The Big Ben Paradox

[Pat Dolan]

I would like to submit this little conundrum to this group for consideration:

Consider a distant observer traveling at .867 c ( gamma=2 ) relative to
the solar system along the line that is collinear with the sun's axis of
rotation. As the clockwork solar system spins beneath him, the distant
observer peers through his powerful telescope at Big Ben in London. In
accordance with special relativity, and after taking relativistic
doppler into account, the distant observer measures Big Ben's little
hand to make one revolution for every two revolutions of his own
wristwatch's little hand. He also observes that Big Ben's little hand
still makes 730.5 revolutions for every revolution that the earth makes
around the sun. From these two observations the distant observer
concludes that in his inertial frame of reference the earth's orbital
velocity is only half the velocity necessary to keep the earth in stable
orbit around the sun.

Will the earth spiral into the sun? If not, why not?

Note: Newtonian gravity is not assumed in this paradox. Invariant
spacetime curvature is assumed to be the cause of the earth's orbit
around the sun.

[Stefan Ram]

Pat Dolan <patd...@comcast.net> writes:
> From these two observations the distant observer
>concludes that in his inertial frame of reference the earth's orbital
>velocity is only half the velocity necessary to keep the earth in stable
>orbit around the sun.
>
>Will the earth spiral into the sun? If not, why not?
>
>Note: Newtonian gravity is not assumed in this paradox. Invariant
>spacetime curvature is assumed to be the cause of the earth's orbit
>around the sun.

I define the total force F on a body with mass m and acceleration
a to be ma (a multiplied by m).

F = ma.

If a := dv/dt, then for a real constant k (not depending on t):
ka=d(kv)/dt (kv = v multiplied by k). If v is being replace by
kv, a is being replaced by ka.

v -> kv: a -> ka.

But this means that when v is being replaced by kv, ma is being
replaced by mka, and F by kF.

v -> kv: ma -> mka, F -> kF.

So, when a moving observer observes a body of mass m that is moving
with speed kv (e.g., k=1/2) where "v" is the speed in the system of
rest, he must conclude that the force, which is F=ma in the system of
rest, is kF=kma, where a is the acceleration in the system of rest,
by the very definition of the force as mass times acceleration.

So, it seems that the force of gravity is also reduced in
the moving system balancing the reduction of speeds.

[Tom Roberts]

On 9/2/23 3:32 AM, Pat Dolan wrote:
> Consider a distant observer traveling at .867 c ( gamma=2 ) relative
> to the solar system along the line that is collinear with the sun's
> axis of rotation. As the clockwork solar system spins beneath him,
> the distant observer peers through his powerful telescope at Big Ben
> in London. In accordance with special relativity, and after taking
> relativistic doppler into account, the distant observer measures Big
> Ben's little hand to make one revolution for every two revolutions of
> his own wristwatch's little hand.

This is wrong -- "after taking relativistic doppler into account" the
observer would measure Big Ben's little hand to rotate at the same rate
as his own wristwatch's little hand. That is, the relativistic
Doppler-shift includes the effect due to relative motion and also the
effect due to "time dilation".

[I ignore the mistake that Big Ben is the nickname for
the Great Bell of the Great Clock of Westminster, and
not the clock itself.]

> He also observes that Big Ben's little hand still makes 730.5
> revolutions for every revolution that the earth makes around the
> sun.

This is correct.

> From these two observations the distant observer concludes that in
> his inertial frame of reference the earth's orbital velocity is only
> half the velocity necessary to keep the earth in stable orbit around
> the sun.

Nope, even if we correct the initial claim to leave "time dilation" in
the observation. SR does not properly handle the gravitation that keeps
earth in its orbit. For that one must use GR, and the components of the
metric in the distant observer's inertial frame are quite different from
those in the solar system rest frame. A correct calculation using the
distant observer's coordinates would model the sun continuing in its
orbit as usual.

IOW: The earth's path through spacetime is independent of the
coordinates used to describe it (this should be obvious). GR includes
this coordinate independence.

> Will the earth spiral into the sun? If not, why not?

It won't. For the simple reason that observations by a distant observer
cannot possibly affect the behavior of the solar system. This OUGHT to
be obvious.

Using GR to model the solar system, one immediately knows that the
predicted orbit of the earth is independent of coordinates, and the
distant observer obtains the same trajectory as an earthbound observer.

