Light travels at approx. 186,000 mps in a vacuum. Let’s say I’m out in space and point my trusty mag-lite, with D-cell batteries, into the black void of space and I click it on for one second. Kind of like a one-time pulsar. Assuming someone/thing on a distant planet were pointing a telescope at the exact spot it needs to be to see my mag-lite flash, would it be able to?
I guess this question falls into a number of areas.
Do the properties of light from any size source (flashlight, star) hold in space? I understand that a flashlight is much too small to be seen from a relatively short distance, but does the light itself behave the same?
Does the power of the source (D-cells vs. a star) dictate the distance light from that source will travel?
Does light dissipate from its source, and can it be determined mathematically the dissipation rate from the power source?
At the core of my question, if I was aiming my flashlight at another space station, exactly 186,000 miles away, would it see the beam from my flashlight? (leave it on for an hour if necessary)
Light behaves better when it is in a vacuum. Light is slowed and bent inside substances. That’s why sticks seem to bend when half in water and why light is slowest in a diamond and why light is separated into colors in a prism.
Not really. It determines from how far away the light can be detected. The energy of the photons will be greater from a stronger source. But both sets of photons will continue forever until they interact with another particle.
Light disperses at a mathematically known rate from a source, going off in all directions. That’s why a coherent beam of light from a laser can be detected at greater distances. It spreads less over a given distance.
I don’t think any present-day equipment could detect it. However, duration leads to greater photon accumulation and time-exposure is exactly how faint stars and objects are detected in space.
Yes, to the extent that all light of the same frequency is the same. Differing frequencies give you the spectrum from red light to blue light and the entire electromagnetic spectrum from radio waves to gamma rays. All photons are identical, though, except for their frequencies (and some quantum properties). Or frequency’s inverse, which is wavelength. Frequency times wavelength equals v, the phase speed. In a pure vacuum v = C, the constant that is the speed of light.
Hold on, there. The energy of photons is directly proportional to their frequency, and a given source will produce a whole spectrum of frequencies, and therefore of energies. A hot source (like a star) will produce a more energetic spectrum than a cooler source like a flashlight, but given the same temperature/spectrum, “more power” merely equates to greater intensity, meaning more photons. The more photons you pump out, the greater the number of photons that travel in a given direction, and the greater the probability that at least one photon will travel a given distance without interacting with something.
I know you know this, but it’s something the OP might not.
For them to see your flashlight, photons from the light have to reach their detecting device. There are two things which diminish the apparent brightness of light, absorption and diffusion. Absorption is when photons hit light absorbing matter and do not make it past. Both the atmosphere and matter in space will absorb your light but this will be a relatively mild effect compared to diffusion.
Imagine shooting buckshot from a shotgun, at close up it will do a lot of damage because all the pellets are packed in tight together. Go further away and the pellets are so diffuse that there’s little chance of any one of them hitting you. Now shoot them several light years and the chances become minuscule.
In addition, there’s no magic marker on the photon that tells the detector that it’s coming from your flashlight. Any detector is going to have many photons hitting it per second and so the only way for it to “detect” your flashlight is if something changes. The more ambient light there is, the greater the chance that your light will be lost among the noise.
Light intensity generally drops with the inverse square of the distance from the light source – it’s rather easy to see why: picture a point light source that’s giving off light uniformly in all directions. Surround that source by a sphere of some radius r. All ‘light rays’ emanating from the source intersect this sphere. Now, picture another sphere centered at the original point source, but with a radius of 2r. This sphere will have a surface four times greater than the first one, and thus, the density of light rays – i.e. the number of intersecting light rays per unit area – will be one fourth that at the first sphere at r. The same reasoning works for ever larger spheres; 3r would amount to an intensity reduction by a factor of 9, 4r by a factor of 16, etc.
Thus, there is no relation between the power of your light source and the distance light travels, since 1/x[sup]2[/sup] never equals 0 for finite x, ignoring the quantized nature of light for the moment.
These two posts both touch on something that I’ve always wondered - does the inverse-square law hold exactly the same for laser light? I know that we cannot build a laser so tight that its light does not spread out at all, but as I understand laser emitters, the starting point of that inverse-square explanation, that your light source is giving off light uniformly in all directions, doesn’t seem to hold. A laser will only emit coherent light in one direction, yes?
Or is that uniformity of direction an illusion, an absorption, say, of any emitted light that doesn’t fall within a particular angular deviation from the desired direction, which would be essentially the same thing as only opening a tiny window up in one of those spheres around the light source and leaving the rest of it opaque?
For that matter, what if we could construct a perfect 3-d parabolic reflector with our light source at the focus of the parabola? Wouldn’t that result in all the light travelling in very nearly the same direction and not suffering inverse-square fading?
If I’ve done my math correctly, a one-watt flashlight bulb emitting light in a 20-degree cone would be as bright as a 21st-magnitude star when viewed from 186,000 miles away. You’d need a fairly large telescope to view this, but it’s doable.
