For Dr. Tom Van Flandern

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21 years 5 months ago #5814 by Abhi
Replied by Abhi on topic Reply from Abhijit Patil
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>

Most physicists are confused about the meaning of "gravitational waves" (GW) because so many recent textbooks have mistakenly implied that these are equivalent to changes in gravitational force (GF). When the definitions are correct, everyone is forced to recognize that GW propagate at speed c but cannot contribute to GF in any way. By contrast, GF is transmitted ftl, according to all six experiments that test this speed.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Look the similarity..

(1) You say that, "When the definitions are correct, everyone is forced to recognize that GW propagate at speed c but cannot contribute to GF in any way."

(2) And I am saying that in Newtonian mechanics, "when the definitions are correct, everyone is forced to recognize that 'decompression waves' propagate at 'speed of sound' but cannot contribute to 'mechanical force' in any way".


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

Is it absolutely necessary? I mean as soon as the person at center cut the band, the whole band and person may fly off even if the band remains stretched at other end. Imagine that the whole band and person in space is moving and waves travelling across band. The stretched band can relax even when the whole band is moving in space.

On the same lines we can consider sun and earth attached by a "gravity band" which is stretched. If the sun cease to exist or "cut" this gravity band, gravitational wave can travel with speed of light on this band. But the moment sun "cut" this gravity band, the entire gravity band and earth can fly off in space. The gravity band near earth may remain stretched until these waves reaches earth.

We can apply same anology at atomic level. I do not see any problem with this.

-Abhi.

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21 years 5 months ago #5825 by Abhi
Replied by Abhi on topic Reply from Abhijit Patil
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>By contrast, gravitation obeys a "transparency principle" whereby the entire interior of any body is just as accessible to gravity as is the surface. So every nucleon within a body is acted upon independently and nearly instantly by gravity without need of any transmission of force from one nucleon to another. Although some parties on your list of names might argue that the transmission of this force occurs at the speed of light, while others (including me) argue that it transmits much faster than the speed of light, no one who is knowledgable in such matters would argue that gravity is transmitted at the speed of sound, the way a mechanical force is.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Let me reflect on this. I always ask people if I have long vertical rod in my hand and we are in state of static equilibrium. Now I leave support at t = 0. With what speed this information will travel at upper end of rod? People always say, with speed of sound. Isn't gravity acting on all electrons and protons in the rod? They say yes. Then why Gravity will wait until decompression wave reaches to every electron and proton in this rod?

I can construct a situation where when I leave the support, immediately the falling end strike to a part of this rod where information regarding release is not reached. This situation will not allow rod to decompress at all and no decompression wave will travel. As there are no decompression waves, there is no information propagating and in such situation this part of rod will refuse to move and act as if some additional mass is placed on it. So even if there is no support, this system of rod will not fall at all. We will see this system of rod just get stucked in space. I have created anti-gravity!!!.

And I am talking about instantaneous propagation of information regarding "removal of force". I am not talking about instantaneous propagation of information regarding "application of force".

Now let me reflect on the special situation. I have explained in another post how law of conservation of energy is violated. Let me explain how Galileo's law ragarding gravatational acceleration is violated. As we know, all bodies fall with same gravitational acceleration regardless of mass of body. Be it feather or apple or any body, all things fall in vaccum with same magnitude of acceleration g.

In our situation extra force = m'g is generated where m' is mass of body and m is mass of system. Let m' = nm. So total mass of system+body = m + nm = (n+1)m. Initially force acting on point M of table is

F(initial force)= mg + m'g = mg + nmg = (n+1)mg.

In our situation extra force = m'g = nmg is generated. So during propagation delay force acting on point M of table is

F'(final force) = (n+1)mg + nmg = (n +1+n)mg = (2n +1) mg.

But mass of system and body = (1+n)m is never changed. Hence during
propoation delay,

Gravitational accelaration = final force / mass of system+body = F' /
(n+1)m = (2n+1)mg / (n+1)m = [(2n+1)/(n+1)]g.

So we find that Gravitational accelearation which is universal constant in our nature is changed. So if the support point of system+body is removed it should fall with accelearation [(2n+1)/(n+1)]g and not g.

In our example, final force acting on point M of table is 300g N due to extra generated force of 100g N. But mass of system+body is always 200 Kg. So we find that gravitationa acceleration in our situation is = 300g/ 200 = (3/2)g = 1.5*9.8 = 14.7 m/s^2.

So the system+body will fall with accelearion 14.7 m/s^2. Was Galileo wrong!!!??

-Abhi.

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21 years 5 months ago #6026 by Abhi
Replied by Abhi on topic Reply from Abhijit Patil

Before you have time to look at this thread, let me write the outline of experiment I am going to propose.
This experiment is initially intended to prove that information regarding separation or release of applied force to body can not propagate through decompression waves with speed of sound.