[I ignore the difficulties the distant observer will face
in obtaining accurate measurements.]

> Note: Newtonian gravity is not assumed in this paradox. Invariant
> spacetime curvature is assumed to be the cause of the earth's orbit
> around the sun.

Hmmmm. This disclaimer does not accurately describe how this is modeled
in GR. Your conclusion about the distant observer calculating earth's
orbit is wrong, and did not properly take into account the coordinate
independence of paths in GR.

Tom Roberts

[William]

On Saturday, September 2, 2023 at 1:32:10=E2=80=AFAM UTC-7, Pat Dolan wrote=
:
> Consider a distant observer traveling at .867 c ( =F0=9D=9B=BE=3D2 ) rela=
tive to the solar system...
> In his inertial frame of reference the earth's orbital velocity is only h=
alf the velocity
> necessary to keep the earth in stable orbit...

Not true, the earth follows a helical geodesic trajectory through spacetime=
, and=20
this helical geodesic is not intrinsically altered by being described in te=
rms of a=20
different system of coordinates (such as the asymptotically flat inertial c=
oordinates
in which the distant observer is at rest). By the way, the *extrinsic* cur=
vature of the=20
earth's trajectory is the same for those two coordinate systems, which may =
be=20
surprising to you if you aren't taking the time component of the trajectory=
into=20
account. (Misner, Thorne, Wheeler illustrates this with a bullet and baseb=
all.)

> Invariant spacetime curvature...

Be careful... as noted above, the *extrinsic* curvature of the trajectory i=
s invariant=20
under Lorentz transformation (which is essentially what you're applying by =
switching
to the asymptotically flat background inertial coordinates in which the dis=
tant high=20
speed object is at rest, superimposed on the mildly curved spacetime surrou=
nding the=20
sun), but the components of the *intrinsic* curvature of spacetime are not =
invariant
under coordinate transformations, they change along with the components of =
the=20
metric as expressed in terms of the different coordinate systems. These th=
ings
are all coordinated so that the invariant intervals are, well, invariant.

> Will the earth spiral into the sun?

No, describing the phenomena in terms of a different coordinate system does=
n't change
the intrinsic phenomena. For example, if you draw two chalk grids on a put=
ting green,=20
and describe the trajectory of a putt going into the hole in terms of one c=
oordinate=20
system, it will also go into the hole when described in terms of the other =
coordinate=20
system. Yes, the ball has different coordinates at the end, but the cup als=
o has different=20
coordinates, so the ball still goes into the cup.=20

The idea that changing the coordinate system used to describe the phenomena=
can=20
somehow change the phenomena is wrong. And no, this does not imply that loc=
al=20
Lorentz invariance has no physical dynamical effects. The dynamical equatio=
ns of=20
physics are locally Lorentz invariant, which is the physical content of spe=
cial relativity.

> Ridiculous! See Einstein's First vs. Kepler's Third, ibid.

I'll assume that by "Einstein's First" you are referring to the principle o=
f special
relativity, i.e., that the equations of physics take the same simple homoge=
neous=20
and isotropic form in terms of every standard system of inertial coordinate=
s, and=20
that by "Kepler's Third" you are referring to Kepler's proposition that the=
squares of=20
the orbital periods of the planets are directly proportional to the cubes o=
f the=20
semi-major axes of their orbits. =20

There's no conflict here, and nothing that makes the above explanation=20
"ridiculous". The principle of relativity is contained in local Lorentz in=
variance,=20
which is clearly satisfied in this situation. It also happens that Kepler'=
s=20
proposition remains satisfied (to the same approximation that it ever was),=
=20
since the angular periods of the helical paths of the planets remain in the=
=20
same proportion to each other in terms of the asymptotic inertial coordinat=
es
in which your distant observer is at rest.

You may be getting confused by trying to apply the Newtonian concepts of=20
instantaneous gravity and Galilean invariance of physical laws, etc., (even
though you we not invoking Newtonian concepts), leading to the quantitative=
=20
Newtonian extrapolation of Kelper's law, relating Newtonian mass to force=
=20
and orbital periods, etc., and pointing out that if all those things were t=
rue, then=20
special relativity would be false. That is correct, but it essentially amo=
unts to=20
saying if special relativity was false then special relativity would be fal=
se. It's=20
a true statement, but it has no meaningful cognitive content.