# Why is there a photon every millimeter on earth coming from a star millions of light years away?

I've been having this confusing thought for so long now it would be amazing if someone could answer me.

Imagine this asterisk * . As you see, from the center point, lines go outwards, just like a sun will emit rays of light in all direction.

BUT, theoretically, there should be a "finite" amount of photons it sends in space, which means that the farther you are from that sun, let's say 1 million light years, the less likely are your eyes of catching the photons emitted from the source?

So does this make the theory of the photon bad? Since we can see a star millions of light years away, no matter where we stand on the ground of this planet. This means that the star can emit at least one photon every millimeter or less millions of light years away so this would mean that it would need an "impossible" density of photons sent at the source on the sun's surface in every direction of the cosmos.

How's this possible?

If you observe a star of radius $R$ from the distance $L$, you will see it as a small disk under the angle $2R/L$ so the solid angle the photons from the star will cover will scale like $(R/L)^2$. That's the percentage of the retina that will be receiving photons: the solid angle measures the "percentage of directions" in which the photons from the star are flying. The number of photons from the star that hit your eyebulb scales like $(R/L)^2$ as well (the dependence $1/L^2$ is what I care about here), because they're divided to all points on the sphere of area $4\pi L^2$, so by dividing these two expressions, you may easily see that the number of photons per unit area of the retina is actually independent of $L$. The star will look smaller as it gets further but the number of photons per unit time that hit a small area of the retina is $L$-independent.

If you're really worried that the star doesn't emit enough photons to satisfy your eyes, note that the Sun emits roughly $4\times 10^{44}$ photons each second. The Earth-Sun distance is roughly 150 million km which is 15 trillion times the radius of the eyebulb. Square it and you will still get that the number of eye-sized areas on the surface of the 1-AU-radius-large sphere is just of order $10^{26}$, still giving you $10^{18}$ photons to each eyebulb per second.

So if you allow 100 photons per eyebulb to be enough to see it, you may still allow the star to be $10^{8}$ times further than the Sun. The sun is 8 light minutes and if multiplied by 100 million, you get something like 200 light years. So with this minimal required number of photons per eyebulb (100 per second), you may see stars up to hundreds of light years away (I can't). Of course, telescopes are collecting starlight from a much larger area than the eyebulb (and they may also patiently collect the photons for a much longer time) so they may see stars much further than that.

• @DavidZaslavsky Lubos is right in the geometrical optics limit due to conservation of brightness. But for all stars (other than the Sun :) the spot size will be limited by other factors (atmospheric distortion, diffraction, aberration, ...) and will be roughly constant (giving a brightness that decreases with distance). – mmc Sep 2 '12 at 16:20
• @DavidZaslavsky Lubos is not saying "the number of photons hitting the eye is constant with distance", he is saying "the number of photons per unit area of [illuminated] retina is actually independent of L". – mmc Sep 2 '12 at 16:42
• Thanks, mmc. @David, I hope that mmc's extra words, which I subscribe to, and the extra terminology helps. BTW I think that not only your discomford is unjustified but even the sentence "pretending that photons are classical objects for a moment" is redundant. I don't have to assume and I didn't assume that photons "were classical": even with the right, fully quantum mechanical photons, I can discuss the statistics of the number of photons absorbed by a region of retina etc. and the results just happen to agree with those in classical physics! – Luboš Motl Sep 3 '12 at 5:55
• Dear David, the number of physical photons in the vacuum is an observable - i.e. a Hermitian operator - much like the position of the electron or its momentum or the total energy in a region. Like any other observable according to quantum mechanics, it is typically in a state that is a superposition of different eigenstates. But the eigenstates form a basis of the Hilbert space. When interactions are considered, "the number of photons" becomes RG-scale-dependent, divergent, and sick, but self-interactions are negligible for states composed of photons/elmg waves only. – Luboš Motl Sep 4 '12 at 6:46
• For all the purposes relevant for the propagation of photons from a star to the Earth, the electromagnetic field is free and solvable. In this approximation, the Hilbert space is a Fock space for bosons and the number of photons $N$ is a well-defined Hermitian operator on the space, much like the number of photons with $p$ in a certain range or solid angle of directions etc. The leading correction to this "free Maxwell field" approximation is the light-light scattering "box diagram" with an electron in the loop - but this correction is very, very tiny. – Luboš Motl Sep 4 '12 at 6:48

## protected by Qmechanic♦Jun 3 '14 at 19:44

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