Galiliean group vs Poincare group and role of mass In what follows I'm trying to generalize an approach from Landau & Lifschits Vol 1 -- unfortunately, the authors use it only in Vol 1 discussing Newtonian mechanics. I was trying to apply the same idea to relativistic case and found some surprising (at least to me!) differences. Although my question arises in physics, I'm more interested in mathematical side of it, that's why I'm asking here on math stack exchange. It may also have a side-effect of not being too rigorous, but I hope it's rigorous enough.
As you probably know, in classical mechanics the Action principle states that equations of motion for a particle are Euler-Lagrange equations for the functional
$$
A[\mathbf{x}(t)] = \int_{t_1}^{t_2} L(t, \mathbf{x}(t), \mathbf{v}(t)) dt
\label{action}
\tag{1}
$$
where $t$ is time, $\mathbf{x}(t)$ is trajectory of the particle and $\mathbf{v}(t) \equiv d\mathbf{x}/dt$ is its velocity vector. For a single free particle Lagrange function $L$ can be determined almost uniquely from symmetries of space, time and the principle of relativity.
From homogeneity of space and time we can assume that $L$ does not explicitly depend on $\mathbf{x}$ and $t$, leaving only dependence on $\mathbf{v}$. Taking isotropy of space into account, we can further rule out dependence on the direction $\mathbf{v}$, thus finally
$$
L = \Lambda(v^2)
$$
Now consider the principle of relativity. Let primed observer be moving with an infinitesimal velocity $\mathbf{u}$ relative to the unprimed one, then the only possible space and time transformation preserving all the symmetries would be:
$$
\left\{ 
  \begin{aligned}
     t' &= t-K\cdot(\mathbf{u}\mathbf{x}) \\ 
     \mathbf{x}' &= \mathbf{x}-\mathbf{u}t \\ 
  \end{aligned}
\right.
\tag{2}\label{transforms}
$$
where $K$ is some constant. Since $K$ has units of $[time]^2/[length]^2$ its absolute value has no physical significance, however its sign does. In Newtonian case, $K=0$. As you've probably guessed in relativity $K=1/c^2$ where $c=299792458 \; meters/second$. Negative value of $K$ corresponds to another case resembling relativity in certain aspects but not realized in Nature.
In order to satisfy the principle of relativity -- i.e. for primed and unprimed observers to have the same form of equations of motion -- integrands in $\ref{action}$ can only differ by a complete differential of some function $F_u(t,\mathbf{x})$:
$$
\Lambda(v'^2)dt'-\Lambda(v^2)dt = dF_u(t,\mathbf{x}) = dt \frac{\partial F_u}{\partial t} + d\mathbf{x}\frac{\partial F_u}{\partial \mathbf{x}} = dt\left\{ \frac{\partial F_u}{\partial t} + \left(\mathbf{v} \frac{\partial F_u}{\partial \mathbf{x}}\right) \right\}\tag{3}\label{relativity}
$$
It is important that this difference is linear in $\mathbf{v}$ since $F_u$ does not depend on $\mathbf{v}$ explicitly.
Calculating $v'^2$ in terms of $v^2$ is straightforward:
$$
\mathbf{v}' = \frac{d\mathbf{x}'}{dt'} = \frac{d\mathbf{x} - \mathbf{u} dt}{dt - K\cdot(\mathbf{u}d\mathbf{x})} = \mathbf{v} - \mathbf{u} + K\cdot(\mathbf{u}\mathbf{v})\cdot\mathbf{v}
$$
$$
v'^2 = v^2 - 2(\mathbf{v}\mathbf{u})\cdot(1-Kv^2)
$$
(all calculations here are up to the first order in infinitesimal relative velocity $\mathbf{u}$). Hence,
$$
\Lambda(v'^2)dt' = \Lambda\left(v^2 - 2(\mathbf{v}\mathbf{u})\cdot(1-Kv^2)\right)\cdot\left( 1-K\cdot(\mathbf{u}\mathbf{v})\right)dt \\
= \Lambda(v^2)dt - (\mathbf{v}\mathbf{u})\cdot\left\{ K\Lambda(v^2) + 2(1-Kv^2)\dot{\Lambda}(v^2) \right\}dt
$$
where by $\dot{\Lambda}$ I denote the derivative of $\Lambda$ by its argument $v^2$.
As we can see the additional term here is already linear in $\mathbf{v}$ and first order in $\mathbf{u}$, so to satisfy $\ref{relativity}$ and not introduce any explicit dependence on $t$ or $\mathbf{x}$, the expression in $\left\{\ldots\right\}$ has to be a constant:
$$
K\Lambda(v^2) + 2(1-Kv^2)\dot{\Lambda}(v^2) \equiv Q
\tag{4}\label{maineq}
$$
In Newtonian case, $K=0$, thus we get a separable equation
$$
2\dot{\Lambda}(v^2) = Q  \implies \Lambda(v^2) = \frac{Qv^2}{2} + C_1
$$
where $C_1$ is integration constant which turns out to be additive and hence can be omitted. Constant $Q$ is called mass and denoted by $M$, so we  get the final expression:
$$
L=\frac{Mv^2}{2}
$$
We also find the explicit form of the transformation $\ref{relativity}$
$$
\Lambda(v'^2)dt'-\Lambda(v^2)dt=-M\cdot(\mathbf{u}\mathbf{v})dt
$$
From Lagrange function we can calculate expressions for energy and momentum
$$
\begin{aligned}
    E &= \frac{Mv^2}{2} \\
    \mathbf{P} &= M\mathbf{v}
\end{aligned}
$$
Their transformation law under $\ref{transforms}$ is
$$
\begin{aligned}
    E' &= E - (\mathbf{u}\mathbf{P}) \\
    \mathbf{P}' &= \mathbf{P} - M\mathbf{u}
\end{aligned}
$$
Note that the mass explicitly enters into transformation law of the momentum.
Now let's turn to the case $K\neq 0$. The solution to $\ref{maineq}$ (d'Alambert equation now) is
$$
\Lambda(v^2) = C_1\sqrt{1-Kv^2} + \frac{Q}{K}
$$
Here's the first surprising result -- roles of $Q$ and $C_1$ switched: now $Q$ is an additive constant and can be omitted, while integration constant $C_1$ can be related to the mass in Newtonian limit
$$
\Lambda(v^2 \to 0) \to C_1 - \frac{C_1 K v^2}{2} \implies C_1=-M/K
$$
so finally
$$
\Lambda(v^2) = -\frac{M}{K}\sqrt{1-Kv^2}
$$
The most interesting part is that by setting $Q=0$ we made action invariant under coordinate transformations $\ref{transforms}$. This simply was not a possibility when $K=0$ (unless action is just zero)!
Another remarkable consequence of this is that owning to invariance of action, energy and momentum form a proper (co-)vector (it can be shown that
$E = -{\partial A}/{\partial t}$ and $\mathbf{P} = {\partial A}/{\partial \mathbf{x}}$), hence their transformation law is goverened by $\ref{transforms}$:
$$
\begin{aligned}
    E' &= E - (\mathbf{u}\mathbf{P}) \\
    \mathbf{P}' &= \mathbf{P} - K\cdot E\mathbf{u}
\end{aligned}
$$
note that no external parameters like mass enter into these equations.
Although the derivation above seems clear, to me it still obscures the reason of these differences behind equation $\ref{maineq}$. So I'm wondering if it's possible to give any "higher-level" (maybe, group theory? my knowledge of it is very poor) explanation.
So my questions are:

