For Dr. Tom Van Flandern

Dr. Tom said above:
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> A rope is so rigid that we cannot observe its elasticity by eye. So let's substitute an elastic band for the rope and repeat your experiment. The revolving person then stretches the elastic band until the centripetal and centrifugal forces are equal. Now, when the band is cut, it does not instantly contract. Instead, it remains stretched for a moment until a compression wave travels from the cut end to the X-end of the band at the speed of sound. The band loses its elasticity and contracts as this compression wave passes.

But this picture requires that the X-end of the band remains stretched for a brief time after the cut happens, until the compression wave arrives. As long as the band near the X-end is stretched, it will continue to exert a force on the attached person or object. So that person X is not free to fly off its previous path until the band relaxes.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Let us fix two elctromagnets at the "center of rubber band". Distance between these two magnets is, say 5 cm. The two magnets are repelling each other. So they remain at fixed distance while the person at center is revolving the rubber band. Now the person at center cut the band. According to you, the band near center starts to contracts first and this compression wave passes across band to other person at X-end.

But in between, at the center of this rubber band, we have fixed two electromagnets. As the compression wave approches the electromagnets, we change the magnetic field in between these two magnets so that the length between two magnets, which was 5 cm does not change at all. (or instead we can increase this length between two magnets i.e. stretch the band!). Now the compression wave is not passing through this "speed breaker" i.e. the band between two magnets is not contracting.

Does this mean that even if source of centripetal force does not exist and only because the band between two magnets has not contracted, the person at X end will keep revolving forever in circular motion as if nothing is happened?

Certainly, you will say, NO. Then established theory of information communication regarding non existence of centripetal force is not working here. We can keep the band stretched. We can stop information reaching to X-end.

Concentrate on the stretched rubber band between two magnets. How the information will "jump" or "skip" these two magnets and stretched rubber band to reach the other X end?

-Abhi.

Let us come to our basic point. I say that when we leave a body so that it falls freely, all the particles which make that body begins to fall at exactly same moment although acceleration of some particles might be different than g.

For example, if I have a vertical rod of mass m with lower end in my hand, then G force mg is acting on my hand. To balance the rod so that it does not fall freely and remains in state of static equilibrium, my hands are exerting upward force of magnitude mg. Now I say that when I leave my hands i.e remove upward force applied to lower end of rod, this information reaches to upper end instantaneously. The molecules at lower end of rod were compressed, so they will fall with aacceleration slightly greater than g as they expand in downward direction. But the acceleration of lower end g+ and upper end g, begins at exactly same moment whatever may be distance between lower end and upper end.

You people say that, as the molecules in lower end are compressed, when I will leave support, these molecules will begin to expand and this information will reach to upper end molecule by molecule with speed of sound. So all the base of your arguments is compressibility (if the support is at lower end of rod) and tension (if the support is at upper end of rod)

Now consider a wooden rod placed vertically on ground. Let its original length in absence of force be L. As we place this wooden rod vertically, it undergoes deformation and its length is decreased by L'. Its lower end is compressed maximun and upper end minimum. Now I take two magnets, M1 and M2. These two magnets are fixed to upper and lower end of this vertical wooden rod. South pole of magnet M1 is at lower end and south pole of other magnet M2 is at upper end. So south-south poles are repelling each other.

Due to this upper magnet M2 fixed to upper end of wooden rod will be pushed upward exerting upward force on upper end of rod. Due to this, rod will undergo tension. Let this upward force acting on upper end of rod be exactly equal to G force. This will cancel the deformation(compressibility)caused by G force and wooden rod regains its original length L.

Now molecules at lower end are NOT compressed. If I remove support from lower end of wooden rod, how this information will propagate to upper end of rod?

Suppose that instead of compression, there is tension at lower end of rod due to repulsive magnetic force exerted by two magnets on each other, then how this information regarding release of rod will propagate to upper end starting from lower end?

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Another my old favourite puzzle for you:

Newtonian physics claims that if the wheel is rolling on ground with velocity v, then velocity of top portion of wheel is v+v = 2v.

