Rigid Bodies. November 5, 2018

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1 Rigid Bodies November 5, 08 Contents Change of basis 3. Einstein summation convention Passive transformation Active transformation Transformations The Levi-Civita tensor Orthogonal transformations 0. Defining property All real, 3-dimensional representations of orthogonal transformations Unitary representations optional 4 3. The special unitary group in dimensions Rotations as SU Infinitesimal SU matrices Finite transformations Rotation of a real 3-vector using SU Spinors Lorentz transformations Rigid Body Dynamics 0 4. Rotating frames of reference Dynamics in a rotating frame of reference Examples 3 5. Lagrange points Case : ϕ Case : sin ϕ Foucault Pendulum odification to gravity The motion of the pendulum oment of Inertia 3 6. Total torque and total angular momentum oment of intertia Example: oment of inertia of a thin rectangular sheet Rotating reference frame and the Euler equation Torque-free motion Torque-free motion of a symmetric rigid body

2 7 Lagrangian approach to rigid bodies Angular velocity vector Action functional for rigid body motion Symmetric body with torque: tops Conserved quantities Solving Potential energy Slowly precessing top Gyroscopes

3 A rigid body is defined as one in which the distance between any two points in the body remains constant. If we think of the body as made up of individual particles, the distances between any two of these particles is fixed. We may equally well consider a continuum approximation. In either case the complete orientation of the body requires six parameters. Beginning from an arbitrary origin and coordinates, we may locate any one point, P, in the body by three coordinates. Picking any other point, P, the distance d between them is fixed, so P lies on a sphere of radius d centered on P, and we may specify the position of P on this sphere by giving two angles. Finally, with the positions of P and P fixed, any third point P 3 lies in a plane with P and P. Since this plane contains the line connecting P and P, the only freedom in specifying P 3 is a single angle, specifying the orientation of this plane. The positions of all further points are fixed by their distances from these three. We therefore require parameters to fully specify the position and orientation of a rigid body. Fix an orthonormal coordinate system, with basis i, j, k, fixed in space, and a second set, i, j, k, fixed in the rigid body. Think of the origin of the second set as P so that the origins of these two systems separated by the position vector R of the point P. The only remaining difference between the basis vectors will be determined by the three angular variables required to specify P, P 3. Change of basis We now seek the relationship between two orthonormal bases with a common origin. The first key fact is that the transformation is linear, and this is immdediate by the definition of a vector basis. Given a set of basis vectors i, j, k, every vector can be expanded as a linear combination v v i + v j + v 3 k Since any other basis is comprised of vectors, the vectors of the new basis i, j, k may be expanded in the old: and conversely, i a i + a j + a 3 k j a i + a j + a 3 k k a 3 i + a 3 j + a 33 k i b i + b j + b 3 k j b i + b j + b 3 k k b 3 i + b 3 j + b 33 k We consider two types of transformation: passive and active. A passive transformation is one in which any given vector remains fixed while we transform the basis. An active transformation is one in which we leave the basis fixed, but transform all vectors. Thus, for a passive transformation and an arbitrary vector v, we expand in each basis: v v i + v j + v 3 k v i + v j + v 3k where the new basis vectors expanded in the old as in Eq. as above. For an active transformation, we consider two vectors v v i + v j + v 3 k v v i + v j + v 3k 3

4 where, since the transformation must still be linear, the components are related by v A v + A v + A 3 v 3 v A v + A v + A 3 v 3 v 3 A 3 v + A 3 v + A 33 v 3 3 Before looking at the relationship between active and passive transformations, we develop some simpler notation.. Einstein summation convention All of this is much easier in index notation. Let the three basis vectors be denoted by ê i, i,, 3, so that ê î ê ĵ ê 3 ˆk and similarly for ê i. Then the basis transformations, Eqs. and, may be written as ê i ê i 3 a ij ê j j 3 b ij ê j j where a ij is the matrix a ij a a a 3 a a a 3 a 3 a 3 a 33 This lets us write three long equations as a single short one. Note that the first index tells us which row, and the second which column. The expansion of the vector v is 3 v v j ê j and an active transformation as 3 v kêk k j 3 j k 3 A kj v j ê k It is easy to see that this will lead us to write 3 j millions of times. The Einstein convention avoids this by noting that when there is a sum there is also a repeated index j, in the cases above. If we are careful never to repeat an index that we do not sum, we may drop the summation sign. Thus, 3 a ij ê j a ij ê j j 3 b ij ê j b ij ê j j 3 v j ê j v j ê j j 4

5 The repeated index is called a dummy index, and it does not matter what letter we choose for it, v j ê j v k ê k as long as we do not use an index that we have used elsewhere in the same expression. Thus, in the basis change examples above, we cannot use i as the dummy index because it is used to distinguish three independent equations: ê a j ê j ê a j ê j ê 3 a 3j ê j Such an index is called a free index. Free indices must match in every term of an expression. Thus, dummy indices tell us that there is a sum, while free indices tell us which equation we are looking at. Since the basis is orthonormal, we know that the dot product is given by ê i ê j δ ij where δ ij is the Kronecker delta, equal to if i j and to 0 if i j. Notice that the expression above represents nine separate equations. If we repeat the index, we have a single equation ê k ê k δ kk 3 Be sure you understand why the result is 3. We can find the relationship between the matrices a ij and b ij, since, substituting one basis change into the other, ê i a ij ê j a ij b jk ê k a ij b jk ê k Taking the dot product with ê m notice that we cannot use any of the indices i, j or k, we have ê i a ij b jk ê k ê m ê i ê m a ij b jk ê k δ mi a ij b jk ê m ê k a ij b jk δ mk a ij b jm and since δ mi δ im is the identity matrix,, this shows that AB so that the matrix B with components b ij is inverse to A, B A This introduces matrix notation, where A and B are entire matrices with components [A] ij a ij [B] ij b ij When we write a matrix product as AB we understand this to be normal matrix multiplication, in which successive columns of b jm are summed with successive rows of a ij. 5

