Vector Boson Propagator and gauge

  • Thread starter ChrisVer
  • Start date
  • #1
ChrisVer
Gold Member
3,337
440

Main Question or Discussion Point

Well I'm trying to understand the difference between these propagators:

[itex] \frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}[/itex]

and

[itex] \frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}[/itex]

My professor told me that they are different gauges, and the from the second you rule out the goldstone modes. Can someone explain it to me a little better or refer me to some textbook? I don't also get the meaning of goldstone modes- if you have a massive vector boson then you don't have goldstone modes [they become the longitudial dofs]

Also I am having one more question. If the first doesn't kill out the goldstone bosons, then can someone -after arriving at the end result- kill them? Maybe by a gauge transformation? and what gauge transformation?
 
Last edited:

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,721
277
Well I'm trying to understand the difference between these propagators:

[itex] \frac{g_{\mu \nu}}{k^{2} - m^2 + i \epsilon}[/itex]

and

[itex] \frac{g_{\mu \nu}+ \frac{ k_{\mu} k_{\nu}}{m^{2}}}{k^{2} - m^2 + i \epsilon}[/itex]

My professor told me that they are different gauges, and the from the second you rule out the goldstone modes. Can someone explain it to me a little better or refer me to some textbook? I don't also get the meaning of goldstone modes- if you have a massive vector boson then you don't have goldstone modes [they become the longitudial dofs]

Also I am having one more question. If the first doesn't kill out the goldstone bosons, then can someone -after arriving at the end result- kill them? Maybe by a gauge transformation? and what gauge transformation?
I am not sure what your prof meant by "killing off the Goldstone modes". The two expressions correspond to different gauges. If you put a parameter ##\xi## in front of the ##k^\mu k^\nu## term, upon summing up any gauge invariant combination of Feynman diagrams (in particular the sum of the diagrams of a given loop order), the ##\xi## dependent terms will cancel out, showing that adding that term does not change anything to the final result. I guess it is possible to think of these extra terms as being associate to the Goldstone modes (like some pieces of the ##g^{\mu \nu}## piece in QED correspond to the longitudinal modes) .
 
  • #3
Avodyne
Science Advisor
1,396
87
You need to look at a QFT text for details; try Srednicki.
 

Related Threads on Vector Boson Propagator and gauge

Replies
1
Views
618
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
8
Views
4K
Replies
6
Views
5K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
1K
Replies
7
Views
1K
  • Last Post
Replies
2
Views
584
  • Last Post
Replies
3
Views
2K
Top