do the particles ever collide in QED

Discussion in 'Physics & Math' started by Arlich Vomalites, Jan 15, 2015.

  1. Farsight

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    3,492
    Yes, on this thread. I've referred to curl before, saying it's also known as rot which is short for rotor, and that you can get a handle on curl as opposed to curvature by thinking of a wire: you bend it into a Fibonacci spiral, it's curved. But if you also bend it orthogonally too, it's curled, like a hair. I tend not to talk much about divergence because fluid analogies and sources and sinks can be misleading. In no sense does an electric or electromagnetic field flow out of a positron into an electron. Nor does wind flow out of an anticyclone into an cyclone. The Falaco soliton is a vortex, but it isn't like the vortex when you empty your bath.

    It's not a proof . It's an explanation of why electrons and positrons move linearly and rotationally the way they do . You couldn't explain it, remember? And now you're carping about adding differential operators? How many times do I have to show you how to combine electric field lines and magnetic field lines to depict the electromagnetic field. Which has a screw nature, and isn't totally unlike the gravitomagnetic field. It ain't magic you know.
     
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  3. Arlich Vomalites Registered Member

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    Do you think that because electron is a fermion, also a spin-1/2 particle, that it means that the electron rotates, that spin-1/2 means a rotating particle?

    I think that in order to get a rotation of 360 degrees, a full circle, you need a spin-1 particle, that is a boson. I think that you are conflating magnetic fields and electric fields. It is the magnetic field which is rotating. You don't have a magnetic field that does only half a circle, a 180 degrees of rotation.

    What is the quantum of the magnetic field?

    The electron has a charge even if it is not rotating. Charge is a fundamental property that particles have, it is not at all a consequence of rotation, or movement. A moving charge produces the magnetic field which is rotating, and a rotating charge produces the magnetic field which is rotating too. Magnetic field is always rotating. But it is wrong to say that rotation or movement causes the charge of the particles. In fact, I think no-one knows where the charge comes from as far as I know.
    The world is just made of positivity and negativity, without that duality the world would not exist. The duality
    that appears as the electron and the positron.
    Where does goodness and evilness come from?

    No. It is still an electron. But it is just not rotating. It has charge, mass, its spin is 1/2, those are the properties of the electron.

    What do you get if the cyclone does only half a circle, a 180 degrees rotation? Do you have a wind or a cyclone?

    The difference between an electron and a positron is the same as the difference between negativity and positivity.

    What is the difference between goodness and evilness? The rotation?
     
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  5. QuarkHead Remedial Math Student Valued Senior Member

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    Carping? Want a little lesson in operator theory? Good, here goes....
    Operators act on vector spaces by definition. If and only if 2 (or more) such operators act on the same vector space (the domain) and have the same image set (the codomain) then they can be added. With this (and a few extra bits and pieces) one says that the space of all such operators is a vector space

    I explained that the set of all vector fields on a given manifold is itself a vector space. But since a vector field is thus a vector, and since the curl operator maps vector fields onto vector fields, and since the divergence operator maps vector fields onto scalar fields, they have different codomains and therefore adding the curl operator to the divergence operator makes no sense

    Leaving aside the fact that we do not know what is meant by a "field line", did you not earlier claim that there exists no electric field, no magnetic field, only the electromagnetic field?

    Did not James R and I explain how the Faraday field strength tensor (at a point) is derived from the electric vector field and the magnetic vector field (both at the same point)?

    And did you not poo-poo our efforts at the time?
     
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  7. tashja Registered Senior Member

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  8. Arlich Vomalites Registered Member

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    Gordon Kane writes in his book Supersymmetry, Squarks, Photinos and the Unveiling of the Ultimate Laws of Nature about the decay of the proton. He writes that it is possible in some GUTs that a proton decays into two positrons and an electron. Correspondingly, an anti-proton could decay into two electrons and a positron,
    now you have an antiproton with a charge= -1 -1 + 1 = -1

    What kind of a reaction could fuse two electrons and a positron, is quite another question.
    But it is possible to imagine some possible reactions which can do that.


    Yes. But we may continue and ask: where does the charge of the antiquarks come from?
    It is possible that the fundamental particles are the electron and the positron, from which the charge comes.
    I am not sure if anyone really knows the origin of the charge, so I can't so more than that.

    I said that it is possible to think that electrons are capable of fusing although they have no inner structure
    similar to protons.

    If you could divide the proton, the nucleus of the hydrogen atom, into two halves, you don't end up having
    hydrogen. By definition the atom is the smallest piece that matter can be divided. The smallest piece of gold is one atom of gold.
    Half of a nucleus of a hydrogen atom is not hydrogen anymore.

    On the other hand, a half of an electron is a half of an electron. Half of an electron does not change into something else. This is because the electron is the most fundamental particle, alongside with the positron.
    Yes, it is true that we can imagine all sorts of things, but I imagine what I said, although to a first order every
    theoretical idea is wrong.

    QED quantizes the electromagnetic field, whose quantum is the photon. The photon is a boson, a massless boson. Correspondingly, one may ask what is the quantum of the electric field. Again it is possible to
    imagine all sorts of things, and I imagine that the quantum of the electric field is a massless fermion.
     
