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It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?

Brandon Enright
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Shookster
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    I think there are two questions here: 1) Why are stars visible even though they are so far away? and 2) Why do stars appears to subtend such a large solid angle even though they are so far away? – Brian Moths Nov 26 '13 at 16:45
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    By this logic, you wouldn't need to step to the side. You could just stand there and they'd blink because so few photons would be reaching your eye. – Brandon Enright Nov 26 '13 at 17:04
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    By this logic, you wouldn't need to step to the side. The Earth is already moving relative to the star much faster than you are moving relative to the Earth. – user Nov 26 '13 at 20:34
  • I want to expand on NowIGetToLearnWhatAHeadIs. You not only has two questions but they ground on the opposite claims: why do we see so few photons that we can filter them individually (we should see much more photons) and, second, why do we see so much photons - we should not see them at all! This question is based on contradicitons and is not a question at all since it answers itself: it is a fact that stars generate that much photons as you see. What is the question? – Val Nov 26 '13 at 21:05
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    possible duplicate of Are there "gaps" in light, or will it hit everywhere? – Matsemann Nov 26 '13 at 23:29
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    A corollary question to this is: Are the photons we receive from stars localized to classical trajectories, or are they spread out? If so, how much? Does the photon's wavefunction look more like the tip of a ray, or more like an expanding half-sphere? – jdm Nov 27 '13 at 15:54
  • @jdm That is interesting. I expect the expectation values would be the same, but is there in any way a measurable effect? – Captain Giraffe Nov 27 '13 at 23:43
  • @CaptainGiraffe Yes you can measure individual photons. – Bill Alsept Jul 22 '21 at 15:33

7 Answers7

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The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons).

You can do the calculation: If you're 10 light-years away from that star, you are nevertheless getting bombarded by 1 million photons per square centimeter in each second.

$$\frac{10^{45}\ \text{photons}/\mathrm s}{4\pi (10 \ \text{lightyears})^2} \approx 10^6\ \text{photons}/(\mathrm{cm^2\ s)}$$

Steve Byrnes
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  • Marginally related: http://physics.stackexchange.com/q/83866/ – John Rennie Nov 26 '13 at 18:51
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    Good answer. I think you could make it better by showing how you arrived at that 1 million photons / sq. cm figure though. – quant Nov 26 '13 at 22:34
  • @ArmanSchwarz - OK, I put the equation – Steve Byrnes Nov 27 '13 at 04:05
  • This and many of the other answers have helped a lot. Thank you all for answering! – Shookster Nov 27 '13 at 04:55
  • @Shookster if this or another response have answered your question please consider marking them as the answer. – quant Nov 27 '13 at 07:26
  • I am not sure of how to mark it as the answer. – Shookster Nov 28 '13 at 04:19
  • Please use the checkmark close to the vote up-down buttons, towards the top left of the answer. – Vaibhav Garg Nov 29 '13 at 07:48
  • The question is At what distance do the photons emitted in one second (and emitted uniformly in all directions) subtend an angle of more than (say) 5 degrees to the stars centre.This distance if more than the radius of our observable universe... will imply one will indeed keep seeing a star even if one steps to the side....of course one assumes that photons emitted in one second for a star of a particular type say our Sun.. – user82412 Jan 22 '16 at 13:07
  • Secondly why don't we face this problem if we consider wave nature of light. – user82412 Jan 22 '16 at 13:11
  • This answer doesn't seem very satisfactory ... what if the star is in another galaxy 10 megaparsecs away? The number of photons would drop dramatically, does that mean we now see it flicker? – Allure Apr 18 '19 at 00:11
  • @Allure if a star is 10 megaparsecs away, it's too dim to be visible at all (with the naked eye). – Steve Byrnes Apr 18 '19 at 17:29
  • @SteveByrnes right, but it's still visible with a telescope - do we see it flicker then? – Allure Apr 18 '19 at 20:11
  • @Allure If you are imaging a star which produces N photons on average per frame, then some frames it will be more than N photons, some frames fewer than N, according to a Poisson distribution. And yes some frames may have zero photons, if N is not too big. I'm not sure that "flicker" is the right technical term for this (it's usually called "shot noise"), but I think it's the same thing you're talking about. – Steve Byrnes Apr 19 '19 at 00:38
  • I think they flicker when enough water molecules (or portions thereof) happen to pass between the path the photons are taking and the terrestrial observor: In space, my understanding is that they don't flicker. – Edouard Jan 10 '22 at 02:38
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Although I agree with all three of the above answers let me present a slightly different perspective on the problem.

