Variation of the Lagrangian expressed as a time derivative of a function

Question

In chapter 4.5 of Jakob Schwichtenberg's Physics from Symmetry, he expresses the variation of the Lagrangian $L = L\left ( q, \dot{q}, t \right )$ with respect to the generalized coordinate $q$ as

$$\delta L = L\left ( q, \dot{q}, t \right ) - L \left ( q + \delta q, \dot{q} + \frac{d\delta q}{dt}, t \right )$$

He then says that letting $\delta L \neq 0$ is fine as long as the action is invariant under the variation:

$$\delta S = \int dt \delta L = 0 \tag{1}$$

He claims that $L$ can differ from another Lagrangian, say, $L^\prime$, by the time derivative of a function $G = G\left ( q, t \right )$. Let's say that

$$L^\prime = L + \frac{dG}{dt}$$

He then goes on to prove that the variation with respect to the new Lagrangian is equal to the variation of the action of the original Lagrangian, namely:

$$\delta S^\prime = \int dt \delta L^\prime = \int dt \delta L + \int dt \frac{d}{dt}\delta G = \delta S + \frac{\partial G}{\partial q}\delta q \Bigg|_\text{(endpoints)} = \delta S \tag{2}$$

where $\frac{\partial G}{\partial q}\delta q \Bigg|_\text{(endpoints)}$ is zero due to the vanishing variation $\delta q$ at the bounds of integration. This proves that $L$ and $L^\prime$ yield the same equations of motion. From this, he posits that the variation in $L$ can be expressed as

$$\delta L = \frac{dG}{dt}\tag{3}$$

implying that the Lagrangian can vary in this fashion while keeping $S$ invariant. However, inserting this expression into $\left ( 1 \right )$ yields

$$\delta S = \int dt \frac{dG}{dt} = G \Bigg|_{\text{(endpoints)}}\tag{4}$$

which is not necessarily zero since there was never any restriction imposed on $G$ to either vanish or have the same value at both integration bounds. Should the variation in $L$ be more accurately expressed as

$$\delta L = \frac{d}{dt}\delta G$$

since it was readily shown in $\left ( 2 \right )$ that the above vanishes when integrated? I can only reason the original form in $\left ( \text{3} \right )$ if I literally interpret $\delta L$ as $\delta L = L^\prime - L$ which I have a hard time justifying since $\frac{d G}{d t}$ is not an infinitesimal, on top of what was shown in $\left ( 4 \right )$.

For what it's worth, I believe the form of $\left ( \text{3} \right )$ due to how it's used immediately after in the book, I just can't seem to justify its form. The only other reference I could find that has the variation of $L$ in this form is this webpage which likewise doesn't explain the form of $\delta L$.

Answer 1

1. OP already seems aware that Ref. 1 on p. 102 implicitly is talking about 2 different types of (infinitesimal) variations:

   • i. Ordinary variations to derive EL equations. These do satisfy boundary conditions.

   • ii. Symmetry variations. These do not satisfy boundary conditions.

   See also e.g. this related Phys.SE post.

2. Ref. 1. appears to use the fact that

   • i. the EL equations are not affected by a boundary term

   to motivate

   • ii. the notion of a quasisymmetry

   and goes on to prove Noether's first theorem for quasisymmetries.

3. Returning to OP's concrete calculation, it is perhaps helpful to say that the function $G$ should be viewed as part of the action in case (i) but not in case (ii).

References:

1. J. Schwichtenberg, Physics from Symmetry, 2nd edition, 2018; p. 102.

Answer 2

To build intuition, you can consider a 1D particle in a uniform force field (eg free fall) with the Lagrangian: $$ L=\frac12\dot x^2-x $$ and equation of motion: $$ \ddot x+1=0 $$ The system is invariant by translation even if $L$ is not: $$ \delta L=-\delta q=\frac d{dt} (-t\delta q) $$ The corresponding conserved quantity predicted by Noether’s theorem is: $$ Q=\dot q+t $$ Another simple example is invariance by Galilean/relativstic boosts of a free particle.

Recall that for the variational principle, the value of the action $S$ is not physical. Only differences in the action of two paths with the same endpoints are physical, and establish the equations of motion. This is why if $\delta S$ is just a boundary term, it is independent of $q$, so the equations of motions are preserved. This property descends to the Lagrangian $L$ which is defined up to a total derivative.

In the previous example: $$ \delta S=-T\delta q $$ which is independent of the bulk trajectory $q$, which is why the equations of motion are unchanged.