# Gravity never zero

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Ivan, Dec 18, 2011.

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1. ### Robittybob1BannedBanned

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gravitational constant G 6.673(10) x 10-11 N*m2/kg2

F = G (M1 * M2)/r^2
solving for r

r = (F/(G* M1*m2) ^ 0.5)

r = 12.02540008 meters

Two Neutrons more than 12 meters apart will not be gravitationally attracted to each other. Could this be right?

I was a bit surprised by that figure, so that means much to my surprise that gravitational attraction does add and would extend indefinitely throughout the Universe with sufficient mass to bend the Space Time.

Could you check this Trippy please?

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3. ### TrippyALEA IACTA ESTStaff Member

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We'll see - I'm about to go home for the day, and it could easily be several hours before I'm able to sit down and do something like that - perhaps you might want to check your basic assumptions with someone like Alphanumeric, Prometheus, or Alex before we move on to the math.

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5. ### Robittybob1BannedBanned

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OK I'll leave it at that, till someone comments.

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7. ### RealityCheckBannedBanned

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Helo again, wiminex.

Forgive me if I seem dense, but the very fact that Trippy USED the qualifier "If they exist" implies that they ARE 'hypothetical' entities. No? So that means he does not treat them as 'a given'.

He then went further and combined that with the statement/requirement that any such Hypothetical entity would have to satisfy certain criteria/limits imposed on that hypothetical by all the available evidence with which it would have to be 'consistent' in order to even 'speculate' on such entities properties/actions to begin with (including "mechanism")..

In short, I am nonplussed as to why you seem to take issue with what he admits is hypothetical and admits is only speculation as to the 'possible' nature/characteristics of such a hypothetical in the context of the conventional scientific/physics paradigm.

I understood perfectly his stance in the matter, and I can find nothing in his statements so far to make me take such a 'combative' tone towards him as you seem determined to pursue.

May I recommend that you ALL go back to 'square one' and sidestep whatever misunderstandings may have prompted the perplexing (to me) 'combative' turn in some of the side-discussions?

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Please understand that I posted this only in the interests of clarity and polite/productive scientific discourse. I am also reading with interest since the subject matter impinges on my own overarching perspective on everything (to be published soon, I hope) which includes a 'graviton-like' physical entity as a possible 'real' Planck Scale entity BUT without the two-way action ascribed to it by QM hypotheses for same.

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Cheers, wiminex, and good to speak with you, and everyone!

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Last edited: Dec 21, 2011
8. ### Robittybob1BannedBanned

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This thread is quite ironic really for I couldn't even spell "Planck" yesterday.
I always been a fan of Heisenberg Uncertainty Principle. But never ever thought I would ever use it in a scientific argument. Cheers.

9. ### James RJust this guy, you know?Staff Member

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Moderator note: 27 off-topic and/or nonsense/rubbish posts have been moved to a separate thread, here:

10. ### wlminexBannedBanned

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. . . .Thanks RC . . . .have a nice Christmas! I look forward to reading your discourse . . .

Last edited: Dec 21, 2011
11. ### Robittybob1BannedBanned

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@ Trippy
Did you get to check the maths? I was going to split the Universe into 2 supermassive Black holes and see how far apart they would have to be before they wouldn't attact each other.
Number of particle in the universe.
Half this number and put each of these quantities into a Black Hole.
Separate them by distance till the Gravitational pull was less than 1 Planck length per second
Take the BHs and separate them further than that, and you have 2 BH in zero gravity.

12. ### OnlyMeValued Senior Member

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Part of the difficulty here lies in the fact that as of yet we have no functional model for quantum gravity. The Planck length is not an aspect of general relativity, the best theory we have for describing gravity. As far as general relativity is concerned, space is smooth and has no smallest dimension or part.., length or distance.

As long as space is smooth, the influence of gravity between any two masses, while it will reach a point where is becomes trivially insignificant, will never be diminished to the point, that abscent other intervening forces, the two gravitating masses will not attract the each other.

13. ### Robittybob1BannedBanned

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You are saying they attract, and therefore move. But I'm saying if the force is too small to make it move by less than it will move by quantum mechanical fluctuations, then it has not moved and it cannot be moved by gravitational attraction.
We can all have opinions but I used principles from Quantum Mechanics, Uncertainty principle, Einstein Space Time bending and Newtonian Gravitational attraction to prove that gravitational attraction can be effectively zero.
I can see your point though for you could say if you placed another neutron beside the ones already there the two Gravitational fields will overlap and the particles at 12 meters will move. So it wasn't zero for they were able to add their strengths.
Taking this to the limit you end up with 2 super massive BHs, and you end up having to keep separating them, to the point where their attraction is incapable of moving them.
For very large bodies like super massive BHs I don't know the separation required for them.

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14. ### TrippyALEA IACTA ESTStaff Member

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The point is (and this is one of the reasons I wanted someone else to check your assumptions) your calculations assume space and time are inherently smooth, because that's what relativity assumes, however, your logic that you're applying effectively asumes space and time are quantized, which contradicts the assumptions implicit in your calculations.

15. ### Robittybob1BannedBanned

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Could you explain that a little more, for all I did was apply the formulas, I definitely didn't consciously think of space and time like you imply.

You maybe right, for this science is rather new to me. As I was saying I couldn't even spell "Planck" yesterday, let alone use the concepts correctly, but I had a go.