6 . Passive transformation Consider a passive transformation from ê j to ê i. Substituting for the relationship between the basis vectors, we have v iê i v i ê i v i b ij ê j v i b ij ê j so that the components of the vector v in the new basis are given by v k v i b ik To prove this formally, we take the dot product of both sides of our equation with each of the three basis vectors, ê k : v iê i ê k v i b ij ê j ê k v iδ ik v i b ij δ jk v k v i b ik Notice how the identity matrix simply replaces one index with another..3 Active transformation What active transformation gives the same components for v as a given passive one? That is, suppose we want an active transformation that takes v v i ê i to v v iêi where v k v ib ik. This requires, 3 v kêk k 3 3 A kj v j ê k j k v k A kj v j v j b jk A kj v j The relationship here is the transpose. Let A have components a ij. Then the transpose of A, called A t, has components a ji : [A] ij a ij [ A t ] ij a ji This allows us to write the relation as v j b jk A kj v j v j [ B t ] kj A kjv j Since, for each value of j and k these are just ordinary numbers, we may write them in any order, so that [ B t ] kj v j A kj v j and therefore B t A 6

7 .4 Transformations We will normally write active transformations, in components, or v k [O] ki v i v Ov in matrix notation. Now consider an active rotation of a vector v to a new vector v. Consider a counterclockwise rotation about the z axis through an angle θ. If v begins as a vector v, 0, 0 along the x-axis then it will end up as v v cos θ, v sin θ, 0. Therefore, the first column of O must be: [O] ij cos θ sin θ 0 If v were initially in the y-direction, 0, v, 0, then it would rotate to a negative x-value, v v sin θ, v cos θ, 0. This becomes the second column of O, and since anything in the z direction is unaffected, the full transformation must be cos θ sin θ 0 [O] ij sin θ cos θ To accomplish the same result with a passive rotation, ê i 3 j a ijê j, we would have rotated the basis clockwise by the same angle. The matrix a ij takes the form a ij Notice that the matrix a ij is the inverse of O..5 The Levi-Civita tensor cos θ sin θ 0 sin θ cos θ In 3-dimensions, we define the Levi-Civita tensor, ε ijk, to be totally antisymmetric, so we get a minus sign under interchange of any pair of indices. We work throughout in Cartesian coordinate. This means that most of the 7 components are zero, since, for example, ε ε if we imagine interchanging the two s. This means that the only nonzero components are the ones for which i, j and k all take different values. There are only six of these, and all of their values are determined once we choose any one of them. Define ε 3 Then by antisymmetry it follows that ε 3 ε 3 ε 3 + ε 3 ε 3 ε 3 All other components are zero. Using ε ijk we can write index expressions for the cross product and the curl. The i th component of the cross product is given by [u v] i ε ijk u j v k 5 7

8 note the two implicit sums on j and k as we check by simply writing out the sums for each value of i, [u v] ε jk u j v k ε 3 u v 3 + ε 3 u 3 v + all other terms are zero u v 3 u 3 v [u v] ε jk u j v k ε 3 u 3 v + ε 3 u v 3 u 3 v u v 3 [u v] 3 ε 3jk u j v k u v u v We get the curl simply by replacing u j by j x j, [ v] i ε ijk j v k 6 If we sum Eqs.5 and 6 with basis vectors e i, where e i, e j, e 3 k, we may write the curl and cross product as vectors: u v [u v] i e i ε ijk u j v k e i v ε ijk e i j v k There are useful identities involving pairs of Levi-Civita tensors. The most general is ε ijk ε lmn δ il δ jm δ kn + δ im δ jn δ kl + δ in δ jl δ km δ il δ jn δ km δ in δ jm δ kl δ im δ jl δ kn 7 This happens because on each ε ijk, the i, j and k must be, and 3 in some order, so any individual term may be written in terms of deltas. For example, ε 3 ε 3 +δ δ δ 33 To check the whole identity explicitly, note that both sides are totally antisymmetric on ijk and on lmn. But in three dimensions there can be only one totally antisymmetric object with three indices, up to an overall multiple. Therefore, since the right side is antisymmetric on ijk and on lmn, it must be proportional to both ε ijk and ε lmn, ε ijk ε lmn λ δ il δ jm δ kn + δ im δ jn δ kl + δ in δ jl δ km δ il δ jn δ km δ in δ jm δ kl δ im δ jl δ kn for some λ. To fix λ we only need to check one component, say ε 3 ε 3 : ε 3 ε 3 λ δ δ δ 33 + δ δ 3 δ 3 + δ 3 δ δ 33 δ δ 3 δ 3 δ 3 δ δ 3 δ δ δ 33 λδ δ δ 33 so we must have λ and the full identity holds. We get a second identity by setting n k and summing, ε ijk ε lmk δ il δ jm δ kk + δ im δ jk δ kl + δ ik δ jl δ km δ il δ jk δ km δ ik δ jm δ kl δ im δ jl δ kk 3δ il δ jm + δ im δ jl + δ im δ jl δ il δ jm δ il δ jm 3δ im δ jl 3 δ il δ jm 3 δ im δ jl δ il δ jm δ im δ jl 8

9 so we have a much simpler, and very useful, relation A second sum gives another identity. Setting m j and summing again, ε ijk ε lmk δ il δ jm δ im δ jl 8 ε ijk ε ljk δ il δ mm δ im δ ml 3δ il δ il δ il 9 Setting the last two indices equal and summing provides a check on our normalization, ε ijk ε ijk δ ii 6 This is correct, since there are only six nonzero components and we are summing their squares. Now we use these properties to prove some vector identities. First, consider the triple product corresponding to the volume of the parallelipiped defined by the three vectors, u v w u i [v w] i u i ε ijk v j w k ε ijk u i v j w k Because ε ijk ε kij ε jki, it is unchanged by cyclically permuting the vectors u v w w u v v w u Each of these gives the same form when written in components, e.g., v w u v i ε ijk w j u k ε ijk u k v i w j ε kij u k v i w j. Renaming the dummy indices gives the original expression. Next, consider a double cross product: [u v w] i ε ijk u j [v w] k ε ijk u j ε klm v l w m ε ijk ε klm u j v l w m Now using our second identity, Eq.8, to replace ε ijk ε klm ε ijk ε lmk δ il δ jm δ im δ jl, and using the Kronecker deltas to change indices, [u v w] i δ il δ jm δ im δ jl u j v l w m δ il δ jm u j v l w m δ im δ jl u j v l w m u j v i w j u j v j w i v i u w w i u v Returning to full vector notation, this is the BAC CAB rule, u v w u w v u v w As one final example of the usefulness of the Levi-Civita symbol, look at the curl of a cross product, [ v w] i ε ijk j [v w] k ε ijk j ε klm v l w m Restoring the vector notation, this becomes ε ijk ε klm j v l w m + v l j w m δ il δ jm δ im δ jl j v l w m + v l j w m δ il δ jm j v l w m + v l j w m δ im δ jl j v l w m + v l j w m m v i w m + v i m w m j v j w i v j j w i v w w v + w v v w v w If you doubt the advantages here, try to prove these identities by explicitly writing out all of the components! 9

