It is often discussed how certain symmetries and conservation laws can be derived from Newton's laws of motion. My question is: can we go the other way? Can Newton's laws of motion be derived only from symmetry assumptions on a dynamical system such as time reversal, time translation, space translation, etc.?

There are clearly more assumptions and definitions needed, e.g. what exactly is meant by a dynamical system (perhaps a state in $\mathbb{R}^n$ and a differential equation stating the dynamics). However, I think there is no need for more "arbitrary" assumptions.

I believe that specifying the symmetries of a system is a more general and natural way of expressing assumptions about it. Practically, it may help with linking classical Newtonian mechanics to dynamical systems that have similar symmetries. Philosophically, it might also illustrate how "arbitrary" these laws are.

  • $\begingroup$ "My question is: can we go the other way?" What do you think? $\endgroup$
    – Bob D
    Feb 17, 2023 at 16:59
  • 2
    $\begingroup$ I suggest taking a look at the discussion in Chapter 1 of Landau and Liftshitz's Mechanics text, where the action of a free particle is derived by imposing symmetry under Galilean boosts. It may not satisfy you completely, but it's a start. $\endgroup$
    – d_b
    Feb 17, 2023 at 17:35
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    $\begingroup$ Related: Deriving the Lagrangian for a free particle $\endgroup$
    – Qmechanic
    Feb 17, 2023 at 17:56
  • $\begingroup$ I expanded my answer, adding discussion of the symmetry properties of F=ma $\endgroup$
    – Cleonis
    Feb 18, 2023 at 8:46

4 Answers 4


Yes, there is a very important theorem in physics called Noether's theorem (or more specifically Noether's 1st theorem) which definitively links symmetries and conservation laws. It states that for every continuous symmetry of the action of a system, there exists a corresponding conserved current. In simpler terms, if the action, which is an integral of the Lagrangian over all possible paths, is invariant under a symmetry group with continuous parameters, then you'll get out a conservation law for that system.

It turns out that time translation symmetry gives energy conservation and spatial translation symmetry gives conservation of momentum. Additionally, for example, rotational invariance gives rise to conservation of angular momentum.

Furthermore, following the formalism of Lagrangian mechanics, you can use the Lagrangian of your system to derive the system's Newtonian equations of motion. Start with a Lagrangian: $$L(q, \dot{q}, t) = T -V$$ Then construct the Euler-Lagrange equation(s): $$\frac{d}{dt}\big (\frac{\partial L}{\partial \dot{q}} \big ) - \frac{\partial L}{\partial q} = 0.$$

  • $\begingroup$ I think this is a good answer, but since the OP asked specifically about Newton's laws, would it be appropriate perhaps to also add how the Euler Lagrange equation (you already mentioned the Lagrangian..) can be shown to basically reproduce them? $\endgroup$
    – Amit
    Feb 17, 2023 at 17:19
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    $\begingroup$ I think this is the opposite of what the OP wants. Starting from the action is equivalent to starting from Newton's laws. Of course once we have the action, we can determine its symmetries and the corresponding conserved quantities. OP would like to know if we can derive the action only by imposing symmetry constraints without knowing ahead of time what the action should be. $\endgroup$
    – d_b
    Feb 17, 2023 at 17:27
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    $\begingroup$ Yes, this is a good point thanks! $\endgroup$
    – klippo
    Feb 17, 2023 at 17:28
  • $\begingroup$ @d_b OP asked how can we recover Newton's laws starting from symmetry. This is done by noting continuous symmetries of the action of a system and following the prescription of Noether's theorem to recover the corresponding conservation law. The conservation law relevant to Newton's laws that OP asked about is conservation of momentum. We also note that the newtonian F=m a is recoverable from the Euler Lagrange equations, though not directly related to symmetries. $\endgroup$
    – klippo
    Feb 17, 2023 at 17:40

In this answer I discuss the following two things:

-Among the multiple symmetries of mechanics is the symmetrical relation between the members of the equivalence class of inertial coordinate systems. (I categorize the property of reciprocity as a form of symmetry.)

-Symmetry properties of F=ma

The equivalance of the set of inertial coordinate systems has been treated by many; I find the treatment by Palash B. Pal particularly effective. The title of the article is 'Nothing but Relativity' Palash B. Pal demonstrates that the constraint of symmetrical relation between the members of the equivalence class of inertial coordinate system is sufficient to narrow down the options to just two: the Galilean transformations and the Lorentz transformations.

