How can I know when the Lagrange multiplier is a constant?

[Coffee_]

Consider a holonomic system where I have $n$ not independent variables and one constraint $f(q1,q2,...,qN,t)=0$. One can rewrite the minimal action principle as:

$\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial q'_i} - \lambda \frac{\partial f}{\partial q_i} = 0$

My question: how can one check if lambda is a constant or a function? Because in general it does not have to be a constant in this reasoning according to my understanding.

Example:

The harmonic oscillator given by the constaint that the distance of the mass to the origin is constant. The small angle approximation is allowed. So here I write out the equations and I indeed do find a harmonic oscillation under the assumption that $\lambda$ is constant. How to show that this is indeed constant? If I try doing this and do some small angle approximations using the tension I eventually find that the $y$-coordinate is constant so I'm probably doing something wrong.

Is the lambda not constant when not using the small angle approximation?

 

[Orodruin]

Consider a holonomic system where I have $n$ not independent variables and one constraint $f(q1,q2,...,qN,t)=0$. One can rewrite the minimal action principle as:

$\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial q'_i} - \lambda \frac{\partial f}{\partial q_i} = 0$

My question: how can one check if lambda is a constant or a function? Because in general it does not have to be a constant in this reasoning according to my understanding.

Example:

The harmonic oscillator given by the constaint that the distance of the mass to the origin is constant. The small angle approximation is allowed. So here I write out the equations and I indeed do find a harmonic oscillation under the assumption that $\lambda$ is constant. How to show that this is indeed constant? If I try doing this and do some small angle approximations using the tension I eventually find that the $y$-coordinate is constant so I'm probably doing something wrong.

Is the lambda not constant when not using the small angle approximation?

This depends on your problem. For a pendulum, lambda is related to the tension in the pendulum, for a mass on a circle lambda is related to the constraining force. Now obviously, if you move with constant speed on a circle, you need a constant (magnitude) centripetal force and your lambda is constant. For the pendulum, the tension changes and so does lambda.

 

[Coffee_]

This depends on your problem. For a pendulum, lambda is related to the tension in the pendulum, for a mass on a circle lambda is related to the constraining force. Now obviously, if you move with constant speed on a circle, you need a constant (magnitude) centripetal force and your lambda is constant. For the pendulum, the tension changes and so does lambda.

Assuming a non constant lambda won't give me a harmonic motion in the x coordinate though .

 

[Orodruin]

Assuming a non constant lambda won't give me a harmonic motion in the x coordinate though .

Why don't you show what you get? In that way it will be easier to judge where you meet a problem.

 

[Coffee_]

Why don't you show what you get? In that way it will be easier to judge where you meet a problem.

http://imgur.com/kLeIwfN

 

[Orodruin]

So, small oscillations imply that you can neglect higher-order terms in the deviation from the equilibrium solution. What does this imply for $\lambda$?

 

[Coffee_]

So, small oscillations imply that you can neglect higher-order terms in the deviation from the equilibrium solution. What does this imply for $\lambda$?

Well $\lambda$ is equal to the tension up to a sign. It seems that one implicitly assumes the tension to be $mg$ in the standard derivation and thus assuming $\lambda$ to be constant?

 

[Orodruin]

Well $\lambda$ is equal to the tension up to a sign. It seems that one implicitly assumes the tension to be $mg$ in the standard derivation and thus assuming $\lambda$ to be constant?

Yes, but you can argue it in a more formal way in the small angle approximation.

 

[stevendaryl]

In general, \lambda is not constant. And in the particular case of a rock on a string being twirled around in a vertical plane in a circle of radius R, the constraint force is the tension, which is NOT constant; the tension is highest at the bottom of the circle, and is lowest at the top of the circle.

The use of Lagrangian multipliers to work with a constraint f(\vec{x}) = 0 is equivalent, at least in the simplest cases, to doing Newtonian physics (F = ma) with an unknown constraint force \vec{F}_c that is in the direction of \nabla f. The magnitude of \vec{F}_c is determined by the criterion that the particle never leaves the surface f(\vec{x}) = 0. If the other forces are sufficient to keep the particle within the surface, then the constraint force is zero. Otherwise, it's whatever magnitude is necessary.

 

[Orodruin]

In general, \lambda is not constant. And in the particular case of a rock on a string being twirled around in a vertical plane in a circle of radius R, the constraint force is the tension, which is NOT constant; the tension is highest at the bottom of the circle, and is lowest at the top of the circle.

It is however constant in the small angle approximation, which is what was currently under discussion.