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EBTX

232 Posts

Posted - 29 Nov 2003 : 17:16:26

From Newton's equation the force of gravity is generally proportional to

(m1 x m2) / r^2

Therefore, algebraically speaking ...
(m1/r) x (m2/r) would indicate that the attenuation of the gravitational field must proceed as 1/r.

If you give it as
m1/(r^2) x m2/(r^2)
we get
(m1 x m2) / r^4

It seems that to use a 1/r^2 attenuation, we must apply it to two masses additively giving ...
m1/r^2 + m2/r^2 = (m1+ m2)/r^2

but, clearly, the masses must be multiplicative when we examine the details of any theory whatsoever. So, for any radiative flux at 1/r^2, the top part of the equation must be multiplicative while the bottom must be additive ... this is in contradiction to mathematical logic.

But, a curved space theory is multiplicative on top and bottom since the field attenuates as 1/r (the curvature relative to some standard of length significantly greater than the radius of curvature).

What does MM have to say about the relationship of mathematical logic and physical logic?

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tvanflandern

USA
2793 Posts

Posted - 29 Nov 2003 : 17:33:54

quote:
Originally posted by EBTX

From Newton's equation the force of gravity is generally proportional to (m1 x m2) / r^2

Yes, but gravitational force is an unobservable abstraction. In all astrophysical problems, we actually observe acceleration, not force. We verify this formula:
a = GM/r^2
It describes the acceleration (a) of any target mass (m) in the field of a source mass (M) at distance (r), and is clearly an inverse square effect. As Galileo showed, the gravitational acceleration of any mass is indepemdent of its own mass.

Then the formula for abstract force (F) is F = m a. That can be combined with the previous formula, if you wish; but it should be clear that there is nothing "inverse r" about the way the mass of the target body enters the formula.

quote:
a curved space theory is multiplicative on top and bottom since the field attenuates as 1/r (the curvature relative to some standard of length significantly greater than the radius of curvature).

There is no "curved space" theory in existence. See p. 32 of MTW. GR calls for "curved spacetime". But "spacetime" is essentially proper time, with no spacial component. See http://metaresearch.org/cosmology/gravity/spacetime.asp.

quote:
What does MM have to say about the relationship of mathematical logic and physical logic?

They can be made consistent when one gets the math right and the physics straight, and carefully adheres to the principles of physics. -|Tom|-

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EBTX

232 Posts

Posted - 30 Nov 2003 : 03:33:00

quote:
As Galileo showed, the gravitational acceleration of any mass is independent of its own mass.

For small bodies this is true. However, if you drop Jupiter on the Earth it will accelerate a good deal more quickly than a bowling ball ;o)

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EBTX

232 Posts

Posted - 30 Nov 2003 : 03:39:39

quote:
Yes, but gravitational force is an unobservable abstraction. In all astrophysical problems, we actually observe acceleration, not force.

I would say that the extent of compression of a spring would amount to the observation of force sans acceleration. Though we don't have any spring scale on the astronomical level, those at our level should be assumed to scale up.

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EBTX

232 Posts

Posted - 30 Nov 2003 : 03:41:30

quote:
There is no "curved space" theory in existence.

In existence ... yes ... mine at least. In the literature ... unfortunately, I agree.

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tvanflandern

USA
2793 Posts

Posted - 30 Nov 2003 : 11:30:42

quote:
Originally posted by EBTX

quote:
[tvf]: As Galileo showed, the gravitational acceleration of any mass is independent of its own mass.
For small bodies this is true. However, if you drop Jupiter on the Earth it will accelerate a good deal more quickly than a bowling ball ;o)

The gravitational acceleration of any target body remains independent of a target body's own mass (m) for target bodies of any mass because the formula describes only the acceleration of the target body: GM/r^2, which is proportional only to the source mass (M).

A separate formula for the acceleration of the source mass is factored by the target body mass: Gm/r^2. The two independent acceleration formulas combined yield the total relative acceleration between any two bodies, which is equal to the sum of their individual, separate accelerations.

For studying whether the force law is inverse linear or inverse square, it is reasonable to adopt target body masses in the huge range wherein their back-acceleration on the source mass is utterly negligible and unobservable. We then plainly see the source mass field drops off with the square of distance, as does any phenomenon that spreads in two dimensions while propagating through a third, such as light intensity.

quote:
quote:
[tvf]: gravitational force is an unobservable abstraction. In all astrophysical problems, we actually observe acceleration, not force.
I would say that the extent of compression of a spring would amount to the observation of force sans acceleration. Though we don't have any spring scale on the astronomical level, those at our level should be assumed to scale up.

