Lagrange multipliers and gauge conditions

[Joey_Joe_Jojo]

Hello, I was hoping someone would be able to clarify a problem I've got. A lagrange multiplier can be introduced into an action to impose a constraint right?

I was wondering what relation lagrange multipliers have to gauge conditions, which are imposed by hand. Am I correct in saying that lagrange multipliers are fields, and hence must be solved for in the resulting field equations arising from an action? They are dynamical?

My question is in regards to the Einstein aether model, in which a vector field u_{a} is introduced into the gravitational action. The term L*(u_{a}u^{a} + 1) is added to the action to make u_{a} a unit timelike vector (L is the lagrange multiplier). But what is the difference between introducing this term into the action and leaving it out, finding the field equations, and then imposing u_{a}u^{a}=-1 by hand?

Thank you for any help,

Ste

 

[Valerie Park]

fanYes, a Lagrange multiplier can be introduced into an action to impose a constraint. The main difference between introducing the term into the action and imposing it by hand is that with the Lagrange multiplier, the constraint is treated dynamically and becomes part of the field equations that must be solved. Gauge conditions are usually imposed by hand as they are not considered dynamical variables. In the case of the Einstein-aether model, the term L*(u_a u^a + 1) enforces the constraint u_a u^a = -1 dynamically, while if you left it out and imposed the same constraint by hand, it would not be treated as a dynamical variable.

 

[sir-pinski]

vo

Hi Stevo,

You are correct in saying that Lagrange multipliers are fields and must be solved for in the resulting field equations arising from an action. They are used to impose constraints on the system and can be thought of as a way to incorporate external information or conditions into the equations of motion.

Gauge conditions, on the other hand, are imposed by hand and are not dynamical fields. They are used to fix the gauge freedom in the system, which is essentially a choice of coordinates or reference frame. Gauge conditions are necessary in certain theories, such as general relativity, to ensure that physically equivalent solutions are not counted multiple times.

In the case of the Einstein aether model, introducing the term L*(u_{a}u^{a} + 1) into the action is equivalent to imposing the constraint u_{a}u^{a}=-1 by hand. The Lagrange multiplier L acts as a sort of "penalty" for violating the constraint and ensures that the field u_{a} remains a unit timelike vector. This approach is often more convenient and elegant than manually imposing the constraint, as it is automatically satisfied in the equations of motion.

I hope this clarifies the relationship between Lagrange multipliers and gauge conditions for you. Let me know if you have any further questions.