It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light and localises it.

The light from the star is not a hail of photons but instead the star is transferring energy to the photon quantum field and this energy spreads out radially and evenly. If you were to describe the light as photons you'd have to say the photons were completely delocalised i.e. they are spread over the whole spherical wavefront and you could not say in which direction the photon was travelling.

As the energy reaches you it can interact with the rhodopsin molecules in your eye and transfer one photon's worth of energy. It's at this point, and only at this point, that the energy is localised into a photon. Even if the star were so dim that it only emitted a few photons worth of energy per second there would still be a finite probability that your eye could interact with it and detect a photon, though that probability would obviously be ludicrously small.

So stepping aside would make little difference because as long as your eye intersected the spherical wavefront somewhere there would still be a finite probability of detecting a photon and therefore seeing the star.

Have a look at my answer to Some doubts about photons for some related arguments.

Community
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John Rennie
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  • Hi John, is this essentially the path integral view of things? That a photon is spread out and takes all possible paths? If so, isn't the radial spread of one photon worth of energy not uniform but instead highly likely to be found in a small area? – Brandon Enright Nov 26 '13 at 18:55
  • @BrandonEnright: the spread of any single photon is spherically symmetric, but if emitted at the surface of a star the $2\pi$ steradians occupied by the star will probably spoil the original symmetry a bit :-) – John Rennie Nov 26 '13 at 18:57
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    @BrandonEnright: you don't need to use the path integral formulation of QFT to come to that conclusion. This is just wave-particle duality. VERY loosely (I hesitate to say this, because it has wrong bits and right bits), things propogate like waves, and they interact like particles. – Zo the Relativist Nov 26 '13 at 18:58
  • @JohnRennie perhaps this needs to be asked as a question then but if a "single photon" is emitted with spherical symmetry how is momentum conserved? I thought each quanta of energy emitted as "a photon" needed to conserve momentum so the probability of it being found in a particular location was based on the uncertainty of the original momentum. Essentially meaning that you're highly likely to find that photon's quanta of energy pretty much in a straight line from wherever the momentum vector was pointing when it was created? – Brandon Enright Nov 26 '13 at 19:01
  • @BrandonEnright: if you insist on thinking of photons you would have to say that the photon and the system that emitted it are in a superposition of states with different momenta. You would only observe the momentum to be in a particular direction when you interacted with the superposition and collapsed it. If you want to pursue this it probably should be a new question rather than hijacking this one. – John Rennie Nov 26 '13 at 19:03
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    This answer is sort of missing the point. It is correct to calculate a starlight photon flux, multiply it by the photon absorption cross-section of rhodopsin, and thereby calculate rhodopsin's photon absorption rate (as if the photons were bullets, randomly distributed in all directions). You are insinuating that there's something wrong with this procedure, but there's not. It gives the right answer! (I mean, the right answer for calculating light absorption rate.) – Steve Byrnes Nov 26 '13 at 21:33
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    @SteveB: yes, both calculations will give the same result for the probability that a photon will enter the eye in a given time. – John Rennie Nov 27 '13 at 07:55
  • @SteveB (and John) I agree that both answers are "right" but John's answer is, for me, the better intuition as it gets away from the "billiard ball" idea of particles: namely, to think of the one, unique electromagnetic field and then the particulate interactions it has with the other quantum fields. See http://physics.stackexchange.com/a/78949/26076 – Selene Routley Nov 28 '13 at 04:00
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    @SteveB there's absolutely nothing wrong with the other answers, though, it's just that the image they tend to bear is kind of baffling when you come to consider, for example, how does a photon know about e.g. Brewster angles when it crosses an interface: thought of as a billiard ball, one tends to think of its interacting "discretely" with only a few molecules at most, but in reality it becomes a quantum superposition of a free photon and excited states of all the molecules in its ME-foreseen field of influence, thus really does see a continuum like the classical field. – Selene Routley Nov 28 '13 at 04:04
  • While both answers are "right" (inasmuch as they are both models), this one feels "more right" :-) – Jim Garrison Oct 28 '15 at 19:30
  • Now for what's probably not the brightest "request for clarification" ever presented: Is the wavefront, in the context of this answer, "spherical" because the star is spherical, because the eye is more-or-less spherical, or because of both of those facts? (As I spent a lot of my life a few mi. from a beach, the front of a wave being spherical is hard to grasp, but I'm making this request because of someone else's question, presented in a paper but not to PSE, about how we can still see the CMB.) – Edouard Jan 10 '22 at 18:32
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The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable.

Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough.

So your question is basically embroiled in observer bias; it assumes that the stars you see are all the stars there are, and it assumes that you could see a single photon if it hit your eye.

Kaz
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Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to blink.

So because you can see the star and it's relatively bright, that means there is enough of a continuous stream of photons reaching the Earth that stepping side to side doesn't change anything. Also, angular resolution isn't quantized so there is never a situation where stepping side to side (while maintaining the same radius from the star) ever changes the probability of receiving a photon.

Brandon Enright
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A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons.

A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer see the source.

gregsan
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    I think the OP is envisioning some sort of gap between "photon streams" which increases as the distance traveled increases in the same way that a slight l/r angle difference at launch time in the firing of two arrows results in an ever-widening gap between them when measure at the target. I feel this is a flaw in his/her conception, but this answer may not address that (?) – horatio Nov 26 '13 at 17:23
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    Seems to answer the part about being able to step to the side and no longer see the star. – Roman Hunde Nov 26 '13 at 17:47
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    A quick safety note: please do not stare into laser beams. – Thomas Nov 28 '13 at 03:54
  • "A laser radiates in only one direction..." is that really true? My understanding is that laser light is unusual only in being coherent - of a single wavelength. The quality of being collimated into a beam (a very small cone) derives from focusing lenses, which can be used with any light source. –  Nov 28 '13 at 07:57
  • @JonofAllTrades I don't think gregsan was talking about that property of the laser, but just referring to any ordinary object that emits light into a cone rather than radiating outwards in every direction (a laser pointer comes to mind..) but the explanation would be valid even for a flashlight (barring atmospheric scattering effects, of course) – Thomas Nov 28 '13 at 08:09
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A very non-technical answer, but in trying to get your head around this, have you thought about the speed of light?

The angle distended by the star on your eyeball (or by your eyeball on the star) is very small. So it seems like a very tiny region of space must be 'full of photons' for the star to be constantly visible, and since the point where you are standing is not special, all similar regions must be equally 'full'. But the region in question is actually a very narrow beam whose length is the speed of light multiplied by the time that images persist in our vision. If the latter is 50ms, the length of the column is 15,000 km - the diameter of the earth. In this there would need to be a few dozen photons for the star to be marginally visible iirc.

Not a rigorous explanation I know, but it might help reconcile your intuition with the science.

jwg
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So starlight propagates spherically and each human eyeball creates localized photons just at the intersection of wavefront and retina. No matter where you are in relation to the star some part of this wavefront will reveal the photon stream. Some kind of sensor that could image the path of all the photons/wave functions as they were emitted would reveal a solid hemisphere of light expanding away from the star...

user89463
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