do the particles ever collide in QED

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

  1. Farsight

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    We've had the 3D plot:

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    OnlyMe put it up. It's like the Williamson / van der Mark electron:

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    Which is like the ring torus:

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    Which you inflate to the spindle-sphere bispinor:

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    See Adrian Rossiter's Antiprism torus animations. Think of this animation as something like "the eye of the storm", surrounded by frame-dragged space. Co-rotating vortices repel, counter-rotating vortices attract.

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    Two of these things with the same chirality will move apart. With the opposite chirality they'll move together. The positron has the opposite chirality to the electron.

    PS: Weren't you going to explain some maths?
     
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  3. paddoboy Valued Senior Member

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    Well let's KISS. A gravitational field is simply warped/curved/twisted spacetime in the presence of mass.

    Light/photons simply follow geodesics in curved spacetime. I told you that before.


    'No, that's just plain wrong. Light'photons follow geodesic paths in curved spacetime. You need to KISS.
    And I would be certain that the great man would totally agree with that.


    Neither are the all the people that see the majority of your claims as hogwash.


    "The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality". – Hermann Minkowski, 1908
     
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  5. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    i suggest listening to your self.
    i also suggest showing me a 3d plot [that's not some fictitious image, which i'm sure is what is coming] of this nonsense.
    now lets see who pipes up.
    yes as soon as you stop diverting with manipulation from your pretending. now show me how to plot these lines in 3d. show me how to use those equations.
    but i suspect you will just respond with some quote or and equation. you will never show me how by your own admission, only because you're such a pretender. but that's just what it appears to be, am i correct ?
     
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  7. Farsight

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    No, you're a troll. You didn't do that math, I gave you a 3D plot and more, and you're still whining. It's back on ignore for you.
     
  8. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    i never said i did that math, this is just once again, another attempt of yours to manipulate and divert. hilarious.
    no, you gave an image, just like i knew you would, only because i'm keen on your pretending.
    ohh my, scary...
    no problem, but everyone else still can read what i say about you. which i know it bothers you.

    so again,
    i suggest listening to yourself.
    now show me how to plot these lines in 3d. show me how to use those equations.
    you will never show me how by your own admission, only because you're such a pretender. but that's just what it appears to be, am i correct ?

    now lets see who pipes up.
     
  9. PhysBang Valued Senior Member

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    2,422
    Ah, the lying continues. You gave no plot, Farsight, you just cut and pasted a picture that you cannot use to do any physics.
     
  10. James R Just this guy, you know? Staff Member

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    39,426
    Farsight:

    Your lack of mathematical knowledge is clearly on display for all to see here.

    The field lines in the above diagram indicate the direction of the field, not the direction of a force.

    To relate the field to the force on a nearby charge, you need to apply \(\vec{F}=q\vec{E}\), where \(q\) is the charge experiencing the force. If \(q\) is positive, the the direction of the force is the same as the field direction. If \(q\) is negative, then the force points in the opposite direction to the field.

    Not true. A positron has a field pointing outwards from the charge, and an electron has a field pointing inwards.

    In the magnetic case, you apply the formula \(\vec{F}=q\vec{v}\times \vec{B}\), taking into account the charge, the velocity and the field.

    There's no problem with the vector description of electric and magnetic fields.
     
  11. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    2,973
    i'm not even sure he understand what chirality is or means.
     
  12. Arlich Vomalites Registered Member

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    91
    There is more than the field. There is also the mass of the electron.

    If you read the paper in my link, it has a concept "massless charge field". So that is what you are talking about,
    you mean that the charge of the electron is the field of the electron. But according to that concept, mass is somewhere else than in the massless charge field.

    The electron is not just the field, because its field is massless. So where is the mass of the electron located?
     
  13. Farsight

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    3,492
    The electron is "just field", and there's field energy in that field. This isn't massless. Who told you it was? Energy that isn't moving at c has a mass-equivalence, a mass. The concept of a massless charge field just doesn't match the hard scientific evidence. You employ pair production to convert a massless field variation, a photon, into an electron and a positron. Each has both mass and charge. There is nothing that has charge that doesn't have mass. And vice-versa. For example a neutron has no net charge, but the magnetic moment betrays the presence of charge, and of course a free neutron decays into an electron, a proton, and an antineutrino.
     
  14. Farsight

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    3,492
    Then you explain \(F(p)=\sum\nolimits_j E_j \epsilon^j \otimes \sum\nolimits_k B_k \epsilon^k= \sum\nolimits_{jk}F_{jk}\epsilon^{jk}\).

    But two inward-pointing fields result in the same motion as two outward-pointing fields. So it just doesn't work. And the clue as to why is in John Jackson's Classical Electrodynamics: "one should properly speak of the electromagnetic field Fμv rather than E or B separately".

    Yes, we know the story. Now, all you've got is two electrons. Or two positrons. You have two fields, both inward or both outward, and in both cases the particles move apart. Why? Why does the charge experience a force in the "other" direction when the charge is negative?

    It is true. I set down two charged particles. They're either both electrons or both positrons or one of each. When you see them move apart you know they're not one of each, but you don't know whether they're electrons or positrons.

    The problem is that one should properly speak of the electromagnetic field Fμv rather than E or B separately. So what's the vector description of Fμv? Which way does the field point?
     
  15. James R Just this guy, you know? Staff Member

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    Farsight:

    What do you need explained, exactly? Are you confused by the notation? The summations? The indices? What's the problem?

    I'm not sure what you're talking about.

    Jackson says that because you can turn an electric field into a magnet field, or vice versa, by changing reference frame.

    Like charges repel. Unlike charges attract. That's just how nature works. There are two types of electric charge, positive and negative. It's just an empirical fact. Nobody knows why there are two types of charge. There just are.

