-1
$\begingroup$

Let any amount of arbitrarily spaced protons appear along a line segment of length $x$ in an otherwise empty universe at $t+0$

What mechanism, if any, could cause the breaking of said alignment between $t+\large\frac{x}{c}$ & $t+\small\infty$?

$\endgroup$
  • 2
    $\begingroup$ The Heisenberg Uncertainty Principle will make that difficult. $\endgroup$ – PM 2Ring Apr 25 '19 at 14:20
  • $\begingroup$ If you allow quantum effects, then the uncertainty principle. $\endgroup$ – Lewis Miller Apr 25 '19 at 14:21
  • 1
    $\begingroup$ What's the point of your question? $\endgroup$ – Bill N Apr 25 '19 at 14:52
  • $\begingroup$ Wouldn't knowing that the position of each proton is still in the exact alinement after they each had an initial effect on each other negate any uncertainty about their velocity ? Otherwise the alinement would've been broken before $t+\large\frac{x}{c}$ $\endgroup$ – pushkin_ Apr 25 '19 at 14:55
  • $\begingroup$ Photons travel as waves (see diffraction), don't have a position until absorbed, and don't have an effect on each other. Your question seems to present photons as little balls, which is incorrect. $\endgroup$ – safesphere Apr 25 '19 at 15:23
0
$\begingroup$

Your question assumes infinitely perfect alignment and exactly zero initial motion of the protons.

Quantum mechanics contradicts both of those assumptions. For all particles there is an inherent minimum uncertainty in position, and a minimum uncertainty in momentum. The more precisely you try to define one, the greater the uncertainty in the other. If you try to define the positions precisely, they are going to have large and random momentums throwing them out of alignment. If you try to get them motionless, the positions will contain significant randomness.

So the answer is that the laws of physics do not allow you to create a perfect alignment in the first place. In fact quantum mechanics says the alignment can't exist at all. The positions and momentum values don't exist - over time they have probabilistic values. Sooner or later a random fluctuation will cause an imbalanced force, which inevitably causes a cascading failure of any unstable arrangement.

| cite | improve this answer | |
$\endgroup$

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