Infinitesimal Rotations

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1 Universit of Connecticut Chemistr Education Materials Department of Chemistr Januar 007 Infinitesimal Rotations Carl W. David Universit of Connecticut, Follow this and additional works at: Recommended Citation David, Carl W., "Infinitesimal Rotations" (007). Chemistr Education Materials..

2 Infinitesimal Rotations C. W. David Department of Chemistr Universit of Connecticut Storrs, Connecticut (Dated: Januar 0, 007) I. SYNOPSIS In earlier readings, educ/, finite rotations and their non-commutative algebra were discussed. In this piece, the infinitesimal rotations used to describe the kinematics of rigid bodies is introduced. II. INTRODUCTION When one considers infinitesimal rotations, one is beginning the discussion of the kinematics of rotation usuall for spatiall etended bodies. Consider the rotation about the -ais, and consider that we do a finite rotation first. Then we would have cos θ sin θ 0 R = sin θ cos θ 0 (.) For a θ = 90 degree rotation, this would ield R = which if it operated on the row vector would ield R = (.) (.) What does this mean? When looking at the R rotation, in the top of Figure, we see that in the new coordinate sstem, the coordinates of the point have changed as shown in the two relevant equations (above), vector. transforms into vector given in Equation. Remember, to see what is going on ou have to position our ee along the rotation ais (the positive part), and look into the origin. We now repeat the procedure about the -ais, the new -ais, R = = (.4) as can be seen in the second part of Figure. That leaves us onl with the -rotation, R = It takes some head turning to see these coordinates in Figure. III. NON-COMMUTATIVITY TO COMMUTATIVITY We know that finite rotations are not commutative, that if we did a rotation about the ais and then about the ais, the result would be different from doing the ais rotation first, then the ais rotation. This non-commutativit is not true when we deal with infinitesimal rotations, amaing as that seems. To begin the discussion we first specialie to infinitesimal rotations. Let s assume that θ is an infinitesimal, almost ero. Then since cos 0 = and, to first order, sin θ θ for θ ver, ver small. We then transform Equation. into its infinitesimal form. Let s write this out in components: = dθ 0 dθ 0 which becomes = dθ 0 dθ = + dθ = dθ + = Tpeset b REVTEX

3 This means, Now = dθ = dθ dθ = ω an angular velocit of rotation. d = ω (.) d = ω (.) (.) Equation. is the î component, while Equation. is the ĵ component. Of course, is not changing, has no time derivative. If we attempt to write this is vector notation, we could write d r = ω r = v (.4) ω r = 0 0 ω = ĵω îω v is the linear velocit of a point during the rotation. What we see here is that the signs are wrong. All right, not wrong, just inconsistent. If one wants to use the form of Equation.4 then one is forced to change the direction of rotation used in Equation.. This corresponds to changing our point of view. Instead of rotating the coordinate sstem, we are rotating the particle. which means = dθ 0 dθ 0 d = ω (.5) d = ω (.6) now consistent with Equation.4 IV. WHY ARE WE NOW COMMUTATIVE? (.7) Wh are these infinitesimal rotations commutative? Take as an eample R and R Reversing the direction of rotation, reconsider the rotation about the -ais, and consider that we do a finite rotation first. Then we would have R = cos ψ sin ψ 0 sin ψ cos ψ and now let s assume that ψ is an infinitesimal, almost ero. Then R dψ 0 dψ In components: = dψ 0 dψ = = +dψ + = dψ Continuing for the ˆk component and for the ĵ component Defining = dψ = dψ dψ ω an angular velocit of rotation. ( ) d ĵ = ω ( ) d ˆk = ω where, of course, is not changing, has no time derivative. If we attempt to write this in vector notation, we could write = d r = ω r = v ĵ ˆk = ĵω + ˆkω î ω 0 0

