Galilean invariance of Lagrangian for non-relativistic free point particle?

Question

In QFT, the Lagrangian density is explicitly constructed to be Lorentz-invariant from the beginning. However the Lagrangian

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

for a non-relativistic free point particle is not invariant under Galilean transformation. This does not ultimately matter because the difference is a total time derivative.

However, is it possible to exhibit a Galilean invariant Lagrangian for a non-relativistic free point particle?

Answer 1

Theorem Schmeorem. A Galilean invariant Lagrangian for any number of classical particles interacting with a potential:

$$ S = \int \sum_k {m_k(\dot{x}_k-u)^2\over 2} + \lambda \dot{u} - U(x_k)\;\;\; dt $$

For any Galilean invariant Lagrangian $L(\dot{x}_k, x_k)$, the Lagrangian

$$ L'(\dot{x}_k,x_k, \lambda, u) = L(\dot{x}_k-u,x_k) + \lambda \dot{u} $$

is explicitly Galilean invariant, and has the same dynamics (assuming the original Lagrangian was Galilean invariant).

The Galilean properties of the x's are as usual. The dynamical variables extend to include $\lambda,u$ which act as Lagrange multipliers. The transformation law for u and $\lambda$ are:

$ x \rightarrow x-vt $

$ u \rightarrow u-v $

$ \lambda \rightarrow \lambda $

And it is trivial to verify that the new Lagrangian is completely invariant. The equation of motion for $\lambda$ just makes $u$ constant, equal to $u_0$, while the equation of motion for $u$ integrates to

$$ \lambda = - \sum_k m_k x_k - M u_0 t $$

up to an additive constant which I have set to zero. This is almost all the equations of motion, but there is one more equation which comes from extremizing the action with respect to $u_0$, which sets

$$ u_0 = \sum_k m_k \dot{x}_k $$

Where the time is unimportant, because this is the center of mass velocity, which is conserved. The Noether prescription in the explicitly Galilean invariant action is trivial--- the conserved quantity associated with Galilean boosts is just $\lambda$, and this is indeed the center of mass position.

Why this works

If you integrate the kinetic energy for the usual free particle action by parts, you get:

$$S = \int \sum_k m\ddot{x}_k x_k + U(x_k) dt$$

This action is Galilean invariant on mass shell, meaning that the non-galilean invariant part is zero when you enforce the equations of motion. This means that adding some additional nondynamical fields should produce a Galilean invariant action off shell, and this is the $\lambda, u$.

Relation to Lorentz transformations

When you perform a Lorentz transformation, the arclength particle action is invariant. But if you fix the origin of the Lorentz transformation on the initial time, the final time is transformed, so the path is not going to the same final time anymore after the transformation. When you take the non-relativistic limit, the final time becomes degenerate with the initial time, but the action cost from shifting the final time does not approach zero.

This means that you need an extra variable to keep track of the infinitesimal bit of final time, and that this extra variable will need a nontrivial transformation law under Galilean transformations.

To find out what this new variable should be, it is always best to consider the analogous thing for rotational invariance. Consider a string at tension with small deviations from horizontal, and let the deviation of the string from horizontal be h(t). The rotationally invariant potential energy is the arclength of the string

$$ U = \int \sqrt{1+h'^2} dx $$

and this is the potential energy which gives the rotationally invariant analog of the wave equation. Once you go to small deviations, the expansion for U gives the usual wave-equation potential energy

$$ U(h) = \int {1\over 2} h'^2 dx $$

and this is no longer rotationally invariant. But it is skew-invariant, meaning adding a constant slope line to h does not change the energy. Except that it does, by a perfect derivative:

$$ U(h + ax) = \int {1\over 2}h'^2 + a h' + {a^2\over 2} dx$$

This is clearly the same exact situation as for the Lorentz invariance turning into Galilean invariance, except using rotational invariance, where everybody's intuition is firm. The additional $a^2\over 2$ energy is due to the quadratic extra length of a rotated string, while the linear perfect derivative $ah'$ integrates to $a (h_f - h_i)$, and this is the amount of reduction/increase in length when you rotate a tilted string.