Reference: Fundamentals of Physics, Halliday, Resnick, Sixth Edition, Page No. 247

I request you to take ride on bicycle on rainy streets. Let velocity of bike + you be v. Now look at the front wheel of bike. Linear velocity of tyre of wheel is v (because that is why bike is moving with velocity v). As you are on rainy street, some mud will stick to tyre of wheel of bike and as tyre is rotating, mud is also roating with tyre.

Now when this mud is thrown in tangential direction from top of tyre, what will be its velocity with relative to you? If you say that velocity of bike will add to velocity of mud, then velocity of mud in the direction of bike with relative to pedestrian should be v + v = 2v and with relative to you, it should be v. That means, mud should escape from tyre of front wheel and travel before your eyes with velocity v. Are you seeing mud escaping and flying before your eyes (STRAIGHT in the direction of motion of bike) with velocity v?

Take high speed camera, take electron microscope or whatever means, even if you find one single molecule of mud flying with velocity v STRAIGHT in direction of motion of bike from TOP of tyre, send your $10/- claim to sciphysics@yahoo.co.in.

Thanks for being with me...

-Abhi.

If I stand on the floor, the forces acting on each atom of my body are primarily (1) a downward acceleration of gravity, the same for all atoms; and (2) a balancing upward acceleration produced by neighboring atoms in my body, and ultimately by the atoms on the floor, whose tensile strength is resisting being crushed by my weight. Even the atoms in the fingertips at my side are receiving this balancing upward acceleration from neighboring atoms to prevent their fall to the floor. My body is therefore in equilibrium.

Now if the floor is a trap door that suddenly opens, I will fall through. Let's examine that process in detail. At the instant the trap door suddenly opens, every atom of my body is still receiving a downward acceleration due to gravity. There is no change in that part of the total force. In addition, every atom is still receiving a balancing upward acceleration from neighboring atoms, except for the atoms on the bottom of my shoes, which will begin to fall because the are no longer in a balanced-force condition. (The upward-pushing floor is absent.)

Now here is the key. Only after the atoms on the bottom of my shoes begin to fall can they release their upward pressure on their neighboring atoms, which can then begin to fall also and to release their upward pressure on their neighbors.

The sudden release of force is like an impulse, which travels through my body at the speed of sound, releasing the upward pressure force as it does so. Although that wave passes through me quickly, it does not do so instantly. So there is a brief period after the trap door opens when parts of my body are not yet accelerating downward. Instead, the slight compression of my body due to its own weight will be released as my feet start to accelerate downward slightly earlier than my head does.

This release of compression is the reverse of what happens in a collision. If a car hits a wall at high speed, its front end stops before its rear end, and the car compresses its length in the direction of motion. Of course, that deceleration is far greater than the 1-g acceleration of gravity on Earth. But the principle is the same. If I were standing on a planet with surface gravity 100 g, I would be a much shorter person (among other problems <img src=icon_smile.gif border=0 align=middle>). If the trap door opened, I would enter free fall, with my feet falling before my head until I had resumed something like my original shape. Decompression waves might have to bounce back-and-forth through me many times before I was fully decompressed.

So acceleration of an entire body does not begin instantly any more than it ends instantly when a crash occurs. This explanation strikes me as logical, understandable, and internally consistent. Moreover, I am unaware of any difference between this explanation and conventional physics. So I guess I agree that conventional physics requires suspended bodies to initially accelerate more slowly than bodies already in free fall.

But I also conclude this is not just conventional physics, but also reality. The difference between the two falling bodies (one just released but still compressed, the other in free fall with its equilibrium shape) is that one has internal upward forces until decompression is complete, and the other does not. It is true that gravity acts almost instantly, and that does not change before or after acceleration begins in any of these examples. But the speed of internal decompression forces is much slower, and cannot happen faster than the speed of sound.

I sometimes find it helpful to carry thought experiments to extremes. So imagine an airless planet having very strong gravity. Atop a tower, suspend an object having a flat, massive base. From the top of the massive base, insert a pole one light-minute tall, with a small sphere attached to the top of the pole. The pole itself consists of 60 sections, each attached to the next by a spring. The goal is to determine how long it is after the base is dropped before the sphere begins to fall.

