The question is hard for me to put into one sentence so please try to completely read the example :

If I had a stick that is $1000 \ \mathrm{Km}$ long and I would push it forward with $1 \ \mathrm{mm}$ in lets say... $10 \ \mathrm{ms}$ (I have completely no idea how long it would take) but let's say I was about to turn on a switch that is $1000 \ \mathrm{Km} + 1 \ \mathrm{mm}$ away from me, how long would it take for me to see the light that is next to the switch to turn on?

Basically the answer would be a calculation based on the distance the light needs to travel in order for it to reach me $+$ that one millisecond. But I cannot understand that that calculation does not consider anything about the stick being pushed, if the stick was $1 \ \mathrm{m}$ and I would perform the same action the calculation would be the same, resulting in the fact that the stick can have any length no matter how far it would still turn on that light as fast as it would if it was $1 \ \mathrm{m}$ long or $1000 \ \mathrm{Km}$ long. Something in my logic way of thinking is telling me that the stick must be slowed down based on the speed of light.

If you can provide an answer to the next example I would fully understand this paradox :

If I have a stick that is $1000 \ \mathrm{Km}$ long and pushing it forward just a little would cause the light in my tower to turn on, $1000 \ \mathrm{Km}$ further away someone is waiting on my push and has the end of the stick placed on the switch of hes light tower. What would the someone see happening if I push that stick ?

There are 3 answers possible :

  1. He(the someone, the 2nd person) will see hes light turn on and then the other.
  2. He would see both lights turn on simontaniously. (my answer, considering the speed of light delay on the moving stick)
  3. He would see the other light turn on 1st and than his. (very unlikely)

The the answer I need is either 1 or 2. I do not understand why it would be 1 but I would love to accept it, in order to accept it I need someone to explain to me why 1 is the answer.

Thank you so much for reading my question and/or leaving a comment or answer !



3 Answers 3


There is no such thing as an incompressible stick. When you push on your end, a compression wave travels down the stick at the speed of sound, which is much slower than the speed of light. The other end of the stick does not move until this compression wave reaches it.

I think the rest of your problems disappear without the existence of an incompressible stick. If you still have any questions remaining, I'm happy to edit this answer.


Here's the answer to your first example:

Though it might take you only 10 ms to push your end of the stick, it would still take a much longer time for the effect of that to reach the other end. This is because there are no perfect rigid bodies. An impulse at one end has to travel at the speed of sound in the material, which is always less than the speed of light. So it will take $l/v_s$ + $l/c$ + 10 ms + 10 ms for you to see the light turn on, where $v_s$ is the speed of sound and $l$ is the length of your stick.

I don't fully understand your second example, but I'm guessing the speed of sound limitation should also make things sensible here.

  • $\begingroup$ +1 good answer :) @Diede, let me add that the answer to your second example is in fact choice 3. The reason is the same as the explanation dbrane has given for your first example, namely that your push travels down the stick at the speed of sound, which is slower than the light from your tower. So the light signal reaches the other person before the push on the stick does, and he sees your light before his own light turns on. $\endgroup$
    – David Z
    Commented Mar 14, 2011 at 17:34
  • $\begingroup$ Thank you sooooo much for these answers, this en lighted me so much ! thanks ! $\endgroup$
    – user2547
    Commented Mar 14, 2011 at 17:59

Does "ACOUSTIMAGNETOELECTRICISM!" by William J. Beaty answer your question?

  • $\begingroup$ That's a fun link! I don't agree with precisely every detail he says, but the passion and novelty he brings to the question is admirable. $\endgroup$
    – Andrew
    Commented Mar 21, 2011 at 1:46

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