Lagrange multipliers in Lagrangian Mechanics

[Gavroy]

Hi

we covered the Lagrange multiplier method in Lagrangian Mechanics and as far as I know, is the physical meaning behind this to be able to solve either some non-holonomic constraints or to get some information about the constraint forces. my problem is, i do not know the physical meaning of those multipliers. what are they telling me and how are they related to the constraint forces?

 

[leinadle]

How comfortable are you with vector calculus? You can visualize Lagrange Multipliers by realizing the constraint is a curve on your surface, and finding where the gradient vectors are parallel to each other finds either the minimum or maximum of your system of equations. Here is a good interactive visualization: http://demonstrations.wolfram.com/TheGeometryOfLagrangeMultipliers/

 

[VantagePoint72]

The Lagrange multiplier by itself has no physical meaning: it can be transformed into a new function of time just by rewriting the constraint equation into something physically equivalent. For example, the constraints $G(x,y) = 0$ for $G(x,y) = x^2 - y^2$ or $G(x,y) = 2x^2 - 2y^2$ are physically equivalent but lead to different multiplier functions when used to the modify the system's Lagrangian. It's like how the electric and magnetic potentials can be gauge transformed in electromagnetism without changing the physics. The physical interpretation as being related to the force required to constrain the motion (for holonomic constraints) only comes when the multiplier is combined with the particular constraining function chosen: $\vec{F(t)} = -\lambda(t) \frac{\partial G}{\partial \vec{x}}$. Neither $\lambda$ nor $G$ by itself is physical; only their combination.

 

[AlephZero]

i do not know the physical meaning of those multipliers. what are they telling me and how are they related to the constraint forces?

The numerical values of the lagrange multipliers ARE the cnstraint forces.

Set up a simple system where you know the answer (e.g. a 3 DOF system of two springs joined together), use LM's to apply some constraints to it (e.g apply a displacement of 1 at one end a displacement of 2 at the other end, and a force on the DOF in the middle) and work it by hand to see what you get.

 

[VantagePoint72]

The numerical values of the lagrange multipliers ARE the cnstraint forces.

This is incorrect. The Lagrange multiplier needs to be combined with the gradient of the constraining function to get the constraint forces. The numerical value of the of the multiplier by itself is not physical, as I've discussed above.

 

[WannabeNewton]

Hi Gavroy. Maybe a simple example will help you so consider the case of a block sliding down a fixed frictionless incline of angle $\theta$. Obviously, the easiest approaches to this would be to either just write down Newton's 2nd law in a convenient coordinate system or to use the generalized coordinate $q$ representing the distance traveled along the incline by the block and just writing down Lagrange's equations. For the second method, the choice of generalized coordinate $q$ will implicitly take into account the constraint which is that the block must stay on the incline during the trajectory before reaching the ground. However, let's do this using the method of Lagrange multipliers.

Using $x$ for the horizontal coordinate and $y$ for the vertical coordinate, we can write out Lagrangian as $L = \frac{1}{2}m\dot{x}^{2} + \frac{1}{2}m\dot{y}^{2} + mgy$ but note that $x,y$ are not independent. They are related by the holonomic constraint $g(x,y) = y - x\tan\theta = 0$. Letting $L' = L - \lambda g$ we simply write down Lagrange's equations as $\frac{\partial L'}{\partial x^{\alpha}} - \frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L'}{\partial \dot{x}^{\alpha}} = 0$. Doing this and combining with the constraint (this is important!) gives us $m\ddot{x} = \lambda \tan\theta$ and $\lambda\cot\theta = mg\cot\theta - m\ddot{x}$ so adding these two gives us $\lambda\sec^{2}\theta = mg\Rightarrow \lambda = (mg\cos\theta)\cos\theta$ . As we know, $N = mg\cos\theta$ is the magnitude of the normal force on the block from the incline so $\lambda$ is the magnitude of the vertical component of the normal force and $\lambda\tan\theta = (mg\cos\theta)\sin\theta = m\ddot{x}$ is the magnitude of the horizontal component of the normal force, as expected.

As you can see this is more work than simply choosing the generalized coordinate $q$ as the distance traveled along the incline, which implicitly takes the constraint into account and deals with a much nicer coordinate system for this problem.

As far as the mathematical nature of lagrange multipliers goes, for holonomic constraints it is very easy to understand in terms of constraint submanifolds of the configuration space and the implicit function theorem. For non-holonomic systems, there are subtleties involved that go into Frobenius' theorem and the tangent bundle of the configuration space but if you want I can direct you to books / resources for these matters.

 

[WannabeNewton]

By the way, just to drive home LastOne's point, note that for the above example we have $\lambda \nabla g = mg\cos\theta[-\sin\theta \hat{x} + \cos\theta\hat{y}]$. If I were to instead write down my constraint as $g(x,y) = y\cos\theta - x\sin\theta = 0$ (using the same coordinate system), which is physically the exact same constraint, I would get a different value for $\lambda$. In particular, I would get $\lambda = mg\cos\theta$ but note that $\lambda \nabla g = mg\cos\theta[-\sin\theta \hat{x} + \cos\theta\hat{y}]$ still holds. In general, if I want to have an unambiguous interpretation of the constraint forces in terms of the Lagrange multipliers, then this would be needed.

Also, just to put the expression in a more recognizable form, rotate the coordinate system clockwise by $\theta$ to get $\begin{pmatrix} \cos\theta & \sin\theta\\ -\sin\theta&\cos\theta \end{pmatrix}\lambda \nabla g = mg\cos\theta\hat{y}'$. This is of course the expression for the normal force if you wrote down Newton's 2nd law in the coordinate system where the horizontal axis is along the incline and the vertical axis is perpendicular to the incline.