Symmetry, groups and gauge theories in the standard model

1. Mar 2, 2005

da_willem

This is my (limited) understanding of particle physics: In particle physics gauge symmetries play an important role. To allow for massive gauge bosons this symmetry is broken. The theory of weak interactions can be derived from a local SU(2) symmetry, and quantumchromodynamics from a local SU(3) symmetry.

I read the structures particles form when sorted by certain quantum numbers can be described by groups too. Eg Y/Tz diagrams baryon plots. What have these structures got to do with the SU(3) groups? As I understand it elements of SU(3) can be written as:

$$U=exp(i \sum_{k=1} ^{k=8} a_k t_k)$$

With the generators t_k [/tex] the Gell-Mann matrices eg. I see the number 8, and also the 3 as in the number of quarks in a baryon, also appearing in the partical structures but can't see the connection between them and this piece of mathematics. So my questions are: 1)how do these particle structures realte to the mathematics of group theory? (note that my knowledge of group theory is very limited) 2) Has the symmetry breaking in gauge theories have anything to do with the broken symmetries (eg the masses of the particles in the multplets differ, the symmetries are not complete) in particle multiplets? Last edited: Mar 2, 2005 2. Mar 2, 2005 marlon The physics of color is not understandable if all one knows is that there are 3 colors. One must really understand something about SU(3). SU(3) is the group of 3 x 3 unitary matrices with determinant 1. This is the symmetry group of the strong force. What this means is that, as far as the strong force is concerned, the state of a particle is given by a vector in some vector space on which elements of SU(3) act as linear (in fact unitary) operators. We say the particle "transforms under some representation of SU(3)". For example, since elements of SU(3) are 3 x 3 matrices (and these matrices can be constructed using the given generators in your formula), they can act on column vectors by matrix multiplication. This gives a 3-dimensional representation of SU(3). The quarks are represented by this 3*1-matrix. The antiquarks can be represented by row vectors because we can multiply a 3*3-matrix with a row vector on the LEFT side of the matrix. The gluons are represented by the socalled adjoint representation which consits out of traceless 3*3matrices. It can be seen that a row of such a matrix represents one quark colour and a colom of such a matrix represents a anti-colour. each gluon is therefore constructed out of a colour-anticolour combination. Given that there are 3 such colours and anticolours, you would expect 9 gluons. However there are only eight . Can you see why ??? ps : you know that the colours are red green and blue and it is the postulate of QCD that the sum of these three represents colour-neutrality !!! This is the main law that needs to be respected : in interactions : the sum of all involved colours must be WHITE First of all, mass itself breaks symmetry because it mixes the two types of chirality which are fundamentally different in nature. So every elementary particle is massless. Read more on this in my journal. I also suggest the text i wrote on the Higgsfield, which is the system that accounts for generated mass after symmetry breaking in QFT. Just look it up in my journal https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=5 ps : the text on elementary particles is on page 3 or 4 regards marlon ps http://www.phys.uu.nl/~thooft/ here you can find a nice course on Lie Groups IN DUTCH Last edited: Mar 2, 2005 3. Mar 2, 2005 dextercioby [/QUOTE] Nice message,Marlon,zero blank... :tongue2: Mine will be longer.I'll invite the OP to read and try as hard as he can to follow the mathematics from the 6-the chapter of P.Ramond's:"Field theory:A Modern Primer",2.ed,1989.The first 3 sections.No need to go to gravity. Daniel. EDIT:The nuclei,particles forum was the next after this one... 4. Mar 2, 2005 marlon ???? Please, be more mature in your posts... Thank you marlon 5. Mar 2, 2005 da_willem Is this the one? it's the only one the university library has. Title FIELD THEORY Author Ramond, P. Edition 2nd ed. Publisher Reading : Benjamin/Cummings, 1981 ISBN 0-8053-7893-6 Descr. 397 blz. Right, my mistake.. Marlon. About the Higgs-field text in jour yournal. To be sure, the ferromagnetism example is just an analogy right?! You text reads as if the Higgs particle has something to do with the aligning of the spins. Marlon, as I read in your journals you know quite a bit about the Higgs mechanism. Personally I totally don't get where all the mass terms come from. I can't find the visual picture you sketch in the equations. I saw the equations in a simple toy model. The Lagrangian yielding the KG equation: $${\cal L}=\partial _\mu \Phi^\dagger \partial ^\mu \Phi -m^2 \Phi^\dagger \Phi$$ was modified by replacing the last term by a potential $$V=\frac{m^2}{2 \phi_0^2}[\Phi^\dagger \Phi -\phi_0^2]^2$$ How does this modify the KG equation? 6. Mar 2, 2005 dextercioby 1.Yes,that's the 1-st edition.I'm asuming that part is unchanged,but one may never know. 2.Compute the NEW equations...Why ask us...?? Daniel. 7. Mar 2, 2005 marlon It is indeed just an analogy Well, you need to understand the philosophy of the Higgs mechanism. The problem is that gauge-invariance is broken when you include a mass term. The trick is NOt to add a mass in order to describe massive particles but you add a potential of a feld with four degrees of freedom. This field (the Higgs field) really mimicks some sort of interaction expressed by the potential V. This interaction (and the associated energy) really mimicks a particle of mass m. marlon 8. Mar 2, 2005 da_willem Aha, that makes sense. Thanks! 9. Mar 2, 2005 marlon 10. Mar 2, 2005 da_willem The trick is to use $$\Phi^\dagger$$ and $$\Phi$$ as the varying fields in the action principle right? Then the EL equations $$\frac{\partial {\cal L}}{\partial \Phi^\dagger} - \partial _\mu (\frac{\partial {\cal L}}{\partial \Phi}) = 0$$ yield $$-\partial _\mu \partial^\mu \Phi -\frac{m^2}{\phi_0^2}[\Phi^\dagger \Phi -\phi_0^2]\Phi =0$$ 11. Mar 2, 2005 marlon I really don't know what dexter told you, but i hope you are not trying to apply the Higgs mechanism here. If so, STOP, because you are wrong. Problem : you are using a mass-term m but that is just the big problem... If you want an illustration, just check http://www.shef.ac.uk/physics/teaching/phy604/ew4.pdf http://www.shef.ac.uk/physics/teaching/phy604/ew5.pdf marlon 12. Mar 2, 2005 da_willem But in the first articles you cite they use the same sort of 'mass' terms in their Lagrangian: $$-(1/2) m^2 \phi^2$$ in the first and $$-\mu^2 \phi ^* \phi$$ in the second example. 13. Mar 2, 2005 marlon yes but what i was trying to say is that this m is NOT mass because it is a non-physical value. Basically the field that you gave (which is correct, i was not trying to say it was wrong) really describes a tachyon. The "mass"value is not physical though, because m is the squareroot of a negative scalar marlon 14. Mar 2, 2005 da_willem I tried to work out the expansion from the ground state $$\phi_0$$: $$-\partial _\mu \partial^\mu \Phi -\frac{m^2}{\phi_0^2}[\Phi^\dagger \Phi -\phi_0^2]\Phi =0$$ $$\Phi=\phi_0+h(x)$$ Filling this in yields: $$(-\partial _\mu \partial^\mu h -2m^2h) -\frac{m^2}{\phi_0^2}[h^3+h^2 \phi_0]=0$$ The first term between brackets is indeed a particle with mass [itex]\sqrt{2}m right?! What about the other terms?

