# Does Galilean relativity constitute a dynamical symmetry or an isometry?

There are many papers which derive the form of the Lorentz transform from elementary symmetry principles (usually homogeneity of spacetime, isotropy of space, and the fact that boosts form a group), e.g. here or here (warning: PDF). These derivations result in an expression for the Lorentz factor of the form $$\gamma(v) = \frac{1}{\sqrt{1 - K v^2}},$$ where $K$ is some non-negative constant. When $K$ is strictly positive, the Lorentz transform is recovered, whereas when $K$ is zero, the Galilean transform is recovered. This style of derivation implies that the Galilean transform and the Lorentz transform occupy similar roles in non-relativistic and relativistic physics, respectively.

However, in relativity, boosts between reference frames (which are described by the Lorentz transform) are part of the isometry group of the Minkowski metric. Galilean boosts, on the other hand, are never (to my knowledge) presented as isometries of the Euclidean metric; in fact, I'm having difficulty showing that a Galilean boost is an isometry.

The other option is that Galilean relativity (i.e. invariance under Galilean boosts) is actually a dynamical symmetry (i.e. a symmetry of the equations of motion, specifically the Euler-Lagrange equations), and not an isometry. Is this the case, and if so, can invariance under Lorentz boosts be similarly conceived as a dynamical symmetry (of the E-L equations for a relativistic field, perhaps)?

When you derive coordinate transformations based on spacetime symmetries, you get 3 results, not just 2. In addition to Lorentz and Galilean, you also get the Euclidean transformation for K<0. This transformation is ruled out, because it makes no physical sense since time is non reversible. Nevertheless it is 1 of 3 mathematical results. Accordingly, the Galilean transformation does NOT correspond to the Eucledian spacetime and therefore cannot be viewed as a result of the symmetries in this spacetime. For example, time is reversible in the Euclidean spacetime, but not reversible in the Galilean transformation. The Galilean transformation corresponds to a degenerative spacetime where the ratio between space and time coordinates is infinitely small (same as the speed of light being infinite). Therefore the Galilean transformation does not reflect a symmetry between space and time, as there isn't any in the Galilean spacetime.

In physics it is straightforward and Galilean and Lorentz transformation are similar, and do correspond to the isometries.

First, since it seems less confusing to you but historically it was much more confusing, the Poincare group (Lorentz boosts plus rotations and translations) transformation all came because of the isometry of the 4D spacetime on which special relativity (SR) resides. The isometries are translation (in all 4 dimensions, i.e., space and time), rotations in space, and boosts which are the uniform motion transformations (to a constant velocity wrt first frame). The conserved quantities are 4 momentum (energy and momentum), angular momentum and for the boosts it's a pretty useless one (so not usually mentioned, it's simply the position of the center of mass at t=0). For the boosts it's not common knowledge, you can see the answers at What is the invariant associated with the symmetry of boosts?, as well as the note there that it is a duplicate for another one that preceded it. The words in the last answer explains it, the math is in the first answer.

For the Galilean transformations it's also energy, 3 momentum, and angular momentum. The uniform motion transformation just gives you the Galilean uniform motion transformation to a prime frame moving with uniform velocity wrt the original inertial frame.

$x\prime$ = x-vt, with K in your equation 0 as you also noted. The conservation law is the same useless one for the conservation of the position of the center of mass at t=0. The Galilean Group is a Group Contraction of the Poincare Group for c= infinity. You can see the Galilean Group and Lie Algebra at https://en.m.wikipedia.org/wiki/Galilean_transformation

BTW, both classical physics and relativistic physics also allow the discrete symmetries for reflection and time inversion. We already know that reflection symmetry is only approximate, as parity is not conserved in the weak interactions. For time reversal we still are not sure, but it seems to be in the microscopic domain, except usually one describes the 4D spacetime as time oriented, i.e., t only increases, which allows us to insure causality.

The Galilean group is not more than the limit of the Poincare group for c going to infinity. No reason to get confused. Neither depends on the equations of motion, both give you Lie algebra generators (eg the Galilean wiki reference noted above has them for the Galilean group) as part of the group (symmetry) structure.

