What does it mean that a Lagrange multiplier is zero?

Question

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 $0$?

Answer 1

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

Answer 2

Consider your Lagrangian:

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

Now, take the transformation:

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

changing variables in your lagrangian:

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

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".

Answer 3

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.