    If your only observation is whether they attract or repel, then you're right: all you know is you have two like charges or two unlike charges. To identify them, you need to do other tests. This isn't a problem for the vector description of the electric field.

    \(F_{\mu \nu}\) is a tensor, not a vector. The "field" in that description has a tensor at every location in space, not a vector. Tensors don't "point" in a particular direction. They are a more complicated mathematical construct than a vector.
     
  16. Arlich Vomalites Registered Member

    Messages:
    91
    I mentioned the model of the electron in the paper in my link, there appears the concept of "massless charge
    field."

    No, the electron is not "just field". The electron has also mass. The mass of the electron is not in the massless
    charge field.

    The paper that I am talking about is a scientific paper. It is written by a scientist.
    What you are telling now does not match the hard scientific evidence.

    There is the massless charge field, it has no mass, but it has charge.

    Also there is a particle that has mass, but no charge. It is called neutrino.
     
  17. QuarkHead Remedial Math Student Valued Senior Member

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    1,740
    Just to put a cat among the pigeons........

    \(F_{\mu \nu}\) is not actually a tensor, it refers to the scalar components of a tensor when expanded on a tensorial basis, which can be represented in matrix form as
    \(F_{\mu \nu} =\begin{pmatrix}0&-E_x &-E_y& -E_z\\E_x&0& B_x & -B_y\\E_y & -B_z & 0 &B_x\\E_z & B_y & B_x &0 \end{pmatrix}\) where \(E_x \equiv \frac{\partial \mathbf{E}}{\partial x}\) etc And assuming (as I did there) that \(\mu\) references the rows and \(\nu\) the columns, it is easy to see that the matrix elements may be separately or severally zero - i.e of a magnetic or electric nature only - but still allowing the notation \( F_{\mu \nu}\) for the components of this tensor

    Note also that we are here talking about a tensor field (Jackson's comment makes no sense otherwise). It is common for physicists to say "a vector" or "a tensor" when in reality they are talking about fields. They are BAD BOYS.

    That said, consider this. Suppose a n-manifold (don't get excited - \(R^n\) for any strictly positive integer n is a manifold) and that X, Y, Z,..... are tensor fields. Then let me write \(\mathfrak{X}\) for the set of all such fields on my manifold. Since \(\alpha X + \beta Y \in \mathfrak{X}\) for arbitrary scalars \(\alpha,\,\,\beta\), it follows that \(\mathfrak{X}\) is a vector space (this is just the definition of a vector space), so that, since elements of a vector space are vectors (obviously) then a tensor field is a vector.

    Which is a lesson in pure math none of you wanted, but is why mathematics is such fun for mathematicians, and so irritating for everyone else
     
  18. Farsight

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    3,492
    I want you to talk us through it and say what each term means. What's the problem?

    You said the arrowheads denoted the direction of the field rather than the force. But you can't explain why the "inward-field" electrons and the "outward-field" positrons repel each other. And we've now established that the field is Fμv and it doesn't point in some direction. The arrowheads don't denote the direction of the field. There is no inward field or outward field. The electron and the positron have opposite chiralities.

    Well it isn't true. The electron's field is the electromagnetic field. It doesn't change just because you moved. Let's just say that you're holding an electron in your hand, and I'm holding another. When your electron is motionless with respect to my electron, you can feel a linear force pushing you away. You say you're in an electric field. Then when you walk past me you also feel your electron swirling in your hand. Then you say you're in a magnetic field too. But what's really there are two electrons with their two electromagnetic fields. Interacting.

    I know why. Electrons and positrons are spin ½ particles, and they have the opposite chirality.

    Yes it is. Because the vector points both ways.

    Now we're getting somewhere. The field is the electromagnetic field. It doesn't point inward, or outward. Strictly speaking, it doesn't really point in any direction. But since North is a whole host of directions and we can understand North, we can also understand the "direction" of the electromagnetic field. It's the same as the direction that this thing is spinning:

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    And it can spin the other way too:

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    Last edited: Jan 21, 2015
  19. Farsight

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    3,492
    I know, I've read Poelz's paper http://arxiv.org/pdf/1206.0620v17.pdf . He's barking up the right tree, but his "massless charge field" is just an electromagnetic wave. A photon. We make electrons and positrons out of photons in pair production. Then when we annihilate the electron with the positron we get two photons.

    It's just field. The field has mass. A photon in a mirror-box adds mass to that system.

    That doesn't mean it's right.

    Yes it does.

    No. Mass and charge go together.

    It's a bit like a photon in that it travels at c, or thereabouts. A photon has an "effective mass" if you slow it down in say glass. The neutrino might be similar. And you've never seen a stopped neutrino.
     
  20. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    one needs to learn and understand phasing.
    except there are , if i remember correctly, six neutrino detectors globally.
     
  21. Farsight

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    Quarkhead: you have a Bx where you should have a Bz. Tsk.
     
  22. PhysBang Valued Senior Member

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    Yet again Faright demonstrates that he does not understand the very basics of SR. The values and directions of electric fields and magnetic fields change depending on the system of coordinates one chooses to use, regardless of whether one is at rest in those system of coordinates or whether anything is at rest in the system of coordinates. That's just the fact of SR as initially written by Einstein in 1905.
    Farsight says this, but he has no theory to back this up. In order to have a theory, he has to be able to produce equations to back up his descriptions and predictions. He have only pictures, most cut-and-pasted from other people.
     
  23. Dr_Toad It's green! Valued Senior Member

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    And not too long ago he stated as fact that photons and electrons have the same spin. I'm sure that the universe changes to suit Farsight's whim of the moment.
     

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