Recent content by btphysics
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Graduate The breaking of the flavour permutational symmetry
Sorry for only answer now, but last weeks i had some personal issues. I am much grateful for your replies, and i have already understand the problem leptrich, you understood my doughts. With this type of matrix next one can generate mass for the 1st generation and even sufficient CP violation.- btphysics
- Post #11
- Forum: High Energy, Nuclear, Particle Physics
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Graduate The breaking of the flavour permutational symmetry
Thanks for the the answer. In that article they give the same explanation, and the matrix is almost equal. I continua to don´t see any clues about what means that expression and moreover under what transformations that matrix is invariant.- btphysics
- Post #3
- Forum: High Energy, Nuclear, Particle Physics
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Graduate The breaking of the flavour permutational symmetry
Hello,I read an article about fo generating mass for the first two generations of quarks by breaking the flavour permutation symmetry S3. In the top of the page 5 of the following article: http://arxiv.org/pdf/hep-ph/9807214v2.pdf , they mention that "Mq1 transforms as the mixed symmetry term...- btphysics
- Thread
- Symmetry
- Replies: 11
- Forum: High Energy, Nuclear, Particle Physics
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Graduate Why Use a Bi-Doublet Scalar Field (2,2) Under SU(2)L x SU(2)R?
The only problem is that in terms of group theory representation, for me, that does not makes sense.- btphysics
- Post #3
- Forum: Beyond the Standard Models
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Graduate Why Use a Bi-Doublet Scalar Field (2,2) Under SU(2)L x SU(2)R?
Hello, why one can use a bi-doublet scalar field (2,2) under SU(2)L x SU(2)R ? In terms of group theory, we should have only triplets (3,1) or (1,3) since 2 x 2=3+1 ? But in left right symmetric models, indeed yukawa coupling are formed with bi-doublet scalars. Best regards- btphysics
- Thread
- Group
- Replies: 3
- Forum: Beyond the Standard Models
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Graduate Cylindrically symmetric line element canonical form
Thanks for your answer. Well, I want a a line element time dependent, and where the plane ( t,ρ) is orthogonal to the ( ø,z) plane . Probably the line element will be of the form of: ds^2= A(t,ρ) ( dt^2-dρ^2)-... I have some doubts on the rest of the expression, as well on the determination...- btphysics
- Post #3
- Forum: Special and General Relativity
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Graduate Bondi's Radiating Line Element: Solve the Mystery!
Thanks of your time with this question. My problem is that Bondi metric has de following form: ds^2= (\frac{V}{r} e^{2\beta}-U^2r^2e^{2\gamma})du^2 + 2e^{2\beta} dudr+ 2Ur^2 e^{2\gamma} du d\theta -r^2 (e^{2\gamma} d\theta^2 + e^{-2 \gamma} sin^2 \theta d \phi^2) All these...- btphysics
- Post #3
- Forum: Special and General Relativity
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Graduate Bondi's Radiating Line Element: Solve the Mystery!
Hello, Some knows how to Bondi deduce his radiating line element? I read some papers and there isn't any hint about hit. Best regards- btphysics
- Thread
- Element Line Line element
- Replies: 2
- Forum: Special and General Relativity
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Graduate Cylindrically symmetric line element canonical form
Hello, What is the most general cylindrically symmetric line element in the canonical form? Best regards.- btphysics
- Thread
- Canonical form Element Form Line Line element Symmetric
- Replies: 3
- Forum: Special and General Relativity
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Chapter 21 Ray D'Inverno Scalar Optics, congruence of null geodesics
Hello, try to use general geodesic equation and then simplify the connection term with the definition of covariant derivative. Read about affine parameters. I'll try to post a solution attempt. ( Ray D'inverno is a nice book has introduction to general relativity aspects).- btphysics
- Post #2
- Forum: Advanced Physics Homework Help