As we know: newtonian mechanics is a limiting case of relativistic mechanics; for non-relativistic velocity the expressions simplify to the expressions for newtonian mechanics..

Part of the set properties that must be granted in order to have special relativity is granting Minkowski spacetime.

The corresponding assumption for newtonian mechanics is the assumption that space-and-time are Euclidean. This includes granting that space and time stand in a particular relation to each other: for newtonian mechanics it must not only be granted that an object released to free motion will move along a straight line; it must also be granted that the object will in equal time intervals cover equal intervals of space.

(Historically its counterpart for the motion of an object being subjected to a central force was formulated earlier: an object in central-force-caused circumnavigating motion will in equal intervals of time sweep out equal areas; Kepler's second law.)

The set of relations between symmetries and corresponding conserved quantities is commonly described using Noether's theorem. It seems to me however, that those relations are not dependent on describing them with Noether's theorem. That is: I don't think Noether's theorem is a necessity to describe those correspondencies, it just so happens that historically that was the mathematical tool involved in the first recognition of those correspondencies.

Symmetry properties of F = ma

Using the standard letters for time, position, velocity and acceleration:

t time
s position
v velocity
a acceleration

I regard the following as an instance of symmetry:

$$ v = \frac{ds}{dt} \tag 1 $$

$$ a = \frac{dv}{dt} \tag 2 $$

Velocity is to position is what acceleration is to velocity.

That symmetry goes a long way.
I start with the case of uniform acceleration:

$$ v = at \tag 3 $$

To find distance covered as a function of time: integrate (3) with respect to the time coordinate:

$$ s = \int_0^t at \ dt = \tfrac{1}{2}at^2 $$

$$ s = \tfrac{1}{2}at^2 \tag 4 $$

Combining the relations (3) and (4) is in effect combining the relations (1) and (2):

(5) is obtained by multiplying both sides of (4) with acceleration $a$, and then applying (3):

$$ as = a\tfrac{1}{2}at^2 = \tfrac{1}{2}a^2t^2 = \tfrac{1}{2}(at)^2 = \tfrac{1}{2}v^2 $$

$$ as = \tfrac{1}{2}v^2 \tag 5 $$

The product $as$ skips the entity in the middle: the velocity $v$. In (5) the product $as$ transforms to the square $v^2$ by transfering a factor $t$ (time)

The position coordinate $s$ is rescaled to velocity by dividing it with the time factor $t$, and the acceleration value is rescaled to velocity by multiplying it with the time factor $t$.

Generalize (5) to accommodate an initial velocity $v_0$, and an initial position $s_0$:

$$ a(s -s_0) = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 \tag 6 $$

For non-uniform acceleration: integration.

The relations (1) and (2) are then used in the following form:

$$ v = \frac{ds}{dt} \qquad \Longleftrightarrow \qquad ds = v \ dt \tag 7 $$

$$ a = \frac{dv}{dt} \qquad \Longleftrightarrow \qquad dv = a \ dt \tag 8 $$

The integral:

$$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 $$

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}{v_0}^2 \tag {9} $$

As we know: integration is addition of rectangular strips.

The integral $\int_{s_0}^s a \ ds$ specifies rectangular strips with width $ds$ and height $a$.

The integral $\int_{v_0}^v v \ dv$ specifies rectangular strips with width $dv$ and height $v$.

The dimensions of the rectangular strips are rescaled; the height is multiplied with the factor $t$ and the width is divided by the factor $t$, that is why the areas are the same.

All of the above is obtained by virtue of the symmetry:

$$ v = \frac{ds}{dt} \tag 1 $$

$$ a = \frac{dv}{dt} \tag 2 $$

We obtain the work-energy theorem by combining $F=ma$ with (9):

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2 $$

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}m{v_0}^2 \tag {10} $$

Historically: the concepts of potential energy and kinetic energy were in circulation before there the work-energy theorem was stated in the modern form. The work-energy theorem clarifies what the concept of energy is: it shows that potential energy and kinetic energy have in common that they both arise from integration with respect to the position coordinate.

For the expression for kinetic energy $ \tfrac{1}{2}mv^2$: diffentiation with respect to the position coordinate recovers $ma$:

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma $$

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = ma \tag {11} $$

And of course: when you differentiate potential energy with respect to the position coordinate you recover the force.

