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Rotations and the Euler angles 1 Rotations

Rotations and the Euler angles1 RotationsConsider two right-handed systems of coordi-nates,XY Zandx1x2x3, rotated arbitrarilywith respect to one another (see Fig. ). Wewould like to be able to link easily the coor-dinates of any vector~Ain the two frames ofreference. Let~eX, ~eY, ~eZbe the unit vectorsfor the axes of the first system, and~e1, ~e2, ~e3the unit vectors for the axes of the secondsystem. Then, by definition:~A=AX~eX+AY~eY+AZ~eZand~A=A1~ e1+A2~e2+A3~e3 XYZxxxAA312 Fig 1. Projection of the same vector~Aonto two different right-handed systems of , we can express one set of projections in terms of the otherone:A1=~e1 ~A= (~e1 ~eX)AX+ (~e1 ~eY)AY+ (~e1 ~eZ)AZA2=~e1 ~A= (~e2 ~eX)AX+ (~e2 ~eY)AY+ (~e2 ~eZ)AZA3=~e1 ~A= (~e3 ~eX)AX+ (~e3 ~eY)AY+ (~e3 ~eZ)AZor, in matrix form: A1A2A3 = ~e1 ~eX~e1 ~eY~e1 ~eZ~e2 ~eX~e2 ~eY~e2 ~eZ~e3 ~eX~e3 ~eY~e3

leading to the rather ugly general formula: Rˆ(φ,θ,ψ) = cosψcosφ−cosθsinψsinφ −sinψcosφ−cosθsinφcosψ sinθsinφ cosψsinφ+cosθcosφsinψ …

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Transcription of Rotations and the Euler angles 1 Rotations

1 Rotations and the Euler angles1 RotationsConsider two right-handed systems of coordi-nates,XY Zandx1x2x3, rotated arbitrarilywith respect to one another (see Fig. ). Wewould like to be able to link easily the coor-dinates of any vector~Ain the two frames ofreference. Let~eX, ~eY, ~eZbe the unit vectorsfor the axes of the first system, and~e1, ~e2, ~e3the unit vectors for the axes of the secondsystem. Then, by definition:~A=AX~eX+AY~eY+AZ~eZand~A=A1~ e1+A2~e2+A3~e3 XYZxxxAA312 Fig 1. Projection of the same vector~Aonto two different right-handed systems of , we can express one set of projections in terms of the otherone:A1=~e1 ~A= (~e1 ~eX)AX+ (~e1 ~eY)AY+ (~e1 ~eZ)AZA2=~e1 ~A= (~e2 ~eX)AX+ (~e2 ~eY)AY+ (~e2 ~eZ)AZA3=~e1 ~A= (~e3 ~eX)AX+ (~e3 ~eY)AY+ (~e3 ~eZ)AZor, in matrix form: A1A2A3 = ~e1 ~eX~e1 ~eY~e1 ~eZ~e2 ~eX~e2 ~eY~e2 ~eZ~e3 ~eX~e3 ~eY~e3 ~eZ AXAYAZ (1)Let us analyze the elements of the 3 3 matrix.

2 By definition~e1 ~eX= cos 1X, where 1 Xis theangle between the two unit vectors~e1and~eX. Similarly, all other elements of this matrix dependonly on the various angles between various sets of axes, but areindependent of the projected vector~A. It follows that for any other vector~B, we will have automatically: B1B2B3 = ~e1 ~eX~e1 ~eY~e1 ~eZ~e2 ~eX~e2 ~eY~e2 ~eZ~e3 ~eX~e3 ~eY~e3 ~eZ BXBYBZ In other words, if we know the 3 3 matrix, then we can find the components of any vector in oneof the systems, if we know them in the continuing, let us introduce some simpler notation.

3 Wewill denote A1A2A3 =~Abody, AXAYAZ =~AXY Zand R= ~e1 ~eX~e1 ~eY~e1 ~eZ~e2 ~eX~e2 ~eY~e2 ~eZ~e3 ~eX~e3 ~eY~e3 ~eZ and therefore we have~Abody= R ~AXY Z. It then follows that:~AXY Z= R 1 ~Abody1where R 1is the inverse of matrix R, and it should be clear that its matrix elements are: R 1= ~eX ~e1~eX ~e2~eX ~e3~eY ~e1~eY ~e2~eY ~e3~eZ ~e1~eZ ~e3~eZ ~e3 If it s not clear, then derive them and check!We can see that the matrix R 1is just the transpose of matrix R( by definition,Mis the transposeofN, , ifmij=njifor alli, j). This property is a consequence of the invariance of thelength of any vector under Rotations .

