For Dr. Tom Van Flandern

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Try thinking about the difference between the center of mass of a "stretchy" device and the various parts that make up the device. As long as the device is hanging by one of its parts, the rest of the device will be distorted by the force of gravity acting on its various parts.

Energy will be stored in the various parts of the device by this distortion.

When the device is dropped, the center of mass will fall smoothly at 9.8 m/Sec^2, but that stored energy will cause some parts of it to oscillate with respect to the center of mass and this oscillatory movement can, under some conditions, cause <b>some of the parts</b> of the device to fall at a rate other than 9.8 m/Sec^2

But, like my previous example with a rotating object, if you average the motion of any point on the device *over a full oscillation* you will get 9.8 m/Sec^2.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Correct. And I agree with this.


<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>When point B touches point G on your device, standard physics does NOT predict that the device stops falling. Point B pushes on point G and point G starts moving,<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Let mass of whole system be m. Hence initially at support point, downward force F = mg is acting and support point is exerting upward force of same magnitude to balance the system.

Now we are NOT releasing the support. We bring ANOTHER body of same mass m and place it at point G. So now force F = mg is acting on point G. Will the point G start falling?

Nope. It will just compress molecule by molecule until this circulate and reach to support point. Now when we leave support, rod ABC is falling and point G is not moving at all. So when point B strikes point G, force mg acts on point G. Now, how point G will know that force mg acting on it is of same system and not of some other body from outside the system?

The information regarding release of support is not reached to point G, hence point G will assume that the force mg acting on it is of some other body from outside the system. It will just begin to compress and NOT falling. There is big difference between compressing and falling. In one second, it may compress by, say 1 cm. But in free fall, in one second, it falls by 4.9 meter in space.

Ask any physicist whether point G will begin to accelerate at exactly same moment we release support. He/she will say NO. Because it is prediction of Newtonian mechanics.

It is strange that I am trying to prove Newtonian mechanics RIGHT regarding instantaneous speed of gravity and I am doing this by trying to prove Newtonian mechanics WRONG regarding speed of propagation of mechanical forces.

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Standard Physics DOES NOT make the prediction you claim.

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It does predict.. officially.

-Abhi.

Abhi,

OK, so you understand about the way oscillating and rotating motion can cause certain parts of a device to be "falling" at a rate that is different than other parts in general, and different than the rate of fall of the center of mass (CM) in particular.

Now you need to realize that you are analyzing a system with several time varying phenomena occuring simultaneously. If you look at just some parts of the device for an arbitrary time interval (one not equal to a complete cycle of oscillation) you are likely to see things that appear to be impossible. <b>Until you take into account</b> the fact that the system is oscillating.

Earlier in this thread Dr Van Flandern suggested replacing the rope in one of your first examples with an elastic band. (that was when I began to realize that I had done an experiment similar to what you were talking about). The reason he made this suggestion is that pressure waves (compression or relaxation) move less rapidly in "springy" materials than in "stiff" materials. This doesn't make much difference as far as analysis goes, but if you are going to do an actual experiment (or try to visualize doing an actual experiment) it can be very helpful. By slowing everything down it will be easier to see what is happening.

In your experimental device, replace the vertical parts AD and CF with slinkies - replace the vertical rod GE with a spring - and replace the horizontal parts ABC and DEF with rubber rods. The gap between B and G doesn't matter, so leave it there. Make it any size you want.

Now hang your modified device from the roof and let it stop wiggeling. The rod ABC is supported along its entire length, so it is not bent. AD and CF are stretched by the weight of DEF and GE. DEF is bent by the weight of GE resting on it. The gap between B and G is (your favorite number goes here. I like 42. What units? How about quark diameters).

Now release ABC.

Rubber rod ABC, under the pull of gravity AND the tension from AD and CF will begin to move down. And it will move down faster than if only gravity were pulling on it.

The upper ends of AD and CF also begin moving down, also faster than if gravity were the only force being applied.

But for a few (milliseconds? microseconds?) the rod DEF and lower ends of AD and CF, still supported by the (now declining but not yet zero) tension in AD and CF, aren't falling yet (or, as the tension drops,they are falling at a rate less than 9.8 m/Sec^2 [g]). Surprize! And this does mean that GE isn't falling yet (or is falling at a rate less than g). In a (short time?), when the pressure waves moving through the device get to the right place in the cycle, GE will be falling faster than if gravity were the only influence. Over a full cycle, GE will fall at an average rate of g.

****************************************************

The CM of the overall device *is* falling from the first instant, because *some parts of it* are falling from the first instant. The acceleration of some of the parts that are falling right now is greater than g. For other parts the rate of fall is equal to or less than g.

At each instant the average rate of fall for ALL of the parts (and the CM) adds up to, would you believe, g?

****************************************************

The cycle continues -

Now B contacts G. GE begins to compress and ABC begins to bend. As G moves down B continues to move down, but now not quite as fast as A and C are moving down.

The movement of G sends a compression wave down GE.

The relaxation waves travelling down AD and CF eventually reach DEF and it begins to un-bend, and to fall. This allows GE to begin to uncompress, and to fall. At first these parts fall slower than g. But the parts near the top are still falling faster than g, so...