We use very long transmission line made of any material in which information regarding release of applied force is supposed to travel in accordance with established knowledge of physics. The force is applied to starting point A of transmission line and "end point B of transmission line is exactly at same level of starting point A" (where force is applied) of transmission line. We apply established knowledge of physics to calculate propagation delay time taken for information regarding release of applied force to travel from starting point A to end point B of transmission line.

If during this propagation delay time, end point B of transmission line does not move in space, then established knowledge of physics is correct. But if end point B of transmission line moves in space during this propagation delay time, then we show that information regarding separation or release of applied force to body does not propagate in accordance with established knowledge of physics.

We use high-speed video camera focused on starting point A and end point B of transmission line and accurate electronic clock in the background. We check every frame of video film in slow motion to see whether end point B of transmission line is displaced in space during propagation delay time or not.

If end point B of transmission line moves in space during this propagation delay time, then we go on proving validity of following principle.

"When the system is in state of static equilibrium and if we release applied force (Gravitational force, mechanical force, magnetic force etc.) to a body, this information regarding release of applied force is transmitted to every point of body instantaneously".

Will you please tell me if there is flaw in at least logic of this experiment?

-Abhi.




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21 years 5 months ago #6220 by Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[tvf]
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>
[Abhi]
Is it absolutely necessary? I mean as soon as the person at center cut the band, the whole band and person may fly off even if the band remains stretched at other end. Imagine that the whole band and person in space is moving and waves travelling across band. The stretched band can relax even when the whole band is moving in space.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I have actually performed this experiment.

During my Junior year studying Physics, through a series of strange events I ended up with a 10 meter rubber band. I rounded up some friends and we went to a grassy court yard behind our appartment to play with it.

We tied one end to a light pole and 6 of us pulled it as far as we could - about 45 meters. When we let it go it took (more than one second but less than two seconds) to hit the pole. The loose end made this really loud buzzing sound as it flapped in the "wind". It sounded a LOT like a giant woopie cushion.

It obviously had enough power to pull one of us. We found a box and a crash helmet and a test pilot. With only 5 of us left to pull, we could only get the elastic to stretch to about 40 meters. Our test pilot took hold of the end of the band and we let go.

The acceleration was impressive. After we figured out how to keep our test pilot from being yanked forward out of the test vehicle, he made some very nice runs. The whole ride lasted about 4 or 5 seconds, but the first 5 meters usually took a little less than one second which means about 1 g initially, dropping of rapidly of course. The test vehicle never quite reached the pole.

On our last run, the band broke next to the knot at the pole. And it broke just before we all released the stretced band. We knew this from the buzzing sound, which startled us and triggered a "reflex release" of the band.

Our test vehicle accelerated until the loose end hit, which took about half a second or so. From a dead stop it moved forward while the loose end was buzzing, 2 or 3 meters total including a little bit of coast.

I didn't think anything about it at the time, because this was what I expected.

I admit the error bars on this one are kind of fuzzy, but even if we can't get reasonable quantitative results from this experiment, it is qualitatively conclusive. And it can be backed up by the following desktop demonstration.

You need a slinky and some stuff to attach to one end. The stuff you attach (the test mass - a toy pickup works great) needs to be substantially more massive than the slinky, so that the center of mass of the slinky can't provide a counter balance to the test mass. IOW, you are trying to simulate a massless slinky.

Pile stuff in the bed of the pickup until you are satisfied. Attach the test mass to one end of the slinky. Grab the free end of the slinky in one hand and the test mass in the other, pull them apart and let them go at the same time.

> Watch the test mass accelerate.

Or, let the free end go first (just barely), then the test mass end.

> Again, watch the test mass accelerate.

It all happens pretty fast, so timing is important. But even without instrumentation you can see what is happening.

Regards,
LB

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21 years 5 months ago #5872 by Abhi
Replied by Abhi on topic Reply from Abhijit Patil
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
On our last run, the band broke next to the knot at the pole. And it broke just before we all released the stretced band. We knew this from the buzzing sound, which startled us and triggered a "reflex release" of the band.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

I am unable to understand fully what you are saying. Did you here the sound of breaking of band when you were still pulling the band? I mean, the band was separated from pole at t = 0, and it take time t for sound to reach you. So were you people still pulling the rope during this time interval t?

Being english my second language, will you please explain "reflex release"?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Our test vehicle accelerated until the loose end hit,<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Accelerated in direction of pole or in opposite direction?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>You need a slinky and some stuff to attach to one end. The stuff you attach (the test mass - a toy pick up works great) needs to be substantially more massive than the slinky, so that the center of mass of the slinky can't provide a counter balance to the test mass. IOW, you are trying to simulate a massless slinky.

Pile stuff in the bed of the pick up until you are satisfied. Attach the test mass to one end of the slinky. Grab the free end of the slinky in one hand and the test mass in the other, pull them apart and let them go at the same time.
> Watch the test mass accelerate.

