You good people recently answered many of my questions about light. I’ve pondered those answers and now have some followup questions:
If light does not travel in a literal sine wave-type fashion, how are we to understand frequency in regard to a single photon traveling through a vacuum? Now, we also know that microwaves can be prevented from leaking out of the mike by a metal mesh, whose holes are smaller than the wavelength of the light waves. But how is the individual photon prevented from flying right through that little hole?
In short, what is the relationship of a photon to the wavelength/frequency?
Photons have no mass, but they are said to have momentum (hence they could push along a solar sail). How is this possible?
As a corrollary to #2 above, a simple thought experiment (based on lots of Star Wars images, I fear) tells me that a powerful enough laser aimed at, say, a coffee cup on the desk would blow it off the desk (or impart not only heat energy but also kinetic energy). Is this impression correct?
How is it that an infra-red remote control or infra-red microphone can work as it does? Namely, why does the signal seem so able to bounce off walls and other objects and make it to the intended receiver? It would seem that if you did the same thing with regular visible light, you’d not get very good reception–but it’s the same light, right?
Dumb relativity question: From the perspective of one of a pair of photons traveling from opposite directions toward the same target, wouldn’t the speed of the approaching photon seem to be c x 2?
1-3. Wave-particle duality. Light behaves like a wave and a photon, depending on how you are making your measurement. It is literally a sinusoidal oscillation of the electromagnetic field and it is a tiny particle.
Walls absorb visible light quite well, but infra-red bounces quite easily I believe. (In any case, if the receiver was looking out for only a very specific wavelength, visible would probably work OK too.)
From a “stationary” observer’s frame, yes, the diference in velocities is 2c. However, there is no way for one photon to seem anything to the other: They cannot “see” each other, nor exchange any information - you could say that they simply cannot be aware of each others’ existence. If two torches flying towards each other at a speed of 0.99c* emit photons at the speed of light, those photons still only go at c in any reference frame.
It would work with visible light as well. At night with the light on, look up at the ceiling. It’s quite obvious that the light is bouncing off the ceiling. And it’s also obvious that if visible light didn’t bounce off stuff, you wouldn’t be able to see anything.
An individual photon isn’t prevented from passing through a hole smaller that its wavelength. It’s a matter of probability. The probability of a photon passing through the hole is:
1 - e[sup]2x/[symbol]l[/symbol][/sup]
where
[symbol]l[/symbol] = wavelength
x = size of the hole.
For example, my microwave has holes 1 mm (0.001 m) in diameter in the mesh.
The microwave operates as 2.45 GHz, so:
[symbol]l[/symbol] = c / f = 3 x 10[sup]8[/sup] / 2.45 x 10[sup]9[/sup] = 0.12 m.
The force that the laser exerts on the coffee cup is somewhere between:
F = P / c, and
F = 2 P / c, where
F = force on the cup
P = power of the laser
The first equation applies if the coffee cup is perfectly black and absorbs all of the photons that strike it. The second equation applies if the cup is perfectly white and reflects them all.
Lets say that the cup is halfway between perfectly black and perfectly white. Then
F = 1.5 P / c.
Let’s try some numbers. The cup weighs 0.1 kg, so we want to apply about 1 N to be able to push it off the table.
1 = 1.5 P / 3 x 10[sup]8[/sup]
P = 200 MW
The Three Mile Island plant can produce about 800 MW. So it would take a quarter of that to push the coffee cup off the table, using light pressure alone. In reality, of course, the photons that the coffee cup absorbed would very quickly heat it and vaporise it.
You’d think so, but addition of speeds doesn’t work like that. Honest.
If two things are coming at you from opposite directions at speeds u and w, then someone sitting on one will see the other going at, not u+v, but (u+v)/(1+uv/c[sup]2[/sup]). So if they’re both going at 0.9c, then the relative speed is (0.9+0.9)/(1+0.9[sup]2[/sup])~0.995c. If they’re both going at c, then the relative velocity is also c.