4 v is, again, the linear velocit of a point during the rotation. However, the signs are wrong. We have run into the same difficult here which we had before, which we address in the same manner as before, we need to use the reversed angle if we are going to keep the vector cross product notation. = dψ 0 dψ Lastl, consider the rotation about the -ais, and consider that we do a finite rotation first. Then we would have R = cos φ 0 sin φ 0 0 sin φ 0 cos φ and now let s assume that φ is an infinitesimal, almost ero. Then R 0 dφ 0 0 dφ 0 In components: This means, = 0 dφ 0 0 dφ 0 = = +dφ + = dφ where, of course, is not changing, has no time derivative. If we attempt to write this is vector notation, we could write d r = ω r = v vecω r = 0 ω 0 Notice, that this last one is in consistent notation, compared to the others. For a composite rotation, about a generalied ais, one would have Ω = 0 ω ω +ω 0 ω ω ω 0 This is known as an antismmetric matri. We also attach a vector to the rotation, a pseudovector, which we will call (as before) ω. ω = ω ω ω Using this composite rotation matri, we have v = ω r At last, we have = dφ = dφ v = ω ω ω dφ = ω an angular velocit of rotation. ( ) d î = ω ( ) d ˆk = ω v = î (ω ω ) + ĵ (ω ω ) + ˆk (ω ω ) so the kinetic energ would be µ v v = µ v ( ω r) = µ ( ω r) ( ω r) which would be

5 4 v v would be µ ( ω + ω ω ω + ω + ω ω ω + ω + ω ω ω ) We would then have, epanding, v v = (ω ω ) + (ω ω ) + (ω ω ) v v = (ω + ω ) (ω ω ) + (ω + ω ) (ω ω ) + (ω + ω ) (ω ω ) which can be rearranged to appear as v v = ω + ω + ω + ω + ω + ω (ω ω ) ω ω ω ω to which we add and subtract ω + ω + ω to obtain KE = µ ( r (ω + ω + ω) (ω + ω + ω ) ) KE = µ ( r ω ω ( ω r) ) = î ( ω ω ω + ω ) + Rearranging, ĵ ( ω ω ω + ω ) + ˆk ( ω ω ω ω ) (5.) V. RIGID BODY KINEMATICS Consider a rigid bod rotating about an arbitrar (in space) ais. The angular velocit is v = ω r If the point has mass m, then the linear momentum of the point during the rotation is m v = m ω r The angular momentum of the point is = r p = r m ( ω r) ( = m r (ω ω ) î + ((ω ω ) ĵ + ((ω ω ) ˆk ) from which we obtain L which epands, in components, to be KE = m ω = ( r ω ( ω r) r ) ω (5.) The equation 5.6 can be rewritten as (ω ω ) (ω ω ) (ω ω ) m = (+((ω ω ) ((ω ω )) î + m = r (ω î + ω ĵ + ω ˆk) (ω + ω + ω )(î + ĵ + ˆk) (5.7) (+((ω ω ) ((ω ω )) ĵ + (+((ω ω ) ((ω ω )) ˆk (5.) = î ( ω ( + ) ω ω ) + ĵ ( ω ( + ) ω ω ) + ˆk ( ω ( + ) ω ω ) (5.4) = î ( ω ( + + ) ω ω ω ) + ĵ ( ω ( + + ) ω ω ω ) + ˆk ( ω ( + + ) ω ω ω ) (5.5) m = r ω ( ω r) r (5.6)

6 5 m = ω îr + ω ĵr + ω ˆkr ω î ω ĵ ω ˆk ω ĵ ω ˆk ω î ω ˆk ω î ω ĵ (5.8) which appears to be î(ω ( + ) ω ω ) + ĵ(ω ( + ) ω ω ) + ˆk(ω ( + ) ω ω ) It is tempting to define an object which allows us to write = m }{{} I ω which looks like a matri, smells like a matri, but in truth is actuall a tensor, the tensor of the moment of inertia. It looks something like: = m r r r }{{} ω ω ω (5.9) The difference between a normal (for us) rotation matri and this construct is that a rotation matri carries coordinates into coordinates, the elements of the matri have no units. Here, on the other hand, the angular momentum on the left hand side, has different units than the ω vector, so the construct must have units which makes the equation itself true. VI. EXTENDING THE ARGUMENT TO MORE THAN ONE POINT PARTICLE Consider that this derivation has been conducted for one particle rotating about a point. For more than one mass point (the situation considered when considering molecules), rotating about the common center of gravit, we will have one such equation for each mass i = m i I }{{} i ω assuming the are all rotating together (rigid bod rotations, our main interest here). For the assembl, we then have = i = ( ) m i I i ω }{{} i i = i m i(ri i ) i m i i i i m i i i i m i i i i m i(ri i ) i m i i i i m i i i i m i i i i m i(ri i ) We see that we ve obtained the tensor of the moment of inertia eplicitl. ω ω ω (6.)

7 6 R R 0-0 R FIG. : Sequential Rotations