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Density LHC vs. Cosmic Rays

  1. Apr 14, 2015 #1
    Hello,

    I have a question regarding the density of the LHC vs. Cosmic Rays and why the big difference doesn't matter for particle research? As a reference we look at ultra-high-energy cosmic rays which are perfect for 'one on one' collisions, but isn't 'dosage' also a parameter to consider?
    --
    In nature there are about a thousand cosmic ray collisions of a few GeV’s (1 GeV= 109 electron Volt) per second per m2. In LHC it are about one 1 billion per second per cm2. That’s 1.000.000 times more for an area which is 10.000 smaller. It is a density difference of 10 billion and unique in the Universe. (see graph)

    The upgraded LHC is going to generate collisions are 1000 times more intense, with energies of 13 TeV (1 TeV= 1012 eV). These collisions are in nature even less frequent per m2 while the density at the LHC of 10 billion per cm2 will be maintained.
    --
    Protons are not only tiny 'stand alone' particles but also wider energy fields connected to the Higgs Field. When cramping a lot of matter/energy in a small box of just a couple millimeters, than there is a different situation in the lab than for cosmic rays in nature whom have a far less dense distribution.

    We see it everywhere in science that dosage plays a key role, except for particle collisions where it is never considered of any importance, why? Are the distances in between (follow up) collisions relatively so spaced out or are there studies of what the limits are of the numbers of collisions one may jam into a particular volume? Also does the 'collision box' gets to be heat up after a while or does everything just fly by? The LHC creates temperatures 105 times hotter than the center of the sun, how much of this energy is spread out through the Higgs Field onto the surrounding matter?

    Regards,

    Michel


    278px-Cosmic-ray_spectrum_with_LHC_luminosity.png
     
  2. jcsd
  3. Apr 14, 2015 #2

    Orodruin

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    I am sorry, I do not see your point. Cosmic ray interactions are relatively infrequent, but you have a lot of target material (i.e., the entire atmosphere). When you do collider physics, you have very tiny cross sectional areas of the beams - you can still crank up the number of collisions, but that is the entire point. This does not a priori mean that the density is so high that it will affect the collisions themselves. The protons at the LHC can still be considered free and the 1-on-1 collision is a very good approximation.

    Temperature is not equivalent to energy. You have to remember that this happens in an extremely tiny region.

    I do not know what your idea of the Higgs field is, but these sentences makes me believe that you have some fundamental misunderstandings in what it is and how it works.
     
  4. Apr 14, 2015 #3
    Yes, my understanding of the Higgs Field is limited.

    I took as a reference the double-slit experiment that can be done with photons as well as protons. So there is the fact that particles can sense the second slit and act accordingly as a wave. This made me wonder to what extend protons at the LHC can still be considered free, because the double slit experiments is done over a distance of a couple of millimeters similar to the collision box at the LHC.

    In chemistry we have reaction–diffusion systems were the concentration of one or more substances distributed in space changes under the influence of two processes: local reactions in which the substances are transformed into each other, and diffusion which causes the substances to spread out over a surface in space.

    This relates to density where for some experiments a reaction can't happen because one substance is too far away from the other. Now taking cosmic rays and the low distribution it shows that no reaction is taking place, but my question here is at what close distance/frequency will you get interaction such as for the double slit. Is there the possibility of superposition of collisions (waves) that we of course do not have with cosmic rays with their low density?

    Similar to a nuclear reactor where the process is controlled by adjusting the number of neutrons produced. Control rods that absorb neutrons are inserted between the uranium fuel rods. Its is the setup of that influences the reaction.

    As the Higgs Field is a scalar field it made me wonder how spread out particles collisions are, like for the double-slit; are it just points flying around or do they diffuse over a wider area and react with the collision-box ... actually the slit-area of the double-slit experiment heats up a little when it bends the light and the photons lose some energy ... or do the magnetic fields of the LHC keep all the particles clear from surrounding matter?
     
