Why is adding gauge fixing directly different from doing so by Lagrange multiplier? For simplicity, we don't use field model.

Direct method

Consider a system $$L(x,\dot x,y,\dot y)=\frac{\dot x^2}{2}+\dot x y+\frac{(x-y)^2}{2} \tag{1} \, .$$ This system has gauge symmetry $$\delta x=f(t),\ \delta y=f(t)-\dot f(t)\tag{2}$$ for arbitrary $f(t)$. Under this transformation, $$\delta L = \frac{d}{dt}(x f+\frac{1}{2}f^2)\tag{3} \, .$$ The Euler-Lagrange equations are: \begin{align} L1 &: \quad \ddot{x}+\dot y- x+ y =0 \tag{4} \\ L2 &: \quad \dot x - x+y = 0 \, .\tag{5} \end{align} We see $$L1= \frac{d}{dt}L2+L2\tag{6}$$ so $(4)$ is not independent of $(5)$ and we need only to solve $(5)$, i.e. $$\dot x =x-y \tag{7} , .$$ We see $y(t)$ is a gauge freedom, and only fixing $y(t)$ we can solve $x(t)$.

Suppose we choose the gauge $y=0$. Then we solve $\dot x-x = 0$ with result $$x= c e^t \, .\tag{8}$$ with constant $c$ determined by initial value.

Lagrange multiplier method

Now let's try with the Lagrange multiplier method, $$L'(x,\dot x , y, \dot y, \lambda)= \frac{\dot x^2}{2}+\dot x y+\frac{(x-y)^2}{2} - \lambda y \, .\tag{9}$$ The Euler-Lagrange equations are \begin{align} \ddot{x}+\dot y- x+ y &= 0 \tag{10} \\ \dot x - x+y-\lambda &= 0 \tag{11} \\ y &= 0 \, . \tag{12} \end{align} Substituting $(12)$ into $(10,11)$ gives $$\ddot{x} - x =0 \quad \text{and} \quad \dot x-x=\lambda$$ therefore $$x = c_1 e^{-t}+c_2 e^t \longrightarrow \lambda = -2 c_1 e^{-t} \, .\tag{13}$$ It's obvious that $(13)$ is different from $(8)$.

Why do these two methods give different results?

Note: $y=0$ is a well-defined gauge condition because for any function $y(t)$ I can always choose gauge transformation $f(t)= c e^t + e^t\int_1^{t} e^{-u}y(u)du $ such that $y = 0$

  • $\begingroup$ I am very happy that you found a simple way to illustrate the question without the added complexity of field theory. This is an excellent example of removing irrelevant complexity so that the heart of the question is clear and so the answers will also be clear and enlightening. $\endgroup$
    – DanielSank
    Commented Dec 13, 2017 at 6:34

2 Answers 2


I) OP's Lagrangian (1) can be rewritten as

$$\begin{align}L ~=~& \frac{\dot{x}^2}{2} + \dot{x}y + \frac{(x-y)^2}{2}\cr ~\stackrel{z\equiv x-y}{=}&~\frac{1}{2}\frac{d(x^2)}{dt}+ \frac{(z-\dot{x})^2}{2}\cr ~\stackrel{w\equiv z-\dot{x}}{=}&~ \underbrace{\frac{1}{2}\frac{d(x^2)}{dt}}_{\text{total time derivative}}+ \frac{w^2}{2}, \end{align}\tag{A}$$ with infinitesimal gauge quasi-symmetry

$$\begin{align} \delta x~=~&f, \cr \delta y~=~&f-\dot{f}, \cr \delta z~=~&\dot{f}, \cr \delta w~=~&0,\cr \delta L~=~&\frac{d(xf)}{dt}. \end{align} \tag{B}$$

Let us take $x$ and $w$ and as the fundamental variables. Remarkably it is then a priori not necessary to impose any boundary conditions (BCs)! The $x$ variable is a gauge degree of freedom. The EL eq. for $w$ is $$ w~\approx~0. \tag{C}$$

A small problem arises: OP's gauge-fixing condition $$ x-\dot{x}-w~\equiv~x-z~\equiv~y~\approx~0 \tag{D}$$ is not transversal to the gauge orbits, i.e. it does not completely fix the gauge, i.e. one may freely add a contribution $ce^t$ to $x$ without leaving the gauge-fixing surface. However, that could in principle be cured by adding one pertinent BC. Then OP's gauge-fixing condition (D) is well-posed.

II) OP's gauge-fixed Lagrangian (9) can be rewritten as

$$\begin{align}L^{\prime} ~=~&L-\lambda y\cr ~=~& \frac{1}{2}\frac{d(x^2)}{dt}+ \frac{w^2}{2}-\lambda(x-\dot{x}-w). \end{align}\tag{E}$$

The EL eqs. read $$\begin{align} \dot{\lambda}+\lambda~\approx~&0,\cr w+\lambda~\approx~&0,\cr x-\dot{x}-w~\approx~&0,\end{align}\tag{F}$$ with solution $$\begin{align} -\lambda ~=~&w~=~ 2 c_1 e^{-t}, \cr x ~=~& c_1 e^{-t}+c_2 e^t ,\end{align}\tag{G} $$ in agreement with OP's eq. (13).

We can now identify the cause of the disagreement with section I: The Lagrange multiplier (which is supposed to be a non-dynamical auxiliary variable) has effectively turned dynamical: Its eom (F) depends on a time-derivative $\dot{\lambda}$. We must choose $c_1=0$. Then agreement with section I is restored.

III) An alternative way of phrasing the problem is that the constraint (D) is effectively non-holonomic, which opens Pandora's box, cf. e.g. this and this Phys.SE posts.


Let me summarize some key points in my opinion:

  • $y$ is not a free variable that can be arbitrarily fixed (like a gauge field) but instead is an auxiliary variable.
  • The Lagrangian you start with is actually a total time derivative.
  • You change the Lagrangian in the Lagrange multipliers case.

Since there are no dynamic term, $g(\dot{y})$ in Eqn.(1), $y$ is an auxiliary variable: You can replace it by its equation of motion, which is $$y=x-\dot{x}$$ After this, you see that your Lagrangian is $$L(x,\dot{x})=\frac{d}{dt}\left(\frac{1}{2}x^2\right)$$ This is the reason why you see a symmetry $$x\to x+f(t)$$ in the first place: No matter how you change $x$, the difference in Lagrangian will be total time derivative as the initial Lagrangian itself is total time derivative.

In general, you cannot add the result of EOM to the Lagrangian: You would get incorrect answers. However, for auxiliary fields which lack dynamical terms, this is possible. In your example, $y$ is not a gauge freedom but is an auxiliary field.

In your Eqn.(9) you change the Lagrangian as you cannot arbitrarily choose $y$. In fact, if you repeat the same calculation with the Lagrangian $$L'(x,\dot{x},y,\dot{y},\lambda)=L(x,\dot{x},y,\dot{y})+\lambda (x-y-\dot{x})$$ you get the same consistent results.


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