Download Euclidean Group and Galilei Group: Definitions and Properties and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
- We can more elegantly define the Euclidean group to be the group of all distance-preserving transformations of Euclidean space. Indeed, if f : Rn^ → Rn^ is a map that preserves distances, that is, ||f (x) − f (y)|| = ||x − y||
for all x, y ∈ Rn, then f (x) = Rx + u
for some (R, u) ∈ E(n).
Proof: We begin with the assumption that f (0) = 0 and thus for any x ∈ Rn, ||f (x)|| = ||x||. Therefore, by showing f is linear, we’ll have shown that f ∈ O(n). Take any x ∈ Rn^ different from zero and any α ∈ R. We have
||αx|| = ||f (αx)|| = ||f (αx) − f (x) + f (x)|| ≤ ||f (αx) − f (x)|| + ||f (x)|| = ||αx − x|| + ||x|| = ||αx||,
and thus equality must occur throughout. However, the triangle inequality says that equality holds if and only if the vectors f (αx) − f (x) and f (x) are linearly dependent. Therefore, f (αx) = βf (x) and thus |α| = |β|. In addition,
|α − 1 |||x|| = ||αx − x|| = ||f (αx) − f (x)|| = ||βf (x) − f (x)|| = |β − 1 |||f (x)|| = |β − 1 |||x||,
from which we can conclude |α − 1 | = |β − 1 | and these together can only happen if α = β. Thus,
f (αx) = αf (x). Now take any x, y ∈ Rn^ such that x 6 = y. Let
z =
f (x + y) − f (x) and w = f (y) −
f (x + y)
and note that
||z|| =
∣f
( (^) x 2
y 2
− f (x)
y 2
x 2
∣f^ (y)^ −^ f
( (^) x 2
y 2
= ||w||.
Note further that
||y − x|| = ||f (y) − f (x)|| = ||z + w|| ≤ ||z|| + ||w||
=
||y − x|| +
||y − x|| = ||y − x||
and thus equality occurs throughout. As before, we must have z = λw for some λ ∈ R and since ||z|| = ||w||, λ = ±1. If z = −w we have f (x) = f (y) which implies ||x − y|| = ||f (x) − f (y)|| = 0 or that x = y contrary to the choice of x and y. Therefore z = w and this says
f (x + y) = f (x) + f (y),
so f is indeed linear and hence f ∈ O(n). For the general case, set f (0) = u and define R(x) = f (x) − u. Then R is also a map that preserves distances and R(0) = 0. From the above, R ∈ O(n) and thus f (x) = Rx + u as desired.
The Galilei Group
Define an element of the Galilei group G(n + 1) to be a triple (f, v, s) where f ∈ E(n), v ∈ Rn^ and s ∈ R. We call f a Euclidean transformation, v a Galilei boost and s a time translation. Any element (f, v, s) ∈ G(n + 1) gives a transformation of (n + 1)-dimensional spacetime
F(f,v,s): Rn+1^ → Rn+
defined by F(f,v,s)(x, t) = (f (x) + vt, t + s)
for all (x, t) ∈ Rn+1. The map F(f,v,s) uniquely determines f, v and s, so we can also think of G(n+1) as a set of maps.
- Given any (f(R,u), v, s), (fR′,u′), v′, s′) ∈ G(n + 1), viewed as maps, we have
F(f(R,u),v,s) ◦ F(f(R′,u′),v′,s′)(x, t) = F(f(R,u),v,s)(R′x + u′^ + v′t, t + s′) = (R(R′x + u′^ + v′t) + u + v(t + s′), (t + s′) + s) = (RR′x + (Ru′^ + u + vs′) + (Rv′^ + v)t, t + (s′^ + s)) = F(f(RR′,Ru′+u+vs′),Rv′+v,s′+s)(x, t)
and since (RR′, Ru′^ + u + vs′) ∈ E(n), Rv′^ + v ∈ Rn^ and s′^ + s ∈ R, G(n + 1) as maps, is closed under composition. For convenience of notation, we’ll henceforth view elements of G(n + 1) as 4-tuples (R, u, v, s) where R ∈ O(n), u, v ∈ Rn^ and s ∈ R and take a binary operation on G(n + 1) defined above, that is (R, u, v, s)(R′, u′, v′, s′) = (RR′, Ru′^ + u + vs′, Rv′^ + v, s′^ + s).
