universe is actually defined by the speed of light and the time since the universe began to exist in a recognisable form, not by the inverse-square law of light intensity falloff. The reason why we can't see stellar objects from beyond that distance (even with the aid of radio-telescopes) is because there are none old enough for their light to have reached us.
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
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I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
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Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
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5 Answers
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The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = fracN4pi r^2.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_texteye = fracN4pi r^2A_texteye.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = fracN4pi r^2A_texteye lt 1$$
Solving for $r$ gives
$$r > sqrtfracNA_texteye4pi$$
Plugging in your numbers gives
$$r > sqrtfrac(10^45)pi(0.005,textrmm/2)^24pi = 4cdot10^19 ,textrmm approx 4000,textrmlight-years$$
This distance is still well within our own galaxy.
edited Jun 17 at 9:33
Loong
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answered Jun 16 at 1:20
Mark HMark H
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The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = fracN4pi r^2.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_texteye = fracN4pi r^2A_texteye.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = fracN4pi r^2A_texteye lt 1$$
Solving for $r$ gives
$$r > sqrtfracNA_texteye4pi$$
Plugging in your numbers gives
$$r > sqrtfrac(10^45)pi(0.005,textrmm/2)^24pi = 4cdot10^19 ,textrmm approx 4000,textrmlight-years$$
This distance is still well within our own galaxy.
edited Jun 17 at 9:33
Loong
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answered Jun 16 at 1:20
Mark HMark H
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$endgroup$
add a comment |
$begingroup$
The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = fracN4pi r^2.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_texteye = fracN4pi r^2A_texteye.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = fracN4pi r^2A_texteye lt 1$$
Solving for $r$ gives
$$r > sqrtfracNA_texteye4pi$$
Plugging in your numbers gives
$$r > sqrtfrac(10^45)pi(0.005,textrmm/2)^24pi = 4cdot10^19 ,textrmm approx 4000,textrmlight-years$$
This distance is still well within our own galaxy.
edited Jun 17 at 9:33
Loong
1,56012 silver badges21 bronze badges
answered Jun 16 at 1:20
Mark HMark H
13.7k3 gold badges32 silver badges45 bronze badges
$endgroup$
The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.
So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by
$$I = fracN4pi r^2.$$
This comes from spreading out the photons over the surface of a sphere surrounding the sun.
The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye:
$$n = IA_texteye = fracN4pi r^2A_texteye.$$
You are looking for the distance beyond which you would see less than one photon per second:
$$n = fracN4pi r^2A_texteye lt 1$$
Solving for $r$ gives
$$r > sqrtfracNA_texteye4pi$$
Plugging in your numbers gives
$$r > sqrtfrac(10^45)pi(0.005,textrmm/2)^24pi = 4cdot10^19 ,textrmm approx 4000,textrmlight-years$$
This distance is still well within our own galaxy.
edited Jun 17 at 9:33
Loong
1,56012 silver badges21 bronze badges
answered Jun 16 at 1:20
Mark HMark H
13.7k3 gold badges32 silver badges45 bronze badges
edited Jun 17 at 9:33
Loong
1,56012 silver badges21 bronze badges
edited Jun 17 at 9:33
Loong
1,56012 silver badges21 bronze badges
edited Jun 17 at 9:33
Loong
1,56012 silver badges21 bronze badges
1,56012 silver badges21 bronze badges
answered Jun 16 at 1:20
Mark HMark H
13.7k3 gold badges32 silver badges45 bronze badges
answered Jun 16 at 1:20
Mark HMark H
13.7k3 gold badges32 silver badges45 bronze badges
answered Jun 16 at 1:20
Mark HMark H
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13.7k3 gold badges32 silver badges45 bronze badges
add a comment |
add a comment |
$begingroup$
There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.
Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).
Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.
But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).
The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.
Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
$endgroup$
2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
|
show 4 more comments
$begingroup$
There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.
Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).
Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.
But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).
The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.
Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
$endgroup$
2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
|
show 4 more comments
$begingroup$
There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.
Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).
Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.
But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).
The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.
Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
$endgroup$
There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.
Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).
Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.
But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).
The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.
Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
edited Jun 16 at 14:31
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
answered Jun 16 at 14:16
StilezStilez
1,9936 silver badges16 bronze badges
1,9936 silver badges16 bronze badges
2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
|
show 4 more comments
2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
2
2
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
$begingroup$
"redshift will make that impossible" -- so what is that redshift? will 500nm photon turn into invisible 2000nm photon or 520nm?
