I understand light consists of photons. Likewise it is a wave with varying intensity in the electric and perpendicular magnetic fields. Say there is a candle lit on the moon. Why wouldn't we be able to see it from Earth? Since light is quantized, would an equal distance in space correspond to a proportional loss in intensity? If so, what are the proportionality constants in the vacuum of space and in 'regular' air?
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2$\begingroup$ All photons propagating in the same sense gives an intense spot. There is no such thing as intensity of a single photon. Individual photon energies are way far to be detectable by eye. $\endgroup$– UKHCommented Aug 7, 2017 at 16:05
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1$\begingroup$ Remember the whole reason we talk about photons is that energy can only be added to or removed from the electromagnetic field in quantized amounts. $\endgroup$– The PhotonCommented Aug 7, 2017 at 16:33
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1$\begingroup$ Very closely related, if not duplicates. physics.stackexchange.com/q/39416 physics.stackexchange.com/q/346565 $\endgroup$– ProfRobCommented Aug 7, 2017 at 16:49
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6$\begingroup$ Possible duplicate of Why does a light wave fade as it travels through space?. Also related: physics.stackexchange.com/q/203972 $\endgroup$– Emilio PisantyCommented Aug 7, 2017 at 19:05
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2$\begingroup$ @UKH Well actually... our rod cells can detect single photons, but the post-processing our eyes use to reduce noise normally inhibits single photon detection. $\endgroup$– PM 2RingCommented Aug 7, 2017 at 20:09
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4 Answers
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A beam of light loses intensity as it travels and spreads. A photon does not. Instead, when dealing with light as photons, the rate of photon absorption/detection decreases as the density of the photon flux decreases. How bright a light is doesn't depend on the intensity of the individual photons (that's constant). It depends on the number of photons from that light source hitting your eye, and as the light becomes more spread-out there are fewer photons from that light source hitting your eye.
In principle, we can detect a candle on the moon, but it's very hard due both to the very small photon flux on a detector, and to the difficulty in distinguishing between a candle photon and a photon emitted by the background.
Experiments such as the lunar laser ranging experiment detect single photons reflected from mirrors on the moon, but they have several things going for them. First, the instantaneous laser power produced by reflection is a good deal higher than the ambient (at a very specific wavelength), and second, since the laser wavelength is well-known the detector can use a very narrow filter to ignore other frequencies. Even so, there are a certain number of false detections which analysis needs to reject.
answered Aug 7, 2017 at 14:38
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16$\begingroup$ Not to mention the difficulty of getting a candle to burn in vacuum :-) $\endgroup$– jamesqfCommented Aug 7, 2017 at 19:23
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12$\begingroup$ If a candle burns in a vacuum and nobody sees it, does it emit light? $\endgroup$ Commented Aug 7, 2017 at 20:26
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$\begingroup$ @jamesqf The moon does have a very thin atmosphere. It's not quite a vacuum, and the atmosphere is thicker than usual "space" (little bits of Hydrogen here and there, with some Helium on occasion). $\endgroup$ Commented Aug 7, 2017 at 20:47
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6$\begingroup$ @wizzwizz4 little bits of hydrogen here and there, with some helium on occasion make it of course much easier to light a wax candle... $\endgroup$ Commented Aug 7, 2017 at 21:10
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1$\begingroup$ @leftaroundabout That was my description of "space". Sorry for the ambiguity. $\endgroup$ Commented Aug 7, 2017 at 21:11
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A good way of thinking about this is to try to work out how many photons reach a detector on Earth from a candle on the moon. I will invent some plausible numbers here.
Let's say the candle has a power of $10\,\mathrm{W}$ and lets say it emits all of its power at $580\,\mathrm{nm}$ (this is yellow light: in real life it emits only a rather small amount of its power as light, but perhaps it emits more total power: in any case this will be OK for now).
The energy of the candle's photons is
$$E = \frac{hc}{\lambda}$$
which is $3.4\times 10^{-19}\,\mathrm{J}$. So at $10\,{W}$, it is emitting about $3\times 10^{19}$ photons per second. We'll assume it does so equally in all directions & further assume that the ones that hit the moon get absorbed (ie I'm not going to count ones reflected from the surface).
Let our detector on Earth have an area $A$, and the distance to the Moon be $4\times 10^8\mathrm{m}$. Assume the Earth's atmosphere is completely transparent. So, the number of photons per second that reach the detector on Earth, $r$ is:
$$r = \frac{3\times 10^{19}}{4\pi \times (4\times 10^8)^2}A$$
(Note I've used the formula for the surface area of a sphere here.) So, if we have a large amateur telescope with a light gathering area of $0.3\,\mathrm{m^2}$ (this is about a 12 inch telescope), it sees photons from the candle at a rate of about $4\,\mathrm{Hz}$: it sees about 4 photons a second.
Apparently the dark-adapted human eye can have a pupil size up to 9mm (but this is rare), and you have two eyes. So your eyes might see, between them, a photon from the candle every couple of minutes.
And this is the case if it emits $10\,\mathrm{W}$ of visible light: in fact it emits far less, which makes it far harder to detect.
This is why you can not see candles on the Moon.
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1$\begingroup$ That's right. The direction of a photon cannot be exactly known; and because of this uncertainty, the probability of detecting a photon at a particular place scales with the number of photons per unit area per second. The inverse square law does the rest... $\endgroup$– FlorisCommented Aug 7, 2017 at 18:08
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1$\begingroup$ 10W optical power from a candle? Maybe if the candle is the size of an ICBM... You could probably scale this calculation down by a factor of about 500 to start getting into realistic territory. It would be more like one photon every two minutes. $\endgroup$– J...Commented Aug 7, 2017 at 18:55
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1$\begingroup$ @J... Yes, my intention was to show that an absurdly bright candle can't be seen, and hence that a realistic one is really seriously below any detection threshold. I've changed the answer slightly at the end to make that clearer. Sorry. $\endgroup$– user107153Commented Aug 7, 2017 at 19:26
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When you emit light from a light source in all direction, the intensity of the light at distance $r$ will decrease proportional to the inverse of the spheres surface ($I(r)\propto\frac{1}{r^2}$). So in therms of waves the intensity is decreased, in terms of photons this is the number of photons you detect per time. This is the same in air, if the light is not absorbed in air.
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In the picture of light being a wave, the intensity is given by the amplitude of the electric field. However, in the particle picture, the intensity is the number of particles per unit time and area (see also Einstein's paper on the photoelectric effect). Therefore, if you move away from a light source, the same number of photons is distributed over a larger area, thus decreasing the intensity. So it is not an effect of an individual photon, but of the distribution of all the photons emitted.
You can now easily see that the proportionality is $I \sim \frac{1}{A} \sim \frac{1}{r^2}$ where $A$ is the area of the sphere and $r$ is the distance from the light source. In air, you would additionally have to account for absorption and scattering of photons by particles in the air, which is highly dependent on the composition (water vapour content, dust, ...), so no universal proportionality constant can be given.
