Yang–Mills and Mass Gap

Discussion in 'Physics & Math' started by Thales, Nov 29, 2017.

  1. NotEinstein Valued Senior Member

    Messages:
    1,255
    I never claimed that you did said that in post #148, so I don't know where you are getting that from?

    Alright, let's take a fresh look.

    I agree so far.

    This part is utter nonsense, because the units don't match.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. hansda Valued Senior Member

    Messages:
    2,424
    Good. You agree to this that \(T\leq t \) as per my post #148.


    Where the units are not matching.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. NotEinstein Valued Senior Member

    Messages:
    1,255
    Only in order to get at least one complete oscillation, as you yourself state. My point is that this is not generally true.

    You wrote: \(f=\frac{t}{T}\)
    \(t\) has units of time.
    \(T\) has units of time.
    \(f\) has units of inverse time.
    The left-hand side thus has units of inverse time, while the right-hand side is unitless. This makes the equation nonsense.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. hansda Valued Senior Member

    Messages:
    2,424
    Yes.

    Why?

    Seems you are not getting the point. To get the numerical value of rpm, you have to use this formula.
     
  8. NotEinstein Valued Senior Member

    Messages:
    1,255
    Because it is physically possible to rotate an object 180 degrees, i.e. half of a full rotation.

    But the formula is nonsense. It's equivalent to asking if you are 5 apples old, or if you are 6 waters high? You seem to not understand the importance of units. Without understanding how units work, one cannot do any meaningful maths in physics at all!
     
  9. hansda Valued Senior Member

    Messages:
    2,424
    We know that massive, spinning particles are generated in pairs from particle photon through symmetry breaking, as in pair production. Here minimum energy of photon will be for one cycle. So, that minimum energy should correlate with minimum mass.
     
  10. NotEinstein Valued Senior Member

    Messages:
    1,255
    Source please.

    Source please.

    That's circular reasoning. You started out by saying that you are generating massive particles. Obviously there will be a minimum mass; their rest mass.
     
  11. hansda Valued Senior Member

    Messages:
    2,424
    You can see wiki article on pair production.
     
  12. NotEinstein Valued Senior Member

    Messages:
    1,255
    You mean this article: https://en.wikipedia.org/wiki/Pair_production ?

    The words "symmetry" and "cycle" are not present on it, thus this article does not back up your statements. Please provide an actual source.
     
  13. hansda Valued Senior Member

    Messages:
    2,424
    Symmetry breaking is known from higgs mechanism
     
  14. NotEinstein Valued Senior Member

    Messages:
    1,255
    Which also isn't mentioned on that page. Are you even trying to back up your statements?
     
  15. hansda Valued Senior Member

    Messages:
    2,424
    You dont know higgs mechanism?
     
  16. NotEinstein Valued Senior Member

    Messages:
    1,255
    Please explain how it is relevant to your original statement, as it is not mentioned on that Wikipedia article.
     
  17. hansda Valued Senior Member

    Messages:
    2,424
    Particle photon is symmetrical. When its symmetry is broken, two assymetric massive particles are generated as in pair production.
     
  18. hansda Valued Senior Member

    Messages:
    2,424
    So, in pair production, the two particles generated can be considered as two half-photons.
     
  19. hansda Valued Senior Member

    Messages:
    2,424
    Thus minimum one photon cycle is necessary for minimum mass generation.
     
  20. hansda Valued Senior Member

    Messages:
    2,424
    So, the minimum mass will be \(E=mc^2=h \) or \(m=\frac{h}{c^2} \) .
     
  21. NotEinstein Valued Senior Member

    Messages:
    1,255
    Please provide a source that states that the photon is symmetrical in the same sense of the word as it is used in "symmetry breaking".

    In what sense are the resulting particle asymmetrical?

    False; electrons are not half-photons. Positrons are not half-photons. They are completely different types of particles.

    Please define how you are using the word "cycle". Please define what you mean by "minimum mass generation".

    The units don't match: the constant of Planck doesn't have units of energy, but energy time. Please correct your equations.

    I see you still haven't managed to provide even a single source for your original statement. Why is that so difficult for you?
     
  22. hansda Valued Senior Member

    Messages:
    2,424
    From my equations \( mr=\frac{4\hbar}{c}\) or \(m=\frac{1}{r}\frac{4h}{2\pi c} \) .

    Comparing these two equations of mass, we can write maximum radius \(r \) will be \(r=\frac{2c}{\pi} \) .
     
  23. NotEinstein Valued Senior Member

    Messages:
    1,255
    That last equation has a radius with units of length per time. A fine demonstration of "garbage in, garbage out".
     

Share This Page