Re: Tom Van Flandern and Newtonian Gravity

From: Tom Van Flandern (tomvf_at_starpower.net)
Date: 08/12/04 

Date: Wed, 11 Aug 2004 20:27:13 -0400

            This replies to Gerald Lasser, Vern, Bilge, Mike, and
Greysky.

"Gerald Lasser" <antispam@nospam-me.com> writes:

>> [tvf]: I do agree with the equations, but only because the force
propagation delays that must exist have been approximated as zero by
assuming infinite propagation speed. Therefore, the electromagnetic
field at any given point is *not* determined only by the conditions on
the past light cone of that point...

> [Lasser]: Again you have given a self-contradictory answer. Maxwell's
equations imply that the electromagnetic field at any given event is
equal to a specific function of the conditions on the past light cone of
that event. They do not involve an infinite propagation speed.
Therefore, you cannot logically both agree with Maxwell's equations AND
disagree with the proposition that the electromagnetic field at any
given event is equal to a specific function of the conditions on the
past light cone of that event.

            Three times now I have given you quotes, citations, and
examples that contradict these assertions. I realize that you are not
saying anything unique to you, and that many "experts" would agree with
you. But I maintain this is wrong for the reasons I have argued. You
cannot counter a reasoned argument with an assertion, however
"authoritative" it might be.

> [Lasser]: Could you please try again to answer the question without
logically contradicting yourself? You either agree with "the math of
current theories" or you do not. Which is it?

            As I said before:

>> [tvf]: I do agree with the equations, but only because the force
propagation delays that must exist have been approximated as zero by
assuming infinite propagation speed.

> [Lasser]: Your premise is false. I just explained to you that,
according to the equations you claim to agree with, electromagnetic
forces are retarded with a propagation speed of c, which is to say, the
electric and magnetic forces at any given event are determined by the
charge and current distribution on the past lightcone of that event.
Maxwell's equations do not assume c is infinite.

            Despite your "explanation" (actually an assertion), I do not
agree. To try to get you to understand why I disagree, I pose the
following problem. Any solution you offer will be enlightening.

            Adopt the physics perspective that light is generated from
disturbances in electric and magnetic potential fields that propagate at
speed c; that forces are related to gradients of potentials; and that
functions and their derivations need not have the same properties. In
particular, recognize that potentials and their gradients need not
propagate at the same speed. Then Coulomb force need not propagate with
speed c just because light does. These premises should be acceptable to
you for discussion purposes because they do not contradict any of your
assertions.

            Operating under these broad premises, please show the
modifications to Maxwell's equations needed to change the propagation
speed of Coulomb force from speed c to infinite speed. (Hint 1:
Propagation delays in potential fields (e.g., any scalar value) affect
light propagation, but not force propagation in any important way. So
ignore those because they are of no interest to this discussion. Hint 2:
There would then be no difference in the equations except for the case
where two charges have a relative transverse motion. Hint 3: IMO, the
force propagation speed is already infinite in Maxwell's equations
because there is no propagation delay term for relative transverse
motion in them. But perhaps you will find some hidden propagation delay
term that has heretofore eluded me and others.)

> [Lasser]: Let me ask you this: Suppose you have two positively charged
particles, held stationary for a long time (by electrically neutral
pliers) at a distance D from each other. They are each subject to a
measurable, palpable, force of repulsion. Now suppose you abruptly move
one of the particles to a slightly different position. I say the
measurable force on the other particle will be totally unaffected until
a time D/c later. What do you say?

            I say the effect will occur much sooner than "D/c later". So
I disagree with you. In physics, experiments are important for deciding
such questions. You have none to back your opinion, only a long-standing
tradition in the field of thinking that way. But I do have an experiment
that backs up my opinion: the Sherwin-Rawcliffe experiment previously
cited. It demonstrated *no lightspeed delay* in the force between
accelerating charges. Charges accelerated jointly in the same direction
respond to each other's instantaneous positions, and not to the
"left-behind potential hill" following acceleration from zero speed.

and "Vern" <vthodge@bealenet.com> writes:

>> [tvf]: One of us completely misunderstood the other. I thought we
were talking about a substitute for gravity, an alternate explanation
for orbital motion. McCarthy is acknowledging ordinary gravity, and
looking for an explanation for angular momentum being so non-random.

