Gravity never zero

[Robittybob1]

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.

A7 1.62E-35 planck length
A8 1.67E-27 Mass
A9 6.67E-11 Gravitational constant

Hi Alphanumeric , I obviuosly went wrong somewhere and in the process of seeing where I went wrong I was wondering if you also have entered some wrong figures
on my Excel the formula went "=SQRT(A9*A8/A7)" giving an answer of 0.083157316 or (83 mm).

So what we are saying is that at a distance of 83 mm 2 neutrons have less attraction than is accounted for by quantum movement.
as the masses are increased (doubled) the distance between will only rise by the square root of that number (2)

Neutrons, Distance, Increase in distance
1,,, 0.083157316,,,
2,,, 0.117602205,,, 1.414213562
4,,, 0.166314633,,, 1.414213562
8,,, 0.23520441,,, 1.414213562
16,,, 0.332629266,,, 1.414213562
32,,, 0.470408819,,, 1.414213562
64,,, 0.665258532,,, 1.414213562
128,,, 0.940817638,,, 1.414213562
256,,, 1.330517064,,, 1.414213562
512,,, 1.881635277,,, 1.414213562
1024,,, 2.661034128,,, 1.414213562
2048,,, 3.763270553,,, 1.414213562
4096,,, 5.322068255,,, 1.414213562
8192,,, 7.526541106,,, 1.414213562
16384,,, 10.64413651,,, 1.414213562

So if this was taken right up to 0.5 X 10^85 for the number of neutons/protons in the SMBH what would the distance be?

 

[AlphaNumeric]

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

It's a little complicated. For example, in a really warped space the ratio Circumference/Diameter might depend on the location of the circle and its radius. However, in nice non-Euclidean spaces circles are indeed circles. For example, if you embed a 2d sphere into 3d Euclidean space and then draw what would appear to a sphere dweeler to be a circle on the sphere it appears to a Euclidean 3d dweeler to be a circle too. This isn't terribly surprising given circles are themselves spheres but it happens to be true in hyperbolic geometry too. Draw a hyperbolic circle in the upper half plane representation of hyperbolic space and you'll obtain a Euclidean circle, though the different metrics disagree on the location of the centre of the circle. In both spherical and hyperbolic geometries the ratio will actually depend on the radius of the circle. This can be seen by considering circles of constant latitude on the Earth. They are circles in the spherical geometry sense, centred on one of the poles. In fact either pole might be a centre but the radius would obviously change. The limit is the circle centred on the North Pole and whose circumfernce is say the Equator but then pushed down to the South pole. It's got zero circumference then so 'pi' = 0 for such a circle! Likewise if you make the radius very small then any crazy geometry (ignoring pathological cases like curvature singularities) will have the ratio being close to 3.141.... because any manifold looks like Euclidean space close up.

 

[Robittybob1]

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?

No way, I am just learning how the read the words let alone do the maths or even have any idea what is being discussed. I have a long way to go before I'll talk the Euclidian language.

 

[Robittybob1]

A7 1.62E-35 planck length
A8 1.67E-27 Mass
A9 6.67E-11 Gravitational constant

So if this was taken right up to 0.5 X 10^85 for the number of neutons/protons in the SMBH what would the distance be?
Neutrons, ,,,,,Distance,
5.00E+84 ,,, 1.85945E+41
That distance in lightyears .... 1.96549E+25
Current size 78 Billion light years (7.7E+10) (Diameter).

Conclusion. If there were just 2 equal SMBH left in the Universe it would need to be a whole lot bigger than it is toay in order for them not to gravitate toward each other.
This is assuming they are stationary wrt each other.


That is surprising to think this, I wonder if the maths is correct?

 

[OnlyMe]

Conclusion. If there were just 2 equal SMBH left in the Universe it would need to be a whole lot bigger than it is toay in order for them not to gravitate toward each other.
This is assuming they are stationary wrt each other.

Only as a discussion of logic....

1. Assuming that the universe, as we currently know it, did in fact begin with the big bang.
2. That the age of the universe, since the big bang is 14.5 billion years, give or take...
3. That we can not assume that we are at the center of the universe and that the light sphere representing what wh can see of the universe is 13.6 billion light years out in all directions.
4. That the propagation velocity of gravity is limited to the speed of light.

