# What would we observe from thompson's lamp thought experiment?

The thought experiment goes like this:

Say there is some circuit which turns a lamp on/off with just a flick of a switch. Say its off; you flick it, it turns on; flick it again it turns off, and so on.

So say you are conducting the experiment for two minutes. When the remaining time halves, you flick the switch; and when that remaining time halves, you flick again...This process goes on infinitely, until two minutes is reached.

Mathematics says that the sum of all those remaining half-times will eventually tend to two minutes. And the outcome i.e. the final state is always unpredictable. We can't really know if its on/off.

Lets not concern about the final state. What would we actually observe towards the end, few micro seconds towards the two minute mark? Wouldn't the lamp appear turned on? What would happen if the time between two flicks tend to zero?

• This video might give some idea: youtube.com/watch?v=u7Z9UnWOJNY – Renae Lider Apr 21 '14 at 20:33
• This is just a restatement of the old Achillies Paradox (or some such name, I remember it had a ancient Greek figure in it). – Olin Lathrop Aug 21 '14 at 18:28

The simplest answer: The lamp would appear on, to us humans.

Actually, this is what happens to any lamp that runs on AC current. Even if you had an LED that can handle AC current, it would still flick on/off fast enough that human eye would not see it flickering. Most lightbulbs, old TVs, and some other gadgets take advantage of the slowness of the human eye.

If we're talking about some robotic observer, or someone (or something) with infinitely fast eyesight, then I'm afraid I don't know the answer, but I'm willing to guess it would still appear on.

If the lamp is incandescent, the heat dissipates too slowly for the on/off cycles to matter after a while. If the light is an LED, then you still have to wait for the electrons to "realize" they don't have to go down the circuit any longer, but you're likely flickering so fast that they still "think" they do. There might be a hole on the other side of the potential drop for the electron to jump down while the light is in it's "off" phase, making the lamp "on" while the switch says its "off." Plus, if you're looking in different parts of the light spectrum, the light is always "on," giving off blackbody radiation.

Eventually your flicker alone needs to break the speed of light to keep up with the "flick-at-half-of-time-remaining" rule, causing other problems. Even a tiny mechanical switch going that fast may cause explosions (due to air resistance), making the existence of the lamp and the circuit a little more uncertain. Certainly, it would produce light, so I'm going with "there is light coming from that circuit."

Letting $T(k)$ be the time at which the $k^\text{th}$ flicker occurs, we have $$T(k)=60\sum_{n=0}^k 2^{-n}.$$ For example, $T(0)=60,T(1)=90,T(2)=105,...,T(\infty)=120$.

The human eye has a response time on the order of 1/30th of a second. Therefore, once $$60\times2^{-k}=\frac{1}{30}\\\Longrightarrow k = \frac{3 \log (2)+2 \log (3)+2 \log (5)}{\log (2)}\approx 10\\\Longrightarrow T(k)=119.941$$ the lamp will appear to be continuously on at low-brightness, and will remain that way for the next 58.59 milliseconds.

## What is the apparent brightness?

Define $L(k)$ as the total time the lamp has been on after $k$ flickers, which gives $$L(k)=60\sum_{n=0}^k\left(\frac{(-1)^n+1}{2}\right)2^n.$$

The brightness $B$ then becomes $$B=\frac{L(\infty)-L(10)}{T(\infty)-T(10)}=\frac{1}{3}.$$

So the lamp will actually appear to be at 1/3 brightness during the last few milliseconds.

• What is this equation? – deostroll Apr 23 '14 at 8:33
• @deostroll: What equation? – DumpsterDoofus Apr 23 '14 at 12:01
• @deostroll: Basically I was just showing that the lamp will appear to flicker on and off faster and faster from time 0 to time 119.94 seconds, and from 119.94 to 120 seconds it will be so fast that to humans the lamp will appear to be glowing at half brightness. – DumpsterDoofus Apr 23 '14 at 12:03
• $$T(k)=60\sum_{n=0}^k 2^{-n}$$? – deostroll Apr 23 '14 at 12:05
• @deostroll: Oh, $T(k)$ is just the time at which the $k^\text{th}$ flicker occurs. The first 10 values ($k$ from 0 to 9) are {60., 90., 105., 112.5, 116.25, 118.125, 119.063, 119.531, 119.766, 119.883}, does that make sense? – DumpsterDoofus Apr 23 '14 at 12:09

The lamp would be "on", but with a reduced luminous flux output. This is very similar to pulse-width modulation, which is used to control the brightness of LEDs in various applications.