# If light does not change speed, how does luminous intensity decrease?

Light is made up of photons. They hit our eyes to create sensation. Suppose two points A and B are in a very long distance apart in vacuum. If we shine a very weak light from point A, practical experience dictates that from point B we would not perceive the same intensity of light (if any). If speed of light is constant, shouldn't all the photons travel from A to B and thus the intensity remain the same?

Note that they are in vacuum, photons cannot hit something and change direction.

• In this case, intensity depends on the number of photons that reach point B, not the on the photon's speed. – Andrei Geanta Aug 31 '18 at 5:54
• @AndreiGeanta Okay, how does the intensity decrease? Shouldn't the number of photons remain the same? – Shuvo Sarker Aug 31 '18 at 5:56
• Even with the best lasers, light spreads out as it travels. Hence fewer photons will hit the target. – hdhondt Aug 31 '18 at 6:07
• Photons in a vacuum can interact with each other. It's just several perturbation terms down the probability series. But that's not the issue here. It's all about beam spreading and packet dispersion. – Carl Witthoft Aug 31 '18 at 15:34
• Maybe "intensity" could be meant as a measure of total energy, so that a redshift also "reduces intensity" as it reduces the power of the signal. – elliot svensson Aug 31 '18 at 16:39

As a beam of light travels, it spreads out so that for each doubling of the distance, the intensity of the light (defined as the number of photons passing through a unit area per second) is decreased by a factor of four. This is a basic consequence of the way that light beams propagate through 3-dimensional space and has nothing to do with the speed of the beam- which does not vary with distance.

So at a point B that is distant from a light source at A, not all of the photons leaving A will hit B. The result is that fewer and fewer photons will arrive at B as the distance between A and B gets larger and larger, causing the intensity of light measured at B to decrease with increasing distance.

• So, this means the photons never actually stop, they keep bouncing off and off? – Shuvo Sarker Aug 31 '18 at 8:49
• +1, but an image such as this one would be beneficial, I think. – pela Aug 31 '18 at 10:11
• @ShuvoSarker photons are elementary point particles described quantum mechanically with a probability wavefunction. They follow geodesic lines (straight lines in flat space) until and if they interact with some other particles or fields. – anna v Aug 31 '18 at 11:52
• Is this law also applicable to a laser beam? – Thomas Ayoub Aug 31 '18 at 12:47
• I assumed OP was asking about noncoherent luminous objects of finite extent where the distance between source and detector is ~orders of magnitude bigger than the size of the source i.e., as defined by a lay person. So I think my answer, as furnished in that context, remains correct. – niels nielsen Aug 31 '18 at 15:49

You do not actually need to use photons here - in fact the often sadly-ignored in many people's minds good ole' classical description of light as an EM wave is sufficient (though that doesn't mean the other way is wrong, but it's important to point it out!).

A light wave, like any wave, has three basic parameters: its amplitude, frequency, and wavelength. (You may also add phase, but phase is really just a reference shift, it does not change the shape of the wave, though it is important when summing waves.) The amplitude of a light wave is the maximum size of its electric field vector, just as the amplitude of a sound wave is the maximum strength of the overpressure relative to the ambient pressure, and the amplitude of a sea wave is the maximum height of a wave crest above the surrounding unperturbed sea level. The amplitude determines the amount of energy, and thus the intensity - what we'd call respectively "brightness", "loudness", and "height", that is transmitted by each wave.

Amplitude has nothing to do with speed - rather speed is what relates frequency and wavelength: $\lambda f = v_w$, where $v_w$ is the speed of the wave. Amplitude ($A$) does not appear in this relation. That means the intensity can go up, down, or stay the same, and have no effect on the speed. Here, of course, $v_w = c$, the speed of light waves in vacuum, or $v_w = \frac{c}{n}$ in a refractive medium with refractive index $n$.

In your scenario, as the waves spread out, their amplitude decreases - just as a sound wave gets less loud the further you go, and a water wave shallows as it travels out from a source of disturbance such as a cast rock. Because speed has nothing to do with amplitude, this can happen without requiring any changes in speed.

In terms of photons, if you want to use them, the intensity is related to the photon number together with frequency. Decreasing the intensity reduces the number of photons, but each photon always travels with speed $c$. In this scenario, the photons become more thinly spread, over a wider region, so a smaller number in each volume of space.

(And yes I'm assuming a "linear" situation in the above, yes caveats about nonlinearity, yes yes, blah blah, but for these purposes, that is sufficient.)

• does our atmosphere have a refractive index too? If yes, what number is it? – undefined Aug 31 '18 at 13:37
• So what causes redshift, then? – elliot svensson Aug 31 '18 at 13:45
• @undefined Wikipedia lists the refractive index of air: 1.000293 but this has to be taken with a grain of salt as it probably changes with density (height), humidity and temperature. – Arsenal Aug 31 '18 at 14:26
• @undefined you really can look that stuff up. And atmospheric refraction is what leads to things like mirages and colored sky at sunset. – Carl Witthoft Aug 31 '18 at 15:36
• @elliotsvensson Mr. Doppler is dismayed at your question. – Carl Witthoft Aug 31 '18 at 15:36

The intensity of each photon does indeed remain the same, but you won't receive all the photons. Even lasers are just very good spotlights; by the time one reaches the moon, the dot would be a few hundred meters wide. You might as well brew your tea bag in a lake.

• Actually, it is the energy of a photon that remains constant. The intensity is something different. – flippiefanus Aug 31 '18 at 12:57
• @flippiefanus True. In physics, "intensity" means power transferred per unit area. It doesn't really make sense to talk about that for single photons. – Fax Sep 3 '18 at 9:33

There are two ways of looking at it:

• Classical: Radiated energy (photons) propagates isotropically in space. This means that it propagates as a spherical shell. By conservation of energy the energy found by integrating along the sphere area should be the same as the initial energy. As the surface area of a sphere is 4πr^2 and the energy is evenly distributed through it that means that energy density on the sphere surface is proportionally inverse to the radius, this is known as the inverse square law. Intensity is defined as power over area unit; therefore it decreases proportionally to the squared distance to the source.
• Quantum mechanics: Photon speed is constant, therefore after a given time its wave function (for probability distribution) will have a spherical shape, this means that the probability of finding a photon when measuring is the same on any point of the surface of the sphere. Therefore the biggest the sphere, less probability you will have of finding a photon. (quantum) Intensity is defined as the number of photons measured in a point in a period of time, therefore it diminishes proportional to the square of distance to the source.

..Adding to Fax's thought: Each photon maintains it's own intensity. If you had a perfect laser beam, it too would maintain it's intensity no matter how far it traveled. Perfect laser beams are not possible of course. The great majority of light that we see originates from a (approximately) central point, so photons are moving further away from each other as they move away from the center. If you considered successive spherical shells surrounding this point (like light being emitted from the sun, or from a candle), you should see that the surface density (and therefore the intensity) of any group of photons in the same shell would fall off as 1/r².

• Is a redder photon less intense than a bluer photon? If so, then redshift (of the expanding universe variety) causes intensity to reduce, right? – elliot svensson Aug 31 '18 at 17:50