So to get a fully tilt-invariant potential energy, you need to add a variable $u$ which is dynamically constrained to equal the total tilt of the string. This variable will distinguish between different rotated versions of the string: rotating the string by itself without rotating the average tilt variable will change the energy--- this is because tilting the horizontal string between 0 and A is not quite the same as the pre-tilted string between 0 and A, the pre-tilted string has a different length. Rotating the total tilt by itself will change the energy, but rotating both does nothing, and this is the encoding of rotatonal invariance.

So you need an average tilt variable to turn explicit rotational invariance to explicit tilt invariance. The total potential energy is then given by the deviations from the average tilt:

$$ U = \int {1\over 2} (h'-u)^2 dx $$

and u transforms as $u-a$ under a tilt by a. This makes the potential energy invariant.

The kinetic energy is given by the time dependence of h, and there must be a Lagrange multiplier to enforce that the total tilt is equal to the average tilt

$$ S = \int {1\over 2} \dot h^2 - {1\over 2} (h'-u)^2 + \beta (u - h') dt dx $$

Where $\beta$ is a global in x Lagrange multiplier for u, forcing it to equal h'. But it does no harm to allow u to vary in x, so long as the Lagrange multiplier enforces that it is constant. The way to do this is to change the Lagrange multiplier term to

$$ - \int \lambda' (u(x) - h'(x)) dx = \int \lambda (u'(x) - h''(x)) $$

But then the equation of motion kills the second term, so you only need a Lagrange multiplier to be:

$$ \int \lambda u'(x)$$

And the equations of motion automatically constrain u to be the average slope. These manipulations have exact analogs in Lorentz transformations, and explain the relation of the explicitly Galilean invariant action to the Lorentz action. The analog of average slope is the center of mass velocity.

Answer 2

Here I would like to expand some of the arguments given in Ron Maimon's inspiring answer. Consider $N$ point particles with positions ${\bf r}_1, \ldots, {\bf r}_N$. The Galilean transformation group is, e.g., explained here. The only transformation, which we will bother to mention explicitly from now on, is the shear transformation

$$\begin{align} t ~\longrightarrow~& t,\cr \qquad {\bf r}_i ~\longrightarrow~& {\bf r}_i- {\bf v}t,\end{align}\tag{1}$$

where ${\bf v}$ is the relative constant velocity of the two reference frames.

1. Let us start from the manifestly Galilean invariant Lagrangian

$$\begin{align} L_1~=~&\sum_{i=1}^{N} \frac{m_i}{2}(\dot{\bf r}_i-{\bf u})^2 + {\bf \lambda} \cdot \dot{\bf u} - V, \cr V~:=~& \sum_{1\leq i<j\leq N}V_{ij}(|{\bf r}_i-{\bf r}_j|),\end{align}\tag{2}$$

where ${\bf u}={\bf u}(t)$ and ${\bf \lambda}={\bf \lambda}(t)$ are a canonical pair of additional variables. The Galilean transformation reads

$$\begin{align} t ~\longrightarrow~& t, \cr {\bf r}_i ~\longrightarrow~& {\bf r}_i- {\bf v}t, \cr {\bf u}~\longrightarrow~& {\bf u}-{\bf v}, \cr {\bf \lambda} ~\longrightarrow~& {\bf \lambda}.\end{align}\tag{3}$$

2. Next let us integrate out the Lagrange multiplier ${\bf \lambda}$. The equation of motion(=eom) for ${\bf \lambda}$ is $\dot{\bf u} \approx 0$. (The $\approx$ sign means in this answer equal modulo eom.) This leaves a zero-mode ${\bf u}_0$, which is independent of $t$. The new Lagrangian

$$ L_2~=~\sum_{i=1}^{N} \frac{m_i}{2}(\dot{\bf r}_i-{\bf u}_0)^2- V\tag{4} $$

is still manifestly Galilean invariant. The Galilean transformation reads

$$\begin{align} t ~\longrightarrow~& t, \cr {\bf r}_i ~\longrightarrow~& {\bf r}_i- {\bf v}t, \cr {\bf u}_0~\longrightarrow~& {\bf u}_0-{\bf v}.\end{align}\tag{5}$$

