Physics, asked by NitishBatham6983, 1 year ago

Is a lagrangian with a background field interaction renormalizable ? If yes, when?

Answers

Answered by RockyAk47

Hey dear here is the answer

By definition, Lagrange multipliers are only coefficients that enter the extremized quantity (action) etc. linearly – and that multiply constraints. In some exceptional cases, an auxiliary field could enter in this way. However, they typically appear in a more complicated way and bilinear terms in the auxiliary fields are a rule rather than an exception. So strictly speaking, they're not Lagrange multipliers. But they are very similar. If no derivatives of these objects appear in the action, they're also "non-dynamical" (not involving time derivatives) and the variation with respect to them implies "non-dynamical" i.e. algebraic equations of motion.

Note that in the normal treatment of extremization, we introduce Lagrange multipliers because we want to extremize the quantity given the assumption that another quantity or other quantities are kept fixed. "Kept fixed" is translated as "conservation laws" into the physics jargon. However, in physics, we rarely consider conserved quantities that are conserved because the conservation law is explicitly written down as an independent constraint. Instead, in physics we usually discover conservation laws nontrivially – the conserved quantity has to be determined by a somewhat non-trivial procedure due to Emmy Noether out of a symmetry. In almost all physical theories, conservation laws are non-trivial consequences of some other, "more elementary" equations of physics.

Hope its help you