$endgroup$
– aaaaaa
Jun 16 at 18:47
4
4
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
$begingroup$
So what about the photons at higher frequencies? Why would they not be shifted from supervisible to intravisible range?
$endgroup$
– Nij
Jun 16 at 19:52
8
8
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
$begingroup$
@Nij That certainly is a good point - but sun-like stars emit the vast majority of their energy in photons no higher than near ultraviolet frequency. If you were looking at one from across the universe, at a redshift of, say, $zge1$, you would only see the part of the radiation that was originally emitted as far ultraviolet or X-rays, and there's very little of that.
$endgroup$
– David Z♦
Jun 17 at 3:49
2
2
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
$begingroup$
The earliest light we see, the CMB, was emitted at about 3000K, which has the peak intensity about 1μm. The scale factor for it is z = 1100, so now we observe it at 2.7K, or peak intensity of about 1mm, which is way out of visible light. The earliest stars started forming around z = 20, so the hottest ones with the temperature of 40,000K would now be seen at 2,000K, which is still visible.
$endgroup$
– Kolmar
Jun 17 at 5:15
1
1
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
$begingroup$
Another problem not mentioned here is that space isn't empty. There are gas atoms and molecules in the "empty" space, which will absorb or scatter some additional photons as they travel toward the observer. I don't know how much of a factor that would be, but I'm pretty sure it would further diminish the distance of visibility.
$endgroup$
– fbitterlich
Jun 17 at 13:15
|
show 4 more comments
$begingroup$
If the sun emits $N=10^45 $ photons per second, an observer at distance $R$ will receive $N/4pi R^2$ photons per second per unit area. If human eye has radius about $r=10^-3 m$, it will receive 1 photon per second if
$$fracNpi r^24pi R^2 =1, $$
$$R sim sqrtN r sim 10^20 msim 1 kpc. $$
This is less than the size of the universe.
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
$endgroup$
add a comment |
$begingroup$
If the sun emits $N=10^45 $ photons per second, an observer at distance $R$ will receive $N/4pi R^2$ photons per second per unit area. If human eye has radius about $r=10^-3 m$, it will receive 1 photon per second if
$$fracNpi r^24pi R^2 =1, $$
$$R sim sqrtN r sim 10^20 msim 1 kpc. $$
This is less than the size of the universe.
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
$endgroup$
add a comment |
$begingroup$
If the sun emits $N=10^45 $ photons per second, an observer at distance $R$ will receive $N/4pi R^2$ photons per second per unit area. If human eye has radius about $r=10^-3 m$, it will receive 1 photon per second if
$$fracNpi r^24pi R^2 =1, $$
$$R sim sqrtN r sim 10^20 msim 1 kpc. $$
This is less than the size of the universe.
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
$endgroup$
If the sun emits $N=10^45 $ photons per second, an observer at distance $R$ will receive $N/4pi R^2$ photons per second per unit area. If human eye has radius about $r=10^-3 m$, it will receive 1 photon per second if
$$fracNpi r^24pi R^2 =1, $$
$$R sim sqrtN r sim 10^20 msim 1 kpc. $$
This is less than the size of the universe.
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
answered Jun 16 at 1:26
atarasenkoatarasenko
4111 silver badge7 bronze badges
4111 silver badge7 bronze badges
add a comment |
add a comment |
$begingroup$
Besides the other errors pointed out in other posts, including the assumption of a 2d universe, the assumption that we can perceive single photons, the assumption that the photon remains of the same wavelength by the time it reaches us, there is still one more rather big factor, namely the assumption that space is completely empty, otherwise we have to take attenuation into account.
This even has a name, interstellar extinction, referring to the extinction of photons via absorption or scattering by interstellar dust or gas along their paths.
[Wikipedia] The amount of extinction can be significantly higher than this in specific directions. For example, some regions of the Galactic Center are awash with obvious intervening dark dust from our spiral arm (and perhaps others) and themselves in a bulge of dense matter, causing as much as more than $30$ magnitudes of extinction in the optical, meaning that less than $1$ optical photon in $10^12$ passes through.
Note that the Milky Way is not flat, and we sit roughly in the middle of one of its spiral arms, so any light has to pass through a significant amount of interstellar matter in our own galaxy before it can reach us. See here for a simple diagram. Beyond that, there is also quite some dust at the edge of our galaxy, and intergalactic dust between galaxies.