The definition of potential energy takes the content of the work-energy theorem to advantage; potential energy is defined such that over time the value of the sum $\Delta E_p$ and $\Delta E_k$ is constant:

$$ \Delta E_p = - \int_{s_0}^s F \ ds \tag {12} $$

$$ - \Delta E_p = \Delta E_k \tag {13} $$

$$ \Delta E_k + \Delta E_p = 0 \tag {14} $$

$$ \frac{d(E_k + E_p)}{dt} = 0 \tag {15} $$

(15) expresses explicitly the symmetry under time translation: if the derivative with respect to time is zero then as you change the time coordinate the value of $(E_k + E_p)$ remains the same.

Going back once more to (1) and (2): I regard that symmetry as the origin of (14)(15).

$$ v = \frac{ds}{dt} \tag 1 $$

$$ a = \frac{dv}{dt} \tag 2 $$


QFT uses a similar symmetry-to-laws approach I'll adapt to your question. The goal is to build the most general Lagrangian compatible with the desired symmetries and some other conditions, such as (i) smooth dependence on phase space coordinates, which demands a quadratic kinetic sector and (ii) not going further than first-order derivatives because they cause a problem we'll try to avoid. Assuming Lagrangians describe the physics might already feel like a cheat, but they're our clearest understanding of how to start from symmetries, so we'll settle for seeing how symmetries motivate a specific Lagrangian from which Newton's second law, in particular, emerges.

Whereas in the QFT of a Lorentz scalar $\phi$ this gives a Lagrangian density of the form $\mathcal{L}=\frac12\partial_\mu\phi\partial^\mu\phi-V(\phi)$, the problem at hand obtains a true Lagrangian because we only want to work with time-dependent particle positions, not spacetime-dependent fields. (The latter don't originate with QFT or even QM, as e.g. electromagnetism or fluid mechanics illustrate.) Hence we get $L=\frac12m\dot{x}^2-V(\vec{x})$, where we've scaled $\vec{x}$ so a mass factor $m$ can appear in the kinetic term. This allows us to easily upgrade to an arbitrarily diverse universe of particles, viz. $L=\frac12\sum_{i=1}^nm_i\dot{x}_i^2-V(\vec{x}_1,\,\cdots,\,\vec{x}_n)$. Our Euler–Lagrange equations are now $\frac{d\vec{p}_i}{dt}=-\nabla_{\vec{x}_i}V$ for $\vec{p}_i:=m_i\dot{\vec{x}}_i$.

Next we need to be careful about an issue that's often much harder to explain from a Lagrangian perspective than a Newtonian one: how do we accommodate non-conservative forces? Short answer, like this; long answer, to help with our present context, the symmetry assumption from which we start has to be weakened from $\frac{\delta S}{\delta\vec{x}_i}=\vec{0}$ to $\frac{\delta S}{\delta\vec{x}_i}=-\vec{Q}_i^p$. This allows us to separate the question of which symmetries appear on the left-hand side from the question of which non-conservative forces appear on the right. In either case,$$\frac{\delta S}{\delta\vec{x}_i}:=\frac{\partial L}{\partial\vec{x}_i}-\frac{d}{dt}\frac{\partial L}{\partial\dot{\vec{x}}_i}.$$This separation again may bother you, but bear in mind the dissipative version of Newton's second law is similar: $\frac{d\vec{p}_i}{dt}=-\nabla_{\vec{x}_i}V-\vec{Q}_i^p$. What's more, dissipative forces are a symptom of our choosing to analyze systems that are physically incomplete, thereby susceptible to something we didn't model. For example, if you describe friction in terms of individual atoms, you no longer need to describe one macroscopic object losing net energy. Unsurprisingly, QFT is so fundamental it obtains conservative Lagrangian densities. (You can invent a dissipative field theory, then quantize its fields, but the real world doesn't seem to work that way.)

We still haven't identified how $\vec{x}_i$ transforms under a different choice of coordinates. The answer must emerge from demanding our ELEs are thereby invariant. So we want to derive Galilean transformations from Newton's second law. That's a subtle issue discussed here.

  • $\begingroup$ The symmetry group for the second law depends on what the force is. For 0 force (free bodies), it's SL(5). (See Derivation of SL(5). For other force laws (e.g. Newton's gravity) you'll get a proper subgroup that may only have Galilei as a subgroup. Instead, Galilei comes first and is used to impose a requirement on what is permissible as a force law. Symmetry a key model-building technique used for designing force laws and other forms of dynamics. $\endgroup$
    – NinjaDarth
    Oct 13, 2023 at 20:08

I'm a fan of Noether and all things related to symmetry! I've given this matter some thought, but it would take a long time to spell it out in detail. So, perhaps a few cursory remarks will suffice for now, maybe with a later edit to refine and add details.