    Last edited: Jan 25, 2015
  9. Farsight

    Messages:
    3,492
    Yes. See The Discovery of Electron Spin by S A Goudsmit:

    "When the day came I had to tell Uhlenbeck about the Pauli principle - of course using my own quantum numbers - then he said to me: "But don't you see what this implies? It means that there is a fourth degree of freedom for the electron. It means that the electron has a spin, that it rotates".

    Note though that the electron isn't some spinning ball. It has intrinsic spin. So does a tornado. It's a tornado because air is rotating. The rotation makes it what it is.

    I'm not conflating anything. Really. And note that the electron is a spin ½ particle. The rotation is different.

    The magnetic field is just one "aspect" of the electromagnetic field, and the quantum of that is the virtual photon. But note that there are no real photons flying back and forth between two magnets.

    No it doesn't. If it isn't rotating, the electron isn't an electron, it's a photon.

    It is. It isn't fundamental. You can create charged particles in pair production. And you can destroy that charge in annihilation.

    Yes, it's a "turn field". It's how you see the electromagnetic field, which is a "twist field", when you're moving through it. Or when it's moving through you.

    I know.

    I think the duality is down to bispinor rotations like this:

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    People doing unto others as they would have others do unto them. Or not, as the case may be.

    Yes. And we create the electron and the positron out of light in pair production. We had something that was moving linearly at c, and then we've got something with spin ½. And a magnetic moment. Which we can diffract. Then after electron-positron annihilation we've got photons again. You don't have to be the brain of Britain to work out that the electron is a wave in a closed path.

    A wind.

    No. But that is the difference between the electron and the positron.
     
  10. Farsight

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    3,492
    I think it makes sense. Here's a depiction of it. Actually it's convergence rather than divergence, but nevermind. We've got convergence then curl then both.

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    I said a charged particle has an electromagnetic field rather than an electric field or a magnetic field. When you combine charged particles in some particular fashion you then see what we call a magnetic field. In that sense a magnetic field exists. Ditto for an electric field.

    You explained nothing.

    I'm still doing it. All you did was put up a mathematical expression with no list of what the various terms meant. It was merely a restatement of an assertion. The explanatory content was nil.
     
    Last edited: Jan 25, 2015
  11. tashja Registered Senior Member

    Messages:
    715
    No, it's not. The electron is a cacti, and the annihilation products are quantum prickly things.

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    Sorry, Farsight. But I couldn't resist. That comment by Prof. Lincoln was hilarious lol.
     
  12. PhysBang Valued Senior Member

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    2,422
    I suppose that's fine if one wants to keep physics to all the results before, say, 1930, but what do we do with all the observations that we have after that point?

    Can you produce an equation for a rotating electron that will reproduce the behavior of quantum mechanical spin?

    This is a nice fantasy, but can you produce the equation that governs a tornado electron that reproduces the behavior of quantum mechanical spin?
     
  13. PhysBang Valued Senior Member

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    2,422
    That, and your response with pictures, is simply crazy talk.

    You are either really too stupid to follow the very simple mathematics QuarkHead laid out for you or you have some serious mental illness that is keeping you from reading and comprehending properly. It's really worrying; for your own health, please see a doctor.
    No, you have stolen mathematical terms that you have picked up from different sources and you throw them around, trying to borrow the weight of authority from them. You can try to run from the meanings of those terms, but it doesn't look good for you.
     
  14. Farsight

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    3,492
    I don't like his answers tashja. They're condescending. The sort of answers you might give to a five-year-old. Hydrogen atoms don't twinkle. Magnets don't shine. We can diffract electrons. It's quantum field theory, not quantum point-particle theory. In QED the electron is an excitation of the electron field. The electron's field is what it is. Those scattering experiments are like hanging out of a helicopter and probing a whirlpool with a bargepole, then when you can't feel anything solid, saying whatever's in the middle of this thing must be really really small. I don't know if you've watched any of the Fermilab videos. Some aren't bad, but by and large they're pitched very low.
     
  15. QuarkHead Remedial Math Student Valued Senior Member

    Messages:
    1,740
    Then there's nothing more to say except that you are WRONG

    Leaving aside what you are calling "curl" in your silly picture seems not have a source, there is is far deeper confusion on your part. I have absolutely no doubt that you will welcome the following enlightenment.

    Suppose 2 functions \(f(x)=2x\) and \(g(x)=x^2\). Then if I say "first f then g" I mean to say "apply the function g to the image of x under f" and write \(g(f(x)) = (2x)^2\). which is easily seen to e different from \(f(g(x)) = 2(x^2)\)

    Note the common notation \(f(g(x)) \equiv (f \circ g)x\)

    This is called "function (or operator, if you prefer) composition". Not addition not multiplication

    So suppose you are willing to drop your assertion that your diagrams refer to operator addition but rather to operator composition, then what your words (but not your silly diagram) seem to say is curl(div). Noting that every student knows that div(curl) = 0, and using the above simple example, tell what is curl(div) for some field of your own choosing
     
  16. Farsight

    Messages:
    3,492
    My silly picture? But Quarkhead, that isn't my picture. Have a look at page 7 of this paper.