Last edited: Mar 2, 2005
15. Mar 2, 2005

marlon

The h³ represents a selfcoupling. This field interacts with itself. The last term represents a coupling between the excitation h and the vaccuumfield. The latter is ofcourse the Higgsfield.

But may i suggest you try calculating this like shown in the first article that i referred to ?

regards
marlon

16. Mar 2, 2005

da_willem

I use the potential from a book "an introduction to the standard model of particle physics" by Cottingham & Greenwood. It is only that they only work with Lagrangian densities and I would like to see how the Higgs mechanism works out in the equations. Switching to another potential would't make things only more difficult for me...

Now to see how the Higss mechanism gives mass to the gauge fields I will have to substitute $$\partial_\mu \rightarrow \partial_\mu -iqA_\mu$$ and $$\partial^\mu \rightarrow \partial^\mu +iqA^\mu$$ before the symmetry breaking (choosing phi real) right?

Last edited: Mar 2, 2005
17. Mar 2, 2005

da_willem

Then I get a KG equation for the Higgs field with mass $\sqrt{2}m$ and a whole bunch of other terms:

$$q^2 A_\mu A^\mu h +iq(\partial_\mu A^\mu -\partial^\mu A_\mu )h$$
Wich are associated with the interaction of the Higgs field with tha gauge field I presume?!

And $$q^2 \phi_0 A_\mu A^\mu+iq \phi_0 \partial_\mu A^\mu$$

How do these terms indicate that the gauge field has acquired mass?

Last edited: Mar 2, 2005
18. Mar 2, 2005

dextercioby

It doesn't matter whether the metric is Bjorken-Drell,or GR type,then this term:
$$iq(\partial_{\mu}A^{\mu}-\partial^{\mu}A_{\mu})h$$

should be ZERO...

Daniel.

19. Mar 2, 2005

da_willem

Should be zero as in I made a mistake and the term should not be there, or is there actually a reason why the term vanishes?!

20. Mar 2, 2005

dextercioby

Yes,because the two terms are the same...?

Daniel.