> [Vern]: If I'm not mistaken, your view is that Newtonian gravity as
modified by GR is correct except that the forces must propagate FTL.

            Not "except". GR is correct, period; *and* the forces in it
already do propagate FTL (i.e., there is zero force propagation delay in
GR, just as in Newtonian gravity).

> [Vern]: Since neither Newton's equations nor GR offer any model or
explanation of what causes gravity and an Aether vortex model does.

            The new book "Pushing Gravity" offers the most complete
explanation for gravity I've ever seen. OTOH, your aether vortex model
offers no explanation whatever for the force of gravity, but at most
might answer questions about the origin of the excess of prograde
angular momentum in the solar system, as McCarthy described in the
message you quoted. But that model requires that normal gravity already
be operating first.

> [Vern]: I posited that an Aether vortex model of gravity solves both
concerns; FTL forces are not needed and a cause is provided (the carrier
medium).

            You have not addressed the issue of how source mass and
target body are linked in the aether vortex model, much less how the
interpretation of the six experiments showing that gravity operates FTL
could change in that model. (All models have force applied locally, but
propagation of new momentum from the source is a different matter.) And
the Le Sage model of gravity in "Pushing Gravity" avoids the five fatal
problems inherent in trying to connect gravity with aether, where
"aether" has its original meaning of the light-carrying medium:

(1) Gravity behaves as if its constituents are particles, not waves, and
has no known uniquely wave properties. In contrast, light has all wave
properties, but no essential particle property.

(2) From the kinetic theory of gases, the mean speed of the constituents
in an aether that serves as the medium for light must be c * sqrt(5)/3.
But the mean speed of gravitons must be no less than 2x10^10 c.

(3) Gravitons interact with one another through collisions (the
explanation for "dark matter"); "photons" do not.

(4) Gravitons can penetrate matter easily, photons cannot.

(5) There are too few photons in the visible universe to produce a force
as strong as gravity.

> [Vern]: You stated reasons why you believe an Aether vortex model is
flawed (highly inclined and retrograde orbits and the orbits or
artificial satellites). Dennis explained why an Aether vortex model
would yield the same observations as the Newtonian or Relativistic model
(if you want to call it that), and mentioned some observations that
offer evidence of an Aether vortex model (the inclination of Uranus and
its moons, etc.).

            We are not getting the same meaning from what Dennis
McCarthy wrote (in your quote). He was explaining angular momentum, not
gravity. What he said works; your interpretation of it does not, as far
as I can see.

and "Bilge" <dubious@radioactivex.lebesque-al.net> writes:

>> [tvf]: I said (paraphrasing) "curved spacetime has nothing to do with
curved space". If you had read my article, you would see the argument
that "spacetime" means proper time, and has no space component.

> [Bilge]: Later in the article, you clearly mention timelike intervals,
so if you are talking about a timelike interval, how can you not be
talking about spacetime?

            Indeed. How can I be making the point that curved spacetime
has nothing to do with space, while making no mention of spacetime? Your
question makes no sense.

>> [tvf]: Then you too would see that your whole set of questions above
and in the remainder of your post make no sense in the context of my
article because you took the meaning of my words in a completely
different sense than was intended.

> [Bilge]: I don't think so, tom. For example, later in that same url,
you state: [tvf]: "In this form, we can see the space-time interval ds
as a purely time-like interval dt that was merely made to look
space-like through multiplying it by c." [Bilge]: And further below
that, you state: [tvf]: "This is an important concept. If the curved
path of a body through space is not caused by a curvature of space, then
clearly an external force is still required to produce and explain the
deviation from straight line motion. And some explanation other than
curved space is needed to understand the equivalence-principle-like
property of gravity."

            And your point is.? Apparently, wherever I use the phrases
"space-like" and "time-like", you jump off the track I was trying to
follow. And this is still happening even after I changed the wording to
clarify the meaning: "a space-like coordinate (meaning one measured in
meters, not to be confused with a 'space-like interval' as used in
general relativity)".

