Assumptions of Physics overview: Classical mechanics and infinitesimal reducibility
AAssumptions of Physics overview:Classical mechanics and infinitesimal reducibility
Gabriele Carcassi, Christine A. Aidala
Physics Department, University of Michigan, Ann Arbor, MI 48109 (Dated: January 7, 2021)We briefly show how classical mechanics can be rederived and better understood as a consequenceof three assumptions: infinitesimal reducibility, deterministic and reversible evolution, and kinematicequivalence. This work is an overview of some of the results of Assumptions of Physics, a projectthat aims to identify a handful of physical principles from which the basic laws can be rigorouslyderived (https://assumptionsofphysics.org).
I. INTRODUCTION
The overall argument (see [1, 2] for more details) canbe summed up in the following diagram that can be usedas a guide throughout this note.The three assumptions lie on the left column. Eachassumption leads to one or two key insights that progres-sively lead to the physical concepts in the middle column.Each of these is then mapped to its corresponding formalframework on the right.
II. INFINITESIMAL REDUCIBILITY
Infinitesimal reducibility assumption: the state ofthe system is reducible to the state of its infinitesimalparts.
That is, giving the state of the whole system isequivalent to giving the state of its parts, which in turnis equivalent to giving the state of its subparts and so on.Under this assumption, the state of the whole is a dis-tribution over the states of the parts. More precisely, let C be the state space for the whole system. We call parti-cle the limit of recursive subdivision. Let S be the statespace for the particles. Then for each c ∈ C we can findone and only one ρ c : S → R that describes the state ofits parts. That is, the state of the whole is a distributionover the infinitesimal parts. If S is a manifold, (cid:82) U ρ c d S gives us the fraction of the system whose parts are in theregion U of the state space.The next step takes the rest of the section. We needto show that S has the structure of phase space, with its conjugate variables. Mathematically, S is a symplec-tic manifold. The key insight is that, on a manifold, ρ c must transform both as a scalar function (the value mustdepend on the point and not on the coordinates) and asa density. Classical phase space is the only space thatallows these invariant distributions.The key concept is to keep track of the unit system,so we need precise terminology. We call state variables aset of quantities ξ a that fully identify a state. That is,we can write each state s ( ξ a ) as a function of the statevariables. We call coordinate q ∈ ξ a a particular variablethat defines a unit. The key problem is to understandhow many coordinates we can have for a given set ofstate variables.We start with the simplest case, where one coordinateis sufficient to define the unit system, which means thefollowing four conditions:1. the state variables can be written as ξ a = { q, k i }
2. we can arbitrarily change coordinate to ˆ q = ˆ q ( q )3. a change of coordinate induces a unique change overthe remaining state variables ˆ k j = ˆ k j ( q, k i )4. the density is the same regardless of the coordinatesused.Now we show that there can only be one k i . Sup-pose we change unit ˆ q = ˆ q ( q ). Call the new unitsˆ ξ b = { ˆ q, ˆ k j } . We have ρ c ( ˆ ξ b ) = ρ c ( s ( ˆ ξ b )) = ρ c ( s ( ξ a )) = ρ c ( ξ a ) = (cid:12)(cid:12)(cid:12) ∂ ˆ ξ b ∂ξ a (cid:12)(cid:12)(cid:12) ρ c ( ˆ ξ b ). Therefore the Jacobian (cid:12)(cid:12)(cid:12) ∂ ˆ ξ b ∂ξ a (cid:12)(cid:12)(cid:12) mustbe equal to 1. Note that, since ˆ q depends only on q , (cid:12)(cid:12)(cid:12) ∂ ˆ ξ b ∂ξ a (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ∂ ˆ q∂q (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ∂ ˆ k j ∂k i (cid:12)(cid:12)(cid:12) . Suppose there is only one variable.Then we would have (cid:12)(cid:12)(cid:12) ∂ ˆ q∂q (cid:12)(cid:12)(cid:12) = 1. But this would mean theunit change cannot be arbitrary. Therefore we must havetwo or more variables and therefore (cid:12)(cid:12)(cid:12) ∂ ˆ k b ∂k a (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ∂q∂ ˆ q (cid:12)(cid:12)(cid:12) . Thisputs a constraint only on the determinant of the transfor-mation. Suppose there are three or more variables. Thisconstraint is not enough to recover the transformationuniquely and therefore ˆ q would not fully define the unitsfor all other state variables. This means there must be two In mathematical terminology, these are the coordinates of themanifold. a r X i v : . [ phy s i c s . c l a ss - ph ] J a n variables: q and a single k . Coordinate independent areasand densities can only be defined on a two-dimensionalmanifold.Now we generalize to multiple coordinates. We