10 Orthogonal transformations. Defining property The squared length of a vector is given by taking the dot product of a vector with itself, v v v An orthogonal transformation is a linear transformation of a vector space that preserves lengths of vectors. This defining property may therefore be written as a linear transformation, such that v Ov v v v v Write this definition in terms of components using index notation. Setting the components of the transformation to [O] ij O ij v we have i O ij v j v v v v v iv i v i v i O ij v j O ik v k v i v i Notice that we change the name of the dummy indices so that we never have more than two indices repeated. Each term in the expression O ij v j O ik v k, for each value of i, j, k is just a real number, so we may rearrange the terms in any order we like. We may also write the dot product as a double sum, v i v i δ jk v j v k. Then v i v i O ij v j O ik v k δ jk v j v k O ij O ik v j v k or 0 O ij O ik δ jk v j v k If, instead of two copies of v i, we had two arbitrary, independent vectors 0 O ij O ik δ jk u j v k we could conclude that O ij O ik δ jk must vanish. However, since v i v j v j v i is symmetric, we need to derive a further general result. Suppose A ij A ji are the components of any antisymmetric matrix, and S ij S ji are the components of an arbitrary symmetric matrix. Consider the trace of the product, We can evaluate this, just from the symmetries, tr AS tr A ij S jk A ij S ji A ij S ji A ij S ij A ji S ij Since dummy indices can be named arbitrarily, we may rename i as j and j as i to write A ji S ij A ij S ji. Then A ij S ji A ji S ji A ij S ji 0 A ij S ji 0 0

11 Therefore, the full contraction of a symmetric matrix with an antisymmetric matrix always vanishes. Returning to our defining property of orthogonal transformations, and recognizing that the matrix 0 O ij O ik δ jk v j v k ij v i v j is symmetric but otherwise arbitrary, we see that we can make no claim about the antisymmetric part of O ij O ik δ jk, since such a contraction vanishes identically in any case. What we can conclude is that the contraction of O ij O ik δ jk with the arbitrary symmetric matrix v j v k requires the vanishing of the symmetric part, O ij O ik δ jk + O ik O ij δ kj 0 This is enough, because O ij O ik O ik O ij δ jk δ kj and we just have two copies of the same thing, O ij O ik δ jk 0 O ij O ik δ jk and since the components of O t are just [O t ] ij O ji, we may write this in matrix notation as O t O This is the defining property of an orthogonal transformation: O t O 0. All real, 3-dimensional representations of orthogonal transformations While we have found one particular rotation matrix, Eq.4, knowing the defining property Eq.0 allows us to find an expression for any orthogonal transformation. Notice that the transpose of the identity is the inverse of the identity, t, so we may consider linear transformations near the identity which also satisfy Eq.0. Let O + ε where the components [ε] ij ε ij are all small, ε ij for all i, j. Keeping only terms to first order in ε ij, the transpose is just O t + ε t The inverse of a transformation near the identity must also be near the identity, so to find the right form for O to first order, set OO + ε + δ + ε + δ εδ + ε + δ

12 To have the inverse requires only δ ε, correct to first order in ε. Now we impose Eq.0, O t O + ε t ε ε t ε so that the matrix ε must be antisymmetric. Every infinitesimal orthogonal transformation differs from the identity by an antisymmetric matrix. Next, we write the most general antisymmetric 3 3 matrix as a linear combination of a convenient basis, ε w i J i w w w 3 w 0 w w 3 w 0 + w w 3 Notice that the components of the three matrices J i are neatly summarized by [J i ] jk ε ijk where ε ijk is the totally antisymmetric Levi-Civita symbol. The matrices J i are called the generators of the transformations. Knowing the generators is enough to recover an arbitrary rotation. Starting with O + ε + ɛn i J i where ɛ is a small parameter and n any unit vector, we may apply O repeatedly, taking the limit O θ lim n On lim n + ɛn ij i n where the limit is taken in such a way that we hold the product nɛ constant, nɛ θ where θ is finite. Quite generally, this limit gives rise to an exponential. Theorem: For any matrix, lim + n ɛn exp θ where the limit is taken so that nɛ θ remains finite. Proof Noting that the calculation does not depend on the size or details of the matrix, we use the binomial expansion, a + b n n n! k0 k!n k! an k b k we have O θ, n lim n On lim + ɛn n lim n n k0 lim n n k0 n! k! n k! n k ɛ k n n n n k + ɛ k k k!

13 ultiplying and dividing by n k, O θ, n lim n k0 lim n k0 k0 n n n n n k + nɛ k k! n k n n n n n k n θ k k! k! θk exp θ We define the exponential of a matrix by the power series for the exponential, which we can compute because we know how to find powers of the matrix. To find the form of a general rotation, we apply the theorem with n i J i. We therefore need to find powers of n i J i. This turns out to be straightforward because of our choice for the three matrices J i. Remembering that n i n j δ ij n n, [n i J i ] jk n i ε ijk [ n i J i ] mn n i ε imk n j ε jkn n i n j ε imk ε jkn n i n j δ ij δ mn δ in δ jm n i n j δ ij δ mn + n m n n k [ n i J i 3] mn n m n n δ mn n m n k δ mk n i ε ikn n m n k n i ε ikn δ mk n i ε ikn n i ε imn [n i J i ] mn The powers come back to n i J i with only a sign change, so we can divide the series into even and odd powers, [ n i J i l] l n m n n δ mn for all l > 0 mn [ n i J i l+] l [n i J i ] mn for all l 0 mn We can now compute the exponential explicitly: [O θ, ˆn] mn [exp θn i J i ] [ mn ] k! θn ij i k k0 [ n i J i 0] mn + δ mn + l mn [ l ] θ l l! n ij i l l l! δ mn n m n n + mn l0 + [ l0 ] θ l+ l +! n ij i l+ l l +! [n ij i ] mn This first sum is cos θ because the l 0 term is missing from the first sum, while the second sum is sin θ, so a general rotation matrix takes the form [O θ, ˆn] mn [exp θn i J i ] mn δ mn + cos θ δ mn n m n n + sin θ [n i J i ] mn mn 3