In fact, the observed low stellar EBL (extragalactic background light) is mainly due to intergalactic opacity, having a much greater effect than obscuration by galaxies. And both of these, together with the comparatively minor correction for finite age and expansion of the universe, appear to account for the observed EBL.
Of course, we also have atmospheric extinction, and you can also calculate roughly how much gets lost by the time it reaches sea level.
answered Jun 18 at 9:57
user21820user21820
1,49911 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Besides the other errors pointed out in other posts, including the assumption of a 2d universe, the assumption that we can perceive single photons, the assumption that the photon remains of the same wavelength by the time it reaches us, there is still one more rather big factor, namely the assumption that space is completely empty, otherwise we have to take attenuation into account.
This even has a name, interstellar extinction, referring to the extinction of photons via absorption or scattering by interstellar dust or gas along their paths.
[Wikipedia] The amount of extinction can be significantly higher than this in specific directions. For example, some regions of the Galactic Center are awash with obvious intervening dark dust from our spiral arm (and perhaps others) and themselves in a bulge of dense matter, causing as much as more than $30$ magnitudes of extinction in the optical, meaning that less than $1$ optical photon in $10^12$ passes through.
Note that the Milky Way is not flat, and we sit roughly in the middle of one of its spiral arms, so any light has to pass through a significant amount of interstellar matter in our own galaxy before it can reach us. See here for a simple diagram. Beyond that, there is also quite some dust at the edge of our galaxy, and intergalactic dust between galaxies.
In fact, the observed low stellar EBL (extragalactic background light) is mainly due to intergalactic opacity, having a much greater effect than obscuration by galaxies. And both of these, together with the comparatively minor correction for finite age and expansion of the universe, appear to account for the observed EBL.
Of course, we also have atmospheric extinction, and you can also calculate roughly how much gets lost by the time it reaches sea level.
answered Jun 18 at 9:57
user21820user21820
1,49911 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Besides the other errors pointed out in other posts, including the assumption of a 2d universe, the assumption that we can perceive single photons, the assumption that the photon remains of the same wavelength by the time it reaches us, there is still one more rather big factor, namely the assumption that space is completely empty, otherwise we have to take attenuation into account.
This even has a name, interstellar extinction, referring to the extinction of photons via absorption or scattering by interstellar dust or gas along their paths.
[Wikipedia] The amount of extinction can be significantly higher than this in specific directions. For example, some regions of the Galactic Center are awash with obvious intervening dark dust from our spiral arm (and perhaps others) and themselves in a bulge of dense matter, causing as much as more than $30$ magnitudes of extinction in the optical, meaning that less than $1$ optical photon in $10^12$ passes through.
Note that the Milky Way is not flat, and we sit roughly in the middle of one of its spiral arms, so any light has to pass through a significant amount of interstellar matter in our own galaxy before it can reach us. See here for a simple diagram. Beyond that, there is also quite some dust at the edge of our galaxy, and intergalactic dust between galaxies.
In fact, the observed low stellar EBL (extragalactic background light) is mainly due to intergalactic opacity, having a much greater effect than obscuration by galaxies. And both of these, together with the comparatively minor correction for finite age and expansion of the universe, appear to account for the observed EBL.
Of course, we also have atmospheric extinction, and you can also calculate roughly how much gets lost by the time it reaches sea level.
answered Jun 18 at 9:57
user21820user21820
1,49911 silver badges17 bronze badges
$endgroup$
Besides the other errors pointed out in other posts, including the assumption of a 2d universe, the assumption that we can perceive single photons, the assumption that the photon remains of the same wavelength by the time it reaches us, there is still one more rather big factor, namely the assumption that space is completely empty, otherwise we have to take attenuation into account.
This even has a name, interstellar extinction, referring to the extinction of photons via absorption or scattering by interstellar dust or gas along their paths.
[Wikipedia] The amount of extinction can be significantly higher than this in specific directions. For example, some regions of the Galactic Center are awash with obvious intervening dark dust from our spiral arm (and perhaps others) and themselves in a bulge of dense matter, causing as much as more than $30$ magnitudes of extinction in the optical, meaning that less than $1$ optical photon in $10^12$ passes through.
Note that the Milky Way is not flat, and we sit roughly in the middle of one of its spiral arms, so any light has to pass through a significant amount of interstellar matter in our own galaxy before it can reach us. See here for a simple diagram. Beyond that, there is also quite some dust at the edge of our galaxy, and intergalactic dust between galaxies.
