Download The rotating frame

Rotating frames
Kevin Gibson
January 3, 2011
Introduction
In a rotating frame of reference objects move with an additional force
usually referred to as a fictitious force. For frames rotating at a constant
angular velocity ω a Centripetal and Coriolis force1 is present and can be
described by
(1a)
F r =m 2 r
cent
┴
(1b)
 cor v =−2 m⋅ ×
F
v
 
r ┴ is the position perpendicular to ω. It is the objective of this
With
piece to develop a space-time description of a rotating frame.
Potential Energy for a Rotating Frame
It can be shown using Lagrangian mechanics that an object in a rotating
frame behaves as if its potential energy is due to a centripetal and Coriolis
force.
2
(2a)
U cent r =−m ⋅r 2┴
2
(2b)
Suppose we choose as a coordinate system a cylindrical coordinate system
(r,θ,z) with θ about the z-axis. Under these conditions the potential energy
terms in (2) combine to become
U cor  r , 
v =2m ⋅ v ×r 
2
U rot r =−m
r 
2m ⋅r⋅v 
2
r 2
U rot r =−m
2 ⋅r⋅P
2
(3)
The relativistic mass times the potential energy for this case (remembering
that Pt = imc) gives
2 r
1 r 2 2
(4)
m⋅U =
⋅ P P − ⋅
 ⋅P
rot
ic
t
2
t
c
1.
Rotating frame scale factors
An observer in a rotating frame could account for the behavior of objects
under the potential energy (3) by the geometry of the space-time of the
region. The next task therefore is to bring in the theory of generalized
gravity2 to obtain the geometry of space-time in a rotating frame. To do so,
rules for using the theory of generalized gravity must be devised.
Symmetry and scale factors
If the physical source of a force on objects has any symmetry, then that
symmetry should be manifested in the geometry of the region. This
suggests the follows rules:
1. If the source of a potential energy is symmetric about some xα (i.e.
symmetric under the transformation xα → - xα or xα → xα + δxα), then the
deviation of the scale factors from the unperturbed geometry must be
independent of xα.
2. There must be a consistency between observers. For example any two
points along a trajectory of a beam of light will have ∆s = 0.
3. If there is no are no other restrictions, hj → Lj.
The last restraint must be employed with care.
The scale factors
For the case of a rotating frame with constant ω, there is no dependence on
θ, t or z (assuming the axis of rotation is along the z-axis). The scale factors
therefore must depend only on r. Moreover the theory of generalized
gravity dictates that, because of the Coriolis force potential energy, there
must be a non-zero htθ and hθt term that should be equal to each other. The
scale factor can now be known to have the form
(5)
ds 2=h ⋅dr 2dz 2h⋅d 22h 2 ⋅dt⋅d h ⋅dt2
r
t
t
Solving for hθ , ht and htθ
The equations for for hθ, ht and htθ are below, obtained from the generalized
gravity equations for ht and htθ respectively.
2h t r  2
(6a)
2−
=
ic
 
c
2.
 
2 r
⋅ht⋅h
ic
h 2t =
(6b)
Borrowing from the Schwartzchild metric for time, postulate that ht has the
general form
(7)
h =ic⋅ 1A⋅r 2
t
Remembering that an object can be said to conserve energy only when hα ≈
Lα, this must occur in this case for small r. We will make the following
approximation for the calculations.
1
(8)
h t≈ic 1 A⋅r 2


2
Equation (5a) now reads
 
 
A
r
2
2−2⋅ 1 ⋅r  =
2
c
A=−
2
(9)
1
2
c
(10)
The general time scale factor is

h t=ic 1−
r ′≡

r 2
r 2
 =ic 1− 
c
r′
(11a)
c

(11b)
Next, substitute ht into (5b). Moreover use the approximation for ht
appropriate to the limit r << r′.
2
2 r
1 r
h =
⋅1−   ⋅ich  
ic
2 r′
2
t
(12)
For the case of r << r′, hθ = Lθ = r.
(13)
This would be true at least for r << r′. Expanding to larger r requires one to
consider speeds at which a beam of light travels if it does so in the xk
direction.
2
h t =2 r
2
The speed of light
Insisting that ds2 = 0 for any two points along the path of light gives us a
constraint to work with. Assuming that a light beam sent along the xk
direction.
3.
2
2
2
2
0=hk⋅dx k h kt⋅dx k⋅dtht⋅dt
(13)
2
Divide by dt2.
2
   
dx
dx
0=h ⋅ k h 2kt⋅ k h2t
dt
dt
2
k
(14)
The derivatives can be converted into speeds for the light beam by including
scale factors.
2
 
 
2
h
h
0= k ⋅ck ′2 kt ⋅c k ′h 2t
∣L k∣
∣L k∣
(15)
ck′ is the velocity if light in the xk direction. The quadratic nature of (15) is
consistent with the velocity of light having two directions.
For some region, the values ck′ indicates something about ht, hkt and hk.
• Light can be at rest in a given region: ht → 0.
• Only one unique nonzero value of ck′ exists: The quadratic term must
vanish, so hk → 0.
• Two nonzero values of ck′ exist: Neither hkt nor hk must vanish, or go to
∞ if ht → 0.
• The solution to ck′ is zero and one nonzero value: ht → 0 and hkt ≠ 0 and
hk ≠ 0.
For the rotating case, as r → r′, cθ′ must have two options. It can be zero (a
non-rotating observer shoots a light beam in the direction of rotation) and
some other nonzero value. The above rules indicate that in this limit ht → 0
(consistent with the above expression for ht) and hθ ≠ 0. Moreover we must
insist that for r << r′ cθ′ ≈ +c, so hθt → 0. The given value for hθt bears out
the second condition. The first condition can also be satisfied if we assume
hθ = Lθ = r for all r.
Solving for hz and hr
To determine hr imagine the following: An observer is at rest in a rotating
frame a distance r from its axis. Another observer is in an inertial frame and
determines the first observer is a distance ri from the axis and is moving at a
speed v = ri∙ω. For the case where ri = r′, v = c. Any two events occurring at
4.
the rotating observer will, by the estimation of the inertial frame observer,
have ∆s = 0, because our rule #2. So relative to this observer
r
(17)
0=−c 2⋅ 1−
⋅ t 2r
[  ]
r′
For this to be true for all δtr, r must also be r′, so the radial distances must be
the same for both rotating and inertial observers.
(18)
h r=1
So the space-time interval for a rotating frame is
c
(19a)
ds 2=dr 2dz 2 r d 22 r 2 ⋅dt d − ⋅dt2
r