The average rate of fall for ALL parts (and the CM) is still g.

And so on ...




As you can see, standard Physics DOES NOT predict that the device will hang in the air, or fall intermittently. But it does predict that some parts of a device, using stored energy, can fall less rapidly (or even rise) for brief periods of time. As long as later on in the cycle that same part is falling more rapidly.


Regards,
LB

First you said this...

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Every atom in the device is subject to the force gravity. As soon as the device is released every atom in it is free to respond to this force by accelerating. And each of them do in fact begin to fall.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
But when I told you that this is not official stand of physicists, you have changed camp and now you are in official camp. But OK, I don't mind.....

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>In your experimental device, replace the vertical parts AD and CF with slinkies - replace the vertical rod GE with a spring - and replace the horizontal parts ABC and DEF with rubber rods. The gap between B and G doesn't matter, so leave it there. Make it any size you want.

Now hang your modified device from the roof and let it stop wiggeling. The rod ABC is supported along its entire length, so it is not bent. AD and CF are stretched by the weight of DEF and GE. DEF is bent by the weight of GE resting on it. The gap between B and G is (your favorite number goes here. I like 42. What units? How about quark diameters).<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Your set up is good. I don't mind since my experiment is not material dependent. But remember you put the gap between B and G equal to "quark diameter".

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> Now release ABC.

Rubber rod ABC, under the pull of gravity AND the tension from AD and CF will begin to move down. And it will move down faster than if only gravity were pulling on it.

The upper ends of AD and CF also begin moving down, also faster than if gravity were the only force being applied

But for a few (milliseconds? microseconds?) the rod DEF and lower ends of AD and CF, still supported by the (now declining but not yet zero) tension in AD and CF, aren't falling yet (or, as the tension drops,they are falling at a rate less than 9.8 m/Sec^2 [g]). Surprize! And this does mean that GE isn't falling yet (or is falling at a rate less than g). In a (short time?), when the pressure waves moving through the device get to the right place in the cycle, GE will be falling faster than if gravity were the only influence. Over a full cycle, GE will fall at an average rate of g.
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If the gap between B and G is euqal to quark diameter, when B covers this gap, it is obstructed by G. Now B has covered distance in space equal to "quark diameter". So it has relaxed by that distance only.

How relaxation in slinkes equal to quark diameter sets wave along slinkies which can carry information regarding release of support? And even if it is carrying information and during the time interval "t" it reaches to point G, then point G will not start falling during this time interval. But point B of rod is now obstructed by point G.

Let mass of whole system be m. So G force equal to mg will act on point G. But point G has no information whether this "mg" is due to its own mass of system or that of other body from outside the system.

For simplicity, suppose rod ABC is not released and system is hanging. Now we bring another body of same mass equal to that of system i.e. m. Now we place this body on point G. So G-force mg will act on point G. spring GE will begin to compress starting from point G. But it is compressing, NOT falling with free fall acceleration.

(1) Why the same situation can not arise when we release rod ABC and point B interacts with point G?

(2) Why can't point G just begin to compress just like when some other body of same mass is placed on it? Afterall the point G has no information during time interval t regarding release of support.

(3) As information propagation begins from rod ABC and point B is not falling with free fall acceleration, leave alone gretaer than g due to stored elastic potential energy, then which part you vizualize falling with free fall acceleration g in the beginning?

(4) And if no part is falling with g, then I prove Newtonian mechanics prediction that things can either hang in space or fall slower than g.


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The CM of the overall device *is* falling from the first instant, because *some parts of it* are falling from the first instant. <hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

We begin from top rod ABC and I said no part can fall with free fall acceleration due to obstruction by point G. Remember, the distance betwen point B and G is just equal to quark diameter. Point G is compressing due to force mg acting on it even though it is unbelieveable that m also includes mass of molecules in spring at point g. But in any case point G is compressing and NOT falling. As rod ABC rests on point G, hence it is coming down with same velocity with which spring is compressing. This velocity obviously depends upon stiffness of spring.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>As you can see, standard Physics DOES NOT predict that the device will hang in the air, or fall intermittently. But it does predict that some parts of a device, using stored energy, can fall less rapidly (or even rise) for brief periods of time. As long as later on in the cycle that same part is falling more rapidly.


Regards,
LB<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

You are standing on table and due to your mass m, G force "mg" is acting on center of table. Standard physics also predict that even if your mass does not change, "10 mg" gravitational force can act on center of table. Wonder if I remove table, with what acceleration you will fall. g or 10g?

I constructed the situation in which your weight can act on table twice(I can make it 100 times or any number of times). And official people said that it can act twice. But whether you will fall with twice acceleration, then "personal" theories props up..

-Abhi.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
As you can see, standard Physics DOES NOT predict that the device will hang in the air, or fall intermittently. .....

Regards,
LB
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Standard physics(so far our situation of falling objects goes) also predict that speed of sound,

c = 0.707*sqrt(Bulk Modulus / Density)

How here we go to prove that...