14 This is the matrix for a rotation through an angle θ around an axis in the direction of n. To see this, let O act on an arbitrary vector v, and write the result in normal vector notation, [O θ, ˆn] mn v n δ mn + cos θ δ mn n m n n + sin θ [n i J i ] mn v n δ mn v n + cos θ δ mn v n n m n n v n + sin θ [n i J i ] mn v n v m + cos θ v m n m n n v n + sin θε imn n i v n Define the components of v parallel and perpendicular to the unit vector n: In terms of these, v v n n v v v n n O θ, ˆn v v + v v n n cos θ sin θ n v v + v cos θ sin θ n v v + v cos θ sin θ n v since n v n v + v n v. This expresses the rotated vector in terms of three mutually perpendicular vectors, v, v, n v. The direction n is the axis of the rotation. The part of v parallel to n is therefore unchanged. The rotation takes place in the plane perpendicular to n, and this plane is spanned by v, n v. The rotation in this plane takes v into the linear combination v cos θ n v sin θ, which is exactly what we expect for a rotation of v through an angle θ according to the form of Eq.4. The rotation O θ, ˆn is therefore a rotation by θ around the axis ˆn. 3 Unitary representations optional 3. The special unitary group in dimensions Every orthogonal group SO n, rotations in n real dimensions may be written as special cases of rotations in a related complex space. For SO 3, it turns out that unitary transformations of a complex, -dimensional space can be used to rotate real 3-vectors. We show that we can write a real, 3-dimensional vector as a complex, traceless hermitian matrix. Consider a linear combinations of traceless, Hermitian matrices, where σ i are the three Pauli matrices, v v σ + v σ + v 3 σ 3 v i σ i σ σ σ i i Prove, as an exercise, that these are the only traceless, Hermitian matrics. This means that there is a one to one correspondence between -dimensional traceless Hermitian matrics and 3-dimensional real vectors, v v i σ i v 3 v iv v + iv v 3 4

15 Notice that linear combinations of these matrices remain traceless and Hermitian. The squared length of the vector v may be written as the negative of the determinant, v det v det 3 v iv v + iv v 3 [ v3 v + iv v iv ] v + v + v3 v These facts give us a second way to describe orthogonal transformations. For the usual real representation of vectors, we have already seen that a real, linear transformation satisfying O t O is an orthogonal transformation. For the matrix representation, we require a similarity transformation, which preserves three properties:. Vanishing trace: v UvU tr v tr UvU 0 We show below that v will be traceless if v is provided the determinant of U is one, det U.. Hermiticity: v UvU U v U U vu For this to reproduce v, we need U U, or equivalently, U U, the condition that U be unitary. 3. Determinant: det v det UvU det U det v det U det v These properties guarantee that v has the form v i σ i for three real numbers v i, if and only if v does. Furthermore, since the determinant is preserved, the vectors v and v have the same length. Therefore, U produces an orthogonal transformation if and only if U is unitary, with determinant. The group of matrices with these two properties is called SU, the group of special i.e, det U unitary U U transformations in -dimensions. 3. Rotations as SU 3.. Infinitesimal SU matrices We now find the general form of an SU transformation. Starting from an infinitesimal rotation, U +ε, we require UU + ε + ε + ε + ε + O ε ε + ε 0 ε ε 5

16 a + b c id which means the generator must be anti-hermitian. Let ε ih, where h is Hermitian, h h c + id a b We also need the determinant of U to be. To first order, with a + b c id + ia + ib ic + d U + i c + id a b ic d + ia ib and a, b, c, d all small, this gives det U + ia + ib ic + d det ic d + ia ib + ia + ib + ia ib ic + d ic d + ia ib + ia a + ab + ib ba + b + c + d + ia ib + ia + ib + second order Dropping the second order terms and canceling the on each side leaves ia0 and therefore a 0. This means that h is traceless, and U may be written as a linear combination of the Pauli matrices, U + iεn σ with ε and n a unit vector. 3.. Finite transformations Finite SU transformations may be found by taking the limit of many infinitesimal transformations. Using our general result [.] where U ϕ, n lim + iεn σk k iϕ exp n σ lim n nε ϕ The exponential is defined as the power series, iϕ U exp n σ k0 k! iϕ n σ so we need to compute powers of the Pauli matrices. For this it is helpful to have the product of any two Pauli matrices, σ i σ j δ ij + iε ijk σ k which you are invited to prove. Then n σ n σ k. 6

17 n i σ i n j σ j n i n j σ i σ j n i n j δ ij + iε ijk σ k n i n j δ ij + iε ijk n i n j σ k n n + i n n σ Higher powers follow immediately. For all k 0,,,... The exponential becomes U iϕ n σ k0 iϕ n σ m+ k ϕ k+ i n σ ia σk k! k0 k k! cos ϕ + in σ sin ϕ k k ϕ k ϕ k + in σ k0 k ϕ k+ k +! There are important things to be gained from the SU representation of rotations. First, it is much easier to work with the Pauli matrices than it is with 3 3 matrices. Although the generators in the - and 3-dimensional cases are simple, the exponentials are not. The exponential of the J i matrices is rather complicated, while the exponential of the Pauli matrices may again be expressed in terms of the Pauli matrices, e iϕ n σ cos ϕ + in σ sin ϕ and this substantially simplifies calculations Rotation of a real 3-vector using SU Now apply this to a 3-vector, written as We have X x σ X x σ U x σ U cos ϕ + in σ sin ϕ x σ cos ϕ in σ sin ϕ x σ cos ϕ cos ϕ + i n σ x σ cos ϕ sin ϕ i x σ n σ cos ϕ sin ϕ + n σ x σ n σ sin ϕ sin ϕ x σ cos ϕ cos ϕ + in ix j [σ i, σ j ] cos ϕ sin ϕ + n σ x σ n σ sin ϕ sin ϕ 7