The starting point would be the Bargmann group, which is the central extension of the Galilei group. But special consideration must be paid to the coordinate representation, since its natural representation lies in five dimensions, not four. The reason for the fifth dimension is that the extension provided by Bargmann also accounts for mass. This means the three components of momentum, the kinetic energy and the mass transform together as a five-vector. To make this compatible with a coordinate representation would require another coordination - fictitious perhaps - to serve as a conjugate to the mass.

The symmetry group has 11 dimensions, instead of just 10. The Noether charges that go with it are the three components each of angular momentum, mass moment and linear momentum, plus the one component each of mass and kinetic energy. The mass moment requires special treatment.

The primary focus is to make this work with bodies of arbitrary composition with the principles that:

Upward And Downard Scalability: What applies to the body applies equally well to its components, and vice versa.
Additivity: All of the Noether charges are additive. No exceptions. (We need to have a conversation about kinetic energy, Γ‰milie du ChΓ’telet; but also about angular momentum.)

The First Law states that for isolated bodies, all Noether charges are preserved. In particular, this means no self-force or other self-interaction. A "body" is defined as any physical system that retains its contents. So, if you think of a system as being a continuum - as Newton did in the first definitions in his Principia - then you are moving the boundary of the "blob", that makes up the body, with the flow of the body.

Self-force dynamics, meaning specifically "radiation", can only be simulated by pretending there is a far-off body on the other end that the body undergoing "self-force" is interacting with - The Absorber.

The Second Law requires that all interactions be body-to-body, that they be part/whole additive both with respect to the body exerting the force and with respect to the body receiving the force. Associated with each Noether charge is an interaction "force" that is proportional to the rate of change of the corresponding Noether charg.

The mass is considered invariable. To account for propulsion therefore requires the same mental gymnastics that are already used in deriving the rocket equation.

The Third Law states that a given set of body-to-body interactions may be considered removed from the context of other interactions that may be taking place and treated as if they were the only interactions. Specifically, any two bodies may be treated as two components of a single isolated body. The additivity of their mutual interactions, taken in conjunction with the absence of any self-interaction by the entire "body" they are the parts of, and by each of the component bodies, then requires that their mutual interactions be equal and opposite to one another.

The function of the Third Law is to enable upward scalability of the First Law and Second Law from components of a body to the body they are components of.

The requirement of additivity imposes a consistency condition on the angular momentum and kinetic energy. They must each be supplied with an "internal" contribution. The reason is that their corresponding expressions, in terms of the body's mass, position and velocity, are quadratic. Their sums have cross-terms. The internal parts of the angular momentum and kinetic energy are there to absorb those extra terms.

The Noether charges corresponding, respectively, to the symmetries of spatial rotation, boost, spatial translation, time translation and translation with respect to the extra coordinate! are the angular momentum $𝐉$, the mass moment $𝐊$, the linear momentum $𝐩$, the kinetic energy $H$ and the mass $m$. The expressions for each, in terms of a body's mass $m$, center of mass position $𝐫$ and velocity $𝐯$ are: $$𝐉 = m𝐫×𝐯 + 𝐒, \quad 𝐊 = m(𝐫 - 𝐯t), \quad 𝐩 = m𝐯, \quad H = \frac{1}{2}m|𝐯|^2 + U, \quad m.$$ The internal contributions are $𝐒$ for angular momentum and $U$ for kinetic energy. The charges are constrained in such a way that: $$\frac{d𝐫}{dt} = 𝐯.$$

