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Neutralization and polarization?

  1. Dec 12, 2003 #1
    I'd like to know:
    -How do two opposite particles neutralize and how does one neutral particle polarize?
    I'm completly neutral about this so go ahead and polarize me.
     
  2. jcsd
  3. Dec 18, 2003 #2
    Well, I can try to explain annihilation to you, I don't want to go into polarization just yet.

    In essence, annihilation is where two opposite particles, meaning that all of their isospin and quantum flavour numbers are opposite in sign and equal in magnitude, come together completely and their quantum numbers add destructively. Take, for example, an electron and positron. The electron has defining quantum numbers;

    [tex]
    Q = -1, L = -1, I_z = -\frac {1}{2}
    [/tex]

    and the positron has;

    [tex]
    Q = +1, L = +1, I_z = +\frac {1}{2}
    [/tex]

    The electric charges [itex]Q[/itex], lepton numbers [itex]L[/itex], and spin numbers S (which are 1/2 for each, taken as opposite to each other) must cancel for the annihilation to occur, which tends to be the case (rather than scattering). The total isospins [itex]I_z[/itex], however, can couple to either 1 or 0 for the resultant photons. So, either both photons will have [itex]I_z = 0[/itex], or one will have [itex]I_z = 0[/itex] and the other [itex]I_z = \pm 1[/itex];

    [tex]
    e^+ e^- \rightarrow \gamma_{|I_z = 0>} \gamma_{|I_z = 0>}
    [/tex]

    or

    [tex]
    e^+ e^- \rightarrow \gamma_{|I_z = 0>} \gamma_{|I_z = \pm 1>}
    [/tex]

    Since the electron and positron have spin 1/2 and spin -1/2 (either way), these couple to zero so that the photons will have spin 1 and spin -1.

    Hopefully this example gives you a taste of the general idea behind annihilation and pair creation (since it is just the reverse of annihilation in most cases).
     
    Last edited: Dec 18, 2003
  4. Dec 19, 2003 #3
    I'm glad

    Although it's not the type of annihilation I've been looking for, I'm glad you remind me of this cause it makes sense if traslated into my lever theory. There I had:
    M1*D1*Q1=-M2*D2*Q2 (explained here: https://www.physicsforums.com/showthread.php?s=&threadid=11214)
    Now in the case of electron and positron M1=M2 and Q1=-Q2 and they stay const => D1=D2 and they change. When D1=D2 becomes 0 then D1/D2=0/0=1 =>
    M1*Q1=-M2*Q2.....(*)
    and the motion in the geometrical space stops but continues into gravitational and electric space. Since D1=D2=0 this means that we have only one particle with M=M1+M2 and charge Q=Q1-Q2 and now the indexes in the equation (*) represent different states of same particle. The charge and the mass of the newly formed particle start to oscillate which creates EM and gravitational waves(or emition of photons). The fact that two photons are released after the annihilation means that also an EM wave is released cause light has both particle and wave nature (not what suits us most in a given situation). Once again I'm glad my theory well coresponds to your view. But I was more interested in the case where after the annihilation there remains only pure void vacuum. Especially, I was interested in the reverse process.
     
  5. Dec 19, 2003 #4
    Re: I'm glad

    Umm... This thing you have is hard for me to understand, mostly because it seems kind of jumbled, but also because of the division by zero thing. I'm not sure what this idea you have is trying to illustrate, especially since it is mixing different properties like mass and charge.


    Actually, two photons are released because it would violate conservation laws and CP rules to release just one. A photon is an EM wave; a wave-packet, more or less. I can't yet see the correspondence between your theory and the distinct annihilation we are discussing.

    You will never find a pure void after an annihilation of two real particles. Now, the "reverse process" does occur and is known as vacuum pair creation. It is a process that creates "virtual" particles through the violation of energy conservation on a local scale for a period of time that is limited by the Heisenberg Uncertainty Principle. Usually, the virtual pairs will come back together and dissappear when their time runs out. This does not occur for real particles. Wherever there is leftover energy from the masses of the annihilating pair, it must be converted into mass or momentum; and momentum often requires a mass to carry it (photons are purely momentum, though).

