bradley j. nartowt nartowt Friday, July 05, 2013, 21:57:17 PHYS 6246 – classical mechanics Dr. Whiting Whiting
show that the motion of a particle particle in the the potential potential field,
V ( r ) kr 1 hr 2 ku hu 2 V (u)
[I.1]
is the same as that of the motion under the kepler potential alone when expressed in terms of a coordinate system rotating or precessing around the center of force. the motion r ( ) from ( r )
dr f ( r )
is cos( 0 )
2
( mku 1) 2 k h u
, while kepler motion is cos( 0 )
2
(1 2 E ) 4 E k h mk
d 2u d 2
u(1 2h m ) k 2
2
d u 2 d
u m 2
dV , du
u( ) u1 cos u2 sin
m 2
2
1 2 E 2
, as
mk
given by goldstein. to subtract these and find the relative motion is to attempt to show work with differentials instead. Use the orbit differential equation
2u 1 mk
1 EA4 EB
Au1 1 EA
~ [const [constant ant]] , so
using [I.1], and,
km 2
( Au1) 2 Bu
2 hm
1 2m
h
[I.2]
2
It is obvious the kepler solution alone (h = 0 in [I.2]) is,
u0 ( ) u10 cos u20 sin mk 2 Using cos t sin t A cos 0 cos t i A sin 0 sin t A Re( e 2
i 0
[I.3]
e i t ) Acos( t 0) , which is proof that
the linear combination of two orthogonal solutions to an ODE is equivalent equivalent to a single phase-shifted sine or cosine, both [I.2] and [I.3] can be represented as,
rGR ( ) Here,
mk 2
r0 cos( 0 ) r0
km 2 2 hm
r0 1
cos( 0 ) r0 (1 )
r K ( )
R cos( 0 ) R
[I.4]
, which is a parameter with units 1/L related to the strength of the kepler potential. For R r 0 1 , that is
the Kepler and GR-orbits start from the same point and ( , ) (3,1.1) 1 we get 100 orbit-cycles, looking as,
[I.5]
1
Note:
3 1.1
means the effects of relativity are (roughly) 1.1 parts in 3 to Newton-potential — Newton-potential — should be on the order of the
speed of light in accordance with the correspondence-principle, so the precession due to GR is very exaggerated in the plots [I.5]
The following analysis is due to Christopher Wagner: Consider to be the precession angle. Consider GR to be the
angle of the orbit suffering general relativity
h 2
r
, and K to be the angle of the orbit due to Kepler potential alone. Then,
GR K . Finally, consider ourselves in to be at the same radial position for both orbits, so r K rGR r ,
[I.6]
The Lagrangian for the potential [I.1] is L 12 m(r
2
2 r 2 GR2 ) kr r h ; equations of motion are mr GR 2
GR
and
mr mr GR2 ( rk2 2r 3h ) . By geometry, GR K , implying2 that GR2 K 2 2 2K . Putting this into the 2
radial equation of motion (as we always do to eliminate ), we get,
mr mr( K 2 2 2 K ) ( rk2
2h r3
) mr3 ( rk2 2
Notice that the Kepler Lagrangian radial equation of motion (h = 0) is mr
m 2 2 r )
2h r3
2
mr3
r k
2
[I.7]
; subtracting this from [I.7], we get
a condition to be satisfied if we are to have motion as if we were in a Kepler frame, like the RHS of [I.6],
0 h 12 m 2 r 2 h 12 m 2
m
h 12 m 2 m1
2
Wikipedia says that the general-relativity potential hr requires h 0 , so replace
m2 ( h m1
)
[I.8]
is attractive, so its negative gradient must be negative. This
h h in [I.8] and write
2 m
( h m1
)
2 m
h 1
m
. The coordinate system must
be moving at this rate in a circle to observe a keplerian elliptical orbit. For negative total energy: show that if the additional potential term is very small compared to the kepler potential, then the angular speed of precession of the elliptical orbit is,
2 mh / ( 2 ) Rewrite the frequency in terms of the Kepler-potential strength
[I.9] mk 2
get an orbit like [I.5] with such slow precession, recall 1.1 1 the Taylor expansion
, and you get 1 2 hk
1 2 . To
0.21 . In general, 1 , so we can use
1 2 2 O( 2 ) . Obviously, the first term in the Taylor expansion gives 1 from the
kepler orbit, so the next order term must be the angular speed of precession that [I.9] is asking for. Using t and normalizing
t to radian-measure by t 2 t ,
2
t 2
2
h
h mk
k
k
The general relativity term [I.1] is not trivially equivalent to a shift in
2
h
m 2
2 mh 2
, which is discussed lengthily in Problem 3.22…
[I.10]
The perihelion o f mercury is observed to precess (after corrections for known planetary perturbations) at the rate of 40” per century. Show that this precession can be accounted for classically if the dimensionless quantity,
h / (ka )
[I.11]
(which is a measure of the perturbing inverse-square potential relative to the gravitational potential) were as small as
7 108 . (the eccentricity of mercury’s orbit is 0.206, and its period is 0.24 years). Putting in the dimensionless [I.11] into [I.10] gives
m ka 2
. Physically,
2
must be related to eccentricity. For our
orbit, the virial theorem can be used to relate total energy to the semimajor axis a as E E 12 V V 12 V , and V is known exactly (not just its average) at the semimajor axis so E 2
angular momentum as
2 mka mka(1 e ) 2
1 k 2 a
E , and e 1 2 mk 2 , which allows us to write the 2
mk E ( e 21 ) kmk/(2a ) ( e 21) mka(1 e2 ) ; effecting all of this in [I.10], 2
2
2 (1 e ) 2
2
2
8 2 7 10
(1 0.206 )(0.24 years) 2
1.91 106
rad year
0.394"/ yr 39.4
" century
[I.12]
