The Lagrangian with Lagrange multiplier in the form
L=T−V+λf(q,˙q,t).
But there are different ways of writing the constraint f=0.
Will that lead to different EOMs?
Let me give an example:
A pendulum with mass m and length ℓ.
We can use let
I=∫t1t0[12m(˙x2+˙y2)−mgy−λ(√x2+y2−ℓ)]dt,
or
I=∫t1t0[12m(˙x2+˙y2)−mgy−λ((x2+y2−ℓ2)2)]dt.
In the first case, we have
m¨x=−λxℓ,m¨y=−mg−λyℓ,
which are the correct EOM.
But for the second case, we have
m¨x=0,m¨y=−mg.
Answer
Let there be given a (configuration) manifold M. Often in physics one assumes that a constraint function χ obeys the following regularity conditions:
χ:Ω⊆M→R is defined in an open neighborhood Ω of the constrained submanifold C⊂M;
χ is (sufficiently1 many times) differentiable in Ω;
The gradient →∇χ is non-vanishing on the constrained submanifold C⊂M.
Here it is implicitly understood that χ vanishes on the constrained submanifold C⊂M, i.e.
C∩Ω = χ−1({0}) := {x∈Ω∣χ(x)=0}.
[Also we imagine that the full constrained submanifold C⊂M is covered by a family (Ωα)α∈I of open neighborhoods, each with a corresponding constrained function χα:Ωα⊆M→R, and such that the constraint functions χα and χβ are compatible in neighborhood overlaps Ωα∩Ωβ.] Since there (locally) is only one constraint, the constrained submanifold will be a hypersurface, i.e. of codimension 1. [More generally, there could be more than one constraint: Then the above regularity conditions should be modified accordingly. See e.g. Ref. 1 for details.]
The above regularity conditions are strictly speaking not always necessary, but greatly simplify the general theory of constrained systems. E.g. in cases where one would like to use the inverse function theorem, the implicit function theorem, or reparametrize χ→χ′ the constraints. [The rank condition (3.) can be tied to the non-vanishing of the Jacobian J in the inverse function theorem.]
Quantum mechanically, reparametrizations of constraints may induce a Faddeev-Popov-like determinantal factor in the path integral.
Example 1a: OP's 1st example (v1) χ(x,y) = x2+y2−ℓ2
Example 1b: OP's 1st example (v3) χ(x,y) = √x2+y2−ℓ
Example 2a: Assume ℓ>0. OP's 2nd example (v1) χ(x,y) = √x2+y2−ℓ2
Example 2b: Assume ℓ>0. OP's 2nd example (v3) χ(x,y) = (x2+y2−ℓ2)2
References:
- M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994; Subsection 1.1.2.
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1 Exactly how many times differentiable depends on application.
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