4 If we denote:~ATbody= A1A2A3 ;~ATXY Z= AXAYAZ then|~A|2=~A ~A=AXAX+AYAY+AZAZ=~ATXY Z ~AXY Z=~ATbody ~Abody(the vector has the same length in any system of coordinates). But~Abody= R ~AXY Z ~ATbody=~ATXY Z RT(this last property can be checked easily using the definition ofthe transposed matrix),and therefore:~ATbody ~Abody= ~ATXY Z RT R ~AXY Z =~ATXY Z RT R ~AXY Zwhich implies that RT R= 1 RT= R property is extremely useful, since it allows us to easily find the inverse of any rotationmatrix, by just taking its Rotation about one axisLet us derive the expression of R3for the case where the axes0 Zand 0x3are parallel, and the sets of axesXYandx1x2arerotated by an angle with respect to one another (see Fig).

5 In this case, we know (see, for instance, discussion of polarcoordinates) that the relationship between the unit vectors is: ~e1= cos ~eX+ sin ~eY~e2= sin ~eX+ cos eY~e3=~eZWe can now compute the various dot products; for instance~e1 ~eX= cos , etc, and we find R3( ) = cos sin 0 sin cos 000 1 123 XYxxxZ Fig 2. Rotation by an angle about theaxisOz= should check that R3( 1) R3( 2) = R3( 1+ 2) meaning that if I rotate first by angle 2followed by a rotation by angle 1(about the same axis!) it s as if I did a single rotation by angle 1+ 2. Which is inverse matrix is then: R 13( ) = RT3( ) = cos sin 0sin cos 0001 = R3( )This makes perfect sense as well; if system 123 is rotated with + with respect to systemXY Z,then systemXY Zis rotated with with respect to 123.

6 As a result, the rotation matrices shouldhave the same form with , and that is precisely what we the same way, we can write down the matrices for Rotations about any other axis. For instance,ifOXand 0x1are kept parallel and we perform a rotation by an angle about them, we find R1( ) = 1000 cos sin 0 sin cos We can now use the fact that any general 3D rotation can be decomposed into a product of 3rotations about 3 different axes, to find the form of a general rotation Euler s anglesWe characterize a general orientation of the body systemx1x2x3with respect to the inertial systemXY Zin terms of the following 3 Rotations :1.

7 Rotation by angle about theZaxis;2. rotation by angle about the newx 1axis, which we will call the line of nodes ;3. rotation by angle about the Rotations are illustrated in the following figure:XYZ1 2 1 2 3 2"331 2"12 We can now write the general rotation matrix that links~Abodywith~AXY Zas the product of the3 Rotations about the corresponding axes: R( , , ) = R3( ) R1( ) R3( ) = cos sin 0 sin cos 000 1 1000 cos sin 0 sin cos cos sin 0 sin cos 000 1 3leading to the rather ugly general formula: R( , , ) = cos cos cos sin sin sin cos cos sin cos sin sin cos sin + cos cos sin sin sin + cos cos cos sin cos sin sin sin cos cos Fortunately, we will never need to use this matrix.

8 All we really need is to be able to write thecomponents of the angular velocity~ in both systems of coordinates. Since~ describes preciselyhow fast the angles vary in time, we have:~ =d~ dt+d~ dt+d~ dt= ~eZ+ ~e1 + ~e3since the three Rotations are about these particular us analyze each contribution to~ .1. ~ =~eZ (with respect toXY Zsystem). Following the Rotations , we find that with respect to123 system, we have:~eZ= cos ~e3+ sin ~e2 = cos ~e3+ sin (sin ~e1+ cos ~e2)and therefore: ~ = sin sin ~e1+ sin cos ~e2+ cos ~e32. ~ =~e1 = (cos ~eX+ sin ~eY) (with respect toXY Z), whereas~e1 = cos ~e1 sin ~e2 ~ = cos ~e1 sin ~e2with respect to ~ =~e3 (with respect to 123), whereas~e3= cos ~e3 sin ~e2 = cos ~eZ sin ( sin ~eX+ cos ~eY) ~ = sin sin ~eX sin cos ~eY+ cos ~eZAdding all three components together, we find that, with respect to the body reference system,~ = (sin sin + cos )~e1+ (sin cos sin )~e2+ (cos + )~e3(2)while with respect to the inertial reference system.

9 ~ = (cos + sin sin )~eX+ (sin sin cos )~eY+ ( + cos )~eZSo if we can solve the EL equations and find how these angles vary in time, we can figure outwhat s the angular speed in either of the two reference Kinetic energy in terms of Euler s anglesLet us choose the CM as the reference point O, and we will choose the principal axes of inertia asthe body reference frame. The total kinetic energy of the object will be:T=12M~V2CM+12~ ICM ~ =12M~V2CM+12 I1 21+I2 32+I3 23 4where 1= sin sin + cos , etc [see Eq. (2)].For an asymmetric top, the general formula is rather complicated, and we will not use it.

10 For asymmetric top withI1=I26=I3, if you put the expressions for 1, 2and 3in and simplify a bit,you find:L=12M~V2CM+I12 2+ 2sin2 +I32 + cos 2 U(~RCM, , , )This is our Lagrangian in terms of our 6 generalized coordinates, namely~RCM, , , .5


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