            So to say it again as plainly as I know how, I demonstrated
that the metric was a measure of proper time, period. However, someone
was sure to object: "but it is measured in meters, not seconds". And my
answer is "True, but that is because it has been multiplied by c to
convert a pure proper time interval measured in seconds into meters."
Apparently, the point you cannot get past is that, to avoid inventing
new terminology, I referred to this units-converted proper time interval
as "space-like" (meaning only that it has the units of space, nothing
more). It is still a pure proper time interval measured in meters
instead of seconds.

            So I have thought hard about how to communicate my intended
meaning, given that the term "space-like" is causing a derailment. What
I came up with is "length-like". It isn't perfect, but the term is now
defined and used in the article. I've deleted all usages of the term
"space-like". For extra clarity, here is the definition again:
"Length-like" is a time interval multiplied by c to give it units of
length. It is therefore equal to the distance a lightwave in vacuum
would travel in the given time interval. But it remains a time interval,
which is why it is only "length-like" and not truly a length.

            The term "time-like" is retained in the article with its
normal relativistic meaning.

> [Bilge]: But a spacetime metric does have something to do with space,
tom. That's why it's called a spacetime metric and not a time metric.

            Okay, finally, we are talking about the substance of the
paper instead of semantics.

> [Bilge]: How about using your same argument to explain the
Schwarzschild geometry: ds^2 = -(1-2m/r)dt^2 + dr^2/(1-2m/r) + r^2
d\Omega^2, rather than one which does not contain any mass?

            Ignoring that you obscure my point by setting c = 1 and
mixing units, my original argument does not change. In fact, I just made
the last equation in the paper the equivalent of yours using my notation
(potential negative, c retained). Look again at the short web version of
the paper: http://metaresearch.org/cosmology/gravity/spacetime.asp. I'm
sure the simple math needed to show the exact equivalence of my equation
and yours is within your range of abilities.

            But note that not one word of the underlying argument
changes when we do this.

>> [tvf]: If the metric contained a plus sign where it now has a minus,
it would measure a space distance.

> [Bilge]: Which only indicates that you think a Euclidean space is
somehow ordained by god as the only "real" geometry and can't imagine a
space with a negative sign in the metric.

            The whole point of the article was to show even the slowest
of relativity-interested people that, with a minus sign in the metric,
spacetime intervals are proper time intervals multiplied by c, and are
*not* intervals in any kind of space, Euclidean or otherwise.

            Of course, you can render my statement false by defining
spacetime/proper time to be some sort of non-Euclidean "space". But then
you would revert us to another semantic argument. My essential point is
no more profound than this: Curved spacetime has nothing to do with
curved space, wherein "space" means the Euclidean 3-space in which
astronomers make observations and measurements such as relativistic
light-bending. Space is not a component of spacetime, whereas velocity
and potential naturally are components of spacetime.

> [Bilge]: What about the fact that the metric also measures spacelike
intervals?

            Take my form of the Schwarzschild metric and introduce the
Euclidean distance traveled as du, where du^2 = dx^2+dy^2+dz^2. Then du
= v dT. Allow dT in the metric to cancel from the denominator of v =
du/dT, then set the remaining dT = 0 in the equation. It then simplifies
to: du/d-tau = c s, where tau is proper time and s = sqrt(1-2 G M/r
c^2). This is an expression for the velocity of a lightwave through a
potential field, and shows that lightwaves slow when passing through a
stronger gravitational potential. It still does not involve space
curvature.

            Now isn't that a simpler way to understand the equation than
the way usually taught?

> [Bilge]: you need to first try to understand what the equations mean,
physically. You give the distinct appearance of either not understanding
what the equations mean or deliberately attempting to misconstrue them
in order to facilitate your argument about forces.

            Understanding the equations physically has been my goal from
the outset. The traditional way of teaching these concepts may be
beautiful and elegant mathematics, but it never made a bit of sense in
physics. Now it does.

and "Mike" <eleatis@yahoo.gr> writes:

> [Mike]: He [tvf] uses the silly rubber *** analogy in an attempt to
convince laymen that a force is required to initiate motion in GR.