*

*is there any way to directly connect coordinate transformations $\ref{transforms}$ to the fact that action can be chosen to be invariant when $K \neq 0 $ and can't be when $K=0$?

*from mathematical perspective, does mass $M$ play a special role in Newtonian case?

 A: *

*In OP's derivation we only have a small beef with OP's order of arguments at one step. It seems a bit unwarranted to a priori assume in eq. (3) that $F_u$ does not depend on ${\bf v}$. After all, the change $\Delta L$
in the Lagrangian under infinitesimal boosts only needs to be a total time-derivative $dF_u/dt$ (so that the infinitesimal boost is a quasi-symmetry). In principle it doesn't matter if $\Delta L$ depends on higher time derivatives. However, the fact that (after a calculation)
$$-\Delta L ~=~\underbrace{[2(1-Kv^2)\dot{\Lambda} +K \Lambda]}_{=:~Q}{\bf u}\cdot{\bf v} \tag{A}$$
doesn't depend on higher time derivatives ${\bf a}$, ${\bf j}$, $\ldots$, (together with the assumption that $F_u$ is a local function) do indeed imply that $F_u$ cannot not depend on time derivatives ${\bf v}, {\bf a}, {\bf j}, \ldots$. This in turn implies that $\Delta L$ is an affine function of ${\bf v}$. Since the expression (A) doesn't dependent on ${\bf x}$ and $t$ as well, we conclude that the quantity $Q$ is a constant. So it turns out that OP's solution for $\Lambda$ is correct & complete.
(Alternatively, by using the method of the second part of my Phys.SE answer here, one derives the following necessary condition
$$ 2(1-Kv^2)\ddot{\Lambda}~=~K\dot{\Lambda},\tag{B}$$
whose full 2-parameter solution precisely coincides with OP's solution.)


*By the way, in the case $K\neq 0$, if we allow the Poincare transformations to be a quasi-symmetry, then we don't have to put the constant $Q$ to zero. (Of course, $Q$ drops out of the EL equation.)


*In the Galilean case $K=0$ it is well-known that Galilean boosts are quasi-symmetries rather than strict symmetries of the Lagrangian. This is intimately related to:

*

*(i) the realization of the corresponding algebras of Noether charges;


*(ii) that the Bargmann algebra, i.e. the centrally extended Galilean algebra is an Inonu-Wigner contraction of the centrally extended Poincare algebra. The Lie algebra & representation theory of the latter are easier.
See also e.g. this Phys.SE post and links therein.
