Department of Physics and Astronomy

The Forbes Group

Galilean Covariance

## Galilean Covariance¶

In his "Dialogue Concerning the Two Chief World Systems", Galileo put forth the notion that the laws of physics are the same in any constantly moving (inertial) reference frame. Colloquially this means that if you are on a train, there is no experiment you can do to tell that the train is moving (without looking outside).

This post will explain formally what Galilean covariance means, clarify the difference between covariance and invariance, and elucidate the meaning of Galilean covariance in classical and quantum mechanics. In particular, it will explain the following result obtained by simply changing coordinates, which may appear paradoxical at first:

Consider a modern Lagrangian formulation of a classical object moving without a potential in 1D with coordinates $x$ and $\dot{x}$, and the same object in a moving frame with coordinates $X = x - vt$ and $\dot{X} = \dot{x} - v$. The Lagrangian and conjugate momenta in these frames are:

\begin{align} L[x, \dot{x}, t] &= \frac{m\dot{x}^2}{2}, & p &\equiv \pdiff{L}{\dot{x}} = m\dot{x},\\ L_v[X, \dot{X}, t] &= \frac{m(\dot{X}+v)^2}{2}, & P &\equiv \pdiff{L_v}{\dot{X}} = m(\dot{X}+v) = p. \end{align}

Perhaps surprisingly the conjugate momentum $P$ is the same in the moving frame whereas Galilean invariance implies that one should have a description in terms of $P = m\dot{X} = p - mv$. The later description exists, but requires a somewhat non-intuitive addition to the Lagrangian of a total derivative.

In [1]:
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## TL;DR¶

Consider a modern Lagrangian formulation of a classical object moving without a potential in 1D. The motion of the particle can be expressed as the trajectory $x(t)$, with velocity $\dot{x}(t)$ where the trajectory $x(t)$ is an extremum $S'[x]= 0$ of the classical action

$$S[x] = \int_{0}^{t_f} L[x, \dot{x}, t] \d{t} = \int_{0}^{t_f} \left(\frac{m\dot{x}^2}{2} - V(x)\right) \d{t}, \qquad x(0) = x_0, \qquad x(t_f) = x_f,$$

constrained to fixed starting $x_0$ and ending $x_f$ positions. A couple of remarks:

1. This should not be taken as fundamental. Instead, one should start with Newton's laws, which are supported by a multitude of experimental results. After finding that the Lagrangian $L[x, \dot{x}, t] = m\dot{x}^2/2 - V(x)$ gives Newton's laws through the principle of extremal action, one can then use this formalism to simplify changing variables, for example. That in this case $L=K-V$ is a difference between kinetic and potential energies is not a fundamental property. One needs to find/check $L$ in each physical context.
2. The form of $L$ is not unique. In particular, the conditions of fixed endpoints means that Lagrangians are equivalent if they differ by a total time derivative:

$$L_1 \equiv L_2 + \diff{}{t}f(x, \dot{x}, t).$$

Varying (using Calculus of Variations) yields the following equations of motion (Euler-Lagrange Equations):

\begin{align} L[x, \dot{x}, t] &= \frac{m\dot{x}^2}{2} - V(x), & p &\equiv \pdiff{L}{\dot{x}} = m\dot{x}. \end{align}

(Note: the second equation here provides the definition of the quantity $p$ which we identify as the "momentum". Momentum is not defined as mass times velocity... it just happens to have this form in this simple case.)

