1 PHYS 705: Classical Mechanics Non-inertial Reference Frames Vectors in Rotating Frames
2 Infinitesimal Rotations From our previous discussion, we have established that any orientation of a rigid body (with one point fixed) can be represented by a single rotation about a particular axis. And this rotation can be parameterized by three parameters (3 Euler angles) Question: Can we represent these three parameters by a vector? Suppose A and B are two such vectors representing two such rotations Then, for them to be vectors, they must commute, i.e. AB BA However, from our previous discussion, finite rotations are not commutative as an operation.
3 Infinitesimal Rotations Recall that if we represent the rotation A by a matrix M A, then the addition of two rotations A and B will be given by a matrix multiplication, i.e., AB M M And, matrix multiplications are not commutative : A B MAMB MBMA Thus, finite rotations CANNOT be represented as vectors! However, as we will demonstrate next. Infinitesimal rotations can be represented by vectors!
4 Infinitesimal Rotations Let consider a general infinitesimal transformation given by the following, x x x x ' i i ij j ij ij j so that the unprimed axes is changed to by an infinitesimal operator. In matrix notations, we have, ' ε ij x i x I ε x Since is assumed to be small, we will show that the sequence of infinitesimal transformation is not important. x ' i
5 Infinitesimal Rotations - Let say we have two infinitesimal transformations:, Op Op 1 2 - Switching the order of the operations, Op Op 2 1 I ε1 I ε Iε 2 2 1 Iε2 I ε1iiε2 ε1ε2 Iε1ε2 O Iε 2 2 2 Iε1 I ε2iiε1ε2ε1 Iε2 ε1o -By neglecting higher order terms, these two expressions are the same since matrix addition is commutative. Infinitesimal transformation is commutative. 2
6 Infinitesimal Rotations - Now, let consider an inverse of an infinitesimal transformation, A Iε It is given simply by: CHECK: - Then, let see what property will have if we want this infinitesimal transformation to be orthogonal. 1 A Iε 1 AA I ε I ε I (to first order) ε - For orthogonal transformations, we need to have A A T 1 - This then gives, A T I ε A I T 1 ε ε T ε ε is an anti-symmetric matrix
7 Infinitesimal Rotations - Further note that, det A det Iε 1 (to first order) An infinitesimal orthogonal transformation corresponds to a proper rotation. ε - Since is anti-symmetric, we can write it generally in component form as, 0 ε d d d 0 3 2 3 1 d 2 1 d d d, d, d (The three quantities can be identified as three independent parameters specifying the infinitesimal rotation. ) 0 1 2 3
8 Infinitesimal Rotations - As an example, we can explicitly write out the infinitesimal Euler rotation, A cos d cos d cos dsin dsin d cos d sin d cos d cos dsin d sin dsin d sin d cos d cos dsin dcos d sin d sin d cos d cos dcos d sin d cos d sin dsin d sin d cos d cos d with cos d cos d cos d 1 sin d d sin d d sin d d This gives
9 Infinitesimal Rotations - As an example, we can explicitly write out the infinitesimal Euler rotation, A 1 d d 0 cos d cos d cos dsin dsin d cos d sin d cos d cos dsin dsin dx sinddx sin, d sin d cos Ad cos sin cos d sin d sin cos d cos dcos d sin d cos d d d 1 d Iε cos dx 1, sin dsin d sin d cos d and keeping cos d 0 d 1 only linear terms And, dω d,0, d d T Note: 1. For notational purpose, we write as a differential but it is NOT an actual differential of a vector. dω dω 2. But, as you will see stands for a vector of differential changes
10 Infinitesimal Rotations - We will now show that these three quantities also form a particular kind of vector. - Under an infinitesimal rotation, the change in the coordinates can be expressed as, - Writing this out explicitly in components, we have, dr r' r Iε rr εr dx1 0 d3 d2x1 dr dx εr d 0 d x 2 3 1 2 dx 3 d2 d1 0 x 3 - The result looks like the cross product between two vectors: dr rdω with x, x, x d, d, d 1 2 3 dr dx x d x d 1 2 3 3 2 dx x d x d 2 3 1 1 3 dx x d x d 3 1 2 2 1 T T r 1 2 3 dω d1, d2, d3
11 d as a Pseudovector - Under an orthogonal transformation B, the anti-symmetric matrix transforms according to the similarity transform: ε' 1 B εb ε - Recall that a regular vector (called polar vector) r, it must satisfy the transformation rule: x ' i bx ij j b bb ij ik jk where is orthogonal, i.e, ij - One can also show that dω must transform as, d' det B b d (*) i ij j (not shown here) - Vectors transforming according to (*) is called an axial vector (pseudovector)