Force is real enough and can be observed and measured for types of force other than gravity. My remark about force being unobservable was qualified to gravitational force in astrophysical problems, where we can observe only acceleration and must infer force.

quote:
quote:
[tvf]: There is no "curved space" theory in existence.
In existence ... yes ... mine at least. In the literature ... unfortunately, I agree.

As MTW comment on p. 32, Riemann spent his entire life in an unsuccessful search for a curved space theory.

In general, the primary reasons for failure of such theories are:
(a) Only a single curvature can be attached to a single point in space at a single instant of time. Yet bodies of different speeds passing through that point all curve by different amounts. And a taut rope passing through it is Euclidean straight.
(b) A stationary body cannot be induced to commence motion by space curvature alone. An external force must act.
(c) Changes of momentum require a source for the new momentum which must itself possess that momentum (i.e., must move) before it can deliver it to a target body. Curved space postulates new momentum ex nihilo, which is a miracle, forbidden in physics.

How does your theory deal with these allegedly fatal objections that stopped all attempts by the minds that preceded you?

(I trust you realize that shooting down your own theory is a good thing. The only way any of us can make real progress in understanding is by falsifying most of the great ideas that occur to us, so as to hone the possibilities for describing nature down to a few solid ones that work. Our worst enemy in achieving that goal is the tendency to fool ourselves.) -|Tom|-

1234567890

360 Posts

Posted - 30 Nov 2003 : 13:33:43

quote:
Originally posted by EBTX

From Newton's equation the force of gravity is generally proportional to

(m1 x m2) / r^2

Therefore, algebraically speaking ...
(m1/r) x (m2/r) would indicate that the attenuation of the gravitational field must proceed as 1/r.

If you give it as
m1/(r^2) x m2/(r^2)
we get
(m1 x m2) / r^4

It seems that to use a 1/r^2 attenuation, we must apply it to two masses additively giving ...
m1/r^2 + m2/r^2 = (m1+ m2)/r^2

but, clearly, the masses must be multiplicative when we examine the details of any theory whatsoever. So, for any radiative flux at 1/r^2, the top part of the equation must be multiplicative while the bottom must be additive ... this is in contradiction to mathematical logic.

But, a curved space theory is multiplicative on top and bottom since the field attenuates as 1/r (the curvature relative to some standard of length significantly greater than the radius of curvature).

What does MM have to say about the relationship of mathematical logic and physical logic?

It's weird how two unrelated individuals can generate practically
identical ideas.

Serviccing synonymic tuberization kerbstone composer scorify? Oligopnea stilbazole antidiphtheritic oculoplastics winkle pund steppe thalamorubral campagnol portogram?

EBTX

232 Posts

Posted - 01 Dec 2003 : 04:05:39

quote:
How does your theory deal with these allegedly fatal objections ...

I shall give them ... but first ... I lost some sleep over the following (actually several hours over two nights ... and I've always said to myself that no idea is worth losing sleep over because it will still be there tomorrow ;o).

From Newton ...

(2M x 2M) / r^2 = (M x M) / (r/2)^2
where M is two different bodies of the same mass. (You don't have subscripts so I didn't use 1 and 2 there)

Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".
_____________________

Now, we double the mass of the asteroids making sure to keep their centers the same distance apart. The force registered on the bathroom scale should now be "400 pounds".
The total inverse square radiative flux on each mass is doubled. We might then conjecture that 2 x 2 on each mass accounts for the quadrupling of the force registered on the bathroom scale.
_______________________

Now we return to the original setup and this time leave the masses alone and put those bodies at half the original distance taking care to keep their centers at that half distance. Then the bathroom scale once again reads "400 pounds".
The total inverse square radiative flux on each mass is quadrupled. If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.
_____________________

So, all inverse square radiative flux theories of gravitation must then be false or, at best, rendered problematic.

Q.E.D.

Serviccing synonymic tuberization kerbstone composer scorify? Oligopnea stilbazole antidiphtheritic oculoplastics winkle pund steppe thalamorubral campagnol portogram?

tvanflandern

USA
2793 Posts

Posted - 01 Dec 2003 : 12:18:00

quote:
Originally posted by EBTX

Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".

As I tried to explain, force and weight are arbitrary and require definition. You personally will weigh a different amount on every planet or moon you visit. In Earth orbit, you are weightless yet falling in a strong gravitational field. You are using "weight" as if it meant "mass" when it does not. If you stick to acceleration and mass, which are always well-defined, no paradoxes will arise.

quote:
If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.