18 Evaluating the products of Pauli matrices, Substituting, in i x j [σ i, σ j ] in i x j iε ijk σ k n x σ n σ x σ n σ n σ x i n j δ ij + iε ijk σ k n σ x n + i x n σ x n n σ + i n σ x n σ x n n σ + in i x n j σ i σ j x n n σ + in i x n j δ ij + iε ijk σ k x n n σ + in x n n x n σ x n n σ x n n n x n σ x n n σ x σ x σ x σ cos ϕ cos ϕ n x σ cos ϕ sin ϕ + x n n σ x σ sin ϕ sin ϕ x σ cos ϕ cos ϕ sin ϕ sin ϕ n x σ cos ϕ sin ϕ + x n n σ sin ϕ sin ϕ x σ cos ϕ n x σ sin ϕ + x n n σ cos ϕ [x cos ϕ n x sin ϕ + x n n cos ϕ] σ [x x n n cos ϕ n x sin ϕ + x n n] σ and equating coefficients, x x x n n cos ϕ n x sin ϕ + x n n x + x cos ϕ n x sin ϕ which is the same transformation as we derived from SO Spinors Because the transformation x UxU is quadratic in U, we find double angle formulas and the real rotation is through an angle ϕ, not ϕ, This means that as ϕ runs from 0 to π, the 3-dim angle only runs from 0 to π, and a complete cycle requires ϕ to climb to 4π. Notice that U and U give the same rotation of x. ore importantly, there is a crucial physical insight. The transformations U act on our hermitian matrices by a similarity transformation, but they also act on some -dimensional complex vector space. Denote a α vector in this space as ψ, so that the transformation of ψ is given by β ψ e iϕ n σ ψ This transformation preserves the hermitian norm of ψ, since ψ ψ ψ U Uψ ψ U U ψ ψ ψ The complex vector ψ is called a spinor, with its first and second components being called spin up and spin down. While spinors were not discovered physically until quantum mechanics, their existence is predictable 8

19 classically from the properties of rotations. Also notice that as ϕ runs from 0 to π, ψ changes by only ψ e iπn σ ψ cos π + in σ sin π ψ ψ so while a vector rotates by π, a spinor changes sign. A complete cycle of an SU transformation therefore requires ϕ to run through 4π. 3.3 Lorentz transformations A simple generlization of SU allows us to describe the Lorentz group. With a general 4-vector given by v α v 0, v, v, v 3 v 0, v we define a Lorentz transformation as any transformation preserving the proper lengths of 4-vectors, s v 0 + v Consider the space of -dim hermitian matrices. A general complex matrix may we written as α γ A β δ If A is hermitian then we require α γ A β δ A ᾱ β γ δ so that α ᾱ a, δ δ b and β γ, where a, b are real and an overbar denotes complex conjugation. We therefore may write any -dimensional Hermitian matrix as a β A β b Introducing four real numbers, x, y, z, t, and setting a t + z, b t z and β x + iy, A takes the form t + z x iy A x + iy t z i 0 t + x + y + z 0 0 i 0 0 t + x σ Hermitian matrices therefore form a real, 4-dimensional space. We choose the identity and the Pauli matrices as a basis for this 4-dim vector space. Now consider the determinant of A, t + z x iy det A det x + iy t z t + z t z x + iy x iy t z x y s This is the proper length of a 4-vector in spacetime, which means that any transformation which preserves the hermiticity and determinant of A is a Lorentz transformation. 9

20 It is now easy to write the Lorentz transformations. The most general, linear transformation of a matrix is by similarity transformation, so we consider any transformation of the form A LAL where we use L on the right so that the new matrix is hermitian whenever A is, L also preserves the determinant provided A LAL L A L LAL A det A det LAL det L det A det L det L det L det L so that det L ±. The positive determinant transformations preserve the direction of time and are called orthochronos, forming the special linear group in complex dimensions, SL, C. Since these transformations preserve τ t z x y, they are Lorentz transformations. 4 Rigid Body Dynamics So far the problems we have addressed make use of inertial frames of reference, i.e., those vector bases in which Newton s second law holds even though we have used the principle of least action, which allows general coordinates. However, it is sometimes useful to use non-inertial frames, and particularly when a system is rotating. When we affix an orthonormal frame to the surface of Earth, for example, that frame rotates with Earth s motion and is therefore non-inertial. The effect of this is to add terms to the acceleration due to the acceleration of the reference frame. Typically, these terms can be brought to the force side of the equation, giving rise to the idea of fictitious forces centrifugal force and the Coriolis force are examples. Here we concern ourselves with rotating frames of reference. 4. Rotating frames of reference Using our expression for a general rotation, it is fairly easy to include the effect of a rotating vector basis. Consider the change, db, of some physical quantity describing a rotating body. We write this in two different reference frames, one inertial and one rotating with the body. The difference between these will be the change due to the rotation, db inertial db body + db rot Now consider an infinitesimal rotation. We found that the transformation matrix for an infinitesimal rotation must have the form O dθ, ˆn dθˆn J where [J i ] jk ε ijk 0

21 and we write dθ so that the rotation is in the counterclockwise positive direction. To check the direction of rotation, we may choose ˆn ˆk and act on a unit vector î in the x direction, [O dθ, ˆn] jk i k δ jk dθn i ε ijk i k i j dθε 3j, 0, 0 + dθ 0,, 0 since ε 3j must have j to be nonzero, and ε 3. The vector acquires a positive y-component, and has therefore rotated counterclockwise, as desired. Suppose a vector at time t, b t is fixed in a body which rotates with angular velocity ω dθ dt n. Then after a time dt it will have rotated through an angle dθ ωdt, so that at time t + dt the vector is In components, Therefore, returning to vector notation, b t + dt O ωt, ˆn b t b j t + dt δ jk ωtn i ε ijk b k t δ jk b k t ωt n i ε ijk b k t b j t ωdt ε kij b k t n i b t + dt b t ωdtb t n Dividing by dt we get the rate of change due to the rotation, db t dt ω b t If, instead of remaining fixed in the rotating system, b t moves relative to the rotating body, its rate of change is the sum of this change and the rate of change due to rotation, db db + ω b t dt dt inertial and since b t is arbitrary, we can make the operator identification expressing the rate of change of any quantity relative to an inertial frame of reference in terms of its rate of change relative to the rotating body, d d + ω dt dt inertial body 4. Dynamics in a rotating frame of reference Consider two frames of reference with a common origin an inertial frame, and a rotating frame. Let r t be the position vector of a particle in the rotating frame of reference. Then the velocity of the particle in an inertial frame, v inertial, and the velocity in the rotating frame, v body, are related by dr dt inertial dr dt body body v inertial v body + ω r To find the acceleration, we apply the operator again, a inertial dv inertial dt + ω r

