1
$\begingroup$

Say I have a box made of inelastic material such that when a particle hits the box, energy is lost through heat. I then put a particle inside of this box and squeeze the box down.

How does this not violate the uncertainty principle or conservation of energy?

You isolate the position of the particle, which leads to an increase of its momentum. However, the particle's collision on the box will reduce its kinetic energy, thus violating the uncertainty principle, or conservation of energy.

$\endgroup$
  • 2
    $\begingroup$ Can you elaborate on just how you would expect the uncertainty principle to be violated? $\endgroup$ – David Z Feb 12 at 4:36
  • $\begingroup$ @DavidZ Well, you isolate the position of the particle, which leads to an increase of its momentum. However, the particle's collision on the box will reduce its kinetic energy, thus violating the uncertainty principle, or conservation of energy. $\endgroup$ – Goldname Feb 12 at 4:42
  • 1
    $\begingroup$ Could you edit the question to reflect that? It'd be useful to make it clear for people who come by and read the question after comments get deleted. $\endgroup$ – David Z Feb 12 at 4:56
  • $\begingroup$ How certain are you about where your box is? $\endgroup$ – Shufflepants Feb 12 at 19:53
0
$\begingroup$

You clarify in a comment:

Well, you isolate the position of the particle, which leads to an increase of its momentum. However, the particle's collision on the box will reduce its kinetic energy, thus violating the uncertainty principle, or conservation of energy.

All laws in physics hold for isolated systems. There is no energy conservation or momentum conservation in systems that are not isolated.

In your description, inelastic, energy and momentum get out of the box so the particle is not isolated within the box. In a real experiment the energy and momentum will leave both from contact with the walls and by radiation due to stray fields on the walls, as black body radiation type. Thus you can apply the $dp.dx$ Heisenberg uncertainty principle only within a $Δ(t)$, as $p$ is continuously diminishing. The most this hypothetical box could squeeze in space would be in the dimensions of the molecules constituting the sides of the box, the final uncertainty in the momentum, and the final inelastic collision: it will then be a part of the molecules of the wall, with the appropriate momentum and position definition.

$\endgroup$
0
$\begingroup$

for the case where the walls are perfectly elastic: as you squeeze down on the particle in the interests of reducing the uncertainty in its position within the box, the particle responds by increasing the uncertainty in its momentum: its bouncing around within the contracting box becomes increasingly violent. Squeeze the box down enough, and the momentum of the particle can become so (momentarily) huge that particle/antiparticle pairs can be briefly produced. At that point, not only do you not know what the momentum of the particle is, but you don't even know how many particles there are in the box at any instant.

For the case where the walls are inelastic, and every bounce of the particle off the walls therefore removes kinetic energy from the particle, you have a refrigerator whose ability to remove kinetic energy from the bouncing particle will tend to zero before you get to absolute zero. This is because chilling down the particle decreases the uncertainty in its momentum (by driving it towards zero) and the particle responds by increasing the uncertainty in its position.

This means that even after refrigerating the particle as much as you wish, there will always be some residual motion left in the particle that you cannot in principle extract, which is referred to as zero-point energy.

$\endgroup$
  • $\begingroup$ When in this scenario do you make a "measurement" of the momentum? Is it when it interacts with the sides of the box? $\endgroup$ – Aaron Stevens Feb 12 at 5:21
  • $\begingroup$ Also worth mentioning: there is uncertainty in the momentum and position of your box. $\endgroup$ – Shufflepants Feb 12 at 19:52
  • $\begingroup$ sure, but the assumption is that the box is far bigger than the particle. $\endgroup$ – niels nielsen Feb 12 at 22:34
  • $\begingroup$ @AaronStevens, the usual way of thinking about that is that the pressure exerted on the walls by the particles comes from their momentum during their collisions with those walls, and that the pressure we then sense is a measure of that momentum. $\endgroup$ – niels nielsen Feb 12 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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