Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Thomson's Lamp paradox:

A mad scientist owns a desk lamp. It begins in the toggled on position. The scientist toggles the lamp off after one minute, then on after another half-minute. After a quarter-minute the lamp is toggled off, then the scientist waits an eigth-minute and turns the lamp on again. The scientist continues toggling the lamp, waiting one-half of the previously waited time between toggles. After a total sum of two minutes of toggling, what is the state of the lamp (on or off)?

The Wiki article states that supertasks are impossible and the lamp is neither on nor off after the two minutes. This does not make sense to me, as this would mean that the lamp is in a superposition of two states.

Proof (1 is on, 0 is off):

$S = \sum \limits_{i=0}^n {(-1)^i}$




$S=\frac1 2$

How can a macroscopic object like a lamp exist in such a state?

share|cite|improve this question
The state is only specified for $t \in [0,2)$. Your sum doesn't make sense. – Umberto P. Jul 17 '13 at 14:35
In reference to your question at the end: How can a macroscopic object like a lamp change states in an infinitely short amount of time? Without the answer to this question, there's not much point in coming up with a physically realizable answer to your question. – Foo Barrigno Jul 17 '13 at 14:38
Theoretically, If you were to consider such a situation, it would appear as if the lamp is both on and off at the same time. Physically, you cannot switch the lamp on and off instantaneously. After a certain point, the lamp will switch states slower than the mad scientist pressing the switch, so you would never get to infinity. – udiboy1209 Jul 17 '13 at 14:38
@Umberto P.: You should consider making that a reply. – Raskolnikov Jul 17 '13 at 15:07
The light will be on twice as long as it is off, so obviously the probability that it will be on at $t=2$ will be 2/3. <grin> – Rick Decker Jul 17 '13 at 15:36

The following tries to avoid physical arguments:

The lamp $L$ has a well defined state $s\in\{0,1\}$ at time $t$ only if there is an $\epsilon>0$ (which may depend on $t$) such that $L$ is in state $s$ during the complete interval $\ ]t-\epsilon,t+\epsilon[\ $. It follows that the lamp is not in a well defined state at any of the switching times $t_n=2-2^{-n}$ $\>(n\geq0)$. Why then should we expect it to be in a well defined state at time $t_\infty=2\ $?

share|cite|improve this answer
Did you mean to specify a relationship between $\epsilon$ and $n$, for example $n>-\log_2\epsilon$ instead of $\>(n\geq0)$? – John Bentin Jul 17 '13 at 17:36
@John Bentin: The instant $t$ is given first. When $t_n<t<t_{n+1}$ for some $n$ you can find an $\epsilon>0$ such that $s$ is constant on $\ ]t-\epsilon,t+\epsilon[\ $. When $t=t_n$ for some $n$ then there is no such $\epsilon$, since the states of the lamp differ immediately to the left and the right of $t$. – Christian Blatter Jul 17 '13 at 18:08

The series $S = \sum \limits_{i=0}^\infty {(-1)^i}$ does not converge, so your calculation makes no sense to begin with (its Cesaro sum does converge to $0.5$, though).

Furthermore, a physical object such as lamp can't be turned off and on at arbitrarily short time intervals. As soon as you reach Planck time (approximately $5\cdot 10^{-44}$ seconds) really weird things are to be expected, though you'd probably be better off asking about it at And that's completely ignoring the fact that it takes some time for the lamp to turn on (or off).

share|cite|improve this answer
After Planck time, physical abd infromational laws don't apply. – Torsten Hĕrculĕ Cärlemän Jul 17 '13 at 15:00
The Cesaro sum actually converges to zero (though I'd rather avoid bumping this question just for that). – tomasz Jan 22 '15 at 18:16

Consider that at the moment when the mad scientist begins to flip the switch infinitely quickly, the switch itself will be moving at an infinite speed back and forth. This is physically impossible since the switch would be unable to travel faster than the speed of light.

If it were somehow possible, the lamp would probably just dim a bit since the light itself would heat up during the on phases, but would not completely turn off during the off-phases.

share|cite|improve this answer

Besides, the switch would break since it would be subjected to unbounded acceleration.

share|cite|improve this answer

Expanding on Umberto P's comment: The state of Thomson's lamp is unspecified or undefined for time $t \geq 2$.

Thomson's lamp is defined to be ON only in the following time intervals:

$[0, 1), [\frac{3}{2}, \frac{7}{4}), [\frac{15}{8}, \frac{31}{16}) \cdots$

Whereas, it is defined to be OFF only in the following time intervals:

$[1, \frac{3}{2}), [\frac{7}{4}, \frac{15}{8}), [\frac{31}{16}, \frac{63}{32}) \cdots$

Notice that no time $t\geq 2$ is in any of these intervals. Therefore, the state of the lamp is undefined or unspecified for time $t \geq 2$.

share|cite|improve this answer

First, while the state of the lamp is defined at every time less than 2 minutes, for any given time, $t$, there are infinitely more switches between $t$ and $2$ minutes, so the state of the theoretical lamp at $2$ minutes is undefined.

Very fast humans can play musical notes at about $20$Hz, so unless our mad scientist is super-human, the light ends off as the scientist hits the 11th iteration and the 12th requires about $34$Hz. Humans are the weak link, not switches.

Once the instantaneous flicker rate reaches $120$Hz (we're using a circuit now), the light is essentially on until we perceive it to be off (which would be a little bit after the 2 minute mark) from the 13th iteration on (the 14th iteration if the switch starts in the off position). Humans fail again because 'healthy' florescent bulbs flicker at $120$Hz, and we don't notice.

By the 39th iteration, the circuit is switching about as fast as WiFi, so if that was the circuit we used, the light finishes off.

Visible light has a maximum frequency of $790$THz, so this whole thing becomes meaningless at the 49th iteration when the light would finish off.

If we instead use ultra fast switching lasers, we can make it about 60 iterations and the light ends on.

Ignoring human and engineering constraints, what about Planck time? After just 150 iterations the switching stops and the light finishes in the on position.

Of course, we still have an infinite number of switches left...

share|cite|improve this answer

The author is trying to conclude that an infinite number of tasks cannot be performed in a finite amount of time because there is a minimum amount of time required to perform any task. However, his argument is flawed. He has created a situation where it is impossible to deduce the state of the lamp and uses that to conclude that infinite tasks are impossible.

Consider a restatement of the problem. Suppose a green particle is moving at a constant speed and travels from point $A$ to point $B$. Halfway between $A$ and $B$ it changes color to red. Three quarters of the way it changes back to green, etc.

What color is the particle when it reaches point $B$?

I haven't told you. I said that it was green at point $A$ and I gave you a rule for determining its color at fractional distances from $A$ to $B$ but I have not defined what the color is at $B$.

Okay, so let's just say that at $B$ the color of the particle is whatever it was last. What color was that? It is undefined because for any time interval before the particle reaches $B$ there are an infinite number of points at which the particle changes color. It is equivalent to asking what the next rational number is after $1$ with the usual linear ordering. There is no next rational number and in this case there is no last color before reaching $B$.

Therefore, I have to tell you what color the particle is at $B$ the same way I have to tell you what state the lamp is in at two minutes.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.