Gravity (over extremely long distances)

[jspstorm]

This question may be a little vague or difficult to answer definatively.

Take an empty universe, completely bereft of all energy and matter, with the exception of two neutrons. These two neutrons are placed a trillion light years apart from each other and each have no velocity with respect to the other.

My question is: Will gravity eventually pull them together?

 

[fatra2]

Yes.

 

[mathman]

Yes and no. Neutrons are radioactive with a half life of about 15 minutes. As a result, they will be long gone before they can come together. However the resultant protons and electrons will eventually come close enough that quantum effects will come into play.

 

[DLuckyE]

Isn't that assuming that there's no lower limit to the strength of gravity? Doens't everything have a lower limit in QM?

That's even besides the fact it'd take a trillion years before they'd even notice the gravity from each other (if it can reach that far)

 

[Lsos]

Yeah and then there's that whole dark energy thing which is supposed to make up most of the universe.

I think it's safe to assume that in this case, we don't know.

 

[DaveC426913]

Without complicating the question beyond the intent of the OP, fatra2 is correct.

As far as we know, gravity has an infinite reach.

 

[pallidin]

Just a thought:
In this case as described by the OP, I think the gravity over that distance would be so extraordinarily weak as to unable to break the inertia of the non-moving neutrons to initiate attractive motion.
Could be wrong, of course, but I tend to agree that there should be some lower limit.

 

[JDługosz]

Inertia is not like static friction.

 

[pallidin]

Inertia is not like static friction.

Understood, but can a stationary bowling ball(for example) move from a stationary position in space if a light feather impacts it slowly?
There is simply not enough force to break inertia.
At least in my thoughts. Could be wrong!

 

[pallidin]

To my understanding, an object at rest tends to stay at rest unless acted upon by outside forces.
Fine, but surely there must be some lower limit which defines at what "force level" the object will start to move.
After all, could I responsibly say that a force of .000000000000000000000000000001 grams will even slightly move a bowling ball in space?

 

[diazona]

Understood, but can a stationary bowling ball(for example) move from a stationary position in space if a light feather impacts it slowly?

Yes. Absolutely.

Inertia is not something that can be "broken". There is no predicted or observed lower limit for the amount of net force that it takes to make an object move. So as far as we can tell, any force, no matter how tiny, applied to any mass, no matter how large, will induce some acceleration.

 

[pallidin]

Ok. Interesting. Thanks.

 

[Mentallic]

Just our of curiosity, could anyone calculate the magnitude with what kind of Newton force we are dealing with between these 2 neutrons? Remember, 2 neutrons and a trillion light years apart

I'd do it myself but I don't know what equations to use.

 

[betel]

The equation would be
F = \frac{G m_n^2}{r^2}
Now we have
m_n = 1.67 \cdot 10{-27}kg
G = 6.67\cdot 10^{-11}N\frac{m^2}{kg^2}
r= 10^{12}ly = 10^{12}\cdot 9.46 \cdot 10^{15}m=9.46\cdot 10^{27}m
Together
F \approx 2\cdot 10^{-120}N

 

[Mentallic]

That's one hell of a force!

Assuming the force stays constant on both neutrons attracting each other, then it would take approx 10⁵³ years for them to collide. At the time of their collision, they'll be slamming into each other at a whopping 200 Planck lengths per second.

But what bothers me is that I've used a simplified version of events with my assumption. Of course the attractive force will increase as they get closer to each other. Anyone know how this could be calculated?

 

[DaveC426913]

Assuming the force stays constant on both neutrons attracting each other,

Why would you assume this? It's like assuming a jumper will reach the ground at the same velocity with which they left the top of the building.

 

[.physics]

I was just astonished to find out how you guys get such brilliant ideas. jspstorm, I liked your query.

Going by the law of physics and usual calculation , the attraction surely look possible though very negligible.But what I feel is that the gravitons responsible for creating gravity that too emitted by a neutron for attracting another neutron even 1 cm away is impossible. Gravitational force has always been used to calculate the force between bodies with large mass.The value of G by definition is found wrt object with huge masses. I don't think this thing practically exists at atomic level though we can prove it theoretically

And one more thing that I strongly believe is that the laws of physics though seem to be applicable in Earth and in nearby galaxies, it might just be a coincidence and different law of physics exists there. Let's take example of an atom - though we can't exactly tell or see if the atomic model is correct, it has been working fairly good for required calculations and reactions. Might be there's some more complex part that has been abstracted to justify a simple model by the atom itself. We are still debating on it aren't we- quantum theory , string theory etc...

So I strongly believe that universe got itself different kind of laws and and it doesn't follow the earthly law of physics. We can't argue that astrophysics is the most mysterious and interesting field of science.

So, the answer for me - NO! if earthly knowledge is applied.

Yes , if there exists any other universal surprise.

 

[Mentallic]

Why would you assume this? It's like assuming a jumper will reach the ground at the same velocity with which they left the top of the building.

Because I couldn't calculate it properly otherwise

 

[betel]

To calculate it properly you would have to use the equations of motion
2m \ddot r = -G\frac{m^2}{r^2}
Or use the easier way of energy conservation
E = 2m \dot r^2 - G\frac{m ^2}{r}=E_0=-G\frac{m^2}{r_0}
Integrating this for r from r_0 to 0 gives
<br /> T=\sqrt{\frac{4}{\pi}}\Gamma\left(\frac{3}{4}\right)^2\frac{1}{\sqrt{G m}} \sqrt{r_0}^3
For r_0=10^{12}ly this is T = 5 10^{60}s \approx 10^{53} y

Of course this calculation is only using Newtonian Gravity and neglecting any effects from Quantum processes or General Relativity. The last has to be taken into account for the last few seconds.

 

[KingNothing]

But what bothers me is that I've used a simplified version of events with my assumption. Of course the attractive force will increase as they get closer to each other. Anyone know how this could be calculated?

I am guessing that you haven't taken calculus 2. Essentially an integral is a way of adding up little bits of an equation evaluated at different points.

In this case, it'd be a matter of finding the time it takes to travel some small distance x based on the radius (distance between the two neutrons), and integrating that over the entire distance.

On a side note, our definition of time is completely invalid in this case, as "earth time" is not the same as "universe time", and if there is no earth, then any measurement of time is invalid.

 

[QuantumPion]