Justification of form $L(v^2)$ of Lagrangian for a free particle in Landau-Lifshitz vol 1

Question

See the screenshot below for Landau's argument on the form of a free particle lagrangian. My question is regarding whether the Lagrangian $L$ of a free particle must only be dependent on $v^2$. In my understanding, the real law of motion is given by the Euler-lagrange equations and not the form of the Lagrangian. So, as long as there is no velocity direction or position dependence in that, we can put faith in those equations.

Now, imagine if $L = q\cdot \dot{q}$. Then, Euler Lagrange give: $$ 0 = \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = \frac{dq}{dt} - \dot{q} = 0 $$ where the last equality follows from definition of velocity. We see that this law predicts nothing and fails to predict the motion of the particle yet there's no mathematical inconsistency is there? How do we reconcile this? What's wrong here? Is it that my example sets the first order variation to zero for all cases and hence we must look at second-order variation or something?

Answer 1

Yes, it is possible that the Lagrangian of a 3D free particle does depend on components and not only on the absolute value of $v$ for a free particle. For instance $$L(v)= mv^2/2 + cv_x+ v_y\cos(ky)+ hv_z/z$$ gives the correct equation for a free particle of mass $m$. (This is a special case of lagrangians which differ from the standard one in view of added total derivatives when you change the values of constants c, k, h, and therefore they produce the same equation of motion. Your proposed Lagrangian is of the same type.)

This Lagrangian, differently of the standard one, depends on the choice of axes we use in our inertial reference frame. This does not seem in agreement with the idea that all choices of axes must be equally acceptable. Also the choice of the origin of the axes should be irrelevant. Your modified Lagrangian is not invariant under that choice.

However these are meta physical requirements, because what we can experimentally control are only the laws of motion (more precisely the family of all possible motions) and not the Lagrangian itself, that it is a theoretical not directly accessible notion.

What physics says is only that the family of solutions is invariant under the full Galileo group and thus under rotations and translations in particular.

However this invariance admits many forms for the Lagranguans which determine the invariant family of solutions. To lift the requirement of invariance at the level of Lagrangians is just matter of convenience and not a physical requirement. It should be clearly declared in any foundational approach and I do not think it is the case in the LL textbook.

I stress that this forced requirement of Galileian invariance fails at the end of day in classical mechanics, revealing its unphysical nature! That is because there is no Lagrangian that is invariant in form under the complete Galileo group!

The argument by LL (which I do not like very much as it is ideologically too far from a safe operationist approach) should be viewed as a simplicity argument and nothing further in my view: The standard Lagrangian is the simplest Lagrangian that gives the right equations of motion in a given inertial reference frame without adding arbitrary terms like dimensional constants, choices of the origin, orientation of axes, and all that.

Answer 2

The Lagrangian with trivial equations of motion you mentioned, $ L = q \dot{q} = \frac{1}{2}\frac{d}{dt} q^2 $, is a total derivative, so the action $$ S = \int_{t_0}^{t_1} L \,dt = \frac{1}{2} (q(t_1)^2 - q(t_0)^2) $$ does not depend on the trajectory.

The reason why we always find a $v^2$ term in the action is because $L = T - V$ in classical physics where $T = \frac{1}{2} m v^2$ is the kinetic energy and $V$ is the potential. This is necessary to recover Newton's law (up to total derivatives or terms not contributing to the equation of motion).

However, it can be the case that the potential $V$ depends on the velocity. For example, the Lorentz force $\boldsymbol{F} = q (\boldsymbol{E} + \boldsymbol{v} \wedge \boldsymbol{B})$ can be derived from a generalized potential $V(\boldsymbol{x},\boldsymbol{v})$, where the force is derived as $\boldsymbol{F} = \frac{d}{dt} \frac{\partial V}{\partial \boldsymbol{v}} - \frac{\partial V}{\partial \boldsymbol{x}}$. In this case the potential reads $$ V = q ( \phi - \boldsymbol{v} \cdot \boldsymbol{A}) $$ where $\boldsymbol{A}$ is the magnetic vector potential $\boldsymbol{B} = \nabla \wedge A$. This potential is velocity dependent, and if you input this in the Euler-Lagrange equation, you recover Newton's law under the Lorentz force. For details on the derivation, look at this Wikipedia page (Lorentz force in terms of potential & Lorentz force and analytical mechanics)

Note also that, from the mathematical point of view, nothing prevents you from condering other types of Lagrangians $L = L(q, \dot{q})$. In particular you can consider minimizations problems, where $L$ has nothing to do with classical physics, but Euler-Lagrangian equations remain useful to minimize the functional $S$. With this formalism you can prove that the straight line is the shortest path between two points, or more interestingly find the shape of a hanging chain which minimizes the potential energy. In the first case $S$ is a length, and in the second case $S$ is the total potential energy of the system. These are two examples where the mathematical framework of minimizing the functional $S[q, \dot{q}]$ has direct applications, whereas in Lagrangian mechanics $S = \int (T-V) dt$ is not an intuitive physical quantity.

Answer 3

The claims about a Lagrangian function in L&L Mechanics are strictly speaking not true in general for all possible Lagrangians, they are an additional symmetry criterion to select "the one and only Lagrangian". This seems to me to be a fool's errand - there is no real need to define the "best" Lagrangian function.

This is because one can describe free particle with e.g. the Lagrangian function

$$ L' = \frac{1}{2}mv^2 + k\mathbf v\cdot \mathbf r $$

or

$$ L'' = \frac{1}{2}mv^2 +bt. $$ These lead to the same equations of motion as the standard Lagrangian function

$$ L = \frac{1}{2}mv^2. $$

Your example with $q\dot{q}$ is a 1D variant of this phenomenon - addition/removal of such term from the Lagrangian does nothing to equations of motion.

These different Lagrangians are fine because they differ from $L$ (which we know is fine) by a total time derivative of a function of coordinates and time only $f(\mathbf r,t)$. Such difference is inconsequential, because the Euler-Lagrange equations are insensitive to such changes of $L$.

Answer 4

1. We interpret the argument in Landau and Lifshitz as that they want to implement Galilean invariance at the level of the Lagrangian rather$^1$ than just at the level of the equations of motion.

2. The Lagrange equations are not changed by scaling the Lagrangian by a non-zero constant or adding a total time derivative. These are trivial modifications and usually not interesting.

3. However a total time derivative in the Lagrangian can potentially play an intertwining/mediating role in implementing quasi-symmetries in Noether's theorem. Most famously for the Galilean boost quasi-symmetry, cf. e.g. this and this Phys.SE post. So we should be more careful.

4. Let us for simplicity assume that the Lagrangian $L(\vec{q},\vec{v},t)$ is of first-order.

5. It turns out that because the vector fields $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial q^i}$ for time and space translations do commute with total time differentiation $$\frac{\mathrm{d}}{\mathrm{d}t}~=~\frac{\partial}{\partial t}+ v^i\frac{\partial}{\partial q^i}+ a^i\frac{\partial}{\partial v^i}+\ldots,$$ we can w.l.o.g. assume that time and space translations are implemented as strict symmetries (rather than just quasi-symmetries) of the Lagrangian.

6. The above shows the Lagrangian is homogeneous wrt. time and space, i.e. that the Lagrangian $L=L(\vec{v})$ only depends on the velocity $\vec{v}$.

7. To argue that the Lagrangian is isotropic in space, i.e. that the Lagrangian $L=L(v^2)$ only depends on the speed, see my Phys.SE answer here.

--

$^1$ See also e.g. this related Phys.SE post.