In fact, the observed low stellar EBL (extragalactic background light) is mainly due to intergalactic opacity, having a much greater effect than obscuration by galaxies. And both of these, together with the comparatively minor correction for finite age and expansion of the universe, appear to account for the observed EBL.
Of course, we also have atmospheric extinction, and you can also calculate roughly how much gets lost by the time it reaches sea level.
answered Jun 18 at 9:57
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answered Jun 18 at 9:57
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answered Jun 18 at 9:57
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answered Jun 18 at 9:57
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add a comment |
add a comment |
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The observable universe is actually defined by the speed of light and the time since the universe began to exist in a recognisable form, not by the inverse-square law of light intensity falloff. The reason why we can't see stellar objects from beyond that distance (even with the aid of radio-telescopes) is because there are none old enough for their light to have reached us.
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
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2
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I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
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– Michael Eric Oberlin
Jun 17 at 8:36
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Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
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– Chromatix
Jun 17 at 11:14
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I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
add a comment |
$begingroup$
The observable universe is actually defined by the speed of light and the time since the universe began to exist in a recognisable form, not by the inverse-square law of light intensity falloff. The reason why we can't see stellar objects from beyond that distance (even with the aid of radio-telescopes) is because there are none old enough for their light to have reached us.
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
$endgroup$
2
$begingroup$
I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
$begingroup$
Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
add a comment |
$begingroup$
The observable universe is actually defined by the speed of light and the time since the universe began to exist in a recognisable form, not by the inverse-square law of light intensity falloff. The reason why we can't see stellar objects from beyond that distance (even with the aid of radio-telescopes) is because there are none old enough for their light to have reached us.
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
$endgroup$
The observable universe is actually defined by the speed of light and the time since the universe began to exist in a recognisable form, not by the inverse-square law of light intensity falloff. The reason why we can't see stellar objects from beyond that distance (even with the aid of radio-telescopes) is because there are none old enough for their light to have reached us.
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
answered Jun 17 at 2:21
ChromatixChromatix
1703 bronze badges
1703 bronze badges
2
$begingroup$
I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
$begingroup$
Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
add a comment |
2
$begingroup$
I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
$begingroup$
Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
2
2
$begingroup$
I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
$begingroup$
I feel that this fails to consider the expansion of the universe in its early days; which was evidently much faster than the speed of light. The current size of the observable universe is approximately 8.8 × 10²⁶ m, but its age is estimated at a little less than 14 billion years; this leads to a maximum distance traveled (given constant space) by light of only 1.3 × 10²⁶ m.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 8:36
$begingroup$
Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
Does this calculation treat "size" as a radius or a diameter, and does it also account for assumed movement of the observed objects in the (considerable) time since their observed emissions occurred? Those details could account for some of the discrepancy.
$endgroup$
– Chromatix
Jun 17 at 11:14
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
$begingroup$
I don't remember at this point, but size or diameter and you have a noticeable discrepancy. Generally it describes COBE, which has always been there. Additionally, this is speed-of-information; where the object goes after the event of emission is irrelevant, as it would have no say in our local reference frame. Respectfully, you aren't working with all of the data or theory.
$endgroup$
– Michael Eric Oberlin
Jun 17 at 17:09
add a comment |
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FWIW, while it's true that you DID do something wrong and the actual answer is much smaller as you'd expect, there's no physical law that says the answer CAN'T be larger than the observable universe.
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– Jahan Claes
Jun 16 at 2:52
6
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@JahanClaes I figured as much - I just assumed I was wrong because if I was right, I'd probably see a LOT more stars from VERY far away every night when I looked up at the sky.
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– Reggie Simmons
Jun 16 at 2:58
25
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I like how this is a pretty good question even if it's based on a simple mistake
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– JollyJoker
Jun 16 at 13:10
2
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That is the relevant fact and the right way to think about your answer, yeah.
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– Jahan Claes
Jun 16 at 15:27
2
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In addition to the several factors mentioned in all the given answers, there are other factors as well. Like, space isn't empty, so attenuation occurs for other reasons than just surface area spread. Also, the photon flux per unit area from the sun's surface is not uniform (and not necessarily static either), so there are some directions from which you might be able to see it just a bit further away... Even with those, this isn't an exhaustive list...
$endgroup$
– twalberg
Jun 17 at 19:02