Long time ago Wheeler proposed an idea that there is only one electron, some kind of a closed(?) four dimensional knot in spacetime, and in one particular time we slice this knot with a plane and observe it as many electrons at different places in space. In this picture a positron is just this same electron, but moving in an opposite direction. The direction itself is arbitrary, by opposite I mean if we trace the whole trajectory and at the place where it intersects this plane again we already turned back.

So for three-dimensional spacetime we would always have an equal amount of electrons and positrons at one time, it shouldn't hold for four dimensions, as we actually observe, right?

The annihilation process would be just particle turning around in time. And we lose two time plane intersections at this moment for this particle knot. How would the photon's "knot" look like? Because there is no antiphoton, I imagine many trajectories for a photon, so there are actually many photons, they are just straight lines in spacetime. It should hold for all particles that are antiparticles itselves.

There is always some confusion of what C, P and T transformations actually do. In this interpretation, C is reflection of particle trajectories in time dimension , P - in space dimensions and T - corresponds to the choice of direction we follow along the trajectory. So CPT just reflects the knot and we traverse it in the opposite direction. It immidiately makes sense!

A very wild thought would be to imagine how these knots would move, because if they were fixed, it would be boring. But it can't change its topology, so everything would allow to change, even the past, but the overall structure would be the same, and it would be reflected on a small scale taking form of quantum effects, but the complex structures as we are wouldn't notice.


closed as off-topic by ACuriousMind, Kyle Kanos, user36790, user10851, John Rennie Oct 24 '15 at 18:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "We deal with mainstream physics here. Questions about the general correctness of unpublished personal theories are off topic, although specific questions evaluating new theories in the context of established science are usually allowed. For more information, see Is non mainstream physics appropriate for this site?." – ACuriousMind, Kyle Kanos, Community, Community, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ there is no antiphoton depends on your meaning, but typically one says that the antiphoton is a photon. $\endgroup$ – Kyle Kanos Oct 24 '15 at 14:23
  • $\begingroup$ @KyleKanos If photons are "not closed, not turning in time" lines within spacetime, then C transformation wouldn't change topology of such lines, it would still be a line, so they would be indistinguishable $\endgroup$ – swish Oct 24 '15 at 14:35
  • $\begingroup$ Yes, so an antiphoton is a photon, just as I said and just as the answers in the question I linked said. My point is that the existence of an antiphoton is a debatable point for those reasons. $\endgroup$ – Kyle Kanos Oct 24 '15 at 14:40
  • $\begingroup$ trajectory in spacetime: but at the same time particles are not totally localized, or sometime largely non-localized (e.g. an electron on its orbital). $\endgroup$ – Fabrice NEYRET Oct 24 '15 at 15:51
  • $\begingroup$ @FabriceNEYRET That's what I meant by moving trajectory, it's not thin, quantum effects make it blurry. $\endgroup$ – swish Oct 24 '15 at 16:20

A particle and its trajectory are two fundamentally different concepts. A particle is a material object, while the trajectory of a particle is the set of spacetime coordinates from one event to the next.

  • $\begingroup$ What exactly would be a contradiction in unifying these concepts? Instead of talking about particles and their trajectories, we speak of particle as manifestation of its trajectory. We look at it not as some thing at some location in spacetime, but also take its whole path into account, its past, future, everywhere it'd been and will be. $\endgroup$ – swish Oct 24 '15 at 17:59
  • $\begingroup$ What is a manifestation? $\endgroup$ – jjack Oct 24 '15 at 18:02
  • $\begingroup$ An English word... Did I use it wrong? :) $\endgroup$ – swish Oct 24 '15 at 19:07
  • $\begingroup$ @swish: What jjack meant is that "a particle" is an abstraction of a physical object, while "a trajectory" is an abstraction of a set of (usually non-existent) measurements on that object. In that sense you are dealing with two levels of abstraction, one less well defined than the other, except for the oversimplified universe of classical mechanics, which simply does not exist. $\endgroup$ – CuriousOne Oct 24 '15 at 19:53
  • $\begingroup$ @CuriousOne Maybe what I'm getting at, is that a trajectory is what actually represents a physical object, and a particle is what we measure and observe, some subset of it. $\endgroup$ – swish Oct 24 '15 at 20:00

Trajectory is a classical concept that lose its meaning in a quantum field theory, where the fundamental objects are omnipresent fields. Particles are only excitations of these fields.

Not to mention that a realistic quantum field theory lives in curved space time, where the concept of particle is ill defined in general.

In summary, no, the "only one electron" can't be considered a realistic model of particles in real world.

EDIT: Little addition for the curved space part: the number of particle in curved space depends on the state of motion of the observer. For instance in the famous Unruh effect, an accelerated observer sees a black body radiation, while an inertial observer sees the Minkowski vacuum. The standard reference for this is Birrell, Davies (Quantum Fields in Curved Space).

  • $\begingroup$ There is always both particle and field representation for every theory, it depends on what are you summing over, field configurations or particle trajectories. Could you elaborate on difficulty of particle definition in curved spacetime, please? $\endgroup$ – swish Oct 24 '15 at 17:10

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