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?

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    $\begingroup$ 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? $\endgroup$ Nov 26, 2013 at 16:45
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    $\begingroup$ 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. $\endgroup$ Nov 26, 2013 at 17:04
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    $\begingroup$ 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. $\endgroup$
    – user
    Nov 26, 2013 at 20:34
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    $\begingroup$ possible duplicate of Are there "gaps" in light, or will it hit everywhere? $\endgroup$
    – Matsemann
    Nov 26, 2013 at 23:29
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    $\begingroup$ 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? $\endgroup$
    – jdm
    Nov 27, 2013 at 15:54

7 Answers 7


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.

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

  • $\begingroup$ Marginally related: physics.stackexchange.com/q/83866 $\endgroup$ Nov 26, 2013 at 18:51
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    $\begingroup$ Good answer. I think you could make it better by showing how you arrived at that 1 million photons / sq. cm figure though. $\endgroup$
    – quant
    Nov 26, 2013 at 22:34
  • $\begingroup$ @ArmanSchwarz - OK, I put the equation $\endgroup$ Nov 27, 2013 at 4:05
  • $\begingroup$ This and many of the other answers have helped a lot. Thank you all for answering! $\endgroup$
    – Shookster
    Nov 27, 2013 at 4:55
  • $\begingroup$ @Shookster if this or another response have answered your question please consider marking them as the answer. $\endgroup$
    – quant
    Nov 27, 2013 at 7:26

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.

  • $\begingroup$ 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? $\endgroup$ Nov 26, 2013 at 18:55
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    $\begingroup$ @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. $\endgroup$ Nov 26, 2013 at 18:58
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    $\begingroup$ 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.) $\endgroup$ Nov 26, 2013 at 21:33
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    $\begingroup$ @SteveB: yes, both calculations will give the same result for the probability that a photon will enter the eye in a given time. $\endgroup$ Nov 27, 2013 at 7:55
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    $\begingroup$ @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. $\endgroup$ Nov 28, 2013 at 4:04

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.


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.


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.

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    $\begingroup$ 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 (?) $\endgroup$
    – horatio
    Nov 26, 2013 at 17:23
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    $\begingroup$ Seems to answer the part about being able to step to the side and no longer see the star. $\endgroup$ Nov 26, 2013 at 17:47
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    $\begingroup$ A quick safety note: please do not stare into laser beams. $\endgroup$
    – Thomas
    Nov 28, 2013 at 3:54
  • $\begingroup$ "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. $\endgroup$
    – user20846
    Nov 28, 2013 at 7:57
  • $\begingroup$ @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) $\endgroup$
    – Thomas
    Nov 28, 2013 at 8:09

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.


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...

  • $\begingroup$ Eyes cannot figure out what a wavefront is. $\endgroup$
    – Karthik
    Apr 18, 2019 at 17:05

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