Now consider the same situation, but described from a frame moving with constant velocity $v$ such that $\dot{X} = \dot{x} - v$. The standard procedure would be to simply change variables $X(t) = x(t) - vt$, yielding the following Lagrangian:

\begin{align} L^{\text{simple}}_v[X, \dot{X}, t] &= \frac{m(\dot{X}+v)^2}{2} - V(X+vt), & P &\equiv \pdiff{L^{\text{simple}}_v}{\dot{X}} = m(\dot{X}+v) = p. \end{align}

Perhaps surprisingly, the conjugate momentum $P$ is the same in the moving frame whereas one typically expects that Galilean invariance would imply that, in the moving frame, one should have a description in terms of $P = m\dot{X} = p - mv$. To recover this, one needs to include an additional time-dependent piece in the transformation:

\begin{align} L^{\text{partial}}_v[X, \dot{X}, t] &= \frac{m(\dot{X}+v)^2}{2} - V(X+vt) - \overbrace{mv\dot{X}}^{\diff{}{t}(mvX)}, & P &\equiv \pdiff{L^{\text{partial}}_v}{\dot{X}} = m\dot{X} = p - mv. \end{align}

Note that we effect this change by adding a total derivative which therefore does not change the equations of motion. This gives us a new "momentum" $P$, which corresponds to our expectations but is no more fundamental or "correct" than $p$ was above. The following are both valid (and different) physical quantities:

• $p$: The momentum in the original non-moving frame.
• $P = p - mv$: The momentum in the new moving frame.

One can continue in this way to define a "full" transformation, adding one more total derivative to make the Galilean covariance manifest:

\begin{align} L^{\text{full}}_v[X, \dot{X}, t] &= \frac{m(\dot{X}+v)^2}{2} - V(X+vt) \overbrace{- mv\dot{X} - \frac{mv^2}{2}}^{-\diff{}{t}\left(mv\dot{X}t + \frac{mv^2t}{2}\right)}, & P &\equiv \pdiff{L^{\text{full}}_v}{\dot{X}} = m\dot{X} = p - mv,\\ &= \frac{m\dot{X}^2}{2} - V(X+vt). \end{align}

The total derivative subtracted here gives an equivalent Lagrangian that has the same form as old Lagrangian, matching our intuition about the Galilean invariance of Newton's law.

## Details¶

### Boosts¶

The focus here will be on boosts: transforming coordinates to frames moving with constant velocity $\vect{v}$:

$$\vect{x} \rightarrow \vect{X} = \vect{x} - \vect{v} t.$$

(The general Galilean transformation also admits an arbitrary translation, but we will ignore this here, treating time-independent translations separately.)

Colloquially, $\vect{x}(t)$ would describe the motion of a particle in a non-moving frame, while $\vect{X}(t)$ describes the motion of the same particle from the perspective of a moving train that is moving past at constant velocity $\vect{v}$ such that the origins of both coordinate systems coincide at time $t=0$. (This latter restriction is what ignoring the arbitrary translations requires.)

### Covariance vs. Invariance¶

The notion of covariance means that the equations governing a process appear the same in different contexts or frames. For example, Newton's second law

$$F = m\ddot{x}$$

is covariant under the Galilean transformation – it still has the same form $F=ma$ in the new frame:

$$F_v = m\ddot{X}.$$

Note, however, that in general, the quantities in Newton's second law are not Galilean invariant. In particular, the force $F_v$ in the new frame will be different that the force $F$ in the old frame on account of the motion:

$$F_v(X, \dot{X}, t) = F(x, \dot{x}, t) = F(X+vt, \dot{X} + v, t).$$

In the case of a free particle, however, $F_x = F_X = 0$, so Newton's equations are both Galilean covariant and the quantities are Galilean invariant.

The confusion in classical mechanics mentioned in the teaser arises from the fact that the Lagrangian is not Galilean invariant. That the same equations of motion (EQM) follow from a different Lagrangian, however, is no surprise because it is well known that two Lagrangians give rise to the same EQM if they differ by a total time derivative:

$$L[x, \dot{x}, t] = L'[x, \dot{x}, t] + \diff{}{t}f(x, \dot{x}, t).$$

This is the case for a Galilean transformation:

$$L_v[X, \dot{X}, t] = L[x, \dot{x}, t] + m\dot{X} + \frac{m v^2}{2} = L[x, \dot{x}, t] + \diff{}{t}\left(-mxv + \frac{m v^2}{2}t\right).$$