12 d as a Pseudovector - Recall that if B is a an orthogonal transformation, det B 1 det B = + 1, then B is called proper, (rotation only, no inversion) det B = -1, then B is called improper (with inversion) - So, we have the following distinction between a polar and axial vector under a improper orthogonal transformation B: V' BV if V is a polar vector V' BV if V is a axial vector (or pseudovector) - Common examples of axial vectors in physics involves the cross product of two polar vectors such as angular velocity, angular momentum,
13 Geometric View of an Infinitesimal Rotations dω We just saw that an inf. rotation leads to an inf. change for : Here is the same situation in the active view: dr rdω (**This was in the passive view) dr r dω r sin r d r dr dr Here, we are rotating the vector (active view) instead of rotating the coordinates (passive view) thus the reverse order of the cross prod. magnitude: direction: rsind RHR r dr dωr (We will take this active view from now on.)
14 Geometric View of an Infinitesimal Rotations Dividing dt from both sides of the equation, we have, dr d dt Ω v r ω dt r d ω Ω dt where is the instantaneous angular velocity ω where lies along the axis of infinitesimal rotation in the time tt, interval in a direction known as the instantaneous axis of rotation. dt (the familiar definition for angular velocity)
15 Rate of Change of a Vector under Rotation x z z ' r R x ' P y ' r ' y Looking at infinitesimal rotation more generally: We are interested in describing vectors in the body (primed) frame as measured in the fixed (unprimed) frame when the rigid body frame is being rotated about a fixed axis. - For simplicity, we also assume (for now) that a point P described by in the body frame does not move wrt the body frame. R ( is assumed to be fixed for now.) r ' Note: Our discussion can be for any vector instead of the specific position vector r ' in the body frame. G
16 Rate of Change of a Vector in under Rotation - However, as viewed from the fixed frame, will be moving (due to ) dω r ' - From our discussion on rotation before, viewed in the fixed frame, dr' dωr' fixed - Or, dividing by dt, we have, dr' dω r ' ω r ' dt dt fixed r ' - More generally, if we now allow to move wrt the body frame as well, dr' dr' ωr' dt dt fixed body
17 Rate of Change of a Vector in under Rotation - Although our discussion was for a given position vector in the body frame, the discussion applies equally well to ANY vector in the body frame. dg dg ωg dt dt fixed body r ' G - So, it is useful to abstract out this operation for any vectors, d d ω dt dt fixed body
18 Rate of Change of a Vector in under Rotation This relation can also be formally derived. The components of a vector measured in the body frame and in the fixed frame are related by an orthogonal transformation G or G( body) AG( fixed) G a G ' i ij j A a ij where is the Euler rotation matrix Reversing the expression between the fixed and body frames, 1 T G a G' a G' a G' i ij j ij j ji j
19 Rate of Change of a Vector in under Rotation As the body moves infinitesimally in time, the components of in the body dg ' j frame will change by and the instantaneous rotation matrix will be given by the infinitesimal rotation matrix, G ' i AIε a ji ji ji So, the components in the fixed frame is then changed by, ' ' G dg G dg i i ji ji j j Keeping terms up to to 1 st order, we have, G dg G' dg' G' i i j j ji j
20 Rate of Change of a Vector in under Rotation Without loss of generality, we pick our fixed frame to be instantaneously coincident with the body frame at time t so that we choose G i G' j at time t BUT, the differential of G in the body and fixed frame are in general not the same and is related by, dg dg ' G ' i j ji j Using the anti-symmetric property of, i.e., dg dg ' G ' i i ij j ε ji ij
21 Rate of Change of a Vector in under Rotation dg dg ' G ' i j ij j Then, recall that we have ε 0 d d d 0 3 2 3 1 d 2 1 d d 0 And using the Levi Civita tensor The last term can be rewritten as, ijk 1, ijk is of forward permutation 1, ijk is of backward permutation 0, otherwise G' d G' d G' ij j ijk k j ikj k j
22 Rate of Change of a Vector in under Rotation So, we have, dg dg ' d G ' i i ikj k j or dg dg dωg fixed body And, finally dg dg ωg dt dt fixed body And the operator equation again, d d ω dt dt fixed body
23 Acceleration in a Rotating Frame - Let use the operator equation to calculate the velocity and acceleration of a point P in the body frame as measured in the fixed frame z z ' r r ' P y ' unprimed wrt fixed frame x R x ' y primed wrt body frame (Here, we also allow R to move as well.)
