# Mass zero field self-energy part?

1. Feb 3, 2006

### Neitrino

Mass zero field can have a self-energy part?

2. Feb 3, 2006

### dextercioby

You mean like a photon's self-energy diagrams...? Of course. The same goes for gravitons and gluons and used to work for neutrinos, too.

Daniel.

3. Feb 3, 2006

### Neitrino

Thks for reply

If massless field suppose fermion field can have self-energy part...then it is infinity and this infinity appears in denominator of full propagator? In massive filed this infinity of self-energy is absorbed by bare infinite mass.
So what does this infinite self-energy does in massless field?

George

4. Feb 3, 2006

### dextercioby

It may go into the renormalization of the coupling constant: electric charge for em interaction, coupling constant for QCD and unfortunately not in the renormalization of Cavendish's constant: G_{N}.

Daniel.

5. Feb 3, 2006

### Neitrino

I also was thinking like that, but as I understand from Ryder or Cheng Li
they begin renormalization of coupling constant from four-ponit particle Green function. I if consider only two-point function... (series of tadpoles)....

6. Feb 3, 2006

### dextercioby

Nope, renormalizations deals only with OPI Green functions. They give the loops and regularization of those integrals is needed.

Daniel.

7. Feb 3, 2006

### Neitrino

Thks dex Im realy weak in renormalizations.
And when considering cutoff of this self-energy part, we make it finite, what to do with assumption that bare mass should be infinite? if it is assumed to be still infinite...... after putting some cutoff self-energy is made finite.

Thks
George

8. Feb 3, 2006

### dextercioby

I'm not into this "cutoff" procedure, i'm more like a "dimensional regularization" fan, simply because it's highly useful in QED, QCD,quantum gravity and so on.

I can still tell you though that no matter what the regularization is, the result should be the same and the renormalization procedure is identical.

The self-energy is still infinite, because you still have to take the limit $\Lambda \rightarrow +\infty$...

Daniel.

9. Feb 3, 2006

### nrqed

In the spirit of effective field theories, all the theories we have now should be cutoff theories, that is they are valid up to some energy scale and one should in principle put a cutoff of the order of that energy scale (a hard cutoff on all momentum integrals is not the bets way to go about it since it breaks all sorts of symmetries, but its a useful way to think about using a cutoff, to get started).

Then the bare constants are simply not infinite. Thats all. The cutoff is never taken to infinity and everything is finite. If a theory is renormalizable, all physical results will be insensitive to the cutoff, as long as all external energies are much below the cutoff used. When energies approach the cutoff, then the theory becomes useless as an effective field theory and one needs to uncover a more fundamental theory.

Hope this makes sense. And helps.

Regards

Pat

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