| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
06 Jun 2007 12:19:16 AM |
| Object: |
Re: Why is the Weinberg angle an angle? |
On Jun 5, 10:18 pm, "Dallas Kennedy" <dkenn...@mathworks.com> wrote:
First, it's not the Weinberg angle. It's the "weak mixing angle" and was
first defined by Glashow in 1961.
Second, it's defined as an acute angle of a right triangle whose sides are
the SU(2)_weak,L gauge coupling "g" and the U(1)_hypercharge gauge coupling
"g'".
|
| g'
theta_W |
---------------------
g
The weak mixing angle theta_W is defined by: tan(theta_W) = g' / g. So it
is angle.
Standard modern treatments of gauge theories and particle physics discuss
these points. See also the Review of Particle Physics 2006.
Hmm. One could define an angle for any two positive quantities g`and g
in this way.
So it does not need to be an angle at all, in the end? Or is there
another reason
that the ratio g'/g is interpreted as the tan of an angle?
(For example, why are g' and g drawn on orthogonal axes? Is there a
reason for this?)
I am just trying to understand - I hope this does not sound obnoxious.
Francois
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| User: "PD" |
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| Title: Re: Why is the Weinberg angle an angle? |
06 Jun 2007 02:02:43 PM |
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On Jun 6, 12:19 am, wrote:
On Jun 5, 10:18 pm, "Dallas Kennedy" <dkenn...@mathworks.com> wrote:
First, it's not the Weinberg angle. It's the "weak mixing angle" and was
first defined by Glashow in 1961.
Second, it's defined as an acute angle of a right triangle whose sides are
the SU(2)_weak,L gauge coupling "g" and the U(1)_hypercharge gauge coupling
"g'".
|
| g'
theta_W |
---------------------
g
The weak mixing angle theta_W is defined by: tan(theta_W) = g' / g. So it
is angle.
Standard modern treatments of gauge theories and particle physics discuss
these points. See also the Review of Particle Physics 2006.
Hmm. One could define an angle for any two positive quantities g`and g
in this way.
So it does not need to be an angle at all, in the end? Or is there
another reason
that the ratio g'/g is interpreted as the tan of an angle?
(For example, why are g' and g drawn on orthogonal axes? Is there a
reason for this?)
I am just trying to understand - I hope this does not sound obnoxious.
Francois
It's not. The mixing matrix is a unitary matrix, and a unitary matrix
can always be represented using an angle, whether that angle is a real
angle or not. It's just shorthand.
Lorentz boosts look like angular rotations, too, almost. Enough to
call give the angle a name: rapidity.
PD
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| User: "Autymn D. C." |
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| Title: Re: Why is the Weinberg angle an angle? |
07 Jun 2007 05:24:05 AM |
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On Jun 6, 12:02 pm, PD <TheDraperFam...@gmail.com> wrote:
It's not. The mixing matrix is a unitary matrix, and a unitary matrix
can always be represented using an angle, whether that angle is a real
angle or not. It's just shorthand.
Lorentz boosts look like angular rotations, too, almost. Enough to
call give the angle a name: rapidity.
Angula (Croocks) are foolish. Use a damned fraction, say, in floating
decimal: .07983 or .31932*.25.
.
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| User: "Autymn D. C." |
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| Title: Re: Why is the Weinberg angle an angle? |
07 Jun 2007 05:29:01 AM |
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On Jun 7, 3:24 am, "Autymn D. C." <lysde...@sbcglobal.net> wrote:
On Jun 6, 12:02 pm, PD <TheDraperFam...@gmail.com> wrote:
It's not. The mixing matrix is a unitary matrix, and a unitary matrix
can always be represented using an angle, whether that angle is a real
angle or not. It's just shorthand.
Lorentz boosts look like angular rotations, too, almost. Enough to
call give the angle a name: rapidity.
Angula (Croocks) are foolish. Use a damned fraction, say, in floating
decimal: .07983 or .31932*.25.
Oh yeah, ratios are best: .54836.
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