24 Acceleration in a Rotating Frame - So, for the velocity vector, we have (as before), dr dr dr' dt dt dt fixed fixed fixed dr dr dr ' dt dt dt fixed fixed body ωr' dr' dr' ωr' dt dt fixed body vel of P rel to vel of rotating vel of P rel to extra piece due to fixed axes frame origin rotating frame rotation of the frame v v v' ωr' f R
25 Acceleration in a Rotating Frame - Differentiate in the fixed frame again, we have dv dvr dv' dr' ω ω r' dt dt dt dt fixed fixed fixed fixed Both v and r are vectors in the body frame but the differentiation is measured in the fixed frame so that we need to apply the operator equation again to expand them.
26 Acceleration in a Rotating Frame dv dvr dv ' dr ' ω ω r ' dt dt dt dt fixed fixd e fixed fixed dv dvr d ' d ' ' ' ' dt f dt v f dt r ω v ω ω r body dt ω r body a a a 2 ωv' ω ωr' ω r' f R r acc of P rel to acc of P rel to fixed axes rotating frame acc of rotating frame origin vel of P in the rotating frame pos of P in the rotating frame
27 Acceleration in a Rotating Frame - To gain some physical insights into the meanings of the various terms, 1. We assume that the body frame is rotating with a constant angular frequency around a fixed axis and the origin of the body frame is not moving as well, i.e., F ma ω 0 and a 0 f Fma 2 m ωv' mω ωr' r R 2. Newton 2 nd Law applies in inertia frames only. That is why we have calculated a f explicitly in the fixed frame and this gives, where F is the net force acting on m as observed in the fixed frame.
28 Acceleration in a Rotating Frame - Rearranging the terms, we can rewrite the equation into the following form, F2 m ωv' mω ωr' mar - Defining F F2 ωv' ωωr' eff m m - We can re-interpret the Newton 2 nd law as an equation applying in the rotating frame, F eff ma r This is F = ma in the rotating frame!
29 Acceleration in a Rotating Frame Feff F2 m ωv' mω ωr' Notes: - F is the actual physical forces in the inertia frame - The extra two terms corresponding to apparent forces (a penalty you pay) for NOT being in an inertial frame. mω ωr' The familiar centrifugal force 2 m ωv' The coriolis force - These two fictitious forces are not associated with any real forces acting on objects in an inertial frame and they are not visible to an inertial observer!
30 Example of Fictitious Forces in a Rotating Frame Just standing on a merry-go-round ω r ' ωr' ω ωr ' - For simplicity, let saying you are standing still (remain stationary relative v ' 0 to the rotating frame) so that there is no coriolis force! - Everything is perpendicular here. The fictitious centrifugal force is mv mωωr m r outward r ' 2 2 ' ' ( ) - Note: this fictitious centrifugal force ~ r so that it gets larger as one moves out and vanishes at the origin.
