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A hypercharge problem

  1. Nov 25, 2007 #1
    Is it true that hypercharge is conserved in the weak ineractions - and whats the difference between normal hypercharge and weak hypercharge? I know how to work out hypercharge but help would be greatly appreciated! Thanks.
     
  2. jcsd
  3. Nov 25, 2007 #2

    blechman

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    WEAK hypercharge is conserved in the electoweak theory, yes. However, when EW theory is broken to QED after the Higgs field gets a vev, then weak hypercharge is no longer conserved (since the Higgs field has Y=1/2, so when it condenses, it breaks that symmetry).

    STRONG hypercharge has absolutely nothing to do with weak hypercharge (except that they're both U(1)'s, but strong hypercharge is global while weak hypercharge is gauged, so even that relationship isn't really there). It's a very unfortunate naming. Strong hyperchage is B+S (baryon + strangeness), up to factors of 1/2. Baryon number is conserved throughout the SM (up to topological effects), while strangeness is violated by the weak interactions (W-boson exchanges), so STRONG hypercharge is violated by EW theory.

    However the STRONG NUCLEAR FORCE (QCD) does conserve STRONG hypercharge, as well as WEAK hypercharge.
     
    Last edited: Nov 25, 2007
  4. Nov 25, 2007 #3
    So how do you define weak hypercharge?

    Also, in isospin, presumeably weak isospin is different to normal isospin. How do you work out the values for weak isospin? Is it similar to normal isospin? Thanks, for any replies.
     
  5. Nov 25, 2007 #4

    blechman

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    I thought you said you knew how to derive these things in your first post?

    Again, if by "normal isospin" you mean "strong isospin" (the thing Heisenberg proposed), then yes, they are different.

    Left handed fermions form weak isospin doublets, and the right-handed fermions are weak isospin singlets. Then, in exact analogy to the strong case, electromagnetic charge is defined as [itex]T^3 + Y[/itex], up to factors of 2. For example: the weak-isospin doublet [itex](u_L,d_L)[/itex] must have hypercharge +1/6 so that the charges of the up and down quarks work out correctly.

    Again, the weak isospin/hypercharge are the same Lie group as the strong case, but the weak symmetries are gauged, while the strong symmetries are global.
     
  6. Nov 25, 2007 #5
    OK, but how do you work out weak hypercharge and weak isospin. (sorry if I seem to know nothing about this, but this is a new area of particle physics for me)
     
  7. Nov 25, 2007 #6

    blechman

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    What do you mean? You use the formula: [itex]Q=T^3+Y[/itex].
     
  8. Nov 25, 2007 #7
    Where Y is normal hypercharge and T3 is weak isospin? (just checking)
     
  9. Nov 25, 2007 #8

    blechman

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    Y is WEAK hypercharge and T3 is WEAK isospin, that's right.
     
  10. Nov 26, 2007 #9
    OK, but how do you work out the weak hypercharge without knowing the weak isospin or vice versa? (normal hypercharge is defined as the baryon number + the strangeness and so on - can you define weak hypercharge in the same way?)
     
  11. Nov 26, 2007 #10
    Also (but please answer my first question first!) what exactly does the third compnent of isospin (strong isospin) describe. I know how to work it out, and I can deal with its equations, but I don't quite know what it actually describes. Is it something to do with the three cartesian co-ordinates (my memory seems to recollect something about this, but it might be totally wrong)?
     
  12. Nov 26, 2007 #11

    blechman

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    The weak isospin gauge group only acts on left-handed fermions, so the left-handed fermions are in a doublet (fundamental) of weak isospin ([itex](u_L,d_L),(e_L,\nu_L)[/itex], similar for each generation), while the right-handed fermions are in singlets. Choosing [itex]T^3[/itex] is purely convention (which fermion comes first in the doublet). In fact, it isn't even always consistent from one paper to the next. But once you fix that convention, your hypercharge is fixed by the electric charge as I mentioned above. That's how you work this stuff out.

    "Isospin" derives its name (and algebraic properties) from quantum mechanical spin angular momentum. If you wish to have a geometric intuition for it, then you can just get rid of the "iso-" prefix. However, this is not a physical angular momentum, so to do this is a bit misleading. In the end, isospin has nothing to do with the usual coordinates: it lives in its own world.

    TECHNICAL: you can think of isospin generators as defining a coordinate basis for a fibre bundle over spacetime. But unless you already understand the words in the last sentence, don't worry about that! :wink:
     
  13. Nov 26, 2007 #12
    OK, so let me get this right - there is no fixed value for weak isospin and isospin has nothin gto do with spin. Also, what is the third component of isospin - is it just a mathematical operator, or can you actually visualize it?
     
  14. Nov 26, 2007 #13

    blechman

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    (1) There is a fixed value for weak isospin (1/2), but whether the third component of weak isospin is +1/2 or -1/2 is a convention.

    (2) Isospin and Spin have absolutely nothing to do with each other other than a historical significance.

    (3)Third component of isospin is just a mathematical operator, not a physical spin, yes. It gets its name from a mathematical similarity to ordinary spin, but not from a physical relationship.
     
  15. Nov 26, 2007 #14
    Thankyou ever so much, you have been very helpful - I now finally begin to understand!
     
  16. Nov 26, 2007 #15

    blechman

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    glad i can help! :smile:
     
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