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16. ### RealityCheckBannedBanned

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Hi everyone.

Very interesting discussion.

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Quick question....

Assuming the universal expansion/recession scenario is as per the standard model; and that the furthest galaxies from us at opposite ends of our observable universal are destined to go beyond the 'observational horizon' due to their eventual 'separation' at greater than lightspeed with respect to their counterparts at the 'other end' of our observable universe, then, since gravity effect is said to propagate at lightspeed, would the gravity of those opposite galaxies on their counterparts become effectively ZERO? In other words, since they would 'drop out' of LIGHT-exchanging 'contact' with each other, they must also 'drop out' of GRAVITY-exchanging 'contact' with each other. No?

Gotta go. Cheers all.

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17. ### Robittybob1BannedBanned

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One would have to think so if the mechanism was a graviton, but what I was getting into trouble over on another forum, and some might already see the implication here, is that even before galaxies have dropped out of sight, they maybe already beyond effective gravitational attraction. I.e. the gravitational attraction even though is still present it is insufficient to create acceleration (movement) at even the 1 Planck length per second ^2.
But if we can see what separation is required in a two SMBH universe, we might see if the distances are astronomical or not. (Does the distance relate to the size of the Universe as we know it?)

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18. ### AlphaNumericFully ionizedRegistered Senior Member

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In a reputable mainstream peer reviewed journal or do you mean you'll upload a text file to some website?

You're solving $Ma = \frac{GM^{2}}{r^{2}}$ for $a = \frac{L}{T^{2}}$ so $\frac{ML}{T^{2}} = \frac{GM^{2}}{r^{2}}$ so $r = \sqrt{\frac{GMT^{2}}{L}}$.

T=1 second, $M \sim 1.6 \times 10^{-27}$ kilos, $L \sim 4 \times 10^{-35}$ metres and $G \sim 1.6 \times 10^{-11}$ whatever. So we have approximately, using 1.6 ~ 10/6 that

$r^{2} \sim \frac{1}{6}\times 10^{-10} \times \frac{1}{6}\times 10^{-26} \times 1^{2} \times \frac{1}{4 \times 10^{-35}} \sim \frac{1}{6 \times 6 \times 4} \times 10^{-10-26+35} \sim \frac{1}{150}10^{-1} \sim 0.000667$ so $r \sim 0.0258$ so about 2.6 cm.

It's a very weak acceleration though so it's not surprising the distance is large. Thermal excitation is enough to completely overwhelm that sort of force.

/edit

Oh and in response to something wlminex said about division by zero (it's now been moved), the issue of division by zero is not arbitrary, it follows specifically from the properties of the mathematical structures considered. You need an additive inverse of the multiplicative identity in a field but you don't need a multiplicative inverse of an additive identity. More basically, you need something which satisfies 1+X=0 but not something which satisfies 0*X=1. wlminex said he thinks X/X=1 should be true for all X, including zero, but it's easy to derive contradictions if you allow that. Rings, groups, fields, modules, division algebras. All of these have slightly different axioms. Some extensions of the set of Reals include an infinity so that 1/0 is allowed by that doesn't mean 0*(1/0) is defined. This is not some concept mathematicians run from, it's something any competent mathematician knows a great deal about and entire research careers are devoted to it. wlminex was once again throwing his 2 cents in and it's only worth 1 cent.

19. ### AlexGLike nailing Jello to a treeValued Senior Member

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Bibbity, when we talk about how much space-time is curved, it's not a measure of the size of the curve, i.e 5 meters, or 1 centimeter, or the Planck length. It's a measure of the difference between a flat, Euclidian geometry and a non-euclidean space.

In an Euclidean geometry, the angles of a triangle equal 180 degrees, and the value of pi is 3.1417....

So to provide a measure of the amount of space/time curvature, (and this is only my own thinking), you should try to quantify the degree of departure from a Euclidean space.

20. ### prometheusviva voce!Registered Senior Member

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A slight note of caution here: pi is always the same value the decimal expansion of which starts 3.14159... because it is defined to be the ratio of the circumference of a circle to it's diameter in Euclidean space. In curved space you may find that this ratio is no longer equal to pi, but that does not mean the value of pi has changed.

Another note: another measure of "how curved" a space is can be given by talking about parallel light rays and perceived distances. (I suspect I may get this wrong) If the space is positively curved, parallel lines meet at some finite distance away (or are at least closer together than when they started), so a light source appears to be further away than it is. If the space has negative curvature, as the parallel lines approach infinity they are further apart than they were to start with, so a light source must appear closer.

It's kind of late and I'm not confident this is correct. Perhaps Robittybob can tell us?

21. ### TrippyALEA IACTA ESTStaff Member

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The formulae are from relativity/newtonian mechanics (more or less).
You are assuming that there is a smallest distance that can be moved - this is (more or less) an assumption given to us by quantum mechnics.
Relativity/Newtonian mechanics do not assume that there is a smallest unit of distance that can be moved.

The two assumptions contradict each other.

At least, that's my understanding anyway.

22. ### AlexGLike nailing Jello to a treeValued Senior Member

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You should use [sarcasm] and [/sarcasm]

23. ### AlexGLike nailing Jello to a treeValued Senior Member

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Right. In non-euclidean space the ratio is different, but it's not pi.