r ≡ 1−
r 2

r′
(19b)
Geodesics
Suppose an object is in the space-time geometry as given above, what would
its path be assuming it is a geodesic? In general relativity a geodesic
trajectory is described by3
d 2 x
dx
dx
(20)
=−∑   ⋅  ⋅ 
d 2
, 
  
d
d
The Γ is called the Christoffel Symbol and λ some parameter. The x
coordinates span all coordinates and include the scale factors. The sum
eliminates any duplicates in indices. The terminology of general relativity
will be altered according to the following:
• Let the xα in general relativity be written as Lα∙xα.
• Let all derivatives of coordinates be rewritten as Lα∙dxα/dλ .
Likewise for all higher order derivatives.
• Do not bother with contra vs. covariant quantities. If a given hαβ has
an ‘i’ in it, this will be taken care of.
With this in mind, the geodesic equation now is
2
[    ]
d x
dx
dx
L⋅ 2 =−∑   ⋅ L  ⋅ L  
d
d
d
,
And the Christoffel Symbol is now
5.
(21a)
   =L⋅∂
[
2
]
[
2
]
[
2
]
h 
h 
h 
L⋅∂ 
−L⋅∂ 
2∣L ∣∣L∣
2∣L ∣∣L ∣
2∣L ∣∣L ∣
(21b)
For the case of the rotating frame, the possible nonzero values of Γ are (for r
<<r´)
r
(22a)
 ttr =− 2
r′
 t  r=−
1
r′
(22b)
r
2
r′
(22c)
 tr =
1
r′
(22d)
 rt =
1
r′
(22e)
 trt =
In the classical limit, v/c << 1 and r <<r´, the geodesic trajectory obeys the
following equations of motion.
d 2 r rc 2
c
(23a)
= [ ]⋅v =2⋅r⋅v
dt 2
r′ 2
r′
d2
r⋅ 2 =⋅vr
dt
(23b)
⋅v r⋅v 
r⋅v r
d 2t ′
c⋅ 2 =−
− 2⋅
 0
2
c
dt
c
(24c)
Critique of geodesic
• First term of (23a) is the centrifugal force term, the second term is part of
the Coriolis force. It is needed so an object at rest in an inertial frame
will move in a circular motion about the z axis with velocity vθ = -ω∙r.
6.
• Equation (23b) says that an object “falling” in the r direction will be
deflected in the –θ direction. This is consistent with the Coriolis
potential that favors motion in this direction (all assuming ω is positive).
• Equation (23c) is important in the classical limit for time to be the same
for all observers.
The proposed scale factors accurately describe the motion of particles in a
rotating frame.
Transformations
Creating a transformation to a rotating frame from an inertial frame (where
the rotating axis is at rest) means the following constraints.
• ∆θr =0 if ∆θi = ω∙∆t.
• The inertial observer, relative to the rotating frame, must rotate at
angular frequency –ω . This is to satisfy the obvious motion of stars
relative to the rotating Earth frame.
Satisfying these constraints means assuming a form for the transformation
r
(25)
r=a⋅i− t i r r=r i z r =z i t r=at i d⋅ i
c
The variables a and d are unit-less quantities. Using these to satisfy the
condition that ds2 for two nearby events must be the same for both observers
(via numerical methods) gives
(26)
a=1 d =0
so
(27)
r= i− t i r r =r i z r=z i t r =t i
A rather Galilean looking transformation!
Quantum wave equation
Applying the rotating frame scale factors to quantum mechanics means the
generalized Klein-Gordon equation4 becomes.
2
r  2
E
r
1 2  r r 
−ℏ 2 r
2
(28)
{ ∂r  ∂r ∂z  ∂  }−
p E=
E − 0 
2m0 r
r
r2
E0
2 E0
2
The tildes signify momentum and total energy (relativistic and potential)
operators.
7.
Summary
A rotating frame can be viewed as a non-orthogonal reference frame and the
path of an object acting under the Centripetal and Coriolis force is a
geodesic of the object through this space-time. Moreover the velocity of a
beam of light in vacuum is not in general an invariant, but can vary with the
geometry.
Works Cited
1. http://en.wikipedia.org/wiki/Fictitious_force
2. Kevin Gibson Theory of Generalized Gravity (unpublished)
http://www.mc.maricopa.edu/~kevinlg/i256/Generalized_Gravity.pdf
3. Wolfgang Rindler, Essential Relativity 2nd ed., p. 130
4. Kevin Gibson Generalizing the Klein-Gordon equation (unpublished)
http://www.mc.maricopa.edu/~kevinlg/i256/Generalized_KG.pdf
8.
Contact information
[email protected]
[email protected]
9.