[Phys-L] Re: Help on a problem from Goldstein

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding the *special* relativistic planetary orbital precession
problem brought up by Leigh:

> Problem 13 from Chapter 6 of Goldstein's "Classical Mechanics"
> reads:
>
>    Show that the relativistic motion of a particle in an attractive
>    inverse square law of force is a precessing ellipse. Compute the
>    precession of the perihelion of Mercury resulting from this
>    effect. ( The answer, about 7" per century, is much smaller than
>    the actual precession of 40" per century which can be accounted
>    for correctly only by general relativity.)
>
> Can someone kickstart my brain with a hint as to what
> approach I should consider to solve this problem?
>
> Thanks,
>
> Leigh

In response Bernard Cleyet wrote:

> My first thought was somewhere I'd heard one could expand and use
> the first and second? terms.  Thereby, adding a cubic term to the
> force. (quad potential, naturellement).  Some theorem, I think, that
> only the linear and quad central force result in no precession.
> That's why some early on suggested gravity was not exactly inverse
> square.
>
> bc, still searching his shelves.

Although it is certainly possible to perturbatively expand the
problem's solution in inverse powers of c^2 and keep the leading
order correction, it is found that when this is done the problem does
not involve a cubic (or any other) correction to the force term.
Rather, the correction occurs in the momentum-dependent terms of
the Hamiltonian.  Nevertheless it is completely unnecessary to
make such a perturbative expansion.  The problem is exactly
solvable.  In fact it is possible to get an exact solution for the
orbit by taking the exact orbit for the *Newtonian* case and
simply redefining the constant parameters and rescale the polar
angle that appear in the orbit so as to include all the SR
relativistic corrections.  After the parameter/variable
mapping/re-definitions the resulting orbit equation is the exact
orbit for the SR problem. Here we assume the force is an attractive
1/r^2 central force law with force constant magnitude G*M*m which
is incorporated into the problem as a static force field obtained
from the negative gradient of an appropriate 1/r potential function
which is the temporal component of a 4-vector potential whose
spatial (magnetic-like) vector potential components happen to
exactly vanish.

In doing this problem we make the following simplifications:
1) The reduced mass effects are ignored by rigidly nailing down the
central center of force (or, equivalently, we imagine the mass ratio
m/M is infinitesimal where m is the mass of the orbiting test
particle and M is the "mass" of the central force appearing in the
force constant coefficient for the attractive 1/r^2 law).
2) Any radiative effects from the orbiting test particle are denied.

Now suppose that we define the following quantities:

G = Newton's universal gravitational constant
[phi] = polar coordinate angle for the orbit
r = polar coordinate radius of the orbit from the force center
E = energy of the orbiting particle where the zero level of this
    energy is set for the particle at rest at r --> [infinity].
    IOW the value of E does *not* include any relativistic rest
    energy the orbiting particle may have.
m = mass of the orbiting particle
h = orbital angular momentum per unit mass of orbiting particle

Suppose we solve the Newtonian version of this problem and find the
orbital ellipse whose orbital parameters are given in terms of the
orbiting particle's values of E & h (i.e the semimajor axis
a = G*M*m/(2*|E|) and eccentricity
e = sqrt(1 - |E|*h^2/(2*m*G^2*M^2))) .  We next make the following
mapping/transcription of variables by direct substitution (where
primed quantities are the corresponding quantities in the special
relativistic version of the problem):

Newton -->  S. R.
r      -->  r'
[phi]  -->  [phi']*sqrt(1 - (G*M)^2/(h'*c)^2)
E      -->  E'*(1 + E'/(2*m*c^2))
h^2    -->  h'^2 - G^2*M^2/c^2
G*M    -->  G*M*(1 + E'/(m*c^2))

After the above substitutions are performed on the Newtonian orbit
the result is the special relativistic orbit.  Because the angle
[phi] is scaled by a constant factor, (ie. sqrt(1 - (G*M)^2/(h'*c)^2)
the resulting orbit in terms of [phi'] no longer describes a closed
ellipse, but is a precessing orbit that no longer closes on itself.
Note that the relativistic energy E' above *also* does *not* include
the orbiting particle's rest energy.  To get the fully relativistic
total energy *including* the rest energy one needs to write
m*c^2 + E'.

As Hugh Logan has already discussed the exact periapsis precession
per orbit is immediately found from the scale factor relating the
Newtonian version of the angle [phi] and the relativistic version
[phi'].  As Hugh already noted the exact precession per orbit is
simply:

[delta-phi'] = 2*[pi]*(1/sqrt(1 - (G*M)^2/(h'*c)^2) - 1)

In the small 1/c^2 limit the leading order value of [delta-phi] boils
down to [pi]*(G*M)^2/(h'*c)^2 .  As also it was previously mentioned
this leading order value is precisely 1/6 of the corresponding
correct value for the *general* relativistic version of the problem.

If we make the replacement for the value of h' above in terms of the
orbit's asymptotic Newtonian semi-major axis a and eccentricity e,
(ie. h'^2 = h^2 = G*M*a*(1 - e^2) ) we get to leading order in 1/c^2:

[delta-phi'] = [pi]*G*M/(c^2*a*(1 - e^2))

in the special relativistic version of the problem.  And we get:

[delta-phi'] = 6*[pi]*G*M/(c^2*a*(1 - e^2))

in the physically correct general relativistic version of the
problem.

It probably ought to be mentioned that not only can the orbital
precession be solved for exactly in the special relativistic version
of the problem, but the precession can *also* be solved for exactly
in the *general* relativistic version as well (I know this because I
was a bit surprised to be able to work it out when I tried my hand at
it).  The exact result requires, however, the evaluation of an
elliptic integral function.  If we use the previous parameter
definitions the *exact* *general* relativistic apsidal precession per
orbit is:

[delta-phi'] = 4*K(k)/sqrt(2*R*sin([theta]+ [pi]/3)/sqrt(3)) - 2*[pi]

where K(k) is the complete elliptic integral of the first kind of
modulus k = sqrt(sin([theta])/sin([theta]+ [pi]/3)) and R is given
by R == sqrt(1 - 12*S^2) and [theta] is given by
[theta] == (1/3)*arccos((1 - S^2*(18 + 108*Q*(1 - Q/2)))/R^3) where
Q == E'/(m*c^2) and S == G*M/(h'*c).

If the above formula for [delta-phi'] is expanded to leading order in
1/c^2 (for the case where the motion is quasi-Newtonian) the result
is:

[delta-phi'] = 6*[pi]*S^2

which is in agreement with the usual expression for the leading order
general relativistic apsidal precession per orbit.

David Bowman
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