My analysis is that, viewed in slow motion, each spring will elongate sequentially one second (or more) after the previous spring, and the sphere will not begin to move until 60 such elongations have occurred, at least a minute after the base starts to fall. To me, this does not contradict the near-infinite speed of propagation of gravitational force. It simply recognizes that the sphere is resting on the pole until the pole starts to fall, which cannot happen very quickly because even a "rigid" rod, pushed at one end, can only communicate a pressure wave up the rod before the other end can start to move.

Anyway, that is the way I analyze this example. -|Tom|-

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
Another of my old favourite puzzles for you:

Newtonian physics claims that if a wheel is rolling on ground with velocity v, then the velocity of the top portion of the wheel is v + v = 2v.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Abhi, it's too late for you to pretend that you are this dumb. You've already demonstrated that you are smart enough to understand this.

So, at the risk of angering the Moderator God, I'm going to publically accuse you of trolling (this means I think you are claiming to believe things that you do not actually believe). There is no penalty for this, so it's not really a big deal.

However, there is an easy way you can convince me that I'm wrong about you. Do this simple experiment and report the results.

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The Experiment - tape a meter stick to the wall, suspend a slinky next to it till it is still, and drop it.

Record this drop with a camera, and review the results frame by frame. Focus on the bottom end of the slinky, and report what you see.

According to standard Physics, the bottom of the slinky will fall very little until the slinky has almost totally relaxed.

According to Abhi Physics, the bottom of the slinky will begin to fall immediately while the slinky is relaxing.

This happens too fast to see which is true without the camera, but even at only 15 frames per second you can easily see which prediction is correct.

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Abhi (and Patrick - I know you are listening, too), it is possible to have many long and interesting discussions in a message board like this without resorting to BS. The USENET groups sci.physics etc. are not moderated, so no matter what you do or say you can keep on doing or saying it as long as people will talk to you (or at you, more likely).

But is that really the kind of attention you want?

I mostly enjoy discussions like this, even when I suspect that the person I'm talking to is playing some kind of game. But it does get old after a while, and once that happens all it takes is saying something that no one could believe to make me loose all interest.

So, IF you want to keep me talking, don't do the BS thing. Just some food for thought.

Regards,
LB

Although I was chased away from this board for asking questions, I come back once in a while to check things and maybe learn something.

To me, it is obvious that Mr. Abhi confuses input and dynamic responce of a system due to that input. In other words,

His system is dynamic and the cause of motion, or input, is the gravity force. The output, or state, of the system is the position
x(t) and this is a dynamic state that can exhibit time constants in various parts of the system, depending on configuration and structure.
Therefore, although you have an impulse input, the output naturally can take some time to reach it's steady state. This has to do with the material configuration of the structure.

Mixing information theory and Newtonian mechanics can lead to several puzzles in the mind of people who do not have a systems input-output perspective of dynamic systems but analyze them only as rigid bodies using D'Alembert's principle, like Abhi is doing. The great contribution of 20th century systems theory was the systems input-output analysis and the tools to model it to explain the behavior of complex systems and motion in a consistent way.

I couldn't see anything in the examples of Abhi leading to a contradiction or inconsistency in the theory. The only thing I can sense is a persistent attempt to "fit" a pre-conceived idea of an inconcistency in some very basic rigid body dynamic problems. Maybe at some level some type of incosistency comes up in Newtonian mechanics but those are very simple examples and have no room for that.

So my conclusion is basically that Abhi:

1. either lacks basic knowledge of mechanics and dynamics or
2. he has discovered of an inconsistency but he tries to present it through trivial examples for the purpose of not revealing the source of it.

Makis

Tom said: <BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> A rope is so rigid that we cannot observe its elasticity by eye. So let's substitute an elastic band for the rope and repeat your experiment. <hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Tom, why not look into a complete closed sphere that's made of an unbreakable, almost infinite elastic membrane. By infolding(s) (=elastic deformations of the medium) it creates 'double, triple ... multi-layerings of the membrane (like dimensional shells of a cigar). In this way the original one inner space in devided in two spaces, ... and by further mainfolds in billions of subsets.

The infolding is a mechanical coupling which can be seen as a pressure valve. Each time the two hyperspaces make together a new unity (micro or macro universe cigars) which has own specific properties (time, degree of stretchability of the membrane, curvatures, etc;). Smaller cigars can be created in the layers of larger cigars if the stretchability of such sub-layer allows that.

... we are all connected