    For your reference, the equation which is useful here for the vacuum pair production relates the energy "borrowed" and the lifetime of the violating condition;

    [tex]
    \Delta E \Delta \tau \leq \frac {\hbar}{2}
    [/tex]

    where [itex]\Delta E[/itex] is the amount of energy the violating pair gains, and [itex]\Delta \tau[/itex] is the time for which the pair exists before dissappearing back into the vacuum.
     
    Last edited: Dec 20, 2003
  6. Dec 21, 2003 #5
    In effort to explain my self

    If dividion by zero is not possible try multiplication. D1/D2 was also (M1/M2)*(-Q1/Q2) which was 1 and doesn't change even if D1=D2=0.
    0/0=1 is equally true as 0=1*0. The initial equation in my research was F1*D1=F2*D2. This is as far as Archimed reached in research of the rules of lever. I have only generalized it to charge and mass using the following equations:
    F1/F2=M1/M2 if masses are same by sign and
    F1/F2=-Q1/Q2 if charges are oposite;
    The equation of Archimed mixes force and distance. This are also different properties of the matter but are mixed. This is just how the lever works. In fact it's only mixing pure numbers cause also
    (D1/D2)*(M1/M2)(-Q1/Q2)=1 (no counter parts)
    Therefore, I don't see why charge and mass cannot be mixed. I wonder how you would describe the functioning of the lever with charge, mass and distance?
    At 1st there was nothing. And then god said: "Let's get bussy!!" - from the Bible
    If I expand only mass then I have to make M1=-m and M2=m on the same side of the equilibrium (which is D1=D2).To establish balance I have to pass one mass on theother side of the equilibrium. When one variable passes equilibrium point then it changes its sing. So if M1=m and M2=m on different side then D1=-D2.
    The conservation of mass would not be violated if you creation of pair of positive and negative mass come as a result of polarization of zero mass. It's the 2nd Newton's law that prevents existance of negative mass while negative mass is in the domain of the real mass.
    Explanation:
    F=ma=m*dV/dt=d(dx/dt)/dt. Integrated indepentdent of F and m gives
    x-x0=delta(x)=V0*t+F*t^2/m. If we observe free fall then V0=0 so
    delta(x)/F=t^2/m.
    Now t^2 is always positive; delta(x)/F also is positive cause displacement is in the same direction of the force causing it . Because of it mass is always positive.
     
  7. Dec 23, 2003 #6
    Re: In effort to explain my self

    Okay, now I see what you're trying to say. At first I thought you were trying to say that charge affected mass by some direct interaction. You are really describing a dipole as a lever, which may be completely valid as long as you avoid the undefined division-by-zero illustration.

    I think this area has been very thoroughly researched as far the basic relations go.


    I think using negative force or torque is valid, but not using negative mass. There is no such thing as negative mass; only mass and energy, mass being a form of energy. Newton's second law says nothing of negative mass; this is not due to suppression by the prescence of positive mass, but simply because negative mass does not exist.
     
  8. Dec 24, 2003 #7
    Re: Re: In effort to explain my self

    the dividion by zero is not really a problem cause if at least one zero appears left at least one should appear right of the equation.

    If this area were toughly reaserched we would have been using the Archimedes equation [tex]F_aD_a=F_rD_r[/tex] instead of Newton's [tex]F_a=-F_r[/tex] cause the Newton's equation is a special case of the Archimedes one when [tex]D_a=-D_r[/tex], thus the Archimedes equation is more general.
    Try to combine the gravity law for [tex]M_1M_2<0[/tex] with [tex]F_i=M_ia_i[/tex] for i=1 to 2.
     
  9. Dec 24, 2003 #8
    Re: Re: Re: In effort to explain my self

    If zero appears on both sides of the equation, take the zero from one side and put it on the other so you can use a limit rather than division by zero on both sides of an equation. Limits are much more valid than leaving zeros on each side.

    By the way, thankyou for using the LaTex typesetting; it made all the difference for me to understand what you are trying to illustrate.

    Archimedes equation is more general only for cases where distance is a generalized coordinate in the problem.


    A case where [itex]M_1M_2<0[/itex]? That can only happen where either [itex]M_1[/itex] or [itex]M_2[/itex] is negative in magnitude, and the other is non-zero and positive. This should not happen! Tell me a documented case where this happens, and I'll look at the reference for myself.
     
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