say twocoordinates are independent if changing the units for onedoes not induce a change of units for the other. Now sup-pose our particle state space S is such that its units arefully defined by n independent coordinates q i . Supposeyou change the first coordinate q without changing theothers. Then we will find a variable k that changes asbefore so that the densities are invariant. Now change thesecond coordinate q in the same way while also fixing k . Then we find a corresponding k . We can proceedin the same way until we exhaust all coordinates, whichmust also mean that there are no state variables left. Wefind that S is 2 n -dimensional, and the state variables are ξ a = { q i , k i } . We define an independent degree of free-dom as the space charted by a pair of such variables.We can dress the result a bit more formally, and showthat S is a symplectic manifold. To characterize marginaldistributions on each degree of freedom, we want to de-fine integrals of the form (cid:82) Σ ρ c ω ( d Σ) such that the den-sity ρ c is invariant. Therefore we need a two-form ω thatassigns an infinitesimal area to each infinitesimal surface,and we need ω to be invariant. Because degrees of free-dom are, at least locally, independent, the total numberof states in a volume is the product of the possible con-figurations of each degree of freedom. This means thevolume form is proportional to ω n . This cannot be de-generate (i.e. it must be nonzero for each infinitesimalvolume) since all regions of phase space must, by def-inition, contain some states. Therefore ω itself cannotbe degenerate. As the degrees of freedom are indepen-dent, the number of states on a surface does not changeif we translate it across independent degrees of freedom.If we imagine a parallelepiped, the integral over the sur-face must be zero (integrals over opposite sides are equaland opposite). Therefore S must come equipped with atwo-form ω that is closed and not degenerate and there-fore S is a symplectic manifold. By convention, we set ω = (cid:126) dq i ∧ dk i = dq i ∧ dp i where p i = (cid:126) k i .Phase space (i.e. a symplectic manifold) is the onlytype of manifold that is able to support coordinate in-variant distributions, which are required to describe aninfinitesimally reducible system. III. DETERMINISTIC AND REVERSIBLEEVOLUTION
Deterministic and reversible evolution assump-tion: given the present state of the system, all future(determinism) and past (reversibility) states are uniquelyidentified.
We first apply the assumptions to the motion of a singleparticle. Let λ : R → S be the evolution over time ofthe state of a particle. Under the assumption, this willbe uniquely identified by the initial state s = λ ( t ) at the initial time t . Secondly, we apply the assumption tothe density in the sense that all the particles that startwith the same initial state must end up in the same finalstate and vice-versa. That is, if ρ ( λ ( t ) , t ) is the densityassociated to the initial particle at the initial time, wemust have that ρ ( λ ( t ) , t ) = ρ ( λ ( t ) , t ): the density mustremain the same throughout the evolution.Now consider the integral (cid:82) Σ ρω ( d Σ). Both the regionand the density will be mapped in time to ˆΣ and ˆ ρ re-spectively. The fraction of the system found in the newregion will have to be the same as the one found in the oldregion. That is, (cid:82) Σ ρω ( d Σ) = (cid:82) ˆΣ ρ ˆ ω ( d ˆΣ). Since both theintegral and the density have to remain constant duringthe evolution, then ω will need to remain the same. Thatis, the areas in phase space must be mapped to areas ofequal size and independent degrees of freedom must bemapped to independent degrees of freedom (or volumeswould not be mapped to equal volumes). The evolutionis a symplectomorphism and corresponds to Hamiltonianevolution. Intuitively, this is the inverse of Liouville’s the-orem: instead of positing Hamiltonian evolution and find-ing conservation of areas and volumes, deterministic andreversible evolution imposes the conservation of areas andvolumes which leads to Hamiltonian evolution.The argument can also be constructed through sta-tistical concepts (i.e. determinism and reversibilitymeans conservation of variance), thermodynamic con-cepts (i.e. determinism and reversibility means only thestate of the system is important for the evolution; thesystem is therefore isolated and must conserve energy)or information theoretic consideration (i.e. under deter-ministic and reversible evolution the amount of informa-tion does not change, so information entropy has to beconserved). IV. KINEMATIC EQUIVALENCE
Kinematic equivalence assumption: the motion ofthe system (i.e. trajectories in physical space-time) isenough to recover its dynamics (i.e. evolutions in statespace) and vice-versa.
