Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – NowIGetToLearnWhatAHeadIs Nov 26 '13 at 16:45
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
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. – Michael Kjörling Nov 26 '13 at 20:34
possible duplicate of Are there "gaps" in light, or will it hit everywhere? – Matsemann Nov 26 '13 at 23:29
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
up vote 57 down vote accepted

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 per second.

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

share|cite|improve this answer
Marginally related: – John Rennie Nov 26 '13 at 18:51
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 B Nov 27 '13 at 4:05
This and many of the other answers have helped a lot. Thank you all for answering! – Shookster Nov 27 '13 at 4:55
@Shookster if this or another response have answered your question please consider marking them as the answer. – quant Nov 27 '13 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.

share|cite|improve this answer
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
@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. – Jerry Schirmer 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
@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 7:55

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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
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
A quick safety note: please do not stare into laser beams. – Thomas Nov 28 '13 at 3:54
"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. – user20846 Nov 28 '13 at 7:57
@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 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.

share|cite|improve this answer

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...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.