- Note that (1E(n), 0 , 0) = (1O(n), 0 , 0 , 0) ∈ G(n + 1). With this binary operation on G(n + 1), it’s clear that (1E(n), 0 , 0) acts as an identity element. Now take any (R, u, v, s) ∈ G(n + 1) and note (R∗, R∗(sv − u), −R∗v, −s) is also a member of this set. We have
(R, u, v, s)(R∗, R∗(sv − u), −R∗v, −s) = (RR∗, R(R∗(sv − u)) + u − sv, R(−R∗v) + v, s − s) = (1O(n), 0 , 0 , 0)
and
(R∗, R∗(sv − u), −R∗v, −s)(R, u, v, s) = (R∗R, R∗u + R∗(sv − u) + −sR∗v, R∗v − R∗v, −s + s) = (1O(n), 0 , 0 , 0).
Thus, (R∗, R∗(sv − u), −R∗v, −s) acts as the inverse to (R, u, v, s).
- Associativity is routine to check, but the details will be omitted. Therefore, G(n+!) is indeed a group. As a set we have G(n + 1) = E(n) × Rn^ × R. However, it is again not the direct product of these groups, but only a semidirect product
we have d = (v, 1) where v ∈ Rn. Putting this together, we see that the total derivative of F exists and is constant with
JF =
R v 0 1
Therefore,
F (x, t) =
R v 0 1
x t
u s
for some (u, s) ∈ Rn+1^ or F (x, t) = (f(R,u)(x) + vt, t + s) ∈ G(n + 1) as desired. The converse is immediate.
The Free Particle
We have just described how the Euclidean group acts on Euclidean space and how the Galilei group acts on Galilean spacetime. Now we will figure out how the Galilei group acts on the phase space of a free particle! Recall that the phase space of a particle in n-dimensional Euclidean space is X = Rn^ × Rn, where a point (q, p) ∈ X describes the particle’s position and momentum. The various subgroups of the Galilei group act on X as follows:
- The translation group Rn^ is a subgroup of E(n) and thus G(n + 1) in an obvious way, and it acts on X as follows: u(q, p) = (q + u, p) u ∈ Rn.
In other words, to translate a particle we translate its position but leave its momentum alone!
- The orthogonal group O(n) is also a subgroup of E(n) and thus G(n + 1) in an obvious way, and it acts on X as follows:
R(q, p) = (Rq, Rp) R ∈ O(n).
In other words, to rotate a particle we rotate both its position and momentum!
- The group of Galilei boosts Rn^ is a subgroup of G(n + 1) in an obvious way, and it acts on X as follows: v(q, p) = (q, p − mv) v ∈ Rn.
In other words, to boost a particle’s velocity by v we subtract mv from its momentum but leave its position alone!
- Finally, the time translation group R is a subgroup of G(n + 1) in an obvious way, and it acts on X as follows: s(q, p) = (q + sp/m, p) s ∈ R.
This is where we are assuming the particle is free: the force on it is zero, so it moves along at a constant velocity, namely p/m.
- Consider the map defined by
(R, u, v, s)(q, p) = (R(q + sp/m) + (u − sv), Rp − mv).
- The above defines an action of G(n + 1) on X Proof: First recall that (1, 0 , 0 , 0) ∈ G(n + 1) is the identity element. We have
(1, 0 , 0 , 0)(q, p) = (1q, 1 p) = (q, p)
for all (q, p) ∈ X as required. Now take any g 1 = (R, u, v, s), g 2 = (R′, u′, v′, s′) ∈ G(n + 1) and note
g 1 (g 2 (q, p)) = (R, u, v, s)(R′(q + s′p/m) + (u′^ − s′v′), R′p − mv′)
= (R[R′(q + s′p/m) + (u′^ − s′v′) + s(R′p − mv)/m] + (u − sv), R(R′p − mv′) − mv) = (RR′(q + (s′^ + s)p/m) + ((Ru′^ + u + vs′) − (s′^ + s)(Rv′^ + v)), RR′p − m(Rv′^ − v)) = (RR′, Ru′^ + u + vs′, Rv′^ + v, s′^ + s)(q, p) = ((R, u, v, s)(R′, u′, v′, s′))(q, p) = (g 1 g 2 )(q, p)
for all (q, p) ∈ X which completes the proof.
