# Why gauge bosons, but no gauge fermions

1. May 2, 2005

### Edgardo

Hello all,

from Marlon's journal, I read the question "DO YOU KNOW WHY FORCE CARRIERS ARE ALWAYS BOSONS ??? WHY DON'T WE HAVE GAUGE FERMIONS ???"

2. May 2, 2005

### dextercioby

Give me an example of a fermionic first class system,field theory,of course...

Daniel.

3. May 2, 2005

### Meir Achuz

Gauge invariance involves correcting problems caused in quantum mechanics by the appearance of the partial derivatives with respect ot x,y,z,t. These partials form a relativistic 4-vector. This requires 4-vector fields so that (in EM)
d/dx-->d/dx-ieA_x, etc. The particle excitations of vector fields have spin one.
So the requirement that gauge particles (the excitations of the gauge fields) must be vector particles follows from the fact that space-time is 4 dimensional.

4. May 2, 2005

### joshuaw

I thought that because the exchange particles must be able to carry integer spin they must be bosons. (Spin flips and all).
Josh

5. May 3, 2005

### dextercioby

Why would they have to carry integer spin...?

Daniel.

6. May 3, 2005

### marlon

good question...

there are several ways to answer this : spin statistics being one of them

marlon

7. May 3, 2005

### vanesch

Staff Emeritus
I like that answer :-) As gauge fields have to appear in the covariant derivative, they have to be vector fields. Nice.

cheers,
Patrick.

8. May 3, 2005

### wangyi

In SuperSymmetry, maybe you will call the Fermions in the same doublet with the Bosons "Gauge Fermions", because they transform together with the gauge bosons, for
example wino or zino (they form chargino and neutrino).

9. May 3, 2005

### marlon

This is indeed a great answer

marlon

10. May 3, 2005

### BlackBaron

"Gauge Fermions" in supersymmetry are generically called "gauginos".
As long as I know, in supersimetric models they contribute with the forces just like regular gauge bosons, but because of the exclusion principle their net effect is not very strong, wich also shows that if we only had "gauge fermions" and no "gauge bosons" the world wouldn't be as we know it at all.

Last edited: May 3, 2005
11. May 3, 2005

### dextercioby

Unless "vector particle" means something else than the quanta of a vector field,that conclusion is incorrect.

Daniel.

12. May 3, 2005

### gptejms

I am not a QFTist and know only a bit of it,so let me ask-----does gravity have to necessarily be a gauge field?If so why?

Last edited: May 3, 2005
13. May 3, 2005

### dextercioby

Gravity is a gauge field.It's not a vector field (spin 1),but a tensor field (spin 2)...It's the idea behind post #11.

Daniel.

14. May 3, 2005

### gptejms

Yeah ok,gravity is a gauge field(only potential differences matter).But some more stupid questions--why does a tensor field have to be spin 2 and vector field spin one?

Last edited: May 3, 2005
15. May 3, 2005

### dextercioby

Aaa,nice question.Group theory.Vectors are $\left(\frac{1}{2},\frac{1}{2}\right)$ irreducible reps of the restricted homogenous Lorentz group and therefore have total spin $\frac{1}{2}+\frac{1}{2}=1$.

Symmetric 2-nd rank tensors (the gravity field $h_{\mu\nu}$ which is the I-st order perturbation expansion of GR metric $g_{\mu\nu}$) are $(1,1)\oplus (0,0)$ irredcible reps of the restricted homogenous Lorentz group and have spin $1+1=2$.

Daniel.

16. May 3, 2005

### gptejms

put it in a language that's more comprehensible

17. May 3, 2005

### dextercioby

I honestly can't."Spin" means group theory.Vectors & tensors mean group theory.That's all there is to it...All u need to understand is the 'coupling' between spin & Lorentz group...

Daniel.

18. May 3, 2005

Staff Emeritus
Quantum spin is a new concept. That means it can't really be explained in terms of older concepts. If you don't want to learn the group theory definitions, you can just accept the fact that the states of spin 1/2 only turn half as fast as the coordinates do when you perform a rotation.

19. May 3, 2005

### Haelfix

Agreed with Dexter.

Although in Supersymmetry the fundamental axioms of Lie algebra are modified (into.. surprise Super Lie Algebra), which is why you can have 'gauge' fermions with the same standard model quantum numbers, and they do indeed participate. However for technical reasons they have to be Majorana fermions, and when you calculate the beta function you end up with a fraction of the contribution as the gauge boson (tho not too small, maybe an order of magnitude less). Of course super symmetry is badly broken, and depending on the mechanism choice, will reduce things further.

20. May 3, 2005

### gptejms

SU(2) you mean--i know that.please complete the argument.