## Classical Mechanics¶

Here is the full set of equations for a 1D system, starting with the Lagrangian, moving through to the Hamiltonian. We will then effect a boost and show the corresponding equations in the new frame:

\begin{gather} L[x, \dot{x}, t] = \frac{m\dot{x}^2}{2} - V(x), \tag{Lagrangian}\\ p = \pdiff{L}{\dot{x}} = m\dot{x}, \tag{Conjugate momentum}\\ \dot{p} = \pdiff{L}{x} = - V'(x) = F, \tag{Euler EqM}\\ H = p\dot{x} - L = \frac{m\dot{x}^2}{2} + V(x) = \frac{p^2}{2m} + V(x), \tag{Hamiltonian}\\ \dot{x} = \pdiff{H}{p} = \frac{p}{m}, \qquad \dot{p} = -\pdiff{H}{x} = -V(x) = F. \tag{Hamiltonian EqM} \end{gather}

### Boosts in the Lagrangian Formalism¶

Now we effect the boost $X = x - vt$, $\dot{X} = \dot{x} - v$. The correct way to change coordinates is to use the same Lagrangian as before:

\begin{gather} L_v[X, \dot{X}, t] = \frac{m(\dot{X}+v)^2}{2} - V(X+vt), \tag{Lagrangian}\\ P = \pdiff{L_v}{\dot{X}} = m(\dot{X}+v) = p, \tag{Conjugate momentum}\\ \dot{P} = \pdiff{L_v}{X} = - V'(X+vt) = F_v, \tag{Euler EqM}\\ H_v = P\dot{X} - L_v = \frac{m(\dot{X}+v)^2}{2} - mv(\dot{X}+v) + V(X+vt) = \frac{P^2}{2m} - Pv + V(X+vt), \tag{Hamiltonian}\\ \dot{X} = \pdiff{H_v}{P} = \frac{P}{m} - v, \qquad \dot{P} = -\pdiff{H_v}{X} = -V'(X+vt) = F_v. \tag{Hamiltonian EqM} \end{gather}

#### Galilean Covariance¶

The previous transformation does not make the Galilean invariance of Newton's laws manifest. In particular, the form of the Hamiltonian is not the same under this transformation, even if the potential is translationally invariant. In particular, for $V(x)=0$ the Hamiltonian becomes:

$$H = \frac{p^2}{2m} \rightarrow H_v = \frac{P^2}{2m} - Pv = \frac{p^2}{2m} - pv = \frac{(p-mv)^2}{2m} - mv^2.$$

Note also that the momentum $P = p$ is invariant under this transformation - it represents the momentum in the rest frame and hence defines a special frame. Together these suggest that to make manifest the Galilean invariance of the theory, one should also shift the momentum $P = p-mv$. Such a transformation is possible, but better expressed as a canonical transformation in the Hamiltonian formulation of classical mechanics. As we shall see, to effect such a transformation, one must also modify the Lagrangian with a total time derivative as alluded to above.

### Canonical Tranformations¶

A boosts can also be implemented via a canonical transformation from a type 3 generating function $S(X, p)$:

$$x = -\pdiff{S}{p}, \qquad P = -\pdiff{S}{X}, \qquad H_v(X, P) = H(x, p) + \pdiff{S}{t}.$$

The difference between a boost and a Galilean transformation can be expressed as:

Simple Boost:

$$S_{v}(X, p) = -p(X+vt),\\ x = X+vt, \qquad P = p, \\ H_v(X, P) = H(X+vt, P) - Pv.$$

Partial Galilean Transformation:

$$S_{I}(X, p) = -p(X+vt) + mvX = -(p-mv)(X+vt) - mv^2 t, \\ x = X+vt, \qquad P = p - mv,\\ H_v(X, P) = H(X+vt, P+mv) - (P+mv)v.$$

