A couple of things (beyond “neat video”):
He doesn’t answer the question about what would happen if you left the light on. You might think this is no big deal, because he correctly says that the light dies out quickly. If you were in a mirrored room such that the average photon trip was 3m (and somehow not interact with you at all), and the mirrors are 99.99% reflective, a photon would reflect 10^8 times a second (i.e. once every 10 nanoseconds), but only reflect 10,000 times on average, so you would expect the room to go dark in less than a millisecond. However, if you keep the light on, you get a build-up of photons for that time. To reach equilibrium, your production rate and loss rate have to be equal, and you only lose 0.01%, or 0.0001 of your photons. If you have just a 1 Watt source of visible light (which would emit around 10^18 photons a second), you need to have 10,000 times as many photons inside to have 1 Watt leaking out.
Put another way, your source is emitting 10^10 photons per bounce interval (10 nanoseconds) but only 10^6 photons leak out. In the next interval, another 10^10 photons are added and 1,000,010 photons leak out (0.0001 from each generation). And so on, with a decaying exponential buildup, until you have 10^14 photons hitting at each bounce, so that 10^10 can escape. That’s when you reach equilibrium.
So your little 1 Watt light gives you a power buildup and you are doing the equivalent of hugging 10 kW of space heaters. Actually the scenario is worse, because your body emits around 100 Watts, in the infrared, so if the mirrors reflected IR you would cook yourself to death. Fast.
The other issue I have is where he says that mirrors flip left and right and not top to bottom. The initial explanation is right — they flip perpendicular to the plane of the mirror, but then he claims that L-R is perpendicular while U-D is parallel, which is nonsense. It’s a plane, so they are both parallel. Mirrors flip front-to-back, i.e. perpendicular. The confusion is that the mental image we have is of someone walking around the mirror to the other side, and that’s not what is going on. It’s a confusion of inversion and rotation, which are two different ways of getting an image like that. There is no left-right flip! Your right hand is still on your right, it’s just that you expect it to be on your left, because of that were a person in the mirror (who has rotated on an axis to look like that), it would be their left hand.
Maybe you’ll like hearing Feynman explain it.
Nice article, but I have one bone to pick.
Re: “Actually the scenario is worse, because your body emits around 100 Watts, in the infrared, so if the mirrors reflected IR you would cook yourself to death. Fast.”
I disagree with a lot of this. You should do the calculation on how much your body emits in the infrared (50 mW per square cm at an emissivity of 1; I’m not sure how big you are). The result is amusing.
But much more important than how much you emit is the difference between how much you emit and how much you absorb, which is determined by the temperature of the emitter and the conditions of the surrounding environment. If I were to surround myself with perfect mirrors, it would be no different than hanging out in a room where the walls were at the temperature of my skin. Folks do that all the time without “cooking themselves to death. Fast.”
In hot weather, the primary cooling method for humans is not radiation, it’s evaporation. Which is why – if I’m properly hydrated – I can hang out in a low-humidity, 100 degree F room without becoming ill. But put humans in 100 degree F rooms at high humidity, and folks start dying.
I’m ignoring blackbody effects, i.e. I assumed losses are transmitted, not absorbed. Mirrors reflect, while walls are absorbing and then emitting, and that happens in both directions. The walls reach thermal equilibrium with you and they also radiate outward. That’s not happening in the mirrored-room case. It’s more like a laser cavity; the power inside is much higher than the emitted power. You are essentially the equivalent of the gain medium, but without the stimulated emission part.
Also, 100 W is a ballpark figure, but it’s about right. Average human surface area is almost 2 m^2 for men, 1.6 m^2 for women. At room temperature, there’s about a net 100W radiation difference. At 100 ºF evaporation becomes more important, but at 72 ºF or thereabouts (room temperature) radiation is the bulk of the energy loss.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.html
I agree with much of what you say. A buck-naked 2-square-meter man at 300 K (and emissivity 1) radiates roughly 1 kW. If you pick the “right” environment for him then the net radiative power output could certainly be 100 W. And I certainly agree that if he’s not sweating, radiation will dominate the energy flow.
But I still claim that if you stick that 300 K dude inside a PERFECT mirrored room and wait, the radiation field will come into thermal equilibrium with the dude. That would be equivalent to being surrounded by 300 K walls, which usually doesn’t result in fast death as long as one can sweat.
Another way of looking at the same thing: if the human body is running at a total power of 100 W (a reasonable power for an awake human at rest) and then the maximum amount of power that could be heating him is 100 W; super-reflective mirrors don’t change energy conservation. Most folks have no problem sweating out 100 W.
But, if the dude in question doesn’t sweat and there’s negligible convection, then he will indeed start to heat up. For a 100 kg dude and 100 W, I’d calculate that he’ll heat up at around 1 degree an hour (assuming his specific heat is essentially that of water). And in that case, I certainly cede the point and stand corrected: he would die of hypothermia in a few hours.
I think you’re confusing heat with temperature. Sure the power of all of the photons bouncing around might be 10 kW, but once you’re exposed to it, you would absorb every one of those photons in a few nanoseconds. That’s not many joules of energy. You’d be fine.
But the energy doesn’t just disappear when you absorb the photons. Your temperature would have to go up as a result. In this sense, ideal mirrors are really, really good insulators.