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Imagine that I pull a stick one meter towards me. And say the stick is 100 (m) long. My friend looks at the stick on the other side. How long will it take before my friend sees the stick to be one meter displaced (ignoring the physiological processes of seeing)? Note that similar questions have been asked here (and there are other variations). It is answered that the signal travels with the speed of sound. Which I can't imagine. In this question something similar is asked to what I have asked here. But in the answers, the velocity of the signal (of which I don't think that it travels with the speed of sound) is not given. And I'm interested in exactly that.
Is the velocity may be the velocity of light for the medium of the stick? Imagine the stick to be made of thick fiberglass.

If I pull the first layer of atoms, doesn't the second layer react after a time it takes for the changing em field surrounding the first layer to reach the second? And just so for the third layer wrt to second, and so on to the last layer? Isn't a propagating pressure different from this?

If I gently pull a stick (say with an acceleration of $0,01(\frac{m}{sec^2}$)) that extends to the moon, will not the stick on the moon start to move in the direction of my pull almost instantaneously? Which means that the other end will start to move after the time it takes for light to reach the moon? I don't see how I introduce a pressure wave by pulling this slow.

If I move the same stick (of one light-second long, extending to the moon) slowly to and fro (with an amplitude of, say, one meter). Will not the stick on the moon start moving to and fro after a light second? If you don't see the same displacement (almost) instantaneously on the moon then the stick must be elongated one meter!

Last edition!
If I hang a bucket stationary on a thousand-kilometer-high balcony (on a thousand-kilometer long rope). Will not the bucket (almost) instantaneously rise one meter too? If this is not so then the rope must have elongated one meter, and I can't imagine accomplishing this by pulling the rope slowly up. If I pull the bucket up in one second, and if it takes $v/L$ (assuming $L$ to be 1000(km) and $v$ to be much less than 1000(km/sec), a good estimate for the propagation speed)) seconds to arrive at the bucket's site, the rope must have elongated by one meter.

Final edition!!
What makes this problem different from the switching on of a light bulb? If I turn the switch the bulb (almost) instantaneously lights up.
Or what if we push marbles through a pipe? If I put an extra marble in a filled pipe, will not instantaneously one drop out on the other side? If the length of the pipe is fixed (and if it's long enough) and if this putting an extra marble in would travel with the speed of sound (for the marbles), wouldn't all marbles temporarily fit in the pipe?

Deschele Schilder
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5 Answers5

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The speed of sound is the speed at which disturbances travel through the medium. When you pull at one end of the stick, you are essentially inducing such a perturbation. Microscopically, this is due to the fact that molecules along the stick react to this only by interactions with its neighbors, and have no idea that you've pulled at the stick until that disturbance reaches its neighbors.

Edit: regarding your edit, changes in pressure correspond to changes in microscopic position. What you're asking about is essentially a macroscopic manifestation of the same phenomenon. To hone your intuition, imagine a 10-foot wide steel cable, the kind that supports a bridge. If an earthquake shakes the whole system, the cable is going to undergo some oscillatory motion. On one hand, these waves travel through the steel cable like some kind of sound wave, but on the other hand, the oscillations might be macroscopically visible and would bear no distinction from the result of a friend shaking one end of the cable really hard.

Deschele Schilder
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jsborne
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    But I'm not referring to oscillatory motion. The atoms in the stick don't start to move due to a pressure I introduce upon pulling (or pressing). Why doesn't the next layer of atoms in the stick (after I pulled the first one) start to move after a time it takes an em disturbance to reach them? – Deschele Schilder May 12 '21 at 15:35
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    Elastic force (like the kind exhibited between atoms in the rod) takes the form $F \sim (\delta x)^2$, and has classical equations of motion that give a characteristic speed of propagation (the fact that there is such a speed emerges from the fact that $x\pm ct$ are the characteristic curves of the PDE; see this). The initial condition need not be oscillatory: we can have the initial condition $f(t=0)$ be a delta function impulse, and the impulse will travel along the rod as $f(x-ct)$. – jsborne May 12 '21 at 16:33
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You seem to believe that when you pull the stick through one meter, momentarily the stick is somehow stretched by one meter at your end before the other end moves to restore the stick's length, which is not the case.

When you start to pull the stick it will elongate microscopically (if the stick was replaced by something more elastic, like a weak spring, you would see it extend).

The effect of the microscopic extension will propagate along the stick at the speed of sound in the stick until a position of equilibrium is reached at which point the force you are applying to the stick will be felt by your friend.

Depending upon the material from which the stick is made, your friend might feel the pull almost instantaneously. The speed of sound in a wooden stick would be around 3 to 4 kilometres per second, so it would travel through a 100m stick in perhaps 0.03s

From that moment on, the stick and your friend are moving in unison (more or less) through the 1m.

Something of the reverse of what I have just described will happen when you apply a force to bring your end of the stick to rest after moving it 1m. The stick will compress somewhat until it and your friend are at rest.