First, as before, we apply the assumption to the parti-cles, which means that for every evolution in phase spacethere should be one and only one trajectory. Note thateach space variable x i is a coordinate, i.e. a state variablethat defines a unit. In fact, the trajectories can be fullydescribed by those units and only those units. So we cansay that q i = x i , each q i will be paired with a conjugate p i and each state { q i , p i } will be mapped to one and onlyone trajectory. At each point x i , then, infinitely manytrajectories must pass, one for each combination of { p i } .Since the trajectories are differentiable in x i , we can de-fine a velocity v i = d t x i . If the equations of motion weresuch that v i = v i ( q i ), then kinematic equivalence wouldfail as the full trajectory would be determined only by q i .So we must have v i = v i ( q i , p i ). The relationship mustbe invertible or kinematic equivalence would fail. At anygiven time, then, we must have the following relationship: x i = q i v j = d t x j = v j ( q i , p k ) (1)Let us call weak equivalence the notion that v j ( q i , p k )must be invertible at every q i and therefore we can write p k = p k ( x i , v j ) as a function of position and velocity.In this case, the Jacobian matrix ∂v j ∂p k must be invert-ible. Since v j = d t q j = ∂H∂p j , the Hessian ∂ H∂p k ∂p j must benonzero everywhere, and therefore must have the samesign which we take to be positive. In this case, and onlyin this case, we can construct a Lagrangian from a Hamil-tonian: L ( x i , v j ) = v k p k ( x i , v j ) − H ( q i ( x i ) , p k ( x i , v j )) (2)These are also exactly the cases where the Lagrangian,using the principle of minimal action, leads to a uniquesolution.Now we look at the whole distribution and how it canbe expressed as a function of position and velocity. Wehave ρ ( q i , p j ) = | J | ρ ( x i , v j ) = (cid:12)(cid:12)(cid:12) ∂v i ∂p j (cid:12)(cid:12)(cid:12) ρ ( x i , v j ) since | J | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂x i ∂q j ∂x i ∂p j ∂v i ∂q j ∂v i ∂p j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) δ ij ∂v i ∂q j ∂v i ∂p j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12) δ ij (cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ∂v i ∂p j (cid:12)(cid:12)(cid:12)(cid:12) − | | (cid:12)(cid:12)(cid:12)(cid:12) ∂v i ∂q j (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ∂v i ∂p j (cid:12)(cid:12)(cid:12)(cid:12) . (3)Note that while the value given by ρ ( q i , p j ) is coordinateindependent, the value given by ρ ( x i , v j ) depends on the choice of coordinate through (cid:12)(cid:12)(cid:12) ∂v i ∂p j (cid:12)(cid:12)(cid:12) . If x i truly sets theunit system by itself, then (cid:12)(cid:12)(cid:12) ∂v i ∂p j (cid:12)(cid:12)(cid:12) must be a function ofposition only. Similar considerations will also hold formarginal distributions (i.e. distributions on a subset ofthe coordinates) which means all components of ∂v i ∂p j mustbe a function of position only. We set: ∂v i ∂p j = 1 m g ij ∂p j ∂v i = mg ji (4)where m is the unit conversion constant between velocityand conjugate momentum while g ij represents the lineardependency.If we integrate, we have: v i = 1 m g ij ( p j − A j ) p j = mg ji v i + A j (5)where A j are arbitrary functions. Note that: v i = d t q i = ∂H∂p i = 1 m g ij ( p j − A j ) (6)We integrate yet again and find: H = 12 m ( p j − A j ) g ij ( p j − A j ) + V (7)where V is another arbitrary function. We recognize thisas the Hamiltonian for massive particles under potentialforces. [1] G. Carcassi, C. A. Aidala, D. J. Baker, and L. Bieri,Journal of Physics Communications2