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What constitutes a particle?

  1. Feb 15, 2005 #1
    De Broglie said that all matter is both a wave and a particle.
    Davisson and Germer proved this in 1927 when they scattered electrons through a single crystal of nickel. :smile:

    Electrons are currently believed to be a fundamental particle and so I can comfortably think of them as having a centralised mass. However I’ve also read that they have managed to diffract Neutrons, Protons and even Helium nuclei and found them all to produce diffraction patterns. :confused:

    Since Neutrons and Protons are made up of quarks and Helium from Protons, Neutrons and Electrons, how is it possible to diffract them? Aren't they just lots of little particles held together by forces? :grumpy:

    What constitutes a particle?
    Electrons are held in "orbit" around a nucleus because of electrostatic forces and so count as part of the same particle, but don't electrostatic forces operate across an infinite distance and so arguably every charged particle is connected to every other charged particle. :yuck:

    Furthermore, would two particles held together by gravity in a vacuum but with no electrostatic bond etc. still diffract as one if they were to simultaneously pass through a slit? :cry:



    Helfen Sie mir bitte!
     
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  3. Feb 15, 2005 #2

    dextercioby

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    And the Scottish physicst George Paget Thomson (the son of Sir James Joseph) proved it,too... :wink:

    That's true...



    Because Luis de Broglie's hypothesis provides the theoretical explanation...Conversly,these experiments confirm the veridicity of Luis de Broglie's hypothesis and hence the veridicity of QM...

    Yes...

    Wrt what theory of physics...?

    True,electromagnetic interaction is both universal and long-ranged...

    What do you mean diffract as one...?Define this notion in the context of QM.

    Daniel.
     
    Last edited: Feb 15, 2005
  4. Feb 16, 2005 #3
    Ok I don't have some of the exact values (such as width of a neutron etc.) but I hope you understand what I mean.

    Take the two particles to be neutrons. They are not held together by any force other than gravity and are travelling through a vacuum at a constant speed relative to the diffraction grating.



    Neutron Rest mass = 1.67 x 10^-27
    Plancks Constant = 6.63 x 10^-34
    Velocity = 1m/s in the direction of the grating
    size of slits = 3.9 x 10^-7 m

    λ=h/ρ
    λ=h/mv
    so long as λ > width of slit then you will get an observable diffraction



    a) if they (the two neutrons) were to pass through the grating at the same time but are both counted as separate particles then:

    λ=h/mv
    λ = (6.63x10^-34) / (1.67x10^-27) x (1)
    λ = 3.97x10^-7 m

    (3.97x10^-7 m) > (3.9x10^-7 m)

    and so they would diffract



    b) However if because of the gravitational attraction they were counted as one particle:

    λ=h/2mv
    λ = (6.63x10^-34) / (2)(1.67x10^-27) x (1)
    λ = 1.985x10^-7

    (1.985x10^-7 m) < (3.9x10^-7 m)

    and so they would not diffract but continue onwards as a particle.



    Realistically they may have to have a different velocity and the grating be a different size for it to work but surely its possible to work out exactly what counts as a single particle and what counts as two separate particles sitting next to each other.
     
  5. Feb 16, 2005 #4

    dextercioby

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    No,no,you didn't get the picture,i'm afraid.Here we don't deal with exact values and strict inequalities,we deal with ORDERS OF MAGNITUDE...For a slit of,let's say,10^{-6}m (one micron),any particle with the (de Broglie's wave) wavelength of the order of magnitude MICRONS will diffract.And if the wavelength is smaller,diffraction patterns will be weaker,and for wavelegths very small wrt the slits,it will not be detectable...I'm not referring to "edge diffraction"...

    Daniel.
     
  6. Feb 16, 2005 #5
    well ok, what if there were a few million neutrons held together by gravity how would you work out the wavelength?
    If someone was to point to a bunch of neutrons and ask you to work out the wavelength would you calculate if for the whole bunch or for each individual neutron?

    whether your dealing with "edge diffraction" or Orders of Magnitude the wavelength is still going to be different and so will the diffraction pattern.
     
  7. Feb 16, 2005 #6

    ZapperZ

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    Again, the same misconception about the double slit experiment appears again.

    Take note that the typical 2-slit interference pattern that you get isn't due to the interference of TWO neutrons on each other. It is due to the interference of ONE neutron with itself! You could shoot just one neutron at a time at the double slit, and STILL, over time, accumulate the idential interference pattern! A single-particle interference pattern is distinctively different than a 2-particle, 3-particle, etc. interference pattern!

    Zz.
     
  8. Feb 16, 2005 #7

    dextercioby

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    The wavelength is computed for each particle.Depends whether the neutrons are held together (like in a nucleus) or act as invidual particles (like ina beam).

    Daniel.
     
  9. Feb 16, 2005 #8

    vanesch

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    When stripped to the bare essentials, I'd say that a particle is defined by a relationship between energy and momentum:

    E^2 = m^2 c^4 + p^2 c^2

    It is the fact that *this* relation comes out of quantum field theory that we say that it describes particles.


    cheers,
    Patrick.
     
  10. Feb 16, 2005 #9

    dextercioby

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    I'd say it's more like BUILT IN.To quantize fields,you have to describe them classically first,which is achived in the context provided by Special Relativity which contains
    [tex] E^2 = m^2 c^4 + p^2 c^2 [/tex]

    Daniel.

    P.S.Maybe i missinterpreted your whole phrase,or just the expression "comes out" which seemed unappropriate to me...
     
  11. Feb 16, 2005 #10
    Zapper - Yes I should have said slit rather than grating


    Dextercioby - "The wavelength is computed for each particle.Depends whether the neutrons are held together (like in a nucleus) or act as invidual particles (like ina beam)."

    That's exactly what I was wondering - what counts as being "held together"


    Vanesch - I'm not sure how E^2 = m^2 c^4 + p^2 c^2 describes what's a particle and what's not (because surely m and p are values of the particle) but that's probably just because I haven't studied relativity yet.
     
  12. Feb 16, 2005 #11

    vanesch

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    My remark was probably not complete...
    Of course I know that the origin of the KG equation is the above relationship translated into time and position derivatives (that's probably the essential step !), and all other relativistic field equations are variations on the same theme. But the fun thing is that when you go through the quantizing mantra (for a free field theory) and you then calculate the common spectrum of the {E,p} operators (they commute so you can do that), then that spectrum has an interesting structure. First of all, because of Lorentz invariance, the eigenvalues of those 4 operators have to make up a 4-vector ; and if you boost them each time so as to make the space-like part 0, you will find that these 4-vectors of eigenvalues take on the following aspect: (E = 0 ; p = 0) ; (E = m ; p = 0) ; (E >= 2m ; p = 0).
    The first one is the vacuum, the second one are 1-particle states and the third one can always be seen as a composition of several 4-momentum vectors which obey E_i^2 = m^2 c^4 + p_i^2 c^2 : we are in the center of mass of a multiparticle system, all with mass m.

    But note that in general, the eigenvectors of (E,p) individually, of the H and P operators, do not have to respect E^2 = m^2 c^4 + p^2 c^2 ! But they can be build up by components who do... that's how I see that particles come out of QFT.

    cheers,
    Patrick.
     
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