The "same logic as above" was that "400 pounds" meant the same as 2 x 2 on each mass at the original distance. If you don't mix up mass and weight, you can't get confused in this step. Weight can be anything without violating Newton's law. It even depends on horizontal speed, as when orbiting. Mass is invariant. -|Tom|-

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1234567890

360 Posts

Posted - 01 Dec 2003 : 14:19:01

quote:
Originally posted by EBTX

quote:
How does your theory deal with these allegedly fatal objections ...

I shall give them ... but first ... I lost some sleep over the following (actually several hours over two nights ... and I've always said to myself that no idea is worth losing sleep over because it will still be there tomorrow ;o).

From Newton ...

(2M x 2M) / r^2 = (M x M) / (r/2)^2
where M is two different bodies of the same mass. (You don't have subscripts so I didn't use 1 and 2 there)

Let the Ms be two spherical asteroids some distance apart with a pole between to keep them from gravitating together. One end of the pole pins a bathroom scale to one asteroid and reads "100 pounds".
_____________________

Now, we double the mass of the asteroids making sure to keep their centers the same distance apart. The force registered on the bathroom scale should now be "400 pounds".
The total inverse square radiative flux on each mass is doubled. We might then conjecture that 2 x 2 on each mass accounts for the quadrupling of the force registered on the bathroom scale.
_______________________

Now we return to the original setup and this time leave the masses alone and put those bodies at half the original distance taking care to keep their centers at that half distance. Then the bathroom scale once again reads "400 pounds".
The total inverse square radiative flux on each mass is quadrupled. If we apply the same logic as above we must conjecture that 4 x 4 on each mass accounts for a 16 fold increase in force (1600 pounds) in contradiction with the 400 which should be registered on the bathroom scale as per Newton.
_____________________

So, all inverse square radiative flux theories of gravitation must then be false or, at best, rendered problematic.

Q.E.D.

Yes, and an equivalent way to look at it would be to
consider M1 and M2 different from one another in Newton's
gravitational force equation. Say the total mass
of the system is 10 (i.e. M1 + M2 = 10). Now, this mass
can be distributed as 4 and 6 ; 2 and 8; 3 and 7; etc. So
the total mass of the system adds up to 10 but when you multiply
the masses, you get a different total each time for the square area of the force.

The flux of gravitational force (or energy) is not conserved in Newton's Force equation because multiplication of masses is used
instead of addition.

EBTX

232 Posts

Posted - 01 Dec 2003 : 15:08:23

quote:
If you don't mix up mass and weight, you can't get confused in this step.

So ... you agree that the scale would read 400 pounds in each case?

If so, we could replace the "force" of gravity with a little rocket at 400 pounds thrust on each of our masses. The acceleration produced should then be twice as great on the small mass as on the doubled mass.

But ... the flux on the smaller mass is 4 times greater on the small mass than on the original mass. And ... it's the flux which is the "rocket". This is the bone of contention. Do you see what I mean?

The problem is the flux at four fold increase. If flux just caused a genuine acceleration, it would be four times greater in the small mass than the doubled mass with a doubled flux propelling it ... not just twice the acceleration.

Or, look at it this way. You have four, 100 pound thrusters on the small mass in the halved distance scenario ... and ... two, 100 pound thrusters on the doubled mass in its scenario ... so the smaller mass will accelerate at four times the rate of the doubled mass.

Thus, an inverse square flux cannot be the "effective cause" of gravity.
_______________

Or ... are you saying that the scales in each case would read something different? If so, would this not constitute a clear violation of Newton's laws?
___________________

And I agree, it is confusing. The closer you look at something, the more it tends to drip through your fingers. I am open to enlightenment ;o)

tvanflandern

USA
2793 Posts

Posted - 01 Dec 2003 : 16:14:59

quote:
Originally posted by EBTX

So ... you agree that the scale would read 400 pounds in each case?

As I understand your setup, which is not well, I agree.

quote:
If so, we could replace the "force" of gravity with a little rocket at 400 pounds thrust on each of our masses. The acceleration produced should then be twice as great on the small mass as on the doubled mass.

No, 200 pounds on each mass for a total of 400 pounds. The total weight measured by the scale should logocally depend on both masses, not just one of them; but it depends on how you choose to define "weight", which is ambiguous until specified. "Weight" normally has meaning only with respect to some one particular body. But if either one of the asteroids were hollow and nearly massless, the reading for the combined forces would drop from 400 to 200 pounds total.

quote:
But ... the flux on the smaller mass is 4 times greater on the small mass than on the original mass. And ... it's the flux which is the "rocket". This is the bone of contention. Do you see what I mean?

