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Suppose we have a mass $m$. We can talk about two of its parameters: The net force applied on it $f(t)$ and its net acceleration $a(t)$.
I want to know whether there is any delay between $f(t)$ and $a(t)$ in the real world.
Newton's equation doesn't include a delay, by asserting that $f(t) = ma(t)$, but in the real world scenario is there any delay?

Ilustrating more:
Suppose before time $t=t_0$ the net force was zero but at time $t=t_0$ the force is non-zero.
At what instant is the acceleration gonna be non-zero? Is it gonna be at $t=t_0$ too? In other words, is there any delay between the information embedded by the "Net force" parameter and the information embedded by the "acceleration" parameter?
Perhaps I'm messing with a more deep problem, namely, whether time is continuous or not, but I'm not sure.

The motivation for this came from thinking that in a resistor, there is probably a delay between $V(t)$ and $i(t)$ in the real-world, even though Ohm's law doesn't include it.

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  • $\begingroup$ No simply because the force is defined by the equation $F=d(mv)/dt$ so if you have no acceleration then you have no force and vice versa. $\endgroup$ – Quantum spaghettification May 11 '15 at 8:01
  • $\begingroup$ Is there a delay? Of course. The center of mass will only move with the average of the mass distribution of the body, which, of course, is compressible, even in classical mechanics. See "speed of sound". $\endgroup$ – CuriousOne May 11 '15 at 10:35
  • $\begingroup$ Ideal resistor have no delay between $V(t)$ and $i(t)$. But real resistors can have parasit capacitances/inductances, that account for that delay. And indeed, you can include those in Ohm's law replacing resistance by impedance. $\endgroup$ – Bosoneando May 17 '15 at 11:33
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In the real world, you push the atoms in the back end of the object, which push the atoms in the next layer, etc. This means that the front end of the object won't start accelerating until $t = t_0 + \Delta x/c_s$, where $\Delta x$ is the length of the object and $c_s$ is the speed of sound in the material of the object.

If you consider the object to be one atom only, you end up having to define when your push starts, since the force you apply ultimately is an electromagnetic force, which has an infinite extend, and thus applies even from far away. When you define your push to start at $t_0$, your acceleration by definition starts here as well.

Because I like drawing, here's a drawing of you pushing an object from far away. For most practical purposes, you can take the acceleration of the object to be zero until you're very close, though. But it is there.

enter image description here

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  • $\begingroup$ The relevant speed here is the speed of sound, not the speed of light. $\endgroup$ – CuriousOne May 11 '15 at 10:36
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    $\begingroup$ True, the speed with which the push propagates turns out to be the speed of sound. My intention was to show that the speed of the signal can never be infinite, since it must always be <c, but of course it's much slower than that. $\endgroup$ – pela May 11 '15 at 11:31
  • $\begingroup$ I edited my answer considering that the signal propagates with the speed of sound, rather than the "less than the speed of light". Thanks, @CuriousOne. $\endgroup$ – pela May 11 '15 at 12:00
  • $\begingroup$ Can you just solve one little remainder question ? Suppose my hand appears from no where at some point in space and starts heading towards the object. Then as soon as it appears, no matter how far from the object, it will emit photons that will eventually hit the object and hence imply some minimal acceleration after some delay based on the speed of light. Is that correct ? Thanks $\endgroup$ – nerdy May 11 '15 at 13:38
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    $\begingroup$ @nerdy: Yes, that's correct. Although since the photons are quantized, if your hand appeared very far from the object, the average flux at the location of the object could be so small that it would take longer before a single photon happened to hit the object. $\endgroup$ – pela May 11 '15 at 18:16
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Suppose before time t=t0 the net force was zero but at time t=t0 the force is non-zero. At what instant is the acceleration gonna be non-zero ? Is it gonna be at t=t0 too ?

Remember that, if we're talking instantaneous measurements, your target can have a non-zero acceleration and still have zero velocity.

As pointed out, in the real world your target is not going to be a perfectly rigid body, but somewhere within that system a force is acting on a particle (at t=0), which will be accelerating (at t=0) even if it's not moving yet (at t=0).

One angle to look at is the simplified case of a force acting on a single fundamental particle (an electron, say) - now we're into the realms of quantum mechanics and asking if wavefunction collapse or equivalent is instantaneous, and/or whether we can even sensibly talk about making a truly instantaneous measurement.

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Newton actually expressed his 2nd law in terms of force and momentum as Joseph wrote in his comment to your question. But considering causality then the second law is better expressed as $$p(t)=\int F(t)dt$$ In other words force comes first, leading to motion.

By considering the case where mass is constant $$ m \frac{d^2}{dt^2}x(t)=F(t) $$ where x is the displacement. Solving for $x$ $$x(t)=\frac{1}{m}\iint F(t)dt $$ By looking at the 2nd law this way, one can only achieve instantaneous displacement if the force is an ideal impulse $$F(t)=A\delta(t)$$ Where $\delta(t)$ is the idealized Dirac delta function which in the real, physical world can never be realized since it requires infinite amplitude. Consider instead the force as a step function, $F(t)=A u(t)$ where A is the magnitude of the force. In this case the displacement leads to a displacement that is parabolic with time $$x(t)=\frac{A}{2m}t^2$$

Note that the mass, $m$ moderates the rate of motion. The larger $m$ is for a given force, the slower the parabolic rate of displacement.

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