The eom for ${\bf r}_i$ are Newton's second law as it should be:

$$\begin{align} m_i \ddot{\bf r}_i~\approx~& -\nabla_i V, \cr i~=~&1,\ldots, N.\end{align}\tag{6} $$

We conclude that

The two Lagrangians $L_1$ and $L_2$ are affirmative answers to the OP's question(v1).

3. Finally let us integrate out the zero-mode ${\bf u}_0$. The eom for ${\bf u}_0$ reads

$$\begin{align} {\bf u}_0 ~\approx~& \frac{\sum_{i=1}^{N}m_i\dot{\bf r}_i}{M}, \cr M~:=~& \sum_{i=1}^{N}m_i. \end{align}\tag{7}$$

The new Lagrangian

$$ L_3~=~\sum_{i=1}^{N} \frac{m_i}{2}\left(\dot{\bf r}_i-\frac{\sum_{j=1}^{N}m_j\dot{\bf r}_j}{M}\right)^2- V \tag{8}$$

is still manifestly Galilean invariant. The eom for ${\bf r}_i$ are Newton's second law with a center-of-mass subtraction:

$$\begin{align} m_i \ddot{\bf r}_i-\frac{m_i}{M}\sum_{j=1}^{N}m_j\ddot{\bf r}_j~\approx~& -\nabla_i V, \cr i~=~&1,\ldots, N.\end{align}\tag{9}$$

This is still consistent with Newton's second law, as we know that the center-of-mass(=CM) of an isolated system must have zero acceleration:

$$\begin{align}\ddot{\bf r}_{CM}~=~&\frac{\sum_{j=1}^{N}m_j\ddot{\bf r}_j}{M}\cr~\approx~&{\bf 0}.\end{align}\tag{10}$$

But $L_3$ does not produce these 3 eom, which determine the CM motion. We conclude that

The Lagrangian $L_3$ is not an answer to the OP's question(v1).

This becomes particularly clear if we choose only one particle $N=1$. Then the Lagrangian $L_3$ vanishes identically $L_3=0$.

Answer 3

The answer is negative. There is no action of the free particle invariant under the Galilean group. In the following, a heuristic explanation will be given and in addition a reference where a more detailed proof is provided.

The basic reason is that the Galilean group cannot be realized on the Poisson algebra of functions on the phase space of the free particle $T^{*}\mathbb{R}^3$ (equipped with the canonical symplectic form). It is only its central extention (Please, see the following Wikipedia page) which can be realized in terms of Poisson brackets. For this central extension, the Poisson brackets between the generators of the boosts $B_i$ and the translations (i.e., the components of the momentum) $P_i$ no longer vanishes but depends on the mass of the particle:

$\{B_i, P_j\} = m\delta_{ij}$.

Since boosts must generate the transformation: $ P_i \rightarrow P_i + m v_i $ on the momentum coordinates via the canonical Poisson brackets, the Boost generators have to be realized as multiples of the position coordinates $Q_i$.

$B_i = m Q_i$

The transformation law of the Boosts on the phase space (which is the manifold of the initial data, thus this realization does not involve time):

$ Q_i \rightarrow Q_i $

$ P_i \rightarrow P_i + m v_i $

$ H(\vec{P}) \rightarrow H(\vec{P}+m\vec{v}) - \vec{P}.\vec{v}-\frac{1}{2}m v^2 $

It is easy to verify that the free particle Hamiltonian is invariant and its transformation satisfies a group law. But this realization still does not make the Lagrangian $ L = \vec{P} .\dot{\vec{Q}} - H$ invariant, because the Cartan-Poincare form: $ \vec{P}.d\vec{Q} $ is not invariant and changes by a total derivative: $m d \vec{v}.\vec{Q}$. Thus the existence of mass prevents the action from being invariant, because of the canonical Poisson bracket and not because of the choice of the dynamics through the particular choice of the Hamiltonian.

