Go Back   Physics Help Forum > College/University Physics Help > Theoretical Physics

Theoretical Physics Theoretical Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Mar 23rd 2014, 10:57 PM   #1
Junior Member
 
Join Date: Dec 2010
Posts: 25
A question about effective action in QFT?

They say that for constant fields the effective potential in the tree approximately is given simply by minus the non-derivative terms in the Lagrangian density.I do not understand why the derivative terms are ''kinematic energy'' so that the potential equalling minimum expectation energy must abandons the kinematic energy(so abandons derivative terms).Because for example in Klein-Gordon Lagrangian the second order derivative(of four-vector) is proportional with square of mass(?) not being kinematic energy.
ndung is offline   Reply With Quote
Old Mar 24th 2014, 08:13 AM   #2
Junior Member
 
Join Date: Dec 2010
Posts: 25
Because of constant field,in effective action it has not derivative term,but I still do not understand why the effective action has only minus non-derivative terms of Lagrangian.B/c quantum effective action also contains the current J corresponding with the value of field.
ndung is offline   Reply With Quote
Old Mar 24th 2014, 08:30 AM   #3
Junior Member
 
Join Date: Dec 2010
Posts: 25
Now I can understand with tree approximation the quantum effective action is coincidence with Lagrangian density.
ndung is offline   Reply With Quote
Old Mar 25th 2014, 09:33 PM   #4
Junior Member
 
Join Date: Dec 2010
Posts: 25
At the moment,I think that this is true for Phi-4 theory
ndung is offline   Reply With Quote
Old Mar 31st 2014, 02:58 PM   #5
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,347
I think I'll be polite and let you know that the lack of response isn't that we are ignoring you. I can't answer you simply because I've never really understood the point of derivative couplings. As far as I can tell a derivative coupling is simply is a new field which we can handle in the usual way. I can't see any reason why that would necessarily become a contribution to the kinetic energy terms, unless you need to actually add a symmetry restoring kinetic energy term that is.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Theoretical Physics

Tags
action, effective, qft, question



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Average Effective Dipole moment zbmay0131 Advanced Electricity and Magnetism 0 Apr 3rd 2013 09:01 AM
Effective Capacitance nickerus Electricity and Magnetism 4 Oct 26th 2010 01:25 AM
Effective potential synclastica_86 Advanced Mechanics 0 Oct 23rd 2009 03:51 AM
Newton's 3rd Law (Law of Action-Reaction) Theoretical Question s3a Kinematics and Dynamics 1 Apr 18th 2009 09:34 PM
action reaction jot321 Kinematics and Dynamics 1 Oct 10th 2008 08:47 AM


Facebook Twitter Google+ RSS Feed