# What does it mean that a Lagrange multiplier is zero?

If I have the following Lagrangian

$$L'=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2$$

I can impose the constraint $$x-y-C=0$$ via a Lagrange multiplier $$\lambda$$, so the new Lagrangian is

$$L=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2+\lambda(x-y-C),$$

whose Euler-Lagrange equations are

\begin{align} \ddot{x}-\lambda&=0\\ \ddot{y}+\lambda&=0\\ x-y-C&=0. \end{align}

Solving this, I have

\begin{align} x(t)&=vt+x_0\\ y(t)&=vt+x_0-C\\ \lambda(t)&=0. \end{align}

What does it mean that the Lagrange multiplier is $$0$$?

• 1. What makes this a physics question rather than a Mathematics question? 2. What about the general interpretation of the value of Lagrange multipliers is not satisfactory in this case? – ACuriousMind Apr 7 at 15:02
• So, is it just that the force needed to keep that constraint is zero? – AFG Apr 7 at 15:07
• Another way of phrasing "the constraint force is zero" is to say "there is a class of initial values for the unconstrained system such that the constraint can be satisfied without the constraint term" – Jerry Schirmer Apr 8 at 14:44
• @JerrySchirmer That is what I was looking for. If you post an answer, I will accept it. – AFG Apr 8 at 16:00
• @JerrySchirmer Oh true! I didn't realize that was yours. Just that I found the comment more general and useful. Thank you! – AFG Apr 8 at 16:18

$$L = \frac{1}{2}{\dot x}^{2} + \frac{1}{2}{\dot y}^{2} + \lambda\left(x - y - C\right)$$

Now, take the transformation:

\begin{align} s &= \frac{1}{2}(x -y - C)\\ r &= \frac{1}{2}(x + y) \end{align}

$$L = {\dot s}^{2} + {\dot r}^{2} + \lambda s$$

Compare this with the unconstrained system:

$$L = {\dot s}^{2} + {\dot r}^{2}$$

which, obviously has as its solution of its EOM:

$$s = At + B, r = Et + F$$

But, note that the constraint equation above only requires $$s=0$$ to be satisfied. This means, that you can satisfy the constraint simply by choosing $$A = B = 0$$ as your initial conditions for the unconstrained equation. Thus, the constraint force is zero, and that's the meaning of why your Lagrange multiplier is zero -- it simply says "you can satisfy the constraints with a choice of initial conditions, alone".

the constraint force is zero, because you don't have potential energy or external forces

put for example external forces towards the x coordained you obtain

$$L=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2+\lambda(x-y-C)+F\,x$$

thus: $$\ddot x=\frac 12 F\\ \ddot y=\frac 12 F\\ \lambda=-\frac 12 F$$

so if $$~\lambda~$$ equal zero it mean, that you don't have external or potential forces

remarks from @BioPhysicist

lets look at the general case , you have

external force components are $$~F_x~,~F_y~$$ and the potential energy is depending on x and y $$~U=U(x,y)~$$

thus

$$L=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2-U(x,y)+\lambda(x-y-C) +F_x\,x+F_y\,y$$

$$\Rightarrow$$

$$\lambda=-\frac 12\,F_{{x}}+\frac 12\,{\frac {\partial }{\partial x}}U \left( x,y \right) +\frac 12\,F_{{y}}-\frac 12\,{\frac {\partial }{\partial y}}U \left( x, y \right)$$ thus $$~\lambda=0~$$ only if

$$U(x,y)=Z(x+y)\qquad \text{and}\\ F_x=F_y=F$$

where $$~Z$$ arbitrary function

• If the additional force was along the line $x-y=C$ then the constraint would still be $0$. So your starting sentence isn't always true. – BioPhysicist Apr 8 at 15:32
• @BioPhysicist I agree with you but this is also special case ? – Eli Apr 8 at 19:26
• Right, but your first sentence makes it seem like any force will do. – BioPhysicist Apr 8 at 19:39
• I wrote potential energy or force – Eli Apr 8 at 21:29
• Your first sentence makes it seem like any potential energy or force will do – BioPhysicist Apr 8 at 22:05

As others have pointed out, it means your constraint force us zero. However, what is lacking is what this means for your system.

At first you might think, "How can the constraint force be $$0$$ if the particle is still being constrained to move on the line $$x-y=C$$? Won't a force be needed to constrain it?" But the thing is, you have just forced the motion to start and remain along the line $$x-y=C$$. So you still have a free particle, but you have just specified the direction that free particle has to move in. The particle still moves along that line at a constant velocity with no force needed to keep it on that line (consistent with Newton's first law).

As Eli points out, including an additional force that is not along the constraint line will give you a non-zero constraint force, as now you need a constraint force to keep the particle along the line in the presence of the external force.