The noninvariance of the Cartan-Poincare form under the boosts is referred to as non-equivariance of the momentum maps associated to the Boosts, which indicates that we cannot redefine the group generators so the Poisson bracket between the Boosts and translation generators vanishes. Please see pages 430-433 and exercise 12.4.6 in "Introduction to mechanics and symmetry", by Marsden and Ratiu for a rigorous proof.

Answer 4

Here we would like to expand some of the arguments given in David Bar Moshe's inspiring answer. In particular, we will argue that$^1$

The natural non-relativistic Lie algebra in Newtonian mechanics is the Bargmann algebra, not the Galilean algebra!

1. We start from the relativistic Hamiltonian Lagrangian $$ L_H ~=~{\bf p}\cdot \dot{\bf x} - H , \tag{1}$$ for a spinless point particle in $n$ spacetime dimensions in the static gauge, cf. e.g. my Phys.SE answer here. Let the Hamiltonian$^2$ $$ \begin{align} H~:=~&p^0c-mc^2\cr ~\stackrel{(3)}{\approx}~&\sqrt{{\bf p}^2c^2+m^2c^4}-mc^2\cr \quad\longrightarrow &\quad H_{\infty}~:=~\frac{{\bf p}^2}{2m}\quad\text{for}\quad c\to \infty \end{align}\tag{2} $$ be the kinetic energy, i.e. the energy $$p^0c~\approx~\sqrt{{\bf p}^2c^2+m^2c^4}\tag{3} $$ minus the rest energy $mc^2$. Note that a constant term $mc^2$ in the Lagrangian does not affect the Euler-Lagrange (EL) equations. We are interested in the non-relativistic limit $c\to \infty$. OP's non-relativistic Lagrangian can be obtained from the non-relativistic Hamiltonian Lagrangian (1) by integrating out the $3$-momenta ${\bf p}$, i.e. by Legendre transformation.

2. The $n(n\!-\!1)/2$ Lorentz generators are $$ \begin{align}J^{\mu\nu}~=~& x^{\mu} p^{\nu} - x^{\nu} p^{\mu}, \cr \mu,\nu~\in~&\{0,1,2,\ldots, n\!-\!1\}.\end{align}\tag{4}$$ We have $$ \frac{p^0}{c} \quad\stackrel{(3)}{\longrightarrow}\quad m\quad\text{for}\quad c\to \infty.\tag{5}$$ The $n\!-\!1$ boost generators are$^3$ $$\begin{align} B^i~:=~&\frac{J^{0i}}{c}\cr ~\stackrel{(4)}{=}~&tp^i-x^i\frac{p^0}{c}\cr \quad\stackrel{(5)}{\longrightarrow}&\quad B_{\infty}^i~:=~tp^i-mx^i\quad\text{for}\quad c\to \infty.\end{align} \tag{6}$$

3. Consider the canonical Poisson brackets $$\begin{align}\{x^{\mu}, x^{\nu}\}~=~&0, \cr \{x^{\mu}, p^{\nu}\}~=~&\eta^{\mu\nu}, \cr \{p^{\mu}, p^{\nu}\}~=~&0, \cr \mu,\nu~\in~&\{0,1,2,\ldots, n\!-\!1\}.\end{align} \tag{7}$$ It is well-known that the Poisson brackets of the Poincare generators $$\begin{align} p^{\mu},J^{\mu\nu},& \cr \mu,\nu~\in~&\{0,1,2,\ldots, n\!-\!1\}.\end{align}\tag{8} $$ reproduce the $n$-dimensional Poincare algebra $iso(n\!-\!1,1)$. The mass $m$ has by definition vanishing Poisson bracket with any element $$ \{m,\cdot\}~=~0. \tag{9}$$

