Must time require particle interaction (including decay)?

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This is really about challenging the understanding of "Time" in the physical sense.

I would like to test the hypothesis that time is merely the manifestation of particles in their interaction to one another. It seems that time may be meaningless (and non-existent) in a one-particle system. If so, doesn't particle decay imply that time is manifested - but if so, then what is it interacting with? ...

First, I am trying to accept the premise that "Time" is merely the manifestation of the sequencing particles which undergo motion. This seems to be easy to fathom when consider:
1) a hypothetical closed but infinite system with a single point particle, for which time is meaningless since nothing is evolving in the system. Even its motion is meaningless, since there is no reference point to reveal motion of the particle. Given it is a point particle, then I assume it can not experience self-rotation, as that would be meaningless.
2) a hypothetical closed system with a two particles. If one particle is a reference point, then the moving position of the second particle can be considered coincident with its history (time and position). So rate of time could be sensed through the speed of the particle, and history of time could be reflected in the progression of positions of the particle.
3) From there, I can extrapolate to a real universe of a multitude of particles. It seems that this model of time is still consistent (if I haven't overlooked anything - YET).

Ah oh, what about particle decay. A particle by itself evidently still experiences, or reveals, time (within itself) as recognized by its characteristic particle decay. As I understand it, a particle will decay without any necessary external influence to induce the decay. Instead, it decays at a intrinsic rate (half life) which is irrespective of its environment. As such, the point-particle system above would still manifest time through its eventual particle decay.

Is it possible that the one-particle system is not really a one-particle system when it comes to decay? Perhaps decay occurs because of some stochastic process amongst its constituent quarks (or strings, etc) as they interact. The stochastic process might lead to decay when probability of a rare "interference" event between the constituents eventually experience the inevitable climax.

QUESTION
1) Are any of these premises flawed with regard to current knowledge?
2) Are the conclusions viable?
 

Answers and Replies

  • #2
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No replies yet - have I posted this in the wrong forum area?
 
  • #3
Vanadium 50
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First, it is not reasonable to expect an answer in under 24 hours. People have other things to do than to patrol the boards, searching for a message from you.

Second, I have absolutely no idea what you are talking about. I hope it's not a personal theory, as those are not something discussed on PF. But I don't understand what you are talking about.
 
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Vanadium 50:

I don't think your comment was necessary. If you don't have time for my question then you didn't need to spend your time criticizing it and me.

Fortunately, I have found most of the contributors on this forum to be informative and pleasant.
 
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  • #5
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I think I found the answer to my question:

http://en.wikipedia.org/wiki/Radioactive_decay

"Such a collapse (a decay event) requires a specific activation energy. In the case of an excited atomic nucleus, the arbitrarily small disturbance comes from quantum vacuum fluctuations."

So, yes, the stochastic process can lead to decay, as the sole subject particle interacts with virtual particles of the (quantum) vacuum state. In that sense, the one-particle system is never really alone; instead it feels and thus interacts with the vacuum state.

As a corollary to the aforementioned premise: time - if defined as a manifestation of particle interaction - must always exist because the vacuum state is never zero and thus will always provide for the presence of some (virtual) particles.
 
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