# Is Galilean boost actually a gauge transformation?

In elementary physics, it is well-known that the Newton's law $$\vec{F}=m\vec{a}$$

is invariant under Galilean transformations. However, Galilean relativity is not introduced in details in ordinary textbooks on Quantum Mechanics and Classical Mechanics. Naively, one usually treats the Galilean group as a spacetime symmetry (i.e global and physical symmetry), just as Lorentzian symmetry in QFT.

In relativistic QFT, one usually applies the Noether's theorem to find conserved quantities associated with these global symmetries, such as energy, momentum and angular momentum etc, because Lorentz transformations are rigid.

It is, however, not the case in a generic Galilean transformation, which is shown below $$t^{\prime}=t+a,\quad\vec{r}^{\prime}=R\vec{r}+\vec{v}t+\vec{b},$$

where $$a\in\mathbb{R}$$, $$\vec{v}$$, $$\vec{b}\in\mathbb{R}^{3}$$, and $$R\in O(3)$$.

In the above equations, a pure Galilean boost clearly depends on time parameter. In other words, the transformation of a classical trajectory in the Lagrangian $$L(\dot{\vec{r}})$$ explicitly depends on $$t$$, and so this is a local (i.e gauge) transformation.

To be specific, I am talking about the following transformation of a trajectory of a classical particle (I am not talking about field theory) $$\vec{r}(t)\rightarrow\vec{r}(t)+\vec{v}t,$$

which clearly depends on the parameter $$t$$. To be even more specific, this transformation takes the form $$\vec{r}(t)\rightarrow\vec{r}(t)+\vec{f}(t),$$

where $$\vec{f}(t)=\vec{v}t$$ is a linear function of $$t$$. In other words, this transformation in Classical Mechanics is not rigid.

To avoid further irrelavent information in the comment section, please notice that the $$c\rightarrow\infty$$ limit of the Lorentz group is not the Galilei group. This was clarified by David Bar Moshe and Qmechanic ♦ here. Unlike the Lorentz group, the Galilei group has a non-trivial central extension, parameterized by the mass of the free Newtonian particle. When one lifts its projective representation, one ends up with the linear representation of its universal central extension.

However, unlike gauge transformations in QED which leaves the Lagrangian density invariant, a Galilean boost changes the Lagrangian of a non-relativistic free particle by a total derivative.

In QFT, there's a huge difference between a physical (global) symmetry and a gauge (local) redundancy. Physical symmetries are in irreducible unitary representations that map states into some other states, while gauge redundancies are in trivial representations that leave the quantum states invariant. So what is going on here with Galilean transformations? It's Eucliean subgroup is clearly physical, but its pure boost seems almost like a gauge transformation.

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– Chris
Apr 20, 2022 at 14:48

The Galilean spacetime is $$\mathbb{R}\times \mathbb{R}^3$$. The Galilean group is the group of all galilean transformations $$(R,\vec{v},\vec{a},s)$$ where $$R\in {\rm SO}(3)$$, $$\vec{v},\vec{a}\in \mathbb{R}^3$$ and $$s\in \mathbb{R}$$, with the property that an event $$(t,\vec{x})$$ is transformed into $$(R,\vec{v},\vec{a},s)\cdot (t,\vec{x})=(R\vec{x}+\vec{v}t+\vec{a}, t+s).\tag{1}$$

Now the distinction between local and global symmetries lies in whether the parameters specifying the transformation are constants or functions. In other words: does the transformation act in the same manner everywhere or it acts differently at each point?

As you can see from (1), the transformation parameters, $$R$$, $$\vec{v}$$, $$\vec{a}$$ and $$s$$ are all fixed. They do not vary from point to point. This is not a local transformation in any way.

Of course $$t$$ appears on the right-hand side of (1), but it is because $$t$$ is part of the data specifying the point upon which the transformation acts. A global symmetry has parameters which do not depend on the point, but of course the result of a particular application to a particular point will depend on the point.

Contrast this to the gauge transformations of QED. Thinking about the action on the gauge potential $$A_\mu(x)$$ it is $$A_\mu(x)\mapsto A_\mu(x)+\partial_\mu \lambda(x)\tag{2}.$$

The parameter that specifies the transformation is $$\lambda(x)$$ and it varies from point to point. (2) is local, (1) is not.

• Can you write down a pure Galilean boost and show me that it does not depend on $t$? Apr 20, 2022 at 4:44
• Let's start from a classical trajectory $\vec{x}(t)$. A pure Galilean boost transforms it to $\vec{x}(t)\rightarrow\vec{x}(t)+\vec{v}t$. Can you show me that this transformation is not $t$-dependent? Apr 20, 2022 at 4:50
• A transformation is specified by a group element, and it is local when the group element depends on the point at which the transformation is applied. Here the group element is specified by $\vec{v}$ and $\vec{v}$ does not depend on $t$.
– Gold
Apr 20, 2022 at 13:57
• To compare to QED consider the global ${\rm U}(1)$ acting on the fermion: $\psi(x)\to e^{i\lambda}\psi(x)$. Infinitesimaly it is $\psi(x)\to \psi(x)+i\lambda \psi(x)$. It is a global ${\rm U}(1)$ because $\lambda \neq \lambda(x)$. Still, $\psi(x)$ does depend on $x$. You saying that $\vec{x}(t)\to \vec{x}(t)+\vec{v}t$ is a local transformation because the shift is $\vec{v}t$ depending on $t$ is like saying that the fermion transformation here is local because the shift is $i\lambda \psi(x)$ depending on $x$ through $\psi$.
– Gold
Apr 20, 2022 at 13:58
• An infinity Galilean boost does not depend on $t$. This Galilean boost is a large transformation. Apr 20, 2022 at 14:41