22 d dt + ω v body + ω r d v body + ω r + ω v body + ω r dt dv + dω dt body dt r + ω dr dt + ω v body + ω ω r dv + dω dt dt r + ω v body + ω ω r body The accelerations are therefore related by a inertial a body + dω dt r + ω v body + ω ω r Since Newton s second law holds in the inertial frame, we have F ma inertial where F refers to any applied forces. Therefore, bringing the extra terms to the left, the acceleration in the rotating frame is given by F m dω dt r mω v body mω ω r ma body This is the Coriolis theorem. We consider each term. The first m dω dt applies only if the rate of rotation is changing. The direction makes sense, because if the angular velocity is increasing, then dω dt is in the direction of the rotation and the inertia of the particle will resist this change. The effective force is therefore in the opposite direction. The second term mω v body is called the Coriolis force. Notice that it is greatest if the velocity is perpendicular to the axis of rotation. This corresponds to motion which, for positive v body, moves the particle further from the axis of rotation. Since the velocity required to stay above a point on a rotating body increases with increasing distance from the axis, the particle will be moving too slow to keep up. It therefore appears that a force is acting in the direction opposite to the direction of rotation. For example, consider a particle at Earth s equator which is gaining altitude. Since Earth rotates from west to east, the rising particle will fall behind and therefore seem to accelerate from toward the west. The final term mω ω r is the familiar centrifugal force that is, the centripetal acceleration treated as a force. For Earth s rotation, ω r is the direction of the velocity of a body rotating with Earth, and direction of the centrifugal force is therefore directly away from the axis of rotation. The effect is due to the tendency of the body to move in a straight line in the inertial frame, hence away from the axis. For a particle at the equator, the centrifugal force is directed radially outward, opposing the force of gravity. The net acceleration due to gravity and the centrifugal acceleration is therefore, g eff g ω r g.035

23 so that the gravitational attraction is reduced by about 3.5%. Since the effect is absent near the poles, Earth is not a perfect sphere, but has an equatorial bulge. 5 Examples 5. Lagrange points Consider the three body problem for gravitation, in which to much more massive bodies move about one another in a circular orbit, with the third, lighter body moving in their combined potential. Assume the motions of all three bodies lie in a single plane. The action for the full system is S ˆt 0 Ẋ + Ẋ + mẋ + G X X + G m X x + G m X x We treat this as a perturbative problem, with m,, so that terms of order m are completely negligible imagine a small rock or satellite in the neighborhood of the earth and the moon. oving to center of mass coordinates, we set R X + X + mx + + m X + X r X X where +. Inverting as we did for the two body problem, and the dominant terms of the action become Ẋ + Ẋ + G X X X R + r X R r Ṙ + µṙ + G r Let our coordinate system be at rest at the center of mass, so the first term drops out. The action is now S ˆt 0 ˆt 0 µṙ + mẋ + G r + Gm r x + G m r x µ ṙ + r θ + r sin θ ϕ + mẋ + G r + Gm r x + G m r x Let the common plane of the motion be specified by θ π. Then for a circular orbit at r r 0, S ˆt The conserved energy is 0 µr ϕ + m ẋ + x ϕ G x + r E µr ϕ + m ẋ + x ϕ G x r 3 + Gm r x + G m r x G m r x G m r x

24 For a circular orbit, we expect G r 0 µr 0 ϕ ω 0 ϕ G r 3 0 ove to a frame rotating with this angular velocity. Since we are looking for stationary points, ϕ x will take this same value, ω 0, so the energy reduces to E mẋ + and effective potential for the satellite is, Gm r0 3 x G m r x G m r x V eff Gm r 3 x G m r x G m r x We need to expand the denominators. Define L r x r + x + rx cos ϕ L r + x rx cos ϕ remembering that x is the vector to m from the center of mass, while r is the separation of and, with ϕ the angle between them. The effective potential is We seek the extrema, V eff Gm r 3 x G m G m L L so we will need Therefore, 0 V eff ˆx x + ˆϕ Gm x ϕ r 3 x G m L L L 3 ˆx x + ˆϕ x ϕ L 3 ˆx x + r cos ϕ L 3 ˆx x + r cos ϕ ˆϕ L L 3 ˆx x r cos ϕ ˆϕ G m L r + x + + ˆϕ rx cos ϕ r sin ϕ r sin ϕ r sin ϕ 0 V eff Gm ˆx r 3 x + G m L 3 x + r cos ϕ + G m L 3 x r cos ϕ +ˆϕ L 3 L 3 Gµmr sin ϕ There are two cases to consider, depending on whether sin ϕ 0 or not. 4

25 5.. Case : ϕ 0 If ϕ 0 then only the ˆx term remains, and L, simplify, Therefore, 0 Gm x + G m r 3 Gm L 3 L r 3 x + L L r + x + rx x + r L x r x + r + G m L 3 x r 0 r 3 xl L + L L r 3 x r x r + x + r x r + x This is a fifth order equation. Expanding the squares, the first term becomes A r 3 x r x r + x r 3 x r rx + x r 3 x µ r + rx + x r4 + µ r3 x + r x µ r3 x 4µ r x rx3 + r x + rx3 + x 4 r 3 x x 4 + rx 3 + r x + µ r 3 x + µ r4 The final pair of terms together become Putting these together, 0 A + B B r x r + x r rx + x r + rx + x 3 r rx + x 3 r rx x + x + r 3 xx5 + r x x + rx r r x3 + µ rx r 5 x + µ rx

26 r 3 x5 + r x4 + r x3 + µ x + x + µ rx + rx r 0 x 5 x x 3 r r r + 3µ x + µ + x r r We consider the much simpler case when, so that µ 0 x 5 r x 3 5 r 6 x r 0 x r 4 x r and we can solve exactly. The first term vanishes if x r 5 6 x r ± ± 4 4 We must have the positive root, so the full equation factors as x x 0 r r + 5 x 4 r Case : sin ϕ 0 In this case, we first solve the ˆϕ equation, 0 ˆϕ If the sine does not vanish then we must have L 3 L 3 L L Gµmr sin ϕ 4 r + x + rx cos ϕ r + x rx cos ϕ rx cos ϕ + rx cos ϕ r r + x cos ϕ r x cos ϕ r Replacing x cos ϕ with this expression, L L L, with. Then the condition reduces to L r + x + rx cos ϕ r + x + r µ r + x 6