Full Galilean Transformation:

$$S_{G}(X, p) = -p(X+vt) + mvX + \tfrac{1}{2}mv^2t = -(p-mv)(X+vt) - \tfrac{1}{2}mv^2t,\\ x = X + vt, \qquad P = p - mv,\\ H_v(X, P) = H(X+vt, P+mv) - (P+mv)v + \tfrac{1}{2}mv^2.$$

#### Hamiltonian¶

How consider a quadratic Hamiltonian with a translational invariance $V=0$:

\begin{align} H(p, x) &= \frac{p^2}{2m} & &\rightarrow & H_v(X, P) &= \begin{cases} \frac{P^2}{2m} - Pv , & \text{(simple boost)}\\ \frac{P^2}{2m} - \frac{mv^2}{2} & \text{(partial Galilean)}\\ \frac{P^2}{2m} & \text{(full Galilean)} \end{cases} \end{align}

With the full Galilean transformation, we see that we have made the Hamiltonian manifestly Galilean invariant - it retains its form with the new coordinates.

#### Lagrangian¶

We can also go back and express the Lagrangian $L_v = P\dot{X} - H_v$. Note: here we consider only quadratic dispersion $H(p, x) = p^2/2m + V(x,t)$ as the general case is complicated by the problem of inverting the dispersion:

Simple Boost: $$P = m(\dot{X} + v),\\ L_v[X, \dot{X}, t] = \frac{m(\dot{X}+v)^2}{2} - V(X+vt) = L[x, \dot{x}, t],\\ L[x, \dot{x}, t] = L_v[X, \dot{X}, t]$$

Partial Galilean Transformation:

$$P = m\dot{X}, \\ L_v[X, \dot{X}, t] = \frac{m\dot{X}^2}{2} - V(X+vt) + \frac{mv^2}{2} = L[x, \dot{x}, t] - \diff{}{t}(mxv - mv^2 t), \\ L[x, \dot{x}, t] = L_v[X, \dot{X}, t] + \diff{}{t}(mXv + mv^2 t)$$

Full Galilean Transformation:

$$P = m\dot{X}, \\ L_v[X, \dot{X}, t] = \frac{m\dot{X}^2}{2} - V(X+vt) = L[x, \dot{x}, t] - \diff{}{t}\left(mxv - \frac{mv^2}{2} t\right),\\ L[x, \dot{x}, t] = L_v[X, \dot{X}, t] + \diff{}{t}\left(mXv + \frac{mv^2}{2} t\right)$$

### Generalized Dispersion (aka Negative Mass)¶

For fun, let's consider a more general problem of a particle with generalized dispersion $E(p)$. The appropriate formalism to start with here is the Hamiltonian framework:

$$H(x, p, t) = E(p) + V(x)\\ \dot{x} = \pdiff{H}{p} = E'(p), \qquad \dot{p} = -\pdiff{H}{x} = -V'(x) = F.$$

To obtain Newton's second law, we differentiate the last equation:

$$\ddot{x} = \dot{p}E''(p) = FE''(p), \qquad F = \frac{1}{E''(p)}\ddot{x} = m(p) a$$

showing that $E''(p) \equiv 1/m(p)$ plays the role of a momentum dependent effective mass.

The Lagrangian formulation of this problem is not very transparent:

$$p = E'^{-1}(\dot{x}), \qquad L(x, \dot{x}) = p\dot{x} - H = \dot{x}E'^{-1}(\dot{x}) - E\Bigl(E'^{-1}(\dot{x})\Bigr) - V(x).$$

This only has the familiar form $L = E - V$ if $\dot{x}E'^{-1}(\dot{x}) - E\Bigl(E'^{-1}(\dot{x})\Bigr) = E\Bigl(E'^{-1}(\dot{x})\Bigr)$ or, going back to the variable $p$:

$$pE'(p) = 2E(p),\qquad \d{\ln E} = 2\d{\ln p}, \qquad E(p) \propto p^2.$$

Thus, the procedure of constructing the Lagrangian as $L = E - V$ only works for quadratic dispersions.

