Are the W bosons charged before symmetry breaking?

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Discussion Overview

The discussion centers around the properties of W bosons in the context of the SU(2) symmetry breaking via the Higgs mechanism. Participants explore whether W bosons possess charge before symmetry breaking and the implications of their mass and charge characteristics in different energy regimes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Francois questions whether W bosons had charge before symmetry breaking, noting that no particle is known to have charge without mass.
  • Some participants argue that the concept of "before" symmetry breaking may not be appropriate, as the SU(2) x U(1) symmetry is believed to have always been broken in the universe's history.
  • There is a suggestion that the question may relate more to theoretical derivations rather than physical states, as W bosons are not considered physical states before symmetry breaking.
  • Francois expresses confusion about the relationship between energy levels and the properties of W bosons, particularly regarding their mass and charge at high energies.
  • Another participant clarifies that at high energies, while the W bosons are still massive, their mass can be approximated as zero for calculations, but their charge cannot be disregarded.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of W bosons' properties before symmetry breaking, with no consensus reached on whether they possess charge or how to conceptualize their characteristics in relation to energy levels.

Contextual Notes

There are unresolved questions regarding the definitions of mass and charge in the context of symmetry breaking and the implications of theoretical models versus physical states.

franoisbelfor
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When the SU(2) symmetry is broken
by the Higgs mechanism,
the W bosons acquire mass
and become the well-known W^+ and W^-
bosons discovered at CERN.

So before the breaking, the Ws had no mass.
Did they have charge?

If yes: No particle is known
without mass but with charge. Are the W
before symmetry breaking the first?

If no: How does charge arise through
symmetry breaking?

Thanks for any help!

Francois
 
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"Before" is maybe not the best word. There was, as far as we know, no time in the universe when the SU(2) X U(1) symmetry was unbroken. The time you are referring two is the time it takes to do the derivation - so your question really is closer to "between steps 5 and 6 in the derivation, do the w's have charge?" Note that these are not physical states.

One can do a similar thing with fermions and treat the left-handed and right-handed chiral projections as separate particles. These are also charged and massless - and also not physical states.
 
Vanadium 50 said:
"Before" is maybe not the best word. There was, as far as we know, no time in the universe when the SU(2) X U(1) symmetry was unbroken. The time you are referring two is the time it takes to do the derivation - so your question really is closer to "between steps 5 and 6 in the derivation, do the w's have charge?" Note that these are not physical states.

One can do a similar thing with fermions and treat the left-handed and right-handed chiral projections as separate particles. These are also charged and massless - and also not physical states.

Thank you! I see my mistake. So at energies far above the symmetry breaking scale, the W is still massive and charged, am I correct? The books sometime give the impression that at high energy, the symmetry is unbroken... I really got something mixed up there.

François
 
franoisbelfor said:
Thank you! I see my mistake. So at energies far above the symmetry breaking scale, the W is still massive and charged, am I correct? The books sometime give the impression that at high energy, the symmetry is unbroken..

Horrible books. It is as trying to learn of God by reading the Bible.

Anyway, yes, the point is that at very high energy the W is still massive but its mass is small for the calculations being done, so it can be taken as zero in order to simplify the calculation. But note that you can not put its charge to zero.
 

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