27 ultiply the ˆx equation by x and substitute for L and x cos ϕ, 0 r 3 x + L 3 x + rx cos ϕ + L 3 x rx cos ϕ r 3 x + L 3 x + x + µr µr so that r 3 x + L 3 x and we have two Lagrange points with nonzero ϕ. 5. Foucault Pendulum L 3 x r 3 x L 6 r 6 µ r + x 3 r 6 µ x 3 + r µ x + r x ± µ r We apply the Coriolis theorem, Eq., to the motion of a pendulum at latitude α. F m dω dt r mω v body mω ω r ma body The rotating frame here is Earth s daily rotation. It is important to have an idea of the relative magnitudes of the various terms. The external force F is due to gravity and tension, both of which are proportional to mg. The second term vanishes because the angular velocity of Earth is extremely close to constant. For the Coriolis and centripetal terms, we need the magnitude of the angular velocity of Earth, which is ω π day day sec Hz The radius of Earth is about m while the velocity of a 0m pendulum is about v gl 9.90 m s. Therefore, the acceleration due to the Coriolis term is of order and the centripetal acceleration is of order vω 9.90 m s Hz m s ω R.0337 m s Therefore, g ω R ωv 7

28 5.. odification to gravity Because the centripetal acceleration does not depend on the motion of the pendulum, we simply regard it as part of the acceleration of gravity. In fact, it is this combined force that determines what we regard as downward, rather than the direction toward the center of mass of Earth. Therefore, if g G R ˆr is the true gravitational acceleration, the acceleration we measure at the surface is g g ω ω r Choose both Cartesian î, ˆk ĵ, and spherical ˆr, ˆθ, ˆϕ coordinates at rest in the Earth, with r R the radius of Earth, θ π α the angle from the north pole, where α is the latitude. These are related by ˆr î sin θ cos ϕ + ĵ sin θ sin ϕ + ˆk cos θ ˆθ î cos θ cos ϕ + ĵ cos θ sin ϕ ˆk sin θ ˆϕ î sin θ sin ϕ + ĵ sin θ cos ϕ The problem is symmetric in ϕ, so we rotate the coordinates so that the pendulum is at ϕ 0. Then the unit normal to Earth s surface in the Cartesian basis is ˆr î sin θ + ˆk cos θ î cos α + ˆk sin α ˆθ î cos θ ˆk sin θ î sin α ˆk cos α With the angular momentum given by ω ωˆk, the centripetal acceleration is ω ω r ωˆk ωˆk Rˆr ωˆk ωˆk R î cos α + ˆk sin α ωˆk ωr cos αĵ ω R cos αî or equivalently, ω ω r ω R cos α ˆr cos α + ˆθ sin α Therefore, the actual downward direction is slightly away from the direction to the center of mass. This makes a fractional correction of about ω R g to the acceleration of gravity. The pendulum, when stationary, will hang in the direction of g and have magnitude g g + ω R cos α + ω 4 R cos α sin α since ω 4 R is only of order The motion of the pendulum g + gω R cos α + ω 4 R cos α g + gω R cos α We now change to Cartesian coordinates centered on the pendulum. Let ˆt be a unit vector opposite the local down, that is, ˆt g g, and let û, ŵ complete an orthonormal basis, pointing to the East and North, respectively. The equation of motion is now T mg mω v mẍ 8

29 where T is the tension and ẍ lies in the uw plane. Let the angle β of oscillation of the pendulum be very small and the supporting cable of very long length L so that the arc of the pendulum is sufficiently close to horizontal motion. To be precise about the position, when the pendulum is displaced through an arc length βl its gain in height is L L cos β Lβ, so the position of the pendulum bob is x Lβˆt + L sin β û cos ϕ + ŵ sin ϕ Lβ û cos ϕ + ŵ sin ϕ 3 where ϕ is the angle in the uw plane measured from the u-axis. The tension is T T cos βˆt + T sin βˆx T ˆT where ˆx is the direction of the displacement from equilibrium given by Eq.3. Decompose the equation of motion in directions parallel and perpendicular to ˆT. There is no acceleration in the along T so, ˆT T mg mω v m ˆT ẍ T mg cos β m ˆT ω v 0 and since ωv is of order 0 4 we may drop it and write Noticing that to first order in β, T mg cos β ˆx ˆT cos βˆt + sin βˆx Lβ + Lβ 0 so the remaining equation is in the horizontal plane. Writing out the remaining equation in the uw plane: Lβˆt + Lβˆx mg sin βˆx mω v mẍ we may write g sin βˆx g βˆx g L Lβˆx. Then defining ω 0 the displacement, the first term becomes g L mg sin βˆx mω 0x To find the Coriolis force, we write the angular velocity of Earth in the new basis, Then ω ˆtω cos θ + ŵω sin θ ˆtω sin α + ŵω cos α ω vˆx ˆtω sin α + ŵω cos α uû + vŵ uŵ vû ω sin α ˆt uω cos α and recognizing βl as the magnitude of 9

30 The ˆt component is a tiny contribution to the tension in the cable, and the tension always adjusts to balance any force applied in its own direction, so we may drop this. Expandint the resulting equations in the ŵ, û basis and cancelling m, we have so we have two coupled equations of motion, ω 0 uû + wŵ uŵ ẇû ω sin α üû + ẅŵ ü ẇω sin α + ω 0u 0 ẅ + uω sin α + ω 0w 0 We may write these as a single complex equation. Let z u + iw. Then multiplying the second by i and adding, z + iżω sin α + ω 0z 0 To decouple the two directions, define the precession velocity, ω p ω sin α, and let z se iωpt This means that z is just s together with a slow rotation in the plane. Finding derivatives, and substituting, This is simple harmonic with z se iωpt ż ṡ iω p s e iωpt z s iω p ṡ e iωpt iω p ṡ iω p s e iωpt s iω p ṡ ω ps e iωpt 0 z + iżω sin α + ω 0z s iω p ṡ ω ps + i ṡ iω p s ω p + ω 0s e iωpt 0 s iω p ṡ ω ps + iω p ṡ + ω ps + ω 0s s + ω 0 + ω p s s s e iωt + s e iωt z s e iωt + s e iωt e iωpt where ω ω0 + ω g p L + ω sin α. For L 0 m and g 9.8 m s, ω 0 while ω 7 0 5, so ωp is neglibible compared to ω0. Let s 0 s 0 and ṡ 0 0, both real, at t 0. Then s r e iϕ e iωt + r e iϕ e iωt r e iωt ϕ + r e iωt+ϕ s 0 r e iϕ + r e iϕ r cos ϕ i sin ϕ + r cos ϕ + i sin ϕ s 0 r cos ϕ + r cos ϕ 0 i r sin ϕ r sin ϕ ṡ 0 iωr e iϕ iωr e iϕ iωr cos ϕ i sin ϕ iωr cos ϕ + i sin ϕ ṡ 0 ωr sin ϕ + ωr sin ϕ 0 iω r cos ϕ r cos ϕ 30