4. It is easy to check that the Poisson algebra with the $n(n\!+\!1)/2\!+\!1$ generators $$\begin{align} H, p^i, J^{ij},& B^i, m, \cr i,j~\in~&\{1,2,\ldots, n\!-\!1\},\cr c~<~&\infty,\end{align}\tag{10}$$ in the non-relativistic limit $c\to \infty$ becomes the Bargmann algebra $$\begin{align}\{ B^i,H\}~\stackrel{(2)+(6)}{=}&~p^i,\cr \{ p^i, B^j\}~\stackrel{(6)+(5)}{=}&~m\delta^{ij}, \cr \ldots,& \end{align}\tag{11}$$ i.e. the centrally extended Galilean algebra. [The ellipsis $\ldots$ in eq. (11) indicates other well-known commutation relations, which we won't repeat here. Unlike the other generators (10), the mass $m$ is not a function of time/a dynamical variable.] The Bargmann algebra is an Inonu-Wigner contraction of $$iso(n\!-\!1,1)\oplus u(1).\tag{12} $$
   Here $iso(n\!-\!1,1)$ is the Poincare algebra (8), while $u(1)$ is an algebra generated by the mass generator $m$, which belongs to the center, cf. eq. (9). Note that the Poisson brackets for the Bargmann algebra continue to hold if we instead use the non-relativistic expressions $H_{\infty}$ and $B_{\infty}^i$. For a related argument, see also Ref. 1.

5. In the rest of this answer we will assume that $c=\infty$. One may check that each Bargmann generator $Q$ satisfies the off-shell identity $$ \{Q,H\} +\frac{\partial Q}{\partial t}~=~0.\tag{13}$$ This means that the Bargmann generator $Q$ generates an infinitesimal quasisymmetry $$\delta ~=~ \{\cdot ,Q\}\epsilon\tag{14}$$ of the Hamiltonian Lagrangian (1), cf. my Phys.SE answer here.

6. Example. The $B^i$ boost generators (6) generate infinitesimal Galilean transformations. They are quasisymmetries $$\begin{align} \{L_H, B^i\}~\stackrel{(1)}{=}~&p_j\frac{d}{dt}\{x^j, B^i\}+\{p_j, B^i\}\dot{x}^j-\{H, B^i\}\cr~\stackrel{(6)}{=}~&\frac{df^i}{dt}, \cr f^i~=~&m x^i,\end{align}\tag{15}$$ with themselves as Noether charges: $$\begin{align} Q^i~=~&\frac{\partial L_H}{\partial \dot{x}^j} \{x^j, B^i\} -f^i\cr ~\stackrel{(1)+(6)+(15)}{=}&~ tp^i-mx^i~\cr\stackrel{(6)}{=}~&B^i. \end{align}\tag{16}$$

References:

1. R. Andringa, E. Bergshoeff, S. Panda & M. de Roo, Newtonian Gravity and the Bargmann Algebra, arXiv:1011.1145, p. 11.

--

$^1$ For OP's initial inquiry into whether it is possible to obtain strict symmetry of the Lagrangian rather than only quasisymmetry, we refer to the other answers.

$^2$ Notations & Conventions: Greek indices $\mu,\nu=0,1,2,\ldots, n\!-\!1 $, denote spacetime indices; while Roman indices $i,j=1,2,\ldots, n\!-\!1 $, (and boldface) denote spatial indices. The $\approx$ symbol denotes an on-shell relation. The Minkowski sign convention is $(−,+,\ldots,+)$. And $x^0\equiv ct.$

$^3$ As usual in canonical quantization, the static gauge-fixing condition $t=\lambda$ (of world-line reparametrization invariance) breaks manifest Poincare symmetry. However, the boost quasisymmetry gets restored in the non-relativistic limit $c\to \infty$.

