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X-Ray tracks in photo film

  1. Aug 6, 2015 #1
    In Understanding Quantum Mechanics, Ch.1, Sec. 5 Roland Omnes says:
    ".. hard x-rays can leave straight-line tracks in photographic emulsion and this is strongly reminiscent of a particle trajectory."

    How can we describe this in terms of the wavefunction / interaction / measurement paradigm? Is is a series of "position measurments" -- one measurement at each grain of the emulsion? And presumably, a grain of emulsion would have to absorb at most a tiny part of the photon's energy if we are to see many more absorptions further along the track. This seems to fly in the face of "all-or-nothing" annihilation of photons. And what of the expected "collapse" that should occur at each grain?
     
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  3. Aug 6, 2015 #2

    ZapperZ

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    I'm not sure I understand your problem here. It appears as if you think "x-ray" beam consists of only ONE photon. It doesn't. There are a gazillion photons in an x-ray beam, the beam has some penetration depth. This means that while there is a fraction of the photons that interact at a certain depth of the material, others pass through unimpeded. And this continues through the thickness of the material until there's no more material, or until there's no more x-ray photons.

    Transpose this process to the photographic plate, and that's why you see a track.

    Zz.
     
  4. Aug 6, 2015 #3
    If the film is exposed long enough, yes there will be "gazillions" of blackened grains / dots (not linear tracks) that will collectively form a dark area on the film. If this is what is meant in the text then it's fine. But I understood it to be talking about a countable number of photons whose individual tracks can be seen, like particle tracks in a cloud chamber.
     
    Last edited: Aug 6, 2015
  5. Aug 6, 2015 #4

    Nugatory

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    That problem was solved by Nevill Mott in 1929. You'll find a bunch of threads here and googling for "Mott problem" or "Mott paradox" will find more, including Mott's original paper.

    A handwaving qualitative and heuristic explanation is that the initial state of the particle is indeed an expanding spherically symmetric wave. However, that wave function collapses when the particle interacts with one of the droplets of suspended liquid; and Mott showed that the subsequent evolution of that post-interaction wave function leads to a state in which there is a very high probability that the next droplet in the interaction will be a little further away in the same general direction. Of course that's for charged particles in a cloud chamber, but a similar analysis works for photons propagating through a light-sensitive emulsion.
     
  6. Aug 6, 2015 #5
    Thank you.
     
  7. Aug 6, 2015 #6
    Pondering over Nugatory's reply led me to wonder if it makes sense to talk of "entanglement" in this context.

    Consider a number of photo grains that happen to lie along a classically possible path (e.g. a radial line going outward from the emitting region). The final state of the grains is significantly correlated -- either you will see a lot of grains on this line that have blackened, or you won't see any blackened grains. On the other hand, if we examine grains that don't lie on the same radial path then there would be no particular correlation among them. So the passage of the x-ray wave has left each radially located set of grains in a correlated (entangled) condition).

    So my question again - is it valid to call this an 'entanglement' scenario?
     
  8. Aug 8, 2015 #7

    Nugatory

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    Entanglement implies correlation, but correlation does not necessarily imply entanglement - it's easy to find correlations that are not caused by entanglement.

    Only to the extent that any measurement can be thought of as an entanglement of the measuring apparatus with the quantum system being measured... And that's one of the ways into the swamp of never-ending interpretational debate.

    A droplet in a cloud chamber or a granule of photosensitive emulsion is not isolated from the environment and quite large enough that decoherence will knock down any superpositions of blackened and not-blackened - and there can be no entanglement without superposition.
     
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