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Unformatted text preview: 5 Rotations 5.1 Rotation matrix The rotation operator ˆ R ( t ) has a matrix R ij ( t ) defined so that if a vector a = a i e i (where e i are the basis vectors in the lab frame) is rotated into vector b = b i e i , then b = ˆ R a , b i = R ij a j The rotated basis vectors e ′ i ( t ) = ˆ R e i satisfy e ′ i ( t ) = e j R ji ( t ) (2) e i = e ′ j ( t ) ( R − 1 ( t ) ) ji (3) To memorize the above, recall that when a → b by the same rotation which takes e i → e ′ i b = e i b i = e ′ j a j e i b i = e i R ij a j = e ′ j a j The requirement of length conservation defines the rotation. It can be expressed as ( ⃗e i ( t ) · e j ( t )) = δ ij . Of course, the labframe vectors also satisfy ( e i · e j ) = δ ij . From (2) we have δ ij = ( e ′ i ( t ) · e ′ j ( t )) = R mi R nj ( e m · e n ) = R mi R nj δ mn = R mi R mj In matrix notation that means R T R = E, or alternatively R − 1 = R T . 5.2 Instantaneous angular velocity Let us consider any vector r ( t ) = c ′ i e ′ i ( t ) = c i ( t ) e i that is fixed in the rotating body ( c ′ i are time independent). At t = 0 the body has not yet rotated, so e ′ i (0) = e i and c i (0) = c ′ i . Rotation R ( t ) transforms vector r (0) = c ′ i e i into r ( t ) = c i e i , so c i ( t ) = R ij ( t ) c ′ j Differentiating w.r.t. time we get ˙ c i = ˙ R ij c ′ j = ˙ R ij R ( − 1) jk c k = X ik c k , where we defined X jk = ˙ R ji ( R − 1 ) ik = ˙ R ji ( R T ) ik , so X = ˙ RR T . First, we prove that X is an antisymmetric matrix. From RR T = E one gets ˙ RR T + R ˙ R T = 0 , ˙ RR T = − ( ˙ RR T ) T , X = − X T . Since X is antisymmetric, it has a form X = x 12 x 13 − x 12 x 23 − x 13 − x 23 with only three independent elements. 7 The velocity of the point described by r can be determined as ˙ r = ˙ e ′ i c ′ i = e j ˙ R ji c ′ i = e ′ j ˙ R ji ( R − 1 ) ik c k = e j X jk c k Note that we made a substitution ˙ e ′ i = e j ˙ R ji . This is only true for constant e j . Later we will study the subject of rotation in rotating reference frame, where e j are timedependent and expressions become more complicated. Now we want to prove that v can be given by the formula v = [ ⃗ω × r ] (4) with some vector ⃗ω . This equation can be translated into a matrix form. Expand ⃗ω = ω i e i ( t ). Then v = [ ω i e i × c j e j ] = ω i c j [ e i × e j ] = ω i c j ϵ ijk e k = ( ϵ ijk ω j c k ) e i which gives v i = ϵ ijk ω j c k = Y ik c k . In matrix form Y = ω 1 − ω 2 − ω 2 ω 3 ω 2 − ω 3 We see that by choosing ω 1 = x 12 , ω 2 = − x 13 , and ω 3 = x 23 we can set Y = X . With this choice of components ω i we have X ik = ϵ ijk ω j The equivalence between X and Y proves formula (4)....
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This note was uploaded on 12/13/2011 for the course PHYS 701 taught by Professor Bazaliy during the Fall '11 term at South Carolina.
 Fall '11
 Bazaliy
 mechanics

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