The part-whole decomposition for two bodies can then be consistently laid out: $$m = m_0 + m_1, \quad 𝐫 = \frac{m_0𝐫_0 + m_1𝐫_1}{m}, \quad 𝐯 = \frac{m_0𝐯_0 + m_1𝐯_1}{m},\\ 𝐒 = 𝐒_0 + 𝐒_1 + \bar{m}\bar{𝐫}Γ—\bar{𝐯}, \quad U = U_0 + U_1 + \frac{1}{2}\bar{m}|\bar{𝐯}|^2,\\ \bar{m} = \frac{m_0m_1}{m}, \quad \bar{𝐫} = 𝐫_0 - 𝐫_1, \quad \bar{𝐯} = 𝐯_0 - 𝐯_1.$$ Despite the additivity of the energy, the mutual dynamics of the two bodies is still captured by the dynamics of the reduced body with the data $\left(\bar{m}, \bar{𝐫}, \bar{𝐯}\right)$, treated as a one-body problem. There is nothing new in saying so, except for the courage of also saying that this is all that's needed! No interaction potentials, beyond what's already there, are required. For instance, if the body-to-body interaction is conservative, then we may write, for their mutual interaction: $$\frac{d𝐩_0}{dt} = 𝐅 = -\frac{d𝐩_1}{dt}, \quad 𝐅\left(\bar{𝐫}\right) = -\frac{βˆ‚V}{βˆ‚\bar{𝐫}}\quadβ‡’\quad \frac{1}{2}\bar{m}|\bar{𝐯}|^2 + V =\text{ constant}, \quad \bar{m}\bar{𝐫}Γ—\bar{𝐯} =\text{ constant}.$$

A similar observation applies to $n+1$ body systems. They can be treated as the $n+1$ components of an isolated body, and the entire system reduced to $n$ fictitious bodies that are subject to internal dynamics comprising their mutual interactions.

The actual composition of the Noether charges is not stipulated; rather it is derived. They are only "charges" in the strictest sense, for isolated bodies, but the term will be used generically to denote the "charges" as time-variable quantities, when the body is interacting.

Each of the charges is treated as a coordinate in the dual Lie algebra. This is the "coadjoint orbit" representation. The action of rotations, boosts, spatial translation and time translations may then be induced from this. The translations by the extra coordinate have no effect - it is a gauge transformation.

Using these transforms, a classification of possible "bodies" may be laid out, with different normal forms derived for each. In particular, the layout of the Noether charges for ordinary bodies - with the internal components $𝐒$ and $U$ - will follow. This is the ultimate justification for including the internal components.

The classes that may be found, along with normal forms for each, include the following:

  • Class Aβ‚€: $m β‰  0$, $m𝐉 + π©Γ—πŠ β‰  𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) β†’ \left(𝐒, 𝟬, 𝟬, U, m\right) ≑ \left(𝐉 + \frac{π©Γ—πŠ}{m}, 𝟬, 𝟬, H - \frac{|𝐩|^2}{2m}, m\right)$.
  • Class A₁: $m β‰  0$, $m𝐉 + π©Γ—πŠ = 𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) β†’ \left(𝟬, 𝟬, 𝟬, U, m\right) ≑ \left(𝟬, 𝟬, 𝟬, H - \frac{|𝐩|^2}{2m}, m\right)$.
  • Class Bβ‚€: $m = 0$, $π©Γ—πŠ β‰  𝟬$:
    $(𝐉, 𝐊, 𝐩, H, M) β†’ (𝟬, 𝝹, 𝝿, 0, 0) ≑ \left(𝟬, \frac{(π©Γ—πŠ)×𝐩}{|𝐩|Β²}, 𝐩, 0, 0\right)$.
  • Class B₁: $m = 0$, $π©Γ—πŠ = 𝟬$, $𝐩 β‰  𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) β†’ (η𝝿, 𝟬, 𝝿, 0, 0) ≑ \left(\frac{𝐉·𝐩𝐩}{|𝐩|^2}, 𝟬, 𝐩, 0, 0\right)$.
  • Class C: $𝐩 = 𝟬$, $𝐊 β‰  𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) β†’ (σ𝝹, 𝝹, 𝟬, U, 0) ≑ \left(\frac{π‰Β·πŠπŠ}{|𝐊|^2}, 𝐊, 𝟬, H, 0\right)$.
  • Class Dβ‚€: $𝐊 = 𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) = (𝐒, 𝟬, 𝟬, U, 0)$.
  • Class D₁: $𝐉 = 𝟬$:
    $(𝐉, 𝐊, 𝐩, H, m) = (𝟬, 𝟬, 𝟬, U, 0)$.

This is meant to be comprehensive: the additivity formula for two class A bodies can be expanded to include additivity formulae for two bodies of these and other classes, the resulting table being somewhat analogous to the table of Klebsch-Gordon coefficients.