  5. Apr 14, 2015 #4

    mfb

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    There is not. The density is way too small to give effects.

    Filling 2*1011 protons in a volume as large as an LHC bunch gives a proton density of roughly 0.000005/nm^3 - that is many orders of magnitude thinner than air. It also means the average distance between protons is of the order of 100 nanometers, or 100 million times the diameter of protons. As result, all those collisions are independent processes in a vacuum. The collision products fly away, the remaining protons fly away, and nothing remains in the collision zone.

    As very rare process (less than one time in a billion), a hadron from a collision can collide with another proton - but that is completely negligible.

    Those experiments are done at energies of milli-electronvolts, 15 orders of magnitude below the LHC energies. The double-slit experiments need such a low energy to make interference visible.

    I agree with Orodruin's post, especially the part about the Higgs. Your questions about it do not make sense.

    The magnets in the LHC keep the protons on a nice circular track, to keep the vacuum the LHC uses pumps.
     
  6. Apr 14, 2015 #5
    How did you get to these numbers? When I do the math, than I take and get:

    Number of protons per bunch: 1.5x1011
    Size of Bunch: 16 microns fully squeezed = 1.6x10-5m2
    Size of a Proton: 0.8418±0.0007 fm (A femtometer is 10-15 meters)

    --

    This roughly gives a density of:
    1x106 protons / nm2
    proton = 1x10-8nm2

    --

    That's 1 proton over a distance of 100 protons.


    btw I don't get your argument of thinner than air ... air can be liquid.


    What is your point? Double-slit is just one electron at a time (size 10-7nm2)and still the interference shows up. Thus the small individual electron can sense the other slit, than the one it passes through, which might be a millimeter away, a distance 10.000 times it's size.
     
    Last edited: Apr 14, 2015
  7. Apr 14, 2015 #6

    mfb

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    The 16 micrometers are RMS in x- and y-direction. The length of the bunches is a few centimeters. Let's assume all protons are within 2 cm x 30 μm x 30 μm, then we get 11000/μm^3 or 0.00001/nm^3. That is a factor 20 more compared to my previous post, I don't remember which values I used there. The conclusion is the same, all collisions happen very far away from other protons.

    A femtometer is a millionth of a meter, and I don't know what your area calculations are supposed to give. You can also see the huge distance difference by the fact that out of 100 billion protons, on average just 20-40 collide (per beam).

    It is thinner than air in the atmosphere by a factor of more than 1000, therefore it is much thinner than liquid air as well.

    The electron does not have an intrinsic size, at least no size has been found. The electron has a wavelength, and this has to be comparable to the slit separation. You need a very low energy to achieve that.
     
  8. Apr 14, 2015 #7
    Alright, depth boosts up the numbers.

    Yes.

    Still it are 6x108 collision/s

    Ok, so the density of protons within a bunch is lower than that of Oxygen within a balloon so to speak.

    But this is my point, you won't be able to light a candle in the higher atmosphere where the air is thin, while liquid air is used for welding. All a matter of density. So cosmic ray collisions happen at a density of 1013 lower than the LHC.

    There's a threshold level between reaction an non-reaction.

    Are you saying that there is no interference for electrons that are more powerful, and that they are less likely to sense the other slit? I thought that only photons would be able to penetrate through walls so to speak at different energy levels. mh. Perhaps they do sense but just fly by ... like how a lazer works by photons flying by that agitate electrons.

    This is where my misconception may be rooted, I thought that all particles would sense the other slit over a distance of just a millimeter ... I guess some particles blow holes in a wall, while other fly through ... and others are in a sense deflected and form interference spots.
     
  9. Apr 14, 2015 #8

    mfb

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    So what?
    I don't see the relation to chemical reactions.
    So what? Both are well-isolated events.
    Not in terms of frequency of reactions. This is not chemistry.