Not really. But with the preceding correction, maybe it will make sense to you. If not, please try to explain your setup again, with special attention to definitions of force and weight, which are ambiguous here. I've never thought about gravity in terms of weight before because it requires a combination of a gravitational force and a non-gravitational force (e.g., a spring in the scales), and requires one to define "weight" for each application. That can get tricky because different laws apply to the non-gravitational force, and because there is no universal menaing to "weight".

If you use Newton's gravitational acceleration law, you can't go wrong because only one mass at a time is involved.

quote:
The problem is the flux at four fold increase. If flux just caused a genuine acceleration, it would be four times greater in the small mass than the doubled mass with a doubled flux propelling it ... not just twice the acceleration.

I don't see why, but feel I may now have completely lost contact with your setup. One double mass means double acceleration for the other mass with no change in its own. And vice versa. So if both masses double, that gives four times the acceleration. What can be ambiguous about that?

quote:
Or, look at it this way. You have four, 100 pound thrusters on the small mass in the halved distance scenario ... and ... two, 100 pound thrusters on the doubled mass in its scenario ... so the smaller mass will accelerate at four times the rate of the doubled mass.

I'm not following you at all now. I can't even guess what you might mean. Is there a typo in this last paragraph? If you meant "double" where you said "four times" in the last line, I might agree, but wonder what the question was.

quote:
Thus, an inverse square flux cannot be the "effective cause" of gravity.

My failure to understand this is surely derivative from my failure to understand the preceding. But don't you have an inate sense of discomfort when you are challenging something that millions of minds before you have thought hard about over centuries and found no fault with? You are still entitled to have your questions answered. It's the presumption in your writing that you have just discovered soemthing that millions have thought hard about and not noticed before that I find ... oblivious.

Clear, unambiguous procedure:

* acceleration of body m is a1 = GM/r^2
* acceleration of body M is a2 = Gm/r^2
* total relative acceleration is a1 + a2
* total relative force depends on definition
* weight depends on definition

These last two depend on whether you want the one-body force or weight (relative to which body?), or the two-body force or weight. Define your terms carefully and you can go there. Otrherwise, be safe and stick with masses and accelerations as everyone else does and no ambiguities will ever arise. -|Tom|-

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EBTX

232 Posts

Posted - 02 Dec 2003 : 14:03:37

quote:
But don't you have an inate sense of discomfort when you are challenging something that millions of minds before you have thought hard about over centuries and found no fault with?

No, not at all. I don't have to worry about my reputation or credibility since I have not got any to start with ;o)

What I am proposing is simple logical raplacement of gravity with rocket propulsion which the scale cannot identify by any means whatsoever.

If you say that a mass of 1 unit produces shadowing effect of 1 unit ... then won't two masses produce 2 units of shadowing effect? Is this not correct by your theory? If it produces 4 units of shadowing, I would indeed be wrong ... but how so? The shadowing effect seems fairly straightforward to me and appears to be additive.

tvanflandern

USA
2793 Posts

Posted - 02 Dec 2003 : 14:27:11

quote:
Originally posted by EBTX

If you say that a mass of 1 unit produces shadowing effect of 1 unit ... then won't two masses produce 2 units of shadowing effect? Is this not correct by your theory?

Doubling the mass doubles the shadow effect. If the two masses are separate, then like two shadows of objects cast by two flashlights, the shadows will sometimes add and sometimes not, depending on location.

A spot equi-distant between two masses feels zero force because the two separate forces cancel.

In general, each mass must be considered separately and the effects of each separate mass then added vectorially (i.e., the sum depends on direction and distance of each). -|Tom|-

Serviccing synonymic tuberization kerbstone composer scorify? Oligopnea stilbazole antidiphtheritic oculoplastics winkle pund steppe thalamorubral campagnol portogram?

EBTX

232 Posts

Posted - 03 Dec 2003 : 04:26:49

quote:
Doubling the mass doubles the shadow effect

Now, if the object mass that the shadow falls on is doubled without changing its area presented to the shadow, will there be any increase in the shadowing effect? Can changing the mass of the shadowed object cause more CGs to hit it from the rear than what is represented by the shadow?

You see, if we double the mass that the shadow falls on, we need to have the same shadowing effect on that added mass as was on the original mass ... (i.e. we need to double the CGs hitting the side away from the shadow of the now doubled object) ... or the shadowing theory breaks down.

Let's say that the shadow represents a shortfall of 42 CGs on the "windward" side of the object it falls on. Now if we double the objects mass that the shadow is falling on, that same shadow must now represent a shortfall of 84 CGs on the now doubled object. How could this be so?