Dear PF...(adsbygoogle = window.adsbygoogle || []).push({});

Please help me with basic question more I think more I get confused...

In O(n) space there are n(n-1)/2 generators...

suppose I have symmetric tensor in O(n) space, it will have n(n+1)/2 independent components... and i am building invariant potential from it (quartic polynomial expression), then I will find which components minimize my potential and will have symmetry breaking.

Broken generators will transform into Goldstone boson.

So initially i have n(n-1)/2 goldstonic degrees of freedom.

But before i minimize my potential i want to diagonalize my tensor...

Does diagonalization consume goldstonic degrees of freedom/consumes generators?

After diagonalisation shall i have again n(n-1)/2 goldstonic degrees of freedom in my bydget or less ?

thank you

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Symmetry breaking/ degrees of freedom

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads for Symmetry breaking degrees | Date |
---|---|

B Electroweak spontaneous symmetry breaking | Oct 15, 2017 |

A Breaking of a local symmetry is impossible, so what about global symmetry... | Oct 2, 2017 |

A How can gauge symmetry be broken? | Jun 27, 2017 |

I What does it mean: "up to total derivatives" | May 9, 2017 |

I Spontaneous symmetry breaking of chiral symmetry | Aug 6, 2016 |

**Physics Forums - The Fusion of Science and Community**