    The wavelength of high-energetic electrons (or protons, or anything else) is way too short for relevant interference effects, yes. That has nothing to do with "flying through walls".
     
  10. Apr 14, 2015 #9
    Isn't the wavelength too short for moving high-energetic particles with a lot of momentum and velocity, but what when they collide and come to a relative standstill; like two billiard balls with the same energy smashing into each other full frontal, doesn't the wavelength spread out again?
     
  11. Apr 14, 2015 #10

    mfb

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    All involved particles in such a collision have high energy, or are very short-living (often both together).
     
  12. Apr 14, 2015 #11
    Ok, there's a lot of energy bundled together, but what about the wavelength of the collision itself? If you look at billiard balls you get a loud 'BANG' when the kinetic energy is converted into heat and sound.
     
  13. Apr 14, 2015 #12

    mfb

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    There is no "wavelength of the collision". Sorry, those questions just do not make sense.
     
  14. Apr 14, 2015 #13
    Than what is the term? When two particles that each have a wavelength collide, than what happens to their wavelengths during a collision?
     
  15. Apr 14, 2015 #14

    mfb

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    Particles do not have "wavelength" as a property that could "do" or "get converted to" something.
    The interesting LHC collisions destroy the protons and create completely new particles. And if you want to calculate their de-Broglie wavelengths (which is questionable on its own), they are of the order of a femtometer or smaller.
     
  16. Apr 14, 2015 #15
    Allright, thanks!
     
  17. Apr 15, 2015 #16
    I went off the deep end after this one ... talking about 'wavelength of particle collisions'.

    Ok, high energy particles don't cause interference patterns, but colliding particles where matter is destroyed should release energy over a wider area, not only in the shape of new particles, no?

    I was mentioning 'wavelength of particle collisions' because for a bomb you have a spherical shock wave spreading out from an explosion. Similar to colliding billiard balls were there's an energy wave that spherically spreads out in the form of heat and sound. There's a conversion of kinetic energy for elastic collisions; electromagnetic force conversion for chemical reactions; and strong nuclear force conversion for nuclear bombs. These newly generated shock-waves spread out through a medium over an area that widely exceeds the place that the particle takes up in space.

    Here's my point regarding the Higgs Field ... Medium (???) ... shockwaves of a pin dropping onto the floor travel through thin invisible air ... and we can hear the collision. Is it perhaps possible that there are shock-waves from particle collisions, that we can't measure with the electromagnetic field, but whom are traveling through the Higgs Field / Vacuum / Medium and reach much further out ... giving all surrounding matter a virtual push?
     
  18. Apr 15, 2015 #17

    Vanadium 50

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    No.

    It's one thing to try and invent new physics to explain things we observe. It's quite another to try and invent new physics to explain things that are not what we observe.
     
    Last edited: Apr 15, 2015
  19. Apr 15, 2015 #18

    mfb

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    You cannot apply concepts of macroscopic reactions like bombs to particle physics. You have completely different situations.

    The bomb has ~10^30 particles, expanding outwards into ~10^30 other particles around it after explosion. Yes, putting two bombs next to each other would matter.

    The proton-proton collisions happen in a perfect vacuum - it does not matter if the next proton is a micrometer or a light year away, both distance scales are massively beyond the scale of the collisions.
    Particles are the only things that can have energy. Particles are the only reaction product of a collision.
    No.
     
  20. Apr 15, 2015 #19
    It's not so much about explaining things we don't observe, it is about looking at a field that that we have proven to exist, in this case the Higgs Field, and superimpose what we know about fields in general.

    http://en.wikipedia.org/wiki/Field_physics

    Either the Higgs Field is dead matter without any real physical properties or it is.
     
  21. Apr 15, 2015 #20
    True, the vacuum has no particles in it, but that doesn't mean that it contains no energy or is without any physical properties.

    http://en.wikipedia.org/wiki/Vacuum_energy
     
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