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I Universe thermodynamic equilibrium & Transactional Interpretation

  1. Nov 23, 2016 #1
    Hello, for those who don't know me I'm layman with some knowledge at popular science level. Discussing with a friend we came to the following point and I would like to ask it here because we had doubts about our reasoning.

    Cramer's Transactional Interpretation proposes that, in a sense, a particle will only be emitted if it "knows that it will be absorbed by a receiver". No advanced confirmation wave = no emission.

    Now the discussion was about the universe heat death reaching thermodynamic equilibrium. We said this will happen when no temperature gradients exist anymore, so no photons will be emitted anymore, not even the faintest radio waves. As far as we know, absolute zero may not be a physical temperature, so let's say thermodynamic equilibrium happens when everything in the universe is at (just say) 0.00001 Kelvin. Presumably because of space expansion, when this will happen matter gravitationally bound structures will be quite far from each other.

    The question is, how can an atom "know" that it may not emit a photon anymore because the rest of the universe is already at its same temperature? As long as it is not itself at absolute 0 K, it might seem that it would still attempt to emit a photon to cool itself down, even if of extremely low energy, the smallest quanta packet possible. In order for heat to flow from hot to cold, the emitter needs to know that it is not the coldest entity in the universe. How does it know that this is the case when the potential receivers are arbitrarily far away from it?

    So we thought that this could be an argument in support of the TI? In this theory an atom will only emit a photon if it "knows" that there is a receiver which will absorb it. So it provides a mechanism for why an atom in a universe in thermodynamic equilibrium will not emit any photon, it will not because it can not get any advanced confirmation wave of reception by anything else in the universe. This would be how it knows that it may not radiate anymore.

    In other interpretations, we could not figure out what would prevent the atom from still emitting the photon. If the emission does not care whether the photon may ever be received or not, it seems that the atom will simply emit it, and time will tell whether it can ever be absorbed or not by any receiver.

    We thought that it is only the TI where we find an explanation for why the atom will not emit the photon, because "it already knows that it can not be received by anything else in the universe".

    I hope I explained our dilemma clearly enough. Thank You for any answers or comments.
  2. jcsd
  3. Nov 23, 2016 #2
    You're misunderstanding "heat death" idea. It postulates a universe at a uniform equilibrium temperature, doesn't have to be (in fact, can't be) absolute zero. If everything's in equilibrium, even at a high temp, no work can be done with that energy. Presumably if universe expands enough it would get all the way down to 0.00001 Kelvin, as you say, even less. Although there would always be some tiny amount of thermodynamic radiation, at equilibrium the emitted radiation will be exactly balanced by absorbed. It has no particular relevance to TI, that I can see.
  4. Nov 23, 2016 #3
    Thanks secur.

    But how would an atom "know" whether it may emit a photon or it may not? If the environment is colder than the atom, radiation may happen. When the environment is at the same temperature as the atom, the atom may not radiate. The "environment" (the nearest atom which could absorb an emitted photon) may be very far from the potentially emitting atom, arbitrarily far away. How does the atom know the temperature of that nearest atom, which may be arbitrarily far away, so that it knows whether it can emit a photon or not?
  5. Nov 23, 2016 #4
    Very interesting point. It does seem to me that TI provides a resolution to the apparent paradox here, i.e., of how an atom knows whether or to emit and in particular how it knows not not to emit at all in a universe in a 'heat death' situation. Of course the interesting thing you point to is the apparent nonlocality: a potential absorber could be arbitrarily distant from a potential emitter and yet the emitter 'knows' it can emit to that absorber. This is best understood in the PTI picture in which quantum systems like atoms are not spacetime objects until and unless they participate in an actualized transaction. See my 2012 CUP book, my 2015 ICP book, and/or https://arxiv.org/abs/1608.00660 and https://arxiv.org/abs/1411.2072 for details.
    Interestingly this is not a new paradox. The whole 'principle of least action' explored by Feynman in his sum-over-paths approach involves a quantum 'knowing' what the path of least action is for any final state. How does it know that? Again PTI provides an explanation. (Note that I'm more recently referring to PTI as RTI because it is also the relativistic extension of the original TI.)
  6. Nov 24, 2016 #5
    Thanks Ms. Kastner. Quite by coincidence I had already read your two papers in the links, it's a honor to have your reply to my question, confirming what I thought.

    To other members (or yourself) I would still like to ask how do other interpretations deal with this issue (if they somehow do).
  7. Nov 25, 2016 #6


    Staff: Mentor


    No. Thermodynamic equilibrium does not mean no radiation; it just means incoming radiation balances outgoing radiation for every object. So objects still emit and absorb photons--they just emit and absorb the same amount of energy in any period of time, so they stay at the same temperature forever.
  8. Nov 25, 2016 #7
    Thanks PeterDonis, I was not aware of that. So if we consider an isolated system containing only two objects at the same temperature, they will engage in a never-ending exchange of photons without any of both objects experiencing any actual physical effect, forever ? In this case I guess my reasoning was completely wrong and any entities above 0 Kelvin will radiate, regardless the temperature of everything else in the universe, even if the temperature of their environment is higher, right?
    Last edited: Nov 25, 2016
  9. Nov 25, 2016 #8


    Staff: Mentor


    Yes. If the temperature of the environment is higher, they will absorb more radiation than they emit, but they will still emit radiation.
  10. Nov 26, 2016 #9
    So the premise of the OP was wrong, thank you, it's clear now.
  11. Dec 6, 2016 #10
    But the point still remains that an atom must 'know' about its surroundings in order to know how much radiation it can absorb/emit. IMHO the direct-action picture of TI accounts for this in the most natural way. And each individual atom does undergo a change in internal state when emitting/absorbing, even if not a change in temperature--the latter being a macroscopic quantity related to the avg. kinetic energy of the constituents.
  12. Apr 23, 2018 #11
    Just an update that the Born Rule has now been explicitly derived in the relativistic transactional interpretation for radiative processes. See my paper with John Cramer: https://arxiv.org/abs/1711.04501
    Comments welcome.
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