• While the Galilean group may indeed be seen as a limit of the Poincare group, the Galilean group is not a special case of the Poincare group, because the Galilean group with the same success may be seen as a limit of the Euclidean group for the infinite c. The important point here is the fact that time is not reversible in the Galilean spacetime for a completely different reason compared to the Minkowski spacetime. The reason is that the latter is hyperbolic. The reason in the Galilean spacetime is the infinite speed of light. A zero can be a limit, but it is neither positive nor negative. Commented Aug 13, 2017 at 20:03
• Very good point @safesphere. To drive the point home, do you know of any formal source showing this? Commented Nov 8, 2023 at 23:30
• @RealPattern See if this helps: philsci-archive.pitt.edu/11265/1/spacetimestructure.pdf Commented Nov 9, 2023 at 3:22
• @safesphere, thanks! Looks good! Is the point you are making in your Aug 13th 2017 comment common knowledge (i.e. generally accepted)? Commented Nov 13, 2023 at 15:55

The Galilean group is a double isometry. It leaves invariant the following two quadratic forms: $$dt^2, \quad \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2,$$ as well as $$dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z} + dt \frac{∂}{∂t}.\tag{1}\label{1}$$ The first two of these correspond, respectively, to a covariant metric and an independent contravariant metric, while the third is an invariant generic to differentiable manifolds. The two metrics are no longer inverses of one another, though their contraction is still a multiple of the identity - namely: the zero multiple. They have zero contraction.

When the Galilean group is centrally extended to the Bargmann group, the resulting group has the following as its invariants: $$dx^2 + dy^2 + dz^2 + 2 dt du, \quad dt,$$ and $$\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 + 2\frac{∂}{∂t}\frac{∂}{∂u}, \quad \frac{∂}{∂u},$$ with ($$\ref{1}$$) extended to and replaced by $$dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z} + dt \frac{∂}{∂t} + du \frac{∂}{∂u}.\tag{2}\label{2}$$

(Edit: For strict equivalence, the linear invariants $$dt$$ and $$∂/∂u$$ should be made quadratic $$(dt)^2$$ and $$(∂/∂u)^2$$, otherwise there will be disagreement on which characterizations allow for time reversal $$t → -t$$, or reversal in the other coordinates. The same with the other linear invariants, below.)

By comparison, the Poincaré group has the following as its invariants $$dt^2 - \frac{1}{c^2}\left(dx^2 + dy^2 + dz^2\right), \quad \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 - \frac{1}{c^2}\left(\frac{∂}{∂t}\right)^2,$$ as well as ($$\ref{1}$$). The contraction of the two metrics is the $$-1/c^2$$ multiple of the identity, so they are conventionally treated as inverses of one another.

Corresponding to the Bargmann group is the (trivial) central extension of the Poincaré group, which has - as its invariants: $$dx^2 + dy^2 + dz^2 + 2 dt du + \frac{1}{c^2}du^2, \quad dt + \frac{1}{c^2}du,\\ \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 + 2\frac{∂}{∂t}\frac{∂}{∂u} - \frac{1}{c^2}\left(\frac{∂}{∂t}\right)^2, \quad \frac{∂}{∂u},$$ and ($$\ref{2}$$).

Of note: the Poincaré group is not the non-relativistic limit of the Bargmann group. It is the extended Poincaré group, which is. At the same time, the Galilean group is not what yields the familiar representations for bodies with mass. Instead, it is the Bargmann group, which does.

The Lie brackets for all of the above-mentioned groups can be succinctly written in vector form as: \begin{align} [Λ,Λ'] &= \left(𝞈×𝞈' - α𝞄×𝞄'\right)·𝐉 + \left(𝞈×𝞄' + 𝞄×𝞈'\right)·𝐊 + \left(𝞈×𝝴' + 𝝴×𝞈'\right)·𝐏\\ &+ \left(𝞄·𝝴'-𝝴·𝞄'\right)M + \left(τ𝞄' - 𝞄τ'\right)·𝐏 \end{align} where $$Λ = 𝞈·𝐉 + 𝞄·𝐊 + 𝝴·𝐏 - τH + ψμ,\\ Λ' = 𝞈'·𝐉 + 𝞄'·𝐊 + 𝝴'·𝐏 - τ'H + ψ'μ,\\ M = μ + αH.$$

Using $$‹⋯›$$ to denote the group generated by $$⋯$$, the following cases apply:

• The Galilean group is $$‹𝐉,𝐊,𝐏,H›$$, with $$μ = 0$$ and $$α = 0$$.
• The Bargmann group is $$‹𝐉,𝐊,𝐏,H,μ›$$, with $$α = 0$$.
• The Poincaré group is $$‹𝐉,𝐊,𝐏,E›$$, with $$E = Mc^2$$ and $$α = 1/c^2$$.
• The extended Poincaré group is $$‹𝐉,𝐊,𝐏,H,μ›$$, with $$α = 1/c^2$$.
The coordinate representations of the groups are given by the following infinitesimal transforms: $$Δ𝐫 = 𝞈×𝐫 - 𝞄t + 𝝴,\quad Δt = -α𝞄·𝐫 + τ,\quad Δu = 𝞄·𝐫 + ψ,$$ where $$𝐫 = (x,y,z)$$, when written in terms of Cartesian coordinates. Using $$Δ(d⋯) = d(Δ⋯)$$, the corresponding transforms on the coordinate differentials can be derived from this, and are given by: $$Δ(d𝐫) = 𝞈×d𝐫 - 𝞄dt,\quad Δ(dt) = -α𝞄·d𝐫,\quad Δ(du) = 𝞄·d𝐫.$$ From this, using the invariant ($$\ref{2}$$), the transforms on the differential operators are given by: $$Δ(∇) = 𝞈×∇ + 𝞄\left(α\frac{∂}{∂t} - \frac{∂}{∂u}\right),\quad Δ\left(\frac{∂}{∂t}\right) = 𝞄·∇,\quad Δ\left(\frac{∂}{∂u}\right) = 0.$$

The infinitesimal rotations and boosts are, respectively, $$𝞈$$ and $$𝞄$$; the infinitesimal spatial and time translations are, respectively, $$𝝴$$ and $$τ$$. The infinitesimal translation $$ψ$$ for the extra coordinate $$u$$ is what appears in the extension of the Galilei group to the Bargmann group.

For the Galilean group, $$u$$ and $$ψ$$ drop out of the picture, and $$∂/∂u = 0$$.

For the extended Poincaré group, the "absolute time" of the Galilei and Bargmann groups lives on as $$s = t + αu$$, whose transforms are: $$Δs = σ = τ + αψ, \quad Δ(ds) = 0.$$ Under conversion from $$(t,u)$$ to $$(s,t)$$, the differential operators convert to: $$\left(\frac{∂}{∂t}\right)_s = \left(\frac{∂}{∂t}\right)_u - \frac{1}{α}\left(\frac{∂}{∂u}\right)_t,\quad \left(\frac{∂}{∂s}\right)_t = \frac{1}{α}\left(\frac{∂}{∂u}\right)_t.$$ Their transforms for the coordinate differentials convert to: $$Δ(∇) = 𝞈×∇ + α𝞄\left(\frac{∂}{∂t}\right)_s,\quad Δ\left(\frac{∂}{∂t}\right)_s = 𝞄·∇,\quad Δ\left(\frac{∂}{∂s}\right)_t = 0.$$

The invariants expressed in terms of $$(s,t)$$ convert to: $$dx^2 + dy^2 + dz^2 - \frac{1}{α}dt^2 + \frac{1}{α}ds^2, \quad ds,\\ \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 - α \left(\frac{∂}{∂t}\right)_s^2 + α\ \left(\frac{∂}{∂s}\right)_t^2, \quad \left(\frac{∂}{∂s}\right)_t,$$ and $$dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z} + ds \left(\frac{∂}{∂s}\right)_t + dt \left(\frac{∂}{∂t}\right)_s.\tag{3}\label{3}$$

The reduction from extended Poincaré to Poincaré corresponds to eliminating $$s$$ by reducing it to the proper time, i.e. by the constraint: $$0 = dx^2 + dy^2 + dz^2 + 2 dt du + α du^2 = dx^2 + dy^2 + dz^2 - \frac{dt^2}{α} + \frac{ds^2}{α},$$ i.e. $$ds^2 = dt^2 - α\left(dx^2 + dy^2 + dz^2\right),$$ and eliminating $$∂/∂s$$ from the picture.

The conversion between $$(s,t)$$ and $$(t,u)$$ is the geometric underpinning to the Foldy-Wouthuysen transform. The extended geometry with all five coordinates $$(x,y,z,t,u)$$, when $$α = 0$$, is the Bargmann geometry. When $$α ≠ 0$$, it has no standard name; but the curved versions of it appear in some papers on 5D cosmology on ArXiv, with the true nature of the geometries that are being dealt with in those papers, unbeknownst to the authors. If I find links, I'll add them in a later edit or comment.

• A great answer at a high level of sophistication (above the paygrade of many on this site) +1 Commented Nov 17, 2023 at 2:57
• Yeah, but I think I need to replace "$dt$" by "$dt^2$", and do the same for the other linear invariants. Otherwise, the two sets of characterizations will disagree on whether time reversal and/or parity reversal are to be included or excluded. Commented Nov 17, 2023 at 11:10
• Well, your posts will serve as an invaluable reference for many when needed for years to come. So sure, please edit them to be consistent and rigorous. Commented Nov 17, 2023 at 14:50