(1) The proerty that determines the extent to which an element of a medium changes in volume when the pressure (force per unit area) on it changes is the "bulk modulus (", defined as

B = -[ p'/(v'/v)]

where v'/v is the fractional change in volume produced by a change in pressure p'.

(2) Density is defined as mass per unit volume. d = m/v

(3) So speed of sound(c) in bulk media is given as

c = sqrt( Elastic Property / Inertial Property) = sqrt( B/d)

(4) All real "rigid" bodies are to some extent elastic, which means that we can change their dimensions slightly by pulling, pushing, twisting or compressing them.

(5) There are three ways in which a solid might change its dimensions and get deformed (a) tensile stress (b) Shearing stress (c) hydraulic stress.

(6) Stress (F/A) = modulus of eleasticity(E) * strain(L'/L).

(7) For tension or compression, the stress on an object is defined as F/A, where F is magnitude of force applied perpendiculerly to the area A on the object.

( If the specimen is long rod and the stress does not exceed the yeild strength, then not only the entire rod but also every section of it experiences same strain.

(9) Due to tensile stress, the rod undergoes deformation and changes in length by L' which is given by

L' = (F*L / A*E)

Where F = force applied, L = length of rod, A = area of rod on which force applied, E = Young's modulus.

(10) We take a rod of length L, area A and mass m. We place it on ground in vertical position. G force F = mg is acting on area A of rod. Due to this force, the rod undergoes deformation and its length is decreased by L' which is given as

L' = (F*L / A*E)

Now when we leave support of this rod so that it falls freely, this information regarding release of support propagate from lower end to upper end molecule by molecule with speed of sound c. So it takes time t = L/c to reach to upper end.

But during this time interval t, lower end is falling with free fall acceleration g. So during time interval t, lower end covers distance h in space which is given as

h = 0.5gt^2

As lower end has fallen by height h until information reaches to upper end, that means the rod has regained its original length L. So the distance h through which lower end has fallen must be equal to deformation L'.

Hence h = L'

(FL / AE) = 0.5gt^2

substituting t = L/c, we get

(FL / AE) = 0.5g(L/c)^2 = 0.5gL^2 / c^2

Hence c^2 = (0.5gL^2*AE) / FL

As F = mg

c^2 = 0.5gL^2AE / mgL

c^2 = 0.5(LA)E / m

Here L*A = length od rod * area of rod = volume of rod = V

Hence c^2 = 0.5V*E /m

c^2 = 0.5 E / (m/v)

As m/v = density of medium =d, we get

c^2 = 0.5 E /d

Hence speed of sound in rod is,

c = sqrt(0.5)*sqrt (E/d)

c = 0.707* sqrt(E/d) = 0.707*sqrt[bulk modulus / density]

We must get standard equation of speed of sound in such situation if deformation L' is equal to height h through which rod falls in time interval t = L/c.

Everything seems to fine except "0.707".

Can you please debug this code" 0.707"?

-Abhi.

Abhi,

Even though I'm pretty sure I now understand what you are saying, because of the language thing it seems like a good idea to do a double check. So ...

It is my understanding that you believe -

1) That in reality your device WILL fall smoothly at 9.8 m/sec^2. (After all, this is what we observe for all dropped objects.)

2) That standard (Newtonian) Physics makes the INCORRECT PREDICTION that your device (the entire device, IOW the CM of your device) will NOT fall smoothly at 9.8 m/Sec^2. According to this incorrect prediction made by standard Physics, the CM of your device will fall faster at times and slower (including zero) at other times.

Please correct me if I'm wrong on either point. (Don't try to justify it here, just correct it.)

----

And a question - would the prediction of standard Physics change, and how would it change, if the gap between G and B in your device is zero?

Regards,
LB

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
1) That in reality your device WILL fall smoothly at 9.8 m/sec^2. (After all, this is what we observe for all dropped objects.)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

It is well proved that all bodies fall with uniform acceleration 9.8 m/s^2. But all the physicists somehow believe that when the body start falling, this information reaches to every other particle in the body with speed of sound.

I am saying that this information regarding falling reaches to every point in body "instantaneously". All though some points may start with different acceleration other than g. But they start accelerating at exactly same moment.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>2) That standard (Newtonian) Physics makes the INCORRECT PREDICTION that your device (the entire device, IOW the CM of your device) will NOT fall smoothly at 9.8 m/Sec^2. According to this incorrect prediction made by standard Physics, the CM of your device will fall faster at times and slower (including zero) at other times.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

In situation like I have given, as the transmission line through which information is supposed to pass remains stretched, no information will pass. We can use magnets so that these lines remains streched or comprssed as the case may be. No contraction. No expansion. No relaxation. And hence no information to other parts of system. Hence system may hang in space without support or fall intermittently.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And a question - would the prediction of standard Physics change, and how would it change, if the gap between G and B in your device is zero?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

If the gap between B and G is zero, then it is no more gap. Physicists will say that it is directly supported by rod ABC and there is tension in rod BE. They will say that when we leave support, this information will reach to point E through compression waves through rod BE.

-Abhi.