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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):

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

$S=1-1+1-1+1...$

$S=1-(1-1+1-1+1...)$

$S=1-S$

$S=\frac1 2$

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

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    $\begingroup$ The state is only specified for $t \in [0,2)$. Your sum doesn't make sense. $\endgroup$
    – Umberto P.
    Jul 17, 2013 at 14:35
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    $\begingroup$ 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. $\endgroup$
    – Foo Barrigno
    Jul 17, 2013 at 14:38
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    $\begingroup$ 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. $\endgroup$
    – udiboy1209
    Jul 17, 2013 at 14:38
  • $\begingroup$ @Umberto P.: You should consider making that a reply. $\endgroup$
    – Raskolnikov
    Jul 17, 2013 at 15:07
  • $\begingroup$ 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> $\endgroup$
    – Rick Decker
    Jul 17, 2013 at 15:36

6 Answers 6

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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\ $?

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  • $\begingroup$ Did you mean to specify a relationship between $\epsilon$ and $n$, for example $n>-\log_2\epsilon$ instead of $\>(n\geq0)$? $\endgroup$
    – John Bentin
    Jul 17, 2013 at 17:36
  • $\begingroup$ @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 $t$. $\endgroup$
    – Christian Blatter
    Jul 17, 2013 at 18:08
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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 physics.se. And that's completely ignoring the fact that it takes some time for the lamp to turn on (or off).

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  • $\begingroup$ After Planck time, physical abd infromational laws don't apply. $\endgroup$
    – Torsten Hĕrculĕ Cärlemän
    Jul 17, 2013 at 15:00
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    $\begingroup$ The Cesaro sum actually converges to zero (though I'd rather avoid bumping this question just for that). $\endgroup$
    – tomasz
    Jan 22, 2015 at 18:16
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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.

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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 $t \geq 2$.

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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.

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I present an alternative explanation:- An event is a change in a system. All events occur in finite time. Say, for example, the Switch Duration Time = 0.001 seconds. Eventually, but before the 2 minutes, the interval between switching will be less that the Switch Duration Time. Therefore, at 2 minutes the lamp will be in the process of switching but this event will be incomplete and the lamp will be neither on or off.

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