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Probing the small and HUP

  1. Oct 26, 2011 #1
    Measuring with great precision the position of a particle, requires high momentum ( and thus energy) of a second particle, that probes the first.

    That is because when you want to guide a particle to a precise position, then you need to give it high momentum, so it has enough leeway to spread out in momentum. (Where the trade-off between precision in position and spread in momentum is of course due to HUP.)

    But the probing particle has high momentum only in one direction, the direction in which it is accelerated. From that would follow that we can only guide the probing particle in one space direction. The other two directions can't be controlled, since in those there is only little leeway in momentum.

    So the question, what happens to the other two space-momentum uncertainty relations in high-energy experiments? (The two relations where space and momentum is orthogonal to the moving direction of the accelerated probing particle.)

    thanks in advanced
     
  2. jcsd
  3. Oct 26, 2011 #2
    Ok, more basic: why do we shoot at the microscopic particle A with a high momentum microscopic particle B to find out more about A?

    What has HUP to do with it, which says nothing about high momentum, but only talks about the spread of momentum?

    And does not HUP apply for all directions in space?
     
    Last edited: Oct 26, 2011
  4. Oct 26, 2011 #3
    Some numbers: if we want to probe the distance 1fm = 10^-15 m, momentum uncertainty is at least 197 MeV.

    So when we shoot a 2 GeV particle at a target, the particle can vary enough in momentum so that HUP allows it to probe distances at 1fm.

    But again, the particle has high momentum only in one direction. What about the other two directions?
     
  5. Oct 27, 2011 #4
  6. Oct 27, 2011 #5

    tom.stoer

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    high energy = short Compton wave length acts like high frequency in a standard microscope; that's the reason for probing smaller structures with higher energy
     
  7. Oct 27, 2011 #6
    humanino and Tom, thanks for answering

    Right, that's one standard explanation that is often given, which is perfectly fine.

    But it is also over and over been told that the need for high-energy experiments to study the microscopic world of elementary particles is directly related to Heisenberg's uncertainty principle.

    I would like to see more clearly how!
     
  8. Oct 27, 2011 #7

    tom.stoer

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    I know this argument but I think it's strange.

    Let Delta x become smaller b/c the system (atom, nucleus, nucleon, quark) becomes smaller; the HUP tells you that Delta p of the constituents increases, which means that in typical nucleons you will have high energy of the quarks, i.e. that they have to be treated relativistically (but of course we know from QCD that there are small-quark-momentum contributions to the nucleon structure). Fine.

    But it tells you nothing about the momentum of the electron via which you want to probe the structure of size Delta-x; this is b/x the HUP is always about a "Delta".
     
  9. Oct 27, 2011 #8

    Bill_K

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    Heisenberg's uncertainty principle is vastly oversold.
    You mean the deBroglie wavelength. The Compton wavelength ħ/mc does not depend on the energy.
     
  10. Oct 27, 2011 #9

    tom.stoer

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    Yes, you are right
     
  11. Oct 27, 2011 #10
    Could you elaborate a bit more?

    And why then do a trillion books and internet sources claim that the need for high-momentum particle accelerators to probe the microscopic world follows directly from the HUP? Are they all wrong?
     
  12. Oct 27, 2011 #11

    tom.stoer

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    They should explain the meaning of 'directly'
     
  13. Oct 27, 2011 #12
    Say you have a particle moving with 1 MeV and another with 1 GeV. Both vary with 197 Mev at scale of 1fm. Does that not imply that we can control and predict the path of the 1 GeV particle better than that of the 1 MeV particle?
     
  14. Oct 27, 2011 #13

    tom.stoer

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    The HUP says nothing about the absolute scale, it is about the Delta only. Delta-x of a quark inside a proton does not change when the proton is accelerated (of course you could argue that it DOES change b/c of Lorentz contraction - 'pancake' - and you are right - but that does not follow directly from with HUP but from Lorentz contraction).

    The problem seems to be that these handwaving arguments seem to be nice a first first sight but that you run into more and more difficulties when asking more detailed question. At a certain point it's better to stop using handwaving arguments and to start to calculate something explicitly. I don't like repairing such arguments.
     
  15. Oct 28, 2011 #14
    But a particle at 1fm has the same uncertainty in momentum, no matter how fast it is flying by, i.e how much momentum it has.

    What does that imply for all the collider experiments?

    You gave an explanation in waves, where high momentum means short waves and an accelerator thus works similar to a light microscope. Could somebody translate it into a particle picture?
     
  16. Dec 14, 2011 #15
    I would like ask again why the need for high momentum particles to study the microscopic world? How is that related to Heisenberg's uncertainty principle which deals with the spread in momentum and position measurements.

    thanks!
     
  17. Sep 5, 2012 #16
    Hi there, I have been worrying about the same thing.

    In my uni it says:


    to probe the nucleus we need Δx<<1fm, this implies Δp<<200MeV
    "This in turn means that the momenta of the particle used as a probe must have a momentum much larger than this, and hence an energy large compared with ≈ 200 MeV."




    So that is saying that Δp<<200MeV means that p>>200MeV

    This was leading me to believe that Δp~p


    However I now think that maybe it is trying (badly) to say that p>>Δp

    Is the reason that you need a high momentum for probing small distances because.... you need a high uncertainty in momentum, and you can't have an uncertainty higher than the momentum you are (within error) creating.

    i.e you can't have momentum of 50MeV with an uncertainty of 200MeV...
    You need to produce a moment that is much larger than the uncertainty in momentum, so that you can ignore the uncertainty

    You need P>>ΔP

    Would anyone like to tell me if this is correct
    Thanks
    Soph
     
  18. Sep 17, 2012 #17
    No ones touching this - that would indicate i've got it wrong or people don't understand my possible explanation haha ah well!
     
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