You’re right in saying that perfectly parallel light would not suffer inverse square attenuation. However, perfectly parallel light is basically impossible to attain – in real life, you’ll always have some divergence. So let’s say your beam has some small opening angle α; then, on a wall in some distance (r), it will produce a spot of radius w = rtanα, illuminating an area of the size A[sub]1[/sub] = πw[sup]2[/sup] = πr[sup]2[/sup]tan[sup]2[/sup]α. Again doubling distance, the illuminated area will be four times as great – A[sub]2[/sub] = π(2r)[sup]2[/sup]tan[sup]2[/sup]α = 4πr[sup]2[/sup]tan[sup]2[/sup]α = 4A[sub]1[/sub], and thus the intensity per unit area again diminished by a factor four. Thus, the inverse square law holds as long as there is some divergence in the beam of light.
Even for a perfect TEM[sub]00[/sub] mode laser, which has the smallest and tightest beam, once you get beyond the “confocal zone” the beam begins to spread very nearly as 1/r[sup]2[/sup], and asymptotically approaches an inverse-square dropoff in intensity the farther from the beam waist you get.
There is no such thing as a truly collimated beam, no matter how you do it. This really isn’t the place to discuss the math of laser propagation (I refuse to use all the codes to write out equations), but consult any good laser text, like Amnon Yariv’s books or Siegman’s, or look up “Gaussian beams” on the internet. The lines of equal relative intensity (relative to the peak intensity, on the center line) form hyperbolae that stay relatively collimated over a characteristic length called the Rayleigh Range or the Confocal Distance (there’s actually a factor of 2 between these two. Sue me.), and beyond that the beam fanbs out and approaches its Far Field Angle. The Rayleigh Range gets smaller (and the Far Field Angle larger) the smaller the Beam Waist is (its smallest dimension anywhere). Diode Lasers have pretty tiny beam waists, simply because of the way they’re mad, so they fan out pretty fast. That’s why you need to use a lens with them to collimate the output.
As for searchlights and other parabolic reflectors, it’s true that a point source placed at the focus will be perfectly collimated – only there’s no such thing as a true point source. Every light source has finite dimensions, so your beam eventually starts to spread. This is actually treated mathematically in at least two books on photometry that I know of.
And lasers are much weirder things than you know. There are other beam shapes besides that hallowed and much-sought-after Gaussian beam, and even in a classic resonator you’ll have higher modes which have larger waists and spread faster. But differently shaped resonator systems can have modes that are not remotely gaussian. There have been planar lasers, in which the “beam” is not emitted as a line along a single axis, but emitted equally in all directions along a plane. And I’ve read one account of a spherical laser, in which the light is emitted into all directions. That seems to go against what most people think characterizes a laser – after all, if the light is going in all directions, how is it different from a light bulb? Answer: coherence, monchromaticity, method of generation, and different photon statistics. And, for advanced students, it’s not ASE, because it required a cavity.
But that only works if you’re doubling your distance from the apparent point source of the light. If you have a 1-ft parabolic reflector, and it’s ground so perfectly that the beam it makes only diverges to 2 ft in diameter after traveling 1 light-year, then it will be only 1.5 ft after half a light-year.
In terms of areas, that means it’s 0.56π ft[sup]2[/sup] after 0.5 light-year, and π ft[sup]2[/sup] after 2 light-years. That’s not a fourfold increase in area after doubling the distance, but a 1.78-fold increase.
If, however, you consider that the beam appears to come from a point source 1 light-year behind your reflector, and use distances from that as the basis for your inverse-square calculation (1.5 ly and 2 ly, respectively, in the above example), it works out. The effect of a real collimated beam is to skew the inverse square law, allowing your light beam to lose less intensity over a distance than it normally would have.
Yes, you’re right. My considerations were only in regard to a beam of vanishing width, equivalent to your ‘apparent point source’ (or viewing the whole situation in a long-distance limit).
So, the dissipation of light from my flashlight would be too great to even make it to a space station 186K miles away in space. If it was possible to do this experiment in a pure vacuum would it still be the case?
Also, pulsars interest me. Is it possible (or probable) that there are pulsars out there that don’t emit the required energy for their light to reach the earth, even though they’ve been cranking away forever? Another way to think about the question is that a pulsar would be able to be seen on some outpost of the solar system (like Pluto), but just not have the energy behind it to reach an earthbound telescope?
I think you misunderstood. Some light from the flashlight would in fact reach the space station, and all of the light that didn’t reach the space station, it’s just because it reached someplace else instead. But in principle, if you have a large enough light-collecting area, or watched for a long enough time (to accumulate more photons), there’s no limit on how far away any given light source could be seen.
There are certainly some pulsars that are far enough away that it would be very difficult or impossible to detect them with any equipment we have, but again, it’s possible, at least in principle, to detect them with a good enough telescope. And I’m not sure what Pluto has to do with anything: From the perspective of any other star, Pluto and Earth are essentially in the same place. The distance between stars is vastly greater than any distance within the solar system.
I did misunderstand. Thanks for clearing this up for me.
I understand that in the vastness of the universe, Pluto and Earth are on top of each other. I was just trying to give a quick example of something that was “further” from our earth-bound telescopes. You answered my question even with my bad example.