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Bohr's response to the Einstein box

  1. Dec 2, 2012 #1
    During one of the Einstein-Bohr debates, Einstein proposed a thought experiment that would prove that one could measure time and energy simultaneously. It's known as the Einstein box:


    Then Bohr came up with a resolution (invoking the gravitational time dilation!) which is described e.g. there on Wikipedia. I also have it in my book, but I don't understand it. I have two questions:

    1. Why isn't the explaination as simple as:

    If the emission time of the photon is measured with uncertainty Δt, the change in vertical momentum that could have been introduced to the photon by gravity is maximally:

    Δp = mg Δt

    (where [itex] m = E/c^2 [/itex])

    On the other hand the uncertainty in the vertical position Δz of the box (and hence, of the photon), corresponds to the uncertainty in the potential energy:

    ΔE = mg Δz

    Then since: [itex] Δz Δp ≥ \hbar [/itex] then also:

    [itex] (\frac{ΔE}{mg})(mg Δt) ≥ \hbar → ΔE Δt ≥ \hbar [/itex]

    2. This must be wrong somehow, otherwise it wouldn't take Bohr a whole day to come up with a solution! :smile: Could you explain to me Bohr's argument in more detail? It seems to me like the gravitational time dilation would be a second order uncertainty, since it's an uncertainty in the uncertainty of the emission time.
  2. jcsd
  3. Dec 2, 2012 #2


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    It is not about the vertical momentum of the photon. You can wait until the box is in equilibrium on the scale, or restrict the photon to a horizontal movement.
    In addition, I don't think it gives a meaningful result to combine those equations like you did.
  4. Dec 2, 2012 #3

    I know, that this is what he proposed, but I can't understand it. Could you explain it to me?

    Why isn't it meaningful?
  5. Dec 2, 2012 #4


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    If you take arbitrary equations for different situations (momentum of the box due to photon emission || energy uncertainty based on some position uncertainty), you cannot expect that a combination gives anything interesting. You need some physical connection between them.
  6. Dec 2, 2012 #5
    No no. Sorry if I didn't explain clearly.

    Δz is the uncertainty of the vertical position of the photon, which we measure, measuring the position of the box.

    Δt is the uncertainty of the emission time of the photon

    Δp is the uncertainty of the vertical momentum of the photon (not of the box), which was introduced by the gravity. Depending on what time within the interval from 0 to Δt, the photon has been emitted, the vertical momentum which it gains lies between 0 and mgΔt

    ΔE is the uncertainty of the overall energy of the photon, which follows from the uncertainty of its vertical position. Of course one might say it's also got an uncertainty in the vertical kinetic energy, but that's ~[itex](Δp)^2 [/itex], so it would be a 2nd order correction.

    So it seems like I'm taking the correct conjugate quantities related to this photon. Am I not?
  7. Dec 2, 2012 #6


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    No, those two uncertainties are different.

    This can be as close to 0 as you like, it does not matter for this setup.


    The height of the photon emission does not matter (at least not in first order).
  8. Dec 2, 2012 #7
    You're right, they're different. Ok, let's forget about this part of that sentence. We can measure z by measuring the position of the box, but Δz depends on the size of the slit, etc.

    The photon falls in the gravitational field, hence it gains vertical momentum because it's accelerated by the gravitational force.

    But [itex]ΔE_{grav}[/itex] is of the same order as Δz (it's mgΔz), what other effects on the energy of the photon would be more significant?


    Let me come back to the Bohr's article: "Discussions with Einstein on Epistemological Problems in Atomic Physics" in which his argument is described (source: http://www.marxists.org/reference/subject/philosophy/works/dk/bohr.htm)

    And then he talks about the gravitational time dilation, which introduces an uncertainty on T.

    If only you could tell me why he assumes, that Δp must "obviously" be smaller than this total impulse...? It seems like this total impulse is the total impulse that gravity can impose on the photon within the time T.
  9. Dec 3, 2012 #8


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  10. Dec 3, 2012 #9
    Ohhh thanks! It's good to know that this argument was wrong. I didn't like it. At all.

    At this point I don't whether I should even bother to try to understand what Bohr said - I still don't get why he needed two Δt's in his derivation. But the explaination with entaglement makes so much more sense and it's so much more elegant...!
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