Understanding Absence of a Microscopic Arrow of Time

[Islam Hassan]

At the particle interaction level, we cannot distinguish a preferred direction of interaction, an arrow of time as they say.

I do not understand this if i) in annihilations, there is a manifest disparity between particles before (massive fermions) and after (photons) an interaction and therefore an inferred directionality and ii) in weak interactions there is an asymmetry in the interaction process.

What am I missing here?


IH

 

[Nugatory]

there is a manifest disparity between particles before (massive fermions) and after (photons) an interaction and therefore an inferred directionality

How so? That reaction is reversible: google for "pair production"/

 

[Dale]

in annihilations, there is a manifest disparity between particles before (massive fermions) and after (photons) an interaction

Annihilations are not the only process. There is also two photon pair production, which is the time reverse of anhilation

 

[Islam Hassan]

To take annihilations, their reversal would I believe require a number of sufficiently energetic photons to converge to a single, very precise microscopic location with very precise microscopic timing. Their combined energy must then be exactly equal to the energy need to produce one given particle only and its antimatter particle only.

If I understand correctly what you are saying is that this process remains possible despite a very low, perhaps infinitesimal probability and that this is why there is no microscopic arrow of time?


IH

 

[Islam Hassan]

Ok, I have read the wiki on pair production...only a single boson is needed to do this. I had thought that annihilation produces many more photons than one, a big burst of photons.

Thanks for the feedback.

What about the weak interaction though?


IH

 

[Orodruin]

Their combined energy must then be exactly equal to the energy need to produce one given particle only and its antimatter particle only.

This is false. As long as the energy is sufficient to be above threshold, any excess energy will go into kinetic energy of the resulting electron and positron.

Ok, I have read the wiki on pair production...only a single boson is needed to do this. I had thought that annihilation produces many more photons than one, a big burst of photons.

A single photon cannot pair produce by conservation of energy and momentum. The energy and momentum is typically taken form a nearby nucleus by exchange of a virtual photon.

 

[Islam Hassan]

This is false. As long as the energy is sufficient to be above threshold, any excess energy will go into kinetic energy of the resulting electron and positron.


A single photon cannot pair produce by conservation of energy and momentum. The energy and momentum is typically taken form a nearby nucleus by exchange of a virtual photon.

Thanks for the clarification. Is pair production the same as annihilation? Or can annihilation produce a multiple photon burst?


IH

 

[anorlunda]

Temperature plays a very big role. At any temperature, an equilibrium will be reached between particle creation and annihilation. The equilibrium point varies with temperature.

Susskind's course on Cosmology, lectures 7 and 8, discuss genesis of particles as a function of temperature during the origin of the universe around the time of the big bang. He also disusses the asymmetries.

Here is lecture 7, watch both 7 and 8.

Is pair production the same as annihilation? Or can annihilation produce a multiple photon burst?

Annihilation of what? Not all particles are alike. Not all decay or creation events are alike.

What about the weak interaction though?

Search for CPT symmetry.

 

[Dale]

Their combined energy must then be exactly equal to the energy need to produce one given particle only and its antimatter particle only.

No. Any additional energy in the photons will become KE in the particle pair.

 

[mfb]

The exact inverse reaction to e.g. $e^- e^+ \to \gamma \gamma$ is $\gamma \gamma \to e^- e^+$. While it is quite easy to get the first reaction the second one is much more challenging for various technical reasons, but it exists and it has been measured in accelerators (with some caveats about the photon sources).

The inverse reaction to $\gamma + N \to e^- e^+ N$ where N is a nucleus would be $e^+ e^- N \to \gamma N$. While possible, it would need the three particles to come together at the same time and with suitable conditions for momentum and energy. That is extremely unlikely.

Edit: Some more.
If you want to reverse a decay, e.g. of the Higgs boson, $H \to \gamma \gamma$, then you have to carefully tune the photon energies for $\gamma \gamma \to H$ (or try often enough with a larger range of energies). To make that more realistic, particle accelerators will often look for final states with more than one particle, e.g. $e^- e^+ \to H Z$ as this doesn't require a tuning that precisely (it also has some other advantages not relevant here). This was the reaction LEP hoped to find, but it turned out that the Higgs was just a few percent too heavy to be produced at LEP.

 

[Demystifier]

What about the weak interaction though?

Even though it violates time-inversion invariance, this violation has nothing to do with the thermodynamic arrow of time, according to which entropy increases rather than decreases.