Marco Ocram
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  • Suppose the stick extends to the moon (and the stick is not too heavy because of that) I pull the stick gently towards me (with a small constant acceleration). will this accelerated motion arrive at the moon after a time that equals the distance to the moon divided by the speed of sound in the stick? I can't imagine. (+1, by the way!) – Deschele Schilder May 12 '21 at 12:49
  • @DescheleSchilder Yes, that's correct. You've stated several times that you don't believe this is correct. Can you state why? Do you think this is too slow? All materials (even solids) can be compressed to a certain extent. This compressibility means pushing on one end does not propagate to the other instantaneously. – BowlOfRed May 12 '21 at 17:00
  • @BowlOfRed I've already found the answer in a comment made by jsborne (below his answer). A pulse travels at the speed of light, $f(x-ct)$. – Deschele Schilder May 12 '21 at 17:05
  • @DescheleSchilder, nothing in that answer suggests the pulse travels at the speed of light. – BowlOfRed May 12 '21 at 17:07
  • @BowlOfRed It's written in a comment below his anwer. – Deschele Schilder May 12 '21 at 17:08
  • @DescheleSchilder The word "light" appears nowhere in the answer or the comment. – BowlOfRed May 12 '21 at 17:16
  • @BowlOfRed Indeed! I thought $c$ referred to the speed of light! If I push a layer of atoms toward another layer, will not the next layer react in the time it takes en em wave to get there? – Deschele Schilder May 12 '21 at 17:20
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    No "c" is the speed of wave propagation in the https://en.wikipedia.org/wiki/Wave_equation. It does not refer to the speed of light in all situations. – BowlOfRed May 12 '21 at 17:23
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    The change in force between any two atoms will happen at the speed of light. But inertia will add significantly longer delays between each point in the chain, and that is the dominant factor and leads to the speed of sound overall propagation. – geshel May 12 '21 at 17:49
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It will be the speed of longitudinal elastic waves in the rod: $$ v_{\rm longitudinal}= \sqrt{\frac Y\rho} $$ where $Y$ is Young's constant, and $\rho$ is the mass density of the material in the rod. I usually call this the "speed of sound" in the rod, although it's not exactly "sound."

mike stone
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  • I can't imagine this to be true. If I pull the stick one meter towards me, doesn't that mean that this one meter pulling towards me must travel at the speed of sound? – Deschele Schilder May 12 '21 at 11:29
  • @DescheleSchilder When you pull a stick - you induce a "pressure wave" to stick molecules medium. This pressure travels to other side of stick with sound wave velocity in medium. – Agnius Vasiliauskas May 12 '21 at 11:32
  • Of course not. But the effect of the pull travels towards the other end at the "speed of sound." – mike stone May 12 '21 at 11:32
  • I merely use the stick to exert pressure on my friend. When will she feel this pressure? – Deschele Schilder May 12 '21 at 11:34
  • @DescheleSchilder, put given longitudinal waves speed and stick length into $t=L/v$ – Agnius Vasiliauskas May 12 '21 at 11:38
  • What is the effect of the pull? – Deschele Schilder May 12 '21 at 11:47
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    @DescheleSchilder It's not easy to see this stuff with your fibreglass pole because the speed of sound in solids is so fast. But you can slow the process down by using a long spring instead of your pole. The intermolecular bonds in the fibreglass are also springy, but they're just a fair bit stiffer than a typical spring. – PM 2Ring May 12 '21 at 15:26
  • @PM2Ring That's the same as the train analogy. But suppose I have an array of atoms, one after another (say they are contained in a cylinder). If I push the first one, isn't the second atom pushed after the amount of time it takes an em wave to reach it? – Deschele Schilder May 12 '21 at 15:44
  • @PM2Ring But what if the long flexible spring (in the video he's referring to) is hanging static on a balcony. What if I slowly move the spring up (with a small upward acceleration)? Or say that a bucket hangs on a piece of rope on a thousand-kilometer-high balcony. I slowly pull the bucket one meter up. Will the bucket rise one meter after a time it takes for sound (pressure waves) to get to the other side? – Deschele Schilder May 12 '21 at 18:00
  • @PM2Ring I get it. But if I stretch a guitar string (which I always tend to do to tight...), won't the stretching occur along the whole string at once? :) – Deschele Schilder May 12 '21 at 18:44
  • Let us continue this discussion in chat. – PM 2Ring May 12 '21 at 18:47
  • @PM2Ring You write "The intermolecular bonds in the fibreglass are also springy" I don't get that. I think you mean that the potential between the atoms is spring-like. There aren't actual springs between them though. When an atom is displaced the next atom will feel the change in this potential after a time it takes for photons to travel to it. And so will the atom next to this atom. The atoms have inertia, but they are almost instantaneously displaced, thereby transmitting a change in the potential to the next atom. Which will almost instantaneously be displaced also, etc. – Deschele Schilder May 12 '21 at 19:30
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this would mean that somewhere between me and her there is a difference of one meter,

Not necessarily. I don't know what material your stick is made of, but the speed of sound in solid materials can be multiple thousands of meters per second. The only way there could be a "difference of one meter" is if you manage to move your end of the stick a full meter in less time than it takes for the "news" that your end is moving to reach the other end. That's maybe only a few tens of milliseconds. You would have to pull extremely hard in order to move your end that fast.

which means that the stick will be elongated (or break).

Yes. If you pull hard enough on your end, you can break the stick. Doesn't matter what it's made of. Every material has some finite tensile strength.

As for "elongated," then that's a yes even if you pull very gently. There is no such thing as a perfectly rigid body.

Solomon Slow
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About $0.25$s.

There are two components in this question:

  • Purely physical - how quickly the other end of the stick responds to a pull, and the answers given above are correct (the excitation travels through the stick with the speed of sound - rather similar to how visual information propagates with the speed of light)
  • Human - how quickly the person holding the stick will feel that the stick is being pulled. Human reaction time is usually taken to be $0.25$s. In this case this time is much longer than the time it takes for the sound wave to travel through $1$m.

Update
My initial answer was for a 1 meter long stick. Since then the question was edited: so, if the stick is sufficiently long, we do need to account for the speed of sound: $$ 0.25 + \frac{L}{c_s} $$

Roger V.
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