# Can a gravitational field be represented as an acceleration? (2nd try)

Discussion in 'Alternative Theories' started by Richard777, Dec 21, 2018.

1. ### Richard777Registered Member

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My previous post has been shown to be seriously incorrect. Thank you for your comments and brutal honesty.
The web page has been revised to include some of your comments.

Second Try;

If so, the wave must approximate Newton gravitational acceleration at significant distances from the associated object. The field strength should be zero at the center of the object. The wave amplitude should be related to the Schwarzschild radius.

The “Newton catastrophe” permits infinite acceleration at the center of a massive object. This error may be avoided if the gravitational wave is zero at the COG.

Can a gravitational field associated with a massive object, be represented as an acceleration wave?

Reference; http://newstuff77.weebly.com 06 A Gravitational Field Equation

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5. ### Mike_FontenotRegistered Member

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I think a far more useful equivalence is between a UNIFORM gravitational field versus an acceleration with no field. That is what Einstein did with his "Equivalence Principle", and he used it to point the way to his general theory of relativity that he completed 10 years later. The Equivalence Principle was useful because his special theory of relativity allowed him to solve the problem of an acceleration with no field for a rotating disk, and his Equivalence Principle then told him that geometry isn't Euclidean when a gravitational field is present. In particular, it told him that for creatures on the disk who regarded the disk to be stationary, the ratio of the disk's circumference to its diameter is NOT equal to pi. This conclusion follows purely from the famous length contraction result of special relativity (combined with the Equivalence Principle).

7. ### exchemistValued Senior Member

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I still cannot see where you get this idea of "Newton catastrophe" from. The gravitational field at the centre of gravity of an object is zero. So the acceleration is zero. Not infinite.

Also a gravitational field is not a periodic phenomenon - it has no time-dependence. So how can a wave, which is intrinsically periodic, possibly model it?

8. ### Neddy BateValued Senior Member

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I'm just taking a wild guess here, but maybe he is thinking of a mass with the physical dimensions of a point? In that case, wouldn't the acceleration due to gravity approach infinite as the distance between two such object approaches zero?

A long time ago I wrote a simple program in QBASIC which was supposed to model gravitational behavior. I had four point masses moving around the computer screen in various ways, and it looked like a ballet of various orbits. One point happened to move past another very closely, and one was ejected quickly off the screen. I zoomed out so I could see it, and it seemed to have reached escape velocity, because no matter how much I zoomed out, it kept moving away from the other three, albeit slower and slower over time. I'm not sure if I had any mistakes in the program, or if this was the correct result of the point masses.

9. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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A wave implies oscillating strength of the gravitational field and this is not seen.
Correct.
There is no amplitude or frequency. If there was one why would it be related to the Schwarzschild radius?
You just said the gravity was zero at the center of mass why do you now say it is infinite????
Newtonian gravity shows the acceleration of gravity is zero at the center of a mass.
There is no error on this in Newtonian gravity. The error is in your understanding.
I don't see how.
You should not spam the site by advertising your page. It will get you banned.

10. ### Mike_FontenotRegistered Member

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I garbled the above post. I'll try again:

In Einstein's original Equivalence Principle scenario, he realized that someone inside a box (and unable to see outside) would not be able to tell whether the box was stationary in a uniform gravitational field, or accelerating at a constant rate with no gravitational field. He then used the basic equivalence between acceleration and a gravitational field without acceleration to show that, in general, geometry in a gravitational field isn't Euclidean. He did this by considering a rotating disk without any gravitational fields present. Special relativity allowed him to analyze that problem, using only his famous length contraction result. He determined that yard sticks laid out around the circumference would be length-contracted, and therefore it took more yard sticks to fill up the circumference than it would if the disk weren't rotating. But yard sticks laid out along the diameter wouldn't be length-contracted (they would be narrower, but not shorter). Then, via his Equivalence Principle, he concluded that for stationary creatures living on the disk who regarded the disk to be non-rotating (and thus without any length-contraction effects on the yard sticks), the ratio of the disk's circumference to its diameter is NOT equal to pi (it's larger), and therefore for those creatures, geometry isn't Euclidean. This result told Einstein something very important about what his general theory of relativity would have to be like. He knew he would have to learn differential geometry to arrive at his theory! (Fortunately, he was helped in that task by his longtime friend Marcel Grossman, who was an expert in differential geometry). It took Einstein 10 more years to complete his general theory, but special relativity and the Equivalence Principle had pointed him in the right direction.