OTOH, if they’re going at a slow speed (eg. anything anything larger than a bacteria ever goes at…) then the addition is almost the same as the way you’re used to, which is why you’re used to it.
Although, as pointed out above, a sufficiently powerful laser will push a cup by light pressure off a table, you can propel the cup with a laser and a great deal less energy – a laser pulse that vaporizes a small layer of the cup will impart an impressive momentum to the cup. I know – I used to work on laser propulsion, asnd we were blasting dime-sized pieces of plastic around with more powerfult than average (but not huge) lasers.
Every now and then you can see reports on the news about Leik Myrabo and his Apollo Lightcraft being propelled by staccatto bursts from a CO2 laser down at White Sands. It’s certainly not being vaporized.
(again) Things get strange when you have two holes, and only one photon.
When a photon hits something, it acts like a point particle. But when it’s in flight, it acts like a probability function, and not like a point particle at all.
In the example above, there was a 1.7% chance that the photon passed through the hole. It’s possible to set up two holes spaced slightly apart, and still have a probability of 1.7% that the photon will get through either one of them.
Let’s say we have a detector that can detect individual photons on the other side of the holes. If the photon passes through one of the holes, it’ll make a mark - a single dot - on the detector.
Now let’s fire a number of photons at the holes, on at a time. After a while, a pattern will show up on the detector. The pattern is an interference pattern of light and dark rings. The pattern tells us that each photon did not pass through a single hole. It’s more like each photon that gets through goes partly through one hole and partly through the other.
I think this is the last question left. The answer is that the equation p = mv is wrong. Well, not exactly wrong, but an approximation. For relatively small speeds, it’s a very good approximation. For a photon, it’s about as bad an “approximation” as you can get. For objects travelling at speeds close to c, the correct formula is p = gammamv, but this still doesn’t work for a photon, since for a photon, m is 0 but gamma is infinite. Yet another equation we can use is E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup], and that equation works for everything, including photons. Since a photon’s mass is zero, this means that p = E/c. And finally, if you want to get quantum mechanical, you can also use the equation p = h/lambda, where lambda is the wavelength. This one also applies to everything, although it’s a little less obvious what it means to have a wavelength of, say, an electron.
That will cause it to accelerate at 10 m/s[sup]2[/sup], or > 1g, which is a bit of overkill. You can probably lower you power requirement by a few orders of magnitude.
Also note that lasers have been used to manipulate atoms for a number of years in laser cooling and trapping. The force a laser can exert on an individual atom is huge, since the atoms can absorb and re-emit photons of order 10[sup]6[/sup] times/second. Even though the individual nudge from photon absorption or emission changes the atom’s velocity by only a few mm/s or perhaps a few cm/s (depending on the wavelength and mass involved), the accelerations can be as high as 1000 g’s or more.
If you do it right you can cool the atoms to microKelvin temperatures, and a few tricks get you to Bose-Einstein condensation and nanoKelvin temperatures.
I think you’re making a common mistake. A photon is not a particle. Photons are “quanta,” which is something very weird. To understand photons, first you have to give up the idea that they’re like tiny bullets racing along. Then what are they? Ah, that’s a separate topic. First stamp out any idea that photons are “particles”, then you can go to the next chapter. Beware of creeping “particle-thinking.” Even the professionals fall into it, rather than consistently imagining photons to be weird entities.
Photons have mass. All energy has mass (remember E=MC^2)
Photons have “zero rest-mass” which means that all their mass is part of their kinetic energy, and if you could bring a photon to a halt, there wouldn’t be anything left. This is quite different than, say, an atom, which contains energy (and mass) even when at relative rest.
I assumed a coefficient of friction between the cup and the table of 1. 1 N would be just sufficient to make the cup move. You lower the laser power by “a few orders of magnitude” and the thing ain’t goin’ nowhere. It ain’t on wheels.