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

1. Mar 20, 2015

### 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?

2. Mar 21, 2015

### Orodruin

Staff Emeritus
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.

3. Mar 21, 2015

### Coffee_

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

4. Mar 21, 2015

### Orodruin

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

5. Mar 21, 2015

### Coffee_

http://imgur.com/kLeIwfN

6. Mar 21, 2015

### Orodruin

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

7. Mar 21, 2015

### Coffee_

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?

8. Mar 21, 2015

### Orodruin

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

9. Mar 22, 2015

### stevendaryl

Staff Emeritus
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.

10. Mar 22, 2015

### Orodruin

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