            I use only the "no magic allowed" axiom of physics. The
rubber *** analogy works as an analogy for explaining why a force is
needed just as well as it worked as an analogy for how GR explains
gravity. But in both cases, the absence of a source for new momentum for
the target body and the absence of a transfer mechanism for this
momentum can be explained in only two ways: (1) a force acts ("force"
being defined as "the time rate of change of momentum"); or (2)
mathematical magic.

            But perhaps you have a third way that no one else has yet
thought of? Before you repeat your misunderstanding of what "spacetime"
is, I suggest you read my article "Does space curve?" (which
demonstrates what spacetime is in physics), and refresh on the physics
definition of "force".

> [Mike]: Thus, he [tvf] uses curved space to argue about curved
spacetime. Obviously wrong.

            Let me get this straight. Bilge and I are having a
discussion about my article in which I demonstrate that curved space
plays no role in GR or in "curved spacetime"; and Bilge disputes me on
that point, maintaining that space and its curvature do play a role in
curved spacetime. But you are now criticizing *me*, not Bilge, for
trying to use curved space in GR?

            If I have that right, I'd be forced to conclude that your
comprehension levels are low. Please tell me that you just got your net
personalities mixed up, or had a brief brain lapse.

> [Mike]: He [tvf] assumes that the object magically appears suddenly on
his silly rubber *** and needs then a cause to move.

            Where did this come from? I was discussing the 3-space
trajectories of real objects such as comets. If we start them from the
stationary point on a linear orbit straight into the Sun, what initiates
motion? And what difference does it make if the comet was injected into
that linear orbit by the passage of a star, or it a spaceship dropped it
off there? The same physics question exists either way.

> [Mike]: The point is that every body in spacetime already exists and
has a trajectory already uniquely defined.

            All physical bodies in orbit have an initial point in time
before which they did not exist in their present form, or were not yet
in orbit. Suppose an asteroid is resting peacefully by itself in deep
space relative to the distant stars, when suddenly a planet ejected from
a distant solar system during a nova explosion comes rushing by and .
Well, I can't say "applies a gravitational force to the asteroid"
because you don't seem to accept the physics definition of force. But
surely something happens to the location of the asteroid in 3-space
relative to the distant stars. Please describe that something and what
makes it happen if there is no force applied. (Hint: In physics,
curvature alone cannot initiate 3-space motion unless a force acts.)

> [Mike]: Another simple fact that Van Flandern does not understand is
that geodesic motion is inertial motion in 4-space where Newton's first
law holds, i.e. no forces are required for it. He argues what happens if
projections in 3-space are taken. The answer is that GR does not hold
any longer and he is left with Newtonian mechanics and the problem of
causality and fictitious forces needed to explain motion in local
non-inertial frames of reference.

            Congratulations! In one sweeping statement, you just wiped
out the entire field of relativistic celestial mechanics, which takes
place in 3-space. I suggest you crack a book and get up to speed. If GR
did not apply in 3-space, where all observations are made, then it would
be an observationally untested theory.

            You almost seem to be arguing that 3-space is not reality.
Are you perhaps a 4-space being, forever unchanging? :-)

and "Greysky" <greyskynospam@sbcglobal.net> writes:

> [Greysky]: Tom, arguing with [Bilge] is like bashing a pebble against
your own head. Your assessment of him is correct. You will never get him
to actually give cogent examples to any argument.

            Thanks for the advice. But I already ignore all the trolling
and attempts to change the subject whenever one of my correspondents is
unable to hold up his/her end of an argument, and I just use any mention
of scientific content in the posts to develop relevant points that might
be of interest to others. It is a procedure I recommend. If nothing
else, staying on topic despite distractions and provocations helps
improve the signal-to-noise ratio in these "sci." newsgroups.

            That way, more people come here for science, not just
entertainment. -|Tom|-

Tom Van Flandern - Washington, DC - see our web site on replacement
astronomy research at http://metaresearch.org