To transform to a frame moving with velocity $v$ with coordinate $X = x - vt$ we use the simple boost $S_{v}(X,p) = -p(X+vt)$:

$$x = -\pdiff{S}{p} = X+vt, \qquad P = -\pdiff{S}{X} = p,\\ H_v(X,P) = H(x,p) + \pdiff{S}{t} = H(x,p) - pv = E(P) - Pv + V(X+vt)\\ \dot{P} = - V'(X+vt), \qquad \dot{X} = E'(P) - v.$$

Note that boosting to a moving frame is equivalent to working with a new effective dispersion relation

$$E_{v}(P) = E(P) - Pv.$$

One might ask, is there a canonical transformation that takes us from the variables $(x, p)$ to $(X=x-vt, P=p-mv = p - v/E''(p))$ as expected? Recall that a canonical transformation must satisfy:

$$\pdiff{P(x,p)}{p} = \pdiff{x(X,P)}{X}, \qquad \pdiff{P(x,p)}{x} = -\pdiff{p(X,P)}{X},\\ 1 + \frac{v E'''(p)}{[E''(p)]^2} = 1, \qquad 0 = 0,$$

Thus, we see that such a transformation will only be canonical if $E'''(p) = 0$ or, again, if $E(p) \propto p^2$ is quadratic. In this case $m(p) = 1/E''(p) = m$ is constant, and we can effect the desired transformation with the type 3 generating function $S(X,p) = -(p-mv)(X + vt)$ which gives:

$$x = -\pdiff{S}{p} = X+vt, \qquad P = -\pdiff{S}{X} = p-mv, \\ H_v(X,P) = H(x,p) + \pdiff{S}{t} = H(x,p) - (p-mv)v = E(P+mv) - Pv + V(X+vt)\\ \dot{P} = - V'(X+vt), \qquad \dot{X} = E'(P+mv) - v.$$

However, there is no link here between $m$ in the transformation and $1/E''(p)$ appearing as the effective mass.

## Quantum Mechanics¶

We now consider the Schrödinger equation:

$$\I\hbar\dot{\psi}(x, t) = -\frac{\hbar^2\psi''(x, t)}{2m} + V(x, t)\psi(x, t) = \left(\frac{\op{p}^2}{2m} + V(x, t)\right)\psi(x, t).$$

We may again implement the change of variables $X = x-vt$. The difference between a boost and a full Galilean transformation is the inclusion of a phase in the transformed wavefunction:

$$\psi_v(X, t) = e^{-\I\phi(X, t)}\psi(x, t) = e^{-\I\phi(X, t)}\psi(X-vt, t)$$

Here are the three transformations:

Simple Boost: $$\hbar\phi(X, t) = 0, \\ \I\hbar\dot{\psi}_v(X, t) = \left(\frac{\op{p}^2}{2m} - \op{p}v + V(X+vt)\right)\psi_v(X, t).$$

Partial Galilean Transformation:

$$\hbar\phi(X, t) = mXv,\\ \I\hbar\dot{\psi}_v(X, t) = \left(\frac{\op{p}^2}{2m} - \frac{1}{2}mv^2 + V(X+vt)\right)\psi_v(X, t).$$

Full Galilean Transformation:

$$\hbar\phi(X, t) = mXv + \tfrac{m}{2}v^2t, \\ \I\hbar\dot{\psi}_v(X, t) = \left(\frac{\op{p}^2}{2m} + V(X+vt)\right)\psi_v(X, t).$$

## Summary¶

When one needs to work in a moving frame, there are two natural choice: a simple boost changing to the variable $X = x-vt$ and a complete Galilean transformation that implements this boost, but also changes the momentum (classical) or phase of the wavefunction (quantum). The latter helps if you are working with a Galilean covariant theory such as with a quadratic dispersion as the resulting Hamiltonian look the same, but if you have a nontrivial dispersion, for example, then just do a simple boost.