31 Therefore, s 0 r cos ϕ + r cos ϕ 0 r sin ϕ r sin ϕ ṡ 0 r sin ϕ + r sin ϕ 0 r cos ϕ r cos ϕ Solving r cos ϕ r cos ϕ s 0 r cos ϕ r sin ϕ r sin ϕ ṡ 0 r sin ϕ If we take the initial velocity to be zero then ϕ 0 and we have Then ϕ 0 and Finally r r cos ϕ s 0 r 0 r sin ϕ ṡ 0 0 r r s 0 r ṡ 0 0 s s 0 e iωt + e iωt s 0 cos ωt and the pendulum simply oscillates along the original displacement. To this, e iωpt adds a very slow precession, so that z u + iv cos ω p t i sin ω p t s 0 cos ωt u s 0 cos ωt cos ω p t v s 0 cos ωt sin ω p t The basic g L oscillation is relatively fast, so seeing it, we separate the motion into two parts oscillation and precession instead of a single looping motion. 6 oment of Inertia 6. Total torque and total angular momentum Fix an arbitrary inertial frame of reference, and consider a rigid body. Consider the total torque on the body. The torque on the i th particle due to internal forces will be τ i N r i F ji j 3

32 where F ji is the force exerted by the j th particle on the i th particle, with F ii 0. The total torque on the body is therefore the double sum, τ internal N i j N r i F ji N i<j j N i< j N r i F ji + r j F ij N r i r j F ji where we use Newton s third law in the last step. However, we assume that the forces between particles within the rigid body are along the line joining the two particles, so we have F ji F ji r i r j r i r j and all the cross products vanish. The total internal torque is thus zero: τ internal 0 Therefore, we consider only external forces acting on the body when we compute the torque. For a macroscopic rigid body it is easier to work in the continuum limit. Let the density at each point of the body be ρ r for a discrete collection of masses, we may let ρ be a sum of Dirac delta functions and recover the discrete picture. The contribution to the total torque of an external force df r acting at position r of the body is dτ r df r and the total follows by integrating this. For a small increment of mass, dm ρ r d 3 x, the force using Newton s second law is df r dm dv dt dv dt ρ r d3 x. Integrating the resulting torque, we have ˆ τ ρ r r dv d 3 x dt ˆ [ ] d dr ρ r r v dt dt v d 3 x Since dr dt v v v 0, and the density is independent of time, τ d ˆ ρ r r v d 3 x dt Compare this to the total angular momentum. The angular momentum dl for a small mass element, dm ρ r d 3 x, is dl ρ r r v d 3 x. Integrating over the whole body gives the total angular momentum, ˆ L ρ r r v d 3 x so the total external torque equals the rate of change of total angular momentum, τ dl dt 4 3

33 6. oment of intertia As it stands, the angular momentum of a rigid body depends on both the velocity of its parts and their arrangement. To describe the motion of a given rigid body, it is useful to separate properties of the body itself from the dynamical properties dependent only on its state of motion. We accomplish this as follows. Suppose the rigid body rotates with angular velocity ω. Then the velocity of any point in the body is ω r, so the angular momentum is ˆ L ρ r r ω r d 3 x ˆ ρ r ωr r r ω d 3 x We would like to extract ω from the integral above, but this requires index notation. Writing the equation in components, ˆ L i ρ r ω i r r i r j ω j d 3 x ˆ ρ r ω j δ ij r r i r j ω j d 3 x ˆ ω j ρ r δ ij r r i r j d 3 x where we may pull ω j out of the integral since it is independent of position. Notice how the use of dummy indices and the Kronecker delta allows us to get the same index on ω j in both terms so that we can bring it outside, keeping the sum on j. Now define the moment of inertia tensor, ˆ I ij ρ r δ ij r r i r j d 3 x 5 which depends only on the shape of the particular rigid body and not on its motion. This moment of inertia tensor is symmetric, I ij I ji The torque equation may now be written as τ i d dt I ijω j We have therefore shown that the angluar momentum is L i I ij ω j 6 where equation of motion in an inertial frame is still τ dl dt. In general, I ij is not proportional to the identity, so that the angular momentum and the angular velocity are not parallel. 6.3 Example: oment of inertia of a thin rectangular sheet For a uniform dimensional rectangle of sides a and b along the x and y axes, respectively, we seek ˆ I ij ρ x, y x δ ij x i x j d x about the center of mass. Since the body is uniform, the density mass per unit area for a thin sheet is constant, ρ ab 33

34 Also note that the center of mass is at R i a, b, so we compute relative to that point, I ij ˆ x R δ ij x i R i x j R j d x ab We compute the integrals one at a time I ˆ x R δ x R x R d x ab ˆb ˆa dy dx x a + y b x ab a ab ab b b b 0 ˆb 0 ˆb 0 0 ˆa dy 0 dy a dx y b y b y b 3 b b b b 3 8 b 3 For I the calculation is similar, yielding I a. The off-diagonal component is I I ˆ x R δ x a y b d ab x ab ab 4ab 0 ˆb 0 dy y b ˆa dx x a 0 [ y b ] b 0 [ b b [ x a ] a 0 ] b [ a 6.4 Rotating reference frame and the Euler equation 0 a ] a 0 Next, look at the equation of motion in a rotating frame of reference. We must replace the time derivative, d d + ω dt dt inertial body 34