Answer 5

Yes. $$S = \int m\frac{|\dot{𝐫}|^2 + 2\dot{u}}2 dt,$$ with the Galilean transform: $$(𝐫,t,u) → \left(𝐫 - 𝐯t, t, u + 𝐯·𝐫 - \frac{|𝐯|^2}2t\right),$$ and the transform: $$(\dot{𝐫},\dot{u}) → \left(\dot{𝐫} - 𝐯, \dot{u} + 𝐯·\dot{𝐫} - \frac{|𝐯|^2}2\right)$$ as corollaries, where $m$ is a Lagrange multiplier. The Euler-Lagrange equations are $$\frac{d(m\dot{𝐫})}{dt} = 𝟬,\quad \frac{dm}{dt} = 0,\quad \frac{|\dot{𝐫}|^2 + 2\dot{u}}2 = 0.$$

But, wait: there's more! A surprise is in store.
This can be deformed to the following: $$S = \int m\frac{|\dot{𝐫}|^2 + 2\dot{u} + α\dot{u}^2}2\frac{dt}{1 + α\dot{u}}.$$ The Euler Lagrange equations are now: $$\frac{d}{dt}\left({\frac{m\dot{𝐫}}{1 + α\dot{u}}}\right) = 𝟬,\quad \frac{d}{dt}\left(m\left(1 - α\frac{|\dot{𝐫}|^2 + 2\dot{u} + α\dot{u}^2}{2(1 + α\dot{u})^2}\right)\right) = 0,\quad \frac{|\dot{𝐫}|^2 + 2\dot{u} + α\dot{u}^2}{2(1 + α\dot{u})} = 0.$$ The last of these equations can be used to simplify the Euler-Lagrange equation for $u$ to: $$\frac{dm}{dt} = 0.$$ Since $$(1 + α\dot{u})^2 = 1 - α|\dot{𝐫}|^2$$ also follows from the last of these Euler-Lagrange equations, then the Euler-Lagrange equation for $𝐫$ can be rewritten as: $$\frac{d}{dt}\left(\frac{m\dot{𝐫}}{\sqrt{1 - α|\dot{𝐫}|^2}}\right) = 𝟬.$$

When $α = 0$, this is the result already derived for Galilean symmetry. When $α > 0$, the result is the mechanics for a relativistic body, where light speed is given by $c = 1/\sqrt{α}$. In that case, the transform is the Lorentz transform - extended to $u$ - which is given in infinitesimal form by $$(Δ𝐫, Δt, Δu) = (-𝞄t, -α𝞄·𝐫, 𝞄·𝐫).$$

The Hamiltonian is given by $$\begin{align} H &= \left(\frac{m\dot{𝐫}}{1 + α\dot{u}}\right)·\dot{𝐫} + \left(m\left(1 - α\frac{|\dot{𝐫}|^2 + 2\dot{u} + α\dot{u}^2}{2(1 + α\dot{u})^2}\right)\right)\dot{u} - \frac{m}{2}\frac{|\dot{𝐫}|^2 + 2\dot{u} + α\dot{u}^2}{1 + α\dot{u}}\\ &= \frac{m}{2}\frac{|\dot{𝐫}|^2 - α\dot{u}^2}{(1 + α\dot{u})^2}. \end{align}$$

On-shell, this is equivalent to: $$H = -\frac{m\dot{u}}{1 + α\dot{u}} = \frac{m}{\sqrt{1 - α|\dot{𝐫}|^2}}\frac{|\dot{𝐫}|^2}{1 + \sqrt{1 - α|\dot{𝐫}|^2}}.$$ For $α = 0$, this is just: $$H = \frac{m|\dot{𝐫}|^2}{2}.$$ For $α = 1/c^2 > 0$, it is: $$H = \frac{m}{a}\left(\frac{1}{\sqrt{1 - α|\dot{𝐫}|^2}} - 1\right) = \frac{mc^2}{\sqrt{1 - |\dot{𝐫}|^2/c^2}} - mc^2.$$