Two SMBHs each located at the edge of "our" visible part of the universe, could be separated by more distance in light years, than the universe is old. Being some 17.2 billion light years apart, give or take, any gravitational interaction between the two would be beyond the propagation time of their individual gravitational fields.

If we were to view the same distances as defined by our light sphere of the observable universe and an accelerating expansion of the universe that even approaches the speed of light, in any one direction, two SMBHs located at opposite extremes relative to our observable horizon, would be experiencing an excellerating expansion of the universe that is greater than the speed of light and thus greater than the propagation velocity of their gravitational fields.., and once again they would be beyond the gravitational influence of each other.

These are both speculative scenarios and there are some arguments that could be made to refute either. The point is that from what we do know of the universe if it does have an age, it also appears that there could exist gravitational masses that are far enough apart that they do not yet interact gravitationally. They would lie outside of the time of light horizon of each other.

If it is accepted that gravity propagates at the speed of light, then there are gravitating bodies that do not interact with one another... But they might in the future...

 

[Robittybob1]

When I considered the two extremes:
1. The two neutron universe
2. The two super massive Black Hole Universe (SMBH)
One had virtually no gravity after 80 mm and the other needed a universe Billions of times billions larger than the one we know to allow the SMBHs to survive.
Now I have read that there needs to be Dark Matter introduced to the structure of the Universe to account for the over abundance of the gravitation that is evident. There didn't seem to be enough matter in the Universe to account for all of it.
We need to try and understand what this Dark Matter is all about.

 

[hansda]

Hi I make this short,

Is it true that gravity always exists and It never goes to zero no matter the distance?

Gravity depends upon mass and distance . Distance can not become infinite but mass can become zero . If the mass gets converted into energy , mass can become zero ; so gravity also can become zero at a very extreme case .

Would not the universe in a "endless" time period, draw it self together again? Make a cycle.. of expansion(big bang) and black holes


note
And even if there is no friction in space would not the universe be able to decelerate and contract itself since the little force never stop to exist, but the speed and the centrifugal force can stop. Sorry for my bad English


Best regards / Ivan Loguin

 

[Dywyddyr]

Gravity depends upon mass and distance . Distance can not become infinite but mass can become zero . If the mass gets converted into energy , mass can become zero ; so gravity also can become zero at a very extreme case .

As usual with you, this is nonsense.
That could only apply if:
A) there were no other masses in this hypothetical universe of yours and
B) if energy in sufficient quantities didn't cause gravity.

Fail.

 

[RealityCheck]

As usual with you, this is nonsense.
That could only apply if:
A) there were no other masses in this hypothetical universe of yours and
B) if energy in sufficient quantities didn't cause gravity.

Hi Dywyddyr, pleased to meet ya.

I hope you don't mind my asking a question prompted by your B) above.

In the Standard Model BBang scenario, the initial conditions are assumed to have been nothing but Energy Quantity/Density at a 'temperature' that did not allow any 'matter/mass' density per se, only 'energy'.

So, given this view, can you elaborate on why (IF there WAS such great energy quantity/density) the inflation/expansion scenario of the BBang could occur at all?

Since, as you suggest, sufficient energy quantity/density would 'cause gravity' which, considering the staggering energy quantity/density at the earliest stages of the BBang, would produce enormous gravitational forces immediately 'space-time' became a physical entity that could be 'curved' by that initial energy quantity/density?

Thanks in advance for your time and trouble in replying.

Back tomorrow. Cheers!

.

 

[Dywyddyr]

Since, as you suggest, sufficient energy quantity/density would 'cause gravity'

Um, it's not exactly a "suggestion" since Einstein showed that this is true - hence the previously unaccounted-for anomaly in the orbit of Mercury.

So, given this view, can you elaborate on why (IF there WAS such great energy quantity/density) the inflation/expansion scenario of the BBang could occur at all?

It's quite simple: gravity didn't "form" as a force until after the BB occurred.

 

[RealityCheck]

.

Hi again, dywyddyr. Just surfing through before logging out, and caught your prompt reply. Thanks!

Um, it's not exactly a "suggestion" since Einstein showed that this is true - hence the previously unaccounted-for anomaly in the orbit of Mercury.


It's quite simple: gravity didn't "form" as a force until after the BB occurred.

Yeah, I know. I was referring to the context of your response B) to handsa in your own words, is all.

My point was that the BBang model 'describes' the inflation/expansion stages after whatever initial conditions obtained regarding the Energy Quantity/Density per se.