    It's Maxwell's picture.

    Gotcha!
     
  17. OnlyMe Valued Senior Member

    Messages:
    3,914
    When you post things like the following,

    Seems like,

    Is an answer that fits right in with your statement/description.
     
  18. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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  19. QuarkHead Remedial Math Student Valued Senior Member

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    1,740
    OK. Since I hold no cows to be sacred, let me rephrase my comment. "Leaving aside Maxwell's silly picture" please answer my substantive question.

    PS Since I never open third party links when posted on a forum, I do not know what sort of argument Maxwell advanced for these pictures. Maybe you might like to explain - in your own words please i.e no quotations (after you have answered my question, of course)
     
  20. James R Just this guy, you know? Staff Member

    Messages:
    39,426
    Farsight:

    You're bluffing again. You didn't understand the maths. That's why you had to ask me. Then you didn't understand my explanation of the maths because you didn't understand some of the concepts I used in the explanation.

    Actually, that equation is the explanation. Why? Because it defines what is meant by "electric field". The observable is the force. The field is a construct.

    In other words, nature is such that like charges repel and unlike charges attract. That gives us \(F\) and the above equation then defines \(E\) for us. See?

    You can't either. If you think you can, give me an experimental test using only the force observable that will tell the difference.

    I already agreed it has an electromagnetic field. That electromagnetic field is simply the electric field for a stationary charge.

    You keep repeating yourself. Assertion without proof meets assertion without proof. Stalemate. See how this works?

    You've linked to some guy who produced a theory of gravity (gravitomagnetism) that is demonstrably incorrect. And regarding his electromagnetic equations, perhaps you can explain why the potentials at time \(t\) depend on an arbitrary past time \(t_r\).

    Maxwell's equations are consistent with electromagnetic waves. In fact, Maxwell himself discovered such waves as a result of formulating his equations.

    You really had no idea where to pitch your explanations when you talk to me, do you? You have no way of judging my level of knowledge. I throw mathematics at you, but since it's all a mystery to you, you think I'm lost in a world of mathematics and don't know the basic physics that precedes the mathematics. So one minute you try to bluff like you have a graduate-level understanding of things like relativity, and the next you think you need to explain school-level basics because I won't know them. The only explanation for such behaviour is that you're not sure what is school-level material and what is graduate level stuff. Because on wikipedia, your main source, it's all mixed in together. You can only cut and paste the pictures and some quotes. You have to skip over the maths because you don't understand it.

    Anyway, back to the topic. When a charged particle goes around the magnetic field lines, the magnetic force on it at each point in its trajectory points radially inward. The force isn't a "twirly" force, or a "twisty" force or a "turn force" or a "screw force". It's just the usual kind of linear force. However, the force varies with position in such a way as to produce circular (or helical) motion.

    I thought as much. In future , don't use terms you don't understand, Farsight. You'll get caught out again.

    Heh.

    My question was a simple one: at a single point in space at which there is (part of) an electromagnetic wave, the electric field is at right-angles to the magnetic field. In such a case, which way does the "electromagnetic field" point?

    And your answer is the mess above. Want to try again?

    You have no idea about the maths I'm used to, not being able to do maths yourself.

    If you can't even get the dimensions of E and B correct, you're a lost cause I'm afraid, Farsight. I shouldn't have to explain such a basic thing to you twice.
     
  21. PhysBang Valued Senior Member

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    2,422
    "The maths is the maths you're used to" is Farsight trying to steal the mathematics from actual physics while sneaking in his own fantasy explanation. James R is correct that since he cannot do the mathematics, he cannot really understand what he is saying when it comes to the various mathematical terms.
     
  22. Farsight

    Messages:
    3,492
    What question? Looking at your previous post:

    Oh nitpick. So I used a + in my picture, the horror. You know full well that I said you combine the radial electric field lines and the concentric magnetic field lines to depict the electromagnetic field lines.

    Huh? You're talking about the divergence of the curl being zero. That's fine for a magnetic field because it's a "turn" field. Curl is also known as rot which is short for rotor. A "rotor" field is something like a turntable. The centre isn't rotating faster than the edge. But there are no pure magnetic fields because that would demand some region of space to be rotating like a roller-bearing, and space isn't like that. You get frame-dragging instead. You perceive a turn field because you are moving through a "twist" field and don't realise it. Or it's moving through you. What do you want me to tell you about the curl of the divergence? I've already said there are no sources and sinks here. The electron is not some bathtub vortex.

    No. And you've just admitted you've never followed up on any of my references to Einstein or Minkowski or Maxwell or anybody or anything else. Ye Gods. No wonder it's so difficult to teach you anything. But I will give you a quote from beneath the picture:

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    "If we subtract from the general value of the vector function σ its value σ0 at the point P, then the remaining vector σ - σ0 will, when there is pure convergence, point towards P. When there is pure curl, it will point tangentially round P; and when there is both convergence and curl, it will point in a spiral manner."

    Compare with my picture:

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  23. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    i also suggest farsight needs to learn and understand modes
     

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