Or, let the free end go first (just barely), then the test mass end.

> Again, watch the test mass accelerate.

It all happens pretty fast, so timing is important. But even without instrumentation you can see what is happening.

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

Two problems here..

(1)Most important thing. How exactly you conclude that you leave the both ends of slinky at exactly same moment. And when I say moment, I mean it. Perhaps you can leave the both ends at same instantaneous moment, if both ends are in your one hand. But in your slinky case, it is not possible.

If you take rubber band and stretch it with your "both" hands, then you can't release the band from both ends at same instantaneous moment. It can happen only once if you try it infinite number of attempts.

(2) Even if I assume that you left both ends at same instantaneous moment, the test mass-toy pick up is heavy. Hence its acceleration is less than free end of slinky. So you visually see the test mass moving slower than free end. If the friction of test mass-toy pick up is greater with ground, you will see the test mass moving more slowly.

Just because you see one mass covering less distance than other mass in same or in opposte directions, it does not mean that both masses does not start accelerating at same instantaneous moment.

I mean if two cars start accelerating from same point in same direction. One car covers 10 meter in one second and other car covers 8 meters in "same" one second, it surely does not mean that both cars didn't start accelerating at "same instantaneous moment".

In your stretched rubber band experiment, I do not see any reason to conclude that when the band broke from pole at t = 0, you were still pulling the band at other end at same instantaneous moment.

The time interval between breaking of band from pole, you hearing the sound, and this information reaching to you through band is very very small considering 40 meter distance from pole and you. I do not think human response system is accurate enough to see what happened in very very small fraction of second.

I am doing this experiment of pulling the rubber band tied to a pole. The band breaks off from pole at t = 0, then at same instantaneous moment this information can reach to me. I am in process of being thrown on ground and during very small fraction of second I here the sound of band breaking off from pole.

What reason is there which makes me to conclude that the information regarding breaking of band didn't reach to me at exactly same moment?

Hope you understood what I am saying. I can explain you that if in certain situations, if information does not travel instantaneously, we end up seeing things just hanging in space without any support at all. There is no need of special costly anti-gravity technic.

-Abhi.

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21 years 5 months ago #5827 by Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[Abhi]
I am unable to understand fully what you are saying. Did you here the sound of breaking of band when you were still pulling the band? I mean, the band was separated from pole at t = 0, and it take time t for sound to reach you. So were you people still pulling the rope during this time interval t?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes, the band broke first. I'm estimating about 1/10 sec for the sound to reach us. Average human reaction time is about 1/4 second - so i'm allowing 1/4 second for us to recognize the sound and 1/4 second for us to release the band. That would leave about 1/4 to 1/2 second for the loose end to reach us.

These are, obviously, estimates. No one was timming anything with a stop watch.

The technique we worked out for "launch" was for us to pull the band past the test vehicle so the pilot could grab it in front of us. This allowed us us to release the band without worrying about being run over by the vehicle. A real concern, considering the high acceleration during the first few meters.

It also gave us a great view of the action.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
Being english my second language, will you please explain "reflex release"?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

If you touch something hot, you instinctively jerk your hand away. This is a "reflex action". Loud noises also cause relfex actions.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
Accelerated in direction of pole or in opposite direction?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Since your theory says "NO acceleration once the band breaks", why does it matter? Either way your theory is falsified. But, the acceleration was directed toward the pole, just like in all of the other runs.

(Nothing else is possible, BTW)




<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
Two problems here..

(1)Most important thing. How exactly you conclude that you leave the both ends of slinky at exactly same moment. And when I say moment, I mean it. Perhaps you can leave the both ends at same instantaneous moment, if both ends are in your one hand. But in your slinky case, it is not possible.

If you take rubber band and stretch it with your "both" hands, then you can't release the band from both ends at same instantaneous moment. It can happen only once if you try it infinite number of attempts.

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I do not conclude that I'm releasing both ends at "exactly" the same moment. I figure that the variation in relaese times is up to a few dozen milliseconds, and that about half the time it would be the free end first, the other half the weighted end goes first.

However, the test mass always accelerates in the same direction, toward the free end.

And, you must have missed the part where I talked about releasing the free (unweighted) end first. If you focus on doing this, it is easy to be sure that you did in fact let the unweighted end go before the weighted end. The test mass still accelerates.

If your theory were correect, all acceleration should vanish as soon as the unweighted end is released. Once both ends are free, ANY acceleration falsifies your theory.

Acceleration clearly continues until the slinky collapses to minimum length. You should do this desktop demonstration for yourself.

But, in all fairness Abhi, until it is repeated with instrumentation and remote release devices on each end, there will be room for doubt. Not much room, but some. We do like to deceive ourselves.

Regards,
LB

PS - one simple and inexpensive way to instrument this experiment would use a meter stick and a camera. Examine the file frame by frame after each run to collect data.


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