3 ms·
It's logical when you think about it. Since particles pop in and out of existence all the time and the time these particles exist are > 0, then it should be pos
by cLeEOGPw 12y ago
It's logical when you think about it. Since particles pop in and out of existence all the time and the time these particles exist are > 0, then it should be possible to move those particles before they disappear. So law of conservation of momentum is not violated, neither any other law of physics. It is just counter intuitive, like most stuff related to quantum world.
- claudius 12y agoI’m not quite convinced. If I understand you correctly, what you’re saying is the following: Suppose we send out one photon ν of momentum p_ν. We hence gain momentum -p_ν. This photon then randomly hits a particle X of a particle-antiparticle pair and transfers its momentum/energy to the particle X, creating/altering it into a particle X'. When X' and ¬X collide again, this momentum/energy is lost and doesn’t hit the opposite wall of the cavity (where it would be transferred onto us again, causing us to lose the previously-gained momentum -p_ν). But this would require that when X' and ¬X annihilate each other, the extra momentum of X' is also annihilated. Intuitively, I’d expect it to be sent out as a photon – with momentum p_ν.
- cLeEOGPw 12y agoHow do you imagine annihilation of the extra momentum?
- claudius 12y agoI don’t.
- cLeEOGPw 12y agoSo why you expect it to happen?
- claudius 12y agoI don’t. That’s why I wrote: > But this would require that when X' and ¬X annihilate each other, the extra momentum of X' is also annihilated. Intuitively, I’d expect it to be sent out as a photon – with momentum p_ν. I.e. annihilation of the extra momentum would be required for this to work on the basis of background fluctuations (in my understanding), but I don’t think that this happens.