- #1

Sandglass

- 5

- 2

So is this mixing angle a symbolic representation of the link between electromagnetism and the weak interaction, derived from the complex calculations of the standard model, or does it have a numerical use in this graph?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- B
- Thread starter Sandglass
- Start date

- #1

Sandglass

- 5

- 2

So is this mixing angle a symbolic representation of the link between electromagnetism and the weak interaction, derived from the complex calculations of the standard model, or does it have a numerical use in this graph?

- #2

ohwilleke

Gold Member

- 2,131

- 1,004

Likewise, the W boson mass, the Z boson mass, and the Weinberg angle are not independent of each other, and the relative magnitudes of those Standard Model constants are functions of the electromagnetic and weak force coupling constants. In other words:

So, while we have five parameters (the Weinberg angle, the electromagnetic coupling constant, the weak force coupling constant, the W boson mass and the Z boson mass), because of electroweak unification, those five parameters don't represent five independent degrees of freedom.

The original W

- #3

Sandglass

- 5

- 2

Since the coupling constant for the weak hypercharge is given by g' cos (theta_w) = e, I naively thought that the weak hypercharge Yw was related to the electric charge Q in the same way by Yw cos(theta_w) = Q. What puzzles me is the fact that the particle graph seems to confirm this with a mixing angle shown in the (Yw, Q) plane.

On the other hand, this cannot be the case since the Gell-Mann-Nishijima formula Q = Yw/2 + T3 cannot be satisfied with g'/e = Yw/Q = cos(theta_w) and g/e = T3/Q = sin(theta_w).

Therefore, is there any theoretical relationship between (g'/e) and (Yw/Q) ?

- #4

Sandglass

- 5

- 2

Does anyone have a reference, either an academic book or a review article that shows this graph, instead of this wiki page ?

- #5

- 22,124

- 13,031

Lee, T.D. (1981). Particle Physics and Introduction to Field Theory.

It's a nice textbook by a Nobel laureate.

- #6

Sandglass

- 5

- 2

I'm new to this and want to make sure I understand: do the letters L and R for particles refer to chirality, not helicity ? The legend is ambiguous.

- #7

ohwilleke

Gold Member

- 2,131

- 1,004

Fermions have chirality. Bosons have helicity. All of the references in the chart with L or R that I see are for fermions, so it is a reference to chirality.do the letters L and R for particles refer to chirality, not helicity? The legend is ambiguous.

- #8

malawi_glenn

Science Advisor

Homework Helper

Gold Member

2022 Award

- 6,210

- 1,706

Not enitirely true. Chirality is a label for how the particle transforms under the Lorentz group.Fermions have chirality. Bosons have helicity. All of the references in the chart with L or R that I see are for fermions, so it is a reference to chirality.

Helicity is never used as a label for anything in these kind of diagrams, since it is not a Lorentz invariant property of the particle.

- #9

- 22,124

- 13,031

Dirac fermions have indeed both chirality and helicity. By definition chirality eigenstates are eigenstates of ##\gamma_5##. Since ##\gamma_5^2=1## the possible eigenvalues of chirality are ##\chi \in \{1,-1 \}##.Fermions have chirality. Bosons have helicity. All of the references in the chart with L or R that I see are for fermions, so it is a reference to chirality.

By definition helicity for a momentum eigenstate is the eigenvalue of the component of the total angular momentum in direction of the momentum, and the corresponding eigenvalues are ##h \in \{1/2,-1/2\}## (with natural units, where ##\hbar=1##).

Only for massless particles ##\chi=2h##, and only in this case helicity eigenstates are Lorentz invariant. For massive particles you can change to another reference frame, where helicity is flipped. That's impossible for massless particles, because you cannot "overtake" the fermion with your new reference frame, because a massless fermion moves with ##c## in all inertial frames.

- #10

ohwilleke

Gold Member

- 2,131

- 1,004

Thanks for the corrections that have clarified the concepts.

- #11

malawi_glenn

Science Advisor

Homework Helper

Gold Member

2022 Award

- 6,210

- 1,706

- #12

Sandglass

- 5

- 2

Thanks for all these clarifications.

- #13

garrett

Gold Member

- 413

- 47

I suspect the magic formula you're looking for is

Since the coupling constant for the weak hypercharge is given by g' cos (theta_w) = e, I naively thought that the weak hypercharge Yw was related to the electric charge Q in the same way by Yw cos(theta_w) = Q. What puzzles me is the fact that the particle graph seems to confirm this with a mixing angle shown in the (Yw, Q) plane.

On the other hand, this cannot be the case since the Gell-Mann-Nishijima formula Q = Yw/2 + T3 cannot be satisfied with g'/e = Yw/Q = cos(theta_w) and g/e = T3/Q = sin(theta_w).

Therefore, is there any theoretical relationship between (g'/e) and (Yw/Q) ?

$$

Q e = Y_W \frac{1}{2} g' \cos{\theta_W} + T_3 g \sin{\theta_W}

$$

The electric charge, weak hypercharge, and weak isospin axes are scaled by their respective coupling constants, and particle charges come in integral (or fractional integral, due to convention) multiples of these. Hope that helps.

Share:

- Last Post

- Replies
- 6

- Views
- 597

- Last Post

- Replies
- 2

- Views
- 567

- Last Post

- Replies
- 1

- Views
- 600

- Last Post

- Replies
- 1

- Views
- 787

- Last Post

- Replies
- 8

- Views
- 692

- Last Post

- Replies
- 6

- Views
- 1K

- Replies
- 2

- Views
- 314

- Replies
- 4

- Views
- 918

- Last Post

- Replies
- 27

- Views
- 3K

- Last Post

- Replies
- 7

- Views
- 2K