31 Example of Fictitious Forces in a Rotating Frame Now, we allow v ' 0 Consider standing in the middle and throwing a rock directly at a fixed target fixed outside the merry-go-around ω View in fixed frame In the fixed inertial frame, the rock experiences no net force and it will go out toward the target along the straight path as shown. - However, as the rocks moves outward, the thrower will rotate. Target fixed in fixed frame To the thrower in the rotating frame, the rock will look like veering off View in rotating frame to right.
32 Example of Fictitious Forces in a Rotating Frame So, within the rotating frame, there is a fictitious Coriolis force v ' ω View in rotating frame 2 ωv' Rule of Thumb: object deflects to the right of v ' MOVIE: http://techtv.mit.edu/videos/3714-thecoriolis-effect https://www.youtube.com/watch?v=dt_xjp77-mk
33 Earth as a Rotating Frame The fact that we live on the surface of the Earth which rotates Means that we are subject to these fictitious forces (Note: For an object on the surface of the Earth, the origin of the body frame is not fixed so that R changes and we need to put back 0 but we will assume Earth to have a fixed rate of rotation, i.e., ) ω ω a R R a a a 2 ωv' ω ωr' f R r
34 Earth as a Rotating Frame To calculate, consider R ω a R It moves in the fixed frame but it is stationary in the body frame. R Considering the rate of change of R in the fixed frame, we have dr dr dt dt fixed body ωr Taking the time derivative as measured in the fixed frame again, we have, dr dr R a ω ωr R dt dt fixed body a ω ωr R
35 Earth as a Rotating Frame Plugging this result into our equation for, we have, a a 2 ωv' ω ω Rr' f r And the effective force in the rotating frame becomes, Letting (gravity) be the only actual physical force, and assuming that we re interested in objects near the surface of the Earth so that a f F F2 m ωv' mω ω Rr' ma eff F mg 0 ω v Feff m g0 mω R 2 m ω ' r R r' R This is the effective gravity g (actual gravity + centrifugal force)
36 Earth as a Rotating Frame Now, let look closer at this effective gravity as a function of latitude on Earth, North Pole ω R - For 90 (North Pole), ωr 0 Then, we have g g 0 - For 0 (Equator), equator 2 gg0 R R ˆ g 0 grˆ South Pole Thus, due to centrifugal effects, gravity g is about 2 0.052ms greater @ the poles than @ equator
37 Earth as a Rotating Frame In general, also notice that a mass on a string doesn t hang straight down: North Pole ω g 0 ω ωr R equator Actual direction of g (way exaggerated) gg ω ωr 0 South Pole HW: find deviation angle as a function of latitude
38 Coriolis Effect on Projectiles: Case 1: At North Pole North Pole R ω 2 m ωv' North Pole ω (out of page) v ' ωv' equator Top View Projectile will defect to the right (as in South Pole the merry-go-round situation).
39 Coriolis Effect on Projectiles: 2 m ωv' Case 2: At Equator sending projectile to the East North Pole ω R equator v ' (into of page) ωv' South Pole Coriolis force is up (from the surface) but it is tiny compared to regular gravity.
40 Coriolis Effect on Projectiles: 2 m ωv' Case 3: At intermediate northern latitudes sending projectile to the East North Pole ω R v ' (into of page) ωv' equator In addition to the up direction, Coriolis force also has a component deflecting the orbit to the right wrt its original direction of South Pole motion (toward South in this case).
41 Coriolis Effect on Projectiles: 2 m ωv' Case 4: Southern latitudes sending projectile to the East North Pole ω The situation is reverse. Coriolis force has a component deflecting the orbit to the left wrt its original direction of motion (toward North in this case). South Pole R equator ωv' v ' (into of page)
42 Coriolis Effect on Projectiles: 2 m ωv' Coriolis effect on weather North Pole ω Low equator Northern hemisphere direction of wind will be defected to the right always resulting into forming a counter-clockwise circulating low pressure cell. South Pole Northern hurricane Southern hurricane