The Euclidean Group Part 4-Advanced Classical and Relativistic Mechanics-Lecture Handout, Exercises of Classical and Relativistic Mechanics

Download The Euclidean Group Part 4-Advanced Classical and Relativistic Mechanics-Lecture Handout and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity! Hence a†1 = a −1 1 but on the real euclidean space Rn we have that a † 1 = a T 1 where a T 1 denotes the transpose and therefore we can make the indentification, a1 = R with R ∈ O(n). Consequently, we have shown that for some pair (R, u) ∈ E(n) the isometry preserving map f(x) | |f(x) − f(y)| = |x− y| can be expressed as: f(x) = Rx+ u Solution to 5: Any element of the Galilei Group defined here as a triple (f, v, s) with f ∈ E(n), v ∈ Rn and s ∈ R gives a transformation of (n+1) dimensional spacetime as: F(f,v,s):Rn+1 → Rn+1. The defined action of the map F(f,v,s) on a spacetime pair (x, t) can be defined by: F(f,v,s)(x, t) = (f(x) + vt, t+ s) For any two elements F(f,v,s) and F ′ (f ′,v′,s′) heretofore simply F and F ′ respectively we want an explicit formula for the composite action F ◦ F ′ := F ′′ of F ′′ on a spacetime pair (x, t). From the definition of how F acts on a spacetime pair we can write unambiguously the composite action as follows: (F ◦ F ′)(x, t) = F (f ′(x) + v′t, t+ s′) = F (R′x+ u′ + v′t, t+ s′) action of f ∈ E(n) on x = (f(R′x+ u′ + v′t) + v(t+ s′), t+ s′ + s) = (R(R′x+ u′ + v′t) + u+ v(t+ s′), t+ s′ + s) = (RR′x+Ru′ +Rv′t+ u+ vt+ vs′, t+ s′ + s)(5) ≡ F ′′(x, t) Now enters ambiguity; however, this is good. We are free to define consecutive additive actions in a manner which ensures that the defined action is self-consistent. Hence if we make the following defining identifications, RR′ := R′′(6) Ru′ + u+ vs′ := u′′and consequently:(7) f ′′(x) := RR′x+Ru′ + u+ vs′ := R′′x+ u′′(8) Rv′ + v := v′′(9) s+ s′ := s′′,(10) equation (5) can now be written as: (RR′x+Ru′ +Rv′t+ u+ vt+ vs′, t+ s′ + s) = (R′′x+ u′′ + v′′t, t+ s′′) = (f ′′(x) + v′′t, t+ s′′) ≡ F ′′(x, t) This defines a self-consistent composition acting on a spacetime pair (x, t) in accordance with the given definition from the onset. 2 docsity.com Solution to 6: Next, we would like the notion of an inverse action F−1. In the set of identifications (6)-(10), if we instead make the alternative identifications, R := R′−1 =⇒ R′′ = I u := R′−1v′s′ −R′−1u′ =⇒ u′′ = 0 v := −R′−1v′ =⇒ v′′ = 0 s := −s′ =⇒ s′′ = 0, (11) then F ≡ F−1 results in (F ◦ F ′)(x, t) = (F−1 ◦ F ′)(x, t) = (x, t). Solution to 7: Lastly, we can now show this defines a group. Recalling that the definition of an algebraic group is a set G, which exhibits under some operation ∗ (taking in two arguments a, b and returning one, c) the properties of closure (∀ a, b ∈ G =⇒ c ∈ G), associativity (∀ a, b, c ∈ G, (a∗ b)∗ c = a∗ (b∗ c)) the existence of an identity (I | aI = Ia = a) and the existence of an inverse (a−1 | aa−1 = a−1a = I). Though in general composition of maps is noncommutative, we want our identity to be well behaved so we check that F−1 ◦ F = F ◦ F−1. Since we have already calculated (F−1 ◦ F ′)(x, t) = (x, t) we need only show that (F ′ ◦ F−1)(x, t) = (x, t). (F ′ ◦ F−1)(x, t) = F ′(f−1(x) + v−1t, t+ s−1) = F ′(R−1x+ u−1 + v−1t, t+ s−1) symbolically = F ′(R′−1x+ (R′−1v′s′ −R′−1u′)−R′−1v′t, t− s′) explicitly = (f ′(R′−1x+R′−1v′s′ −R′−1u′ −R′−1v′t) + v′(t− s′), t− s′ + s′) = (R′(R′−1x+R′−1v′s′ −R′−1u′ −R′−1v′t) + u′ + v′t− v′s′, t) = (R′R′−1x+R′R′−1v′s′ −R′R′−1u′ −R′R′−1v′t+ u′ + v′t− v′s′, t) = (x+ v′s′ − u′ − v′t+ u′ + v′t− v′s′, t) = (x+ v′s′ − v′s′ + v′t− v′t+ u′ − u′, t) = (x, t) Hence, our set has an inverse. The associativity of composition ensures that our maps as defined are associative. The identity element is found immediately. It is simply the object FI(x, t) := (fI(x) + vIt, t+ sI), where fI(x) := RIx+ uI, so that RI := I and uI := 0, vI := 0 and lastly sI := 0. As a consequence of Rn vector spaces forming a group under vector addition and O(n) itself being a group under multiplication it follows that our maps F defined by the triplet (f, v, s) is thus itself a group having satisfied all the definitions above. Solution to 9: We want to find the action of our previous quadruple, instead now on the phase space X. We thus define the following reasonable action and then show it can be made self-consistent. The definition of the action is as follows, given any element (q, p) ∈ X : (u, s, v, R)(q, p) := (Rq + u+ s(Rp+mv) m ,Rp+mv), with the defining successive application of two such maps g and g′ as (gg′)(q, p) := g′′(q, p) where the quadruple g′′ = (u′′, s′′, v′′, R′′) is given in terms of g and g′ as follows: (u′′, s′′, v′′R′′) := (u+Ru′ − s′v, s′ + s, v +Rv′, RR′). 3 docsity.com