So my question relates to that stage where 'space-time' is created during that BBang process. And as far as I can gather, the effective separation in 'space-time' of the energy content was very small, and hence interaction distances for gravity effects were also over short distances in 'space-time' at that stage. And since the energy quantity/density at that initial stage of inflation/expansion was supposed to be vast, I am unclear as to why the BBang process 'continued' rather than being immediately 'reversed' by the still-close proximity and vastly dense/quantity energy content which, as you point out, should have 'curved' space-time and so manifesting EXTREME gravity commensurate with that initial energy quantity/density?

I hope this provides more context for the question asked?

Thanks again for your time and trouble, Dywyddyr. Catch ya later!

.

 

[Dywyddyr]

So my question relates to that stage where 'space-time' is created during that BBang process. And as far as I can gather, the effective separation in 'space-time' of the energy content was very small, and hence interaction distances for gravity effects were also over short distances in 'space-time' at that stage. And since the energy quantity/density at that initial stage of inflation/expansion was supposed to be vast, I am unclear as to why the BBang process 'continued' rather than being immediately 'reversed' by the still-close proximity and vastly dense/quantity energy content which, as you point out, should have 'curved' space-time and so manifesting EXTREME gravity commensurate with that initial energy quantity/density?

Er, good question. "I dunno" is the short answer. Possibly, since inflation was supposed at one stage to be exponential, then it's as simple as "gravity couldn't "keep up" with the expansion".

 

[Robittybob1]

Er, good question. "I dunno" is the short answer. Possibly, since inflation was supposed at one stage to be exponential, then it's as simple as "gravity couldn't "keep up" with the expansion".

Isn't more like it takes Mass to have Gravity, and Mass is possibly related to the Higgs particle which gives particles mass. And the BB had to expand and cool somewhat before particles formed, so gravity as you said had "not been formed yet". Good question though.

 

[Dywyddyr]

Isn't more like it takes Mass to have Gravity

I take it you missed, entirely, the post where I stated:
"energy in sufficient quantities causes gravity".

 

[Robittybob1]

I take it you missed, entirely, the post where I stated:
"energy in sufficient quantities causes gravity".

Well I will believe you if you can show me where this is proven.

 

[Dywyddyr]

Well I will believe you if you can show me where this is proven.

Oh, you also missed this bit:
"Einstein showed that this is true - hence the previously unaccounted-for anomaly in the orbit of Mercury."

You might have heard of him?

 

[Robittybob1]

Oh, you also missed this bit:
"Einstein showed that this is true - hence the previously unaccounted-for anomaly in the orbit of Mercury."

You might have heard of him?

I had noted that bit and thought at the time you must have that connection wrong. I would have looked into what Einstein was proving when he correctly calculated the precession of Mercury, if I had ever thought you would be using it as part of an argument against me. I just don't recall it was "energy density". So you show me exactly what Einstein was proving.

 

[Dywyddyr]

I had noted that bit and thought at the time you must have that connection wrong.

Then your education is flawed.

How about E = mc[sup]2[/sup]?
Enough energy acts as a mass. Mass exerts gravity...

I would have looked into what Einstein was proving when he correctly calculated the precession of Mercury, if I had ever thought you would be using it as part of an argument against me. I just don't recall it was "energy density". So you show me exactly what Einstein was proving.

Roughly and simply, that close to the Sun there's enough energy to exert extra gravitational pull on Mercury.

 

[Robittybob1]

Then your education is flawed.

How about E = mc[sup]2[/sup]?
Enough energy acts as a mass. Mass exerts gravity...


Roughly and simply, that close to the Sun there's enough energy to exert extra gravitational pull on Mercury.

It is E = M C^2 not E = M OK!

There has to be something about "Relativity" in your answer.

Just looking at it. Einstein knew about the error and fitted his equations to fit the case, rather than the other way around.
He then was surprised that the equations worked. It was more good luck by the sound of it.

 

[Dywyddyr]

It is E = M C^2 not E = M OK!

Um:

How about E = mc[sup]2[/sup]?

There has to be something about "Relativity" in your answer.

Er, it IS GR.

Just looking at it. Einstein knew about the error and fitted his equations to fit the case, rather than the other way around.

Yeah?
And how do you know this?

He then was surprised that the equations worked. It was more good luck by the sound of it.

Bollocks.