In genric frames, their representations are, respectively

  • Class Aβ‚€: $(𝐉, 𝐊, 𝐩, H, m) = (m𝐫×𝐯 + 𝐒, m(𝐫 - 𝐯t), m𝐯, \frac{1}{2}m|𝐯|^2 + U, m)$.
  • Class A₁: $(𝐉, 𝐊, 𝐩, H, m) = (m𝐫×𝐯, m(𝐫 - 𝐯t), m𝐯, \frac{1}{2}m|𝐯|^2 + U, m)$.
  • Class Bβ‚€: $(𝐉, 𝐊, 𝐩, H, m) = (η𝝿 + 𝐫×𝝿, 𝝹 - 𝝿t, 𝝿, H, 0)$ - version A.
  • Class Bβ‚€: $(𝐉, 𝐊, 𝐩, H, m) = (𝝹×𝐯 + 𝐫×𝝿, 𝝹 - 𝝿t, 𝝿, 𝝿·𝐯, 0)$ - version B.
  • Class B₁: $(𝐉, 𝐊, 𝐩, H, m) = (η𝝿 + 𝐫×𝝿, -𝝿t, 𝝿, H, 0)$.
  • Class C: $(𝐉, 𝐊, 𝐏, H, m) = (σ𝝹 + 𝝹×𝐯, 𝝹, 𝟬, U, 0)$.
  • Class Dβ‚€: $(𝐉, 𝐊, 𝐩, H, m) = (𝐒, 𝟬, 𝟬, U, 0)$.
  • Class D₁: $(𝐉, 𝐊, 𝐩, H, m) = (𝟬, 𝟬, 𝟬, U, 0)$.

The Class A bodies have time-like worldlines, while the Class B bodies have synchronous worldlines and play the role of "instantaneous-action-at-a-distance-impulses". A continuum of such bodies is one that is smeared out in time as a time-density, and takes on the form of a continuously-acting action-at-a-distance force.

The Class C and D bodies are homogeneous and all-pervasive. In the case of Class D, the components may be considered as disembodied versions of the "internal" components of the Class A bodies. Thus, a Class A body may be treated as "pure" if it is free of internal components, else it may be considered as the joining of a pure Class A body and Class D body. A similar observation applies Class C and Class Dβ‚€, in relation to Class D₁ and the internal energy $U$.

In modular representation theory, the "pure" representations are normally called "indecomposeable". I'm not entirely sure the ones I'm calling "pure" actually are indecomposeable. There may be further decomposition for the cases I listed as "pure".

What I don't have clear answers to are:

  • Are there restrictions on what kinds of body-to-body interactions may occur, if it is required that the forces be expressible as functions of the Noether charges?
  • Can the 100%-additivity rule be used to create a relativistic version of this which undercuts the no-interaction theorem? Recall that the theorem states that in many body dynamics, the Noether charges must be strictly additive; particularly: no non-additive potential energy term $V$ may occur. We're undercutting this.
  • Howver, there is a price to pay. To consistently maintain the correspondence limit to non-relativistic theory requires adding an 11th dimension to the PoincarΓ© group. The reduction from it to the PoincarΓ© group is not trivial. Part of what was the central charge, when Galilei was extended to Bargmann, after relativization, and then after reduction, finds its way into the unextended PoincarΓ© group, while part of what is eliminated by this reduction is what was already present in the unextended Galilei group before it was lifted to Bargmann and relativizied. This needs to explored in further detail.
  • $\begingroup$ The part/whole scalability idea and "everything is part of an isolated body" idea are interesting and it can be used for treating bodies with variable composition. This could be used for a better analysis of the rocket equation, as you suggested, and I'll give it a try on a "rocket equation" question. $\endgroup$
    – NinjaDarth
    Oct 13, 2023 at 20:01
  • 1
    $\begingroup$ About the no-interaction theorem: if you deform the Bargmann group, by "relativizing" it, then the 11th generator becomes the mass and the symmetry group is a one-dimensional extension of the Poincaré group. Dropping down to Poincaré, however, the central charge remains intact. Instead, what goes away is the very "internal energy" (U) that you're using to escape non-additivity! So, it's definitely a culprit. But it may not be enough to escape the no-interaction theorem. (BTW, you misspelled Poincaré.) $\endgroup$
    – NinjaDarth
    Oct 13, 2023 at 20:30
  • 1
    $\begingroup$ I corrected the error. Thank you. I will used the idea of the internal energy being the thing that goes away, in another reply. $\endgroup$ Oct 15, 2023 at 3:13

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