11. ### Q-reeusValued Senior Member

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Forgotten or never read what I wrote last part back here?:
That Richard777 never responded indicates mental illness. That thread being yet still in Physics & Maths speaks of an upstairs SciForums illness.
This current thread, based on an edited only marginally less absurd concoction than his Dec 17 2018 pdf version, will remain here in Physics & Maths how long?
Maybe I should have used reverse psychology and praised the original as brilliant and worthy of it's spot.

12. ### Neddy BateValued Senior Member

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Interesting. I knew Einstein had described the closed room being uniformly accelerated upward, but I did not know he also went on to describe the rotating disk to be equivalent to gravity, in general. Do you happen to remember which of his papers or books might have had this in it?

13. ### Neddy BateValued Senior Member

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Thanks, I had missed that, so you already beat me to it! I agree with your assessment, but since nothing else is going on in this section, we can try to make the best of it for now.

Q-reeus likes this.
14. ### Q-reeusValued Senior Member

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Neddy Bate wrote:
Some confusion by MF over conceptual originality it seems. Not Einstein: https://en.wikipedia.org/wiki/Ehrenfest_paradox

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16. ### exchemistValued Senior Member

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Yes indeed. Q-reeus also made this point earlier - as I see has now been pointed out. But the poster speaks of "the centre of a massive object", not the artificial (and physically impossible) scenario of mass concentrated at a mathematical point.

The whole thing seems to be daft.

17. ### Q-reeusValued Senior Member

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Yes Einstein extended Ehrenfest's strictly SR based analysis, to give it a more general geometric significance as covered in that Wikipedia article.

18. ### Neddy BateValued Senior Member

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Wow, cool. So let's assume a non-rotating sphere of some mass. Let's also assume the mass of the object is sufficient to provide a gravitational acceleration similar to a hypothetical mass-less, (non-gravitational), object of the same size which is rotating at some rate.

Does relativity say that the Ehrenfest idea applies equally well to both the non-rotating gravitational object with mass, and the rotating non-gravitational object with no mass?

In other words, is the non-Euclidean circumference-to-diameter ratio altered by the same factor?

Last edited: Dec 23, 2018
19. ### Q-reeusValued Senior Member

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Non-Euclidean in both cases but the specifics are necessarily different. In GR, for any gravitating mass having axial symmetry, only the radial component of spatial metric will be a function of Newtonian potential. That is, only along a radial direction will (accumulated) distance be different to the non-gravitating limit M -> 0.
[I'm only confident above is strictly correct when confined to the mid-plane of symmetry for a cylindrical mass - of which a circular disk is a limiting case.]

Which characteristic is exactly opposite of the spinning disk case where only the circumferential distance is different to the non-spinning limit omega -> 0.
Of course in practice non-uniform elastic strain owing to centripetal acceleration will exceed by many orders of magnitude the infinitesimal purely relativistic corrections.

Last edited: Dec 23, 2018
20. ### Mike_FontenotRegistered Member

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It was in his popular book "Relativity", originally published in 1916, and revised in 1952, and now published by Crown in paperback. I always recommend that little book as the best starting place for anyone wanting to learn relativity theory. It's out of favor now by a lot of people who consider Einstein to have been naive and unsophisticated by modern standards. But I don't agree ... what he had that no one else had was uncommon intuition, and the ability to question things that "everyone knows is true".

21. ### Neddy BateValued Senior Member

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Thank you, very interesting. I was wondering whether the gravitational non-Euclidean effects were similarly-analogous or oppositely-analogous to the rotating case. And you have answered it clearly, (assuming you are correct, which I assume you are).

Intuitively, to me, it makes sense that it would be opposite because the creatures living on the rotating disk have a type of gravity which tends to make objects fall away from the center, whereas the creatures living on the non-rotating mass have a type of gravity which tends to make objects fall toward the center. Do you suppose that is one of the reasons why it is opposite, or is that more of a coincidence?

Last edited: Dec 23, 2018
22. ### Neddy BateValued Senior Member

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Thank you. A long time ago I had read the 1920 version, Relativity: The Special and General Theory. I recall the part about the non-Euclidean geometry in both the rotating and gravitational cases, but thought Einstein was maintaining a difference, rather than an equivalence, between the two types of 'gravity' that result.

23. ### Mike_FontenotRegistered Member

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I think he used the term "equivalence" strictly to mean that all measurements would be the same in the two cases. For example, the lengths and widths of the two groups of yard sticks (those along the circumference, and those along the diameter) would be the same in both scenarios. I think he definitely considered the two scenarios to be different: in one case, you have acceleration with no gravitational field, and in the other case, you have a gravitational field, and no acceleration.

I also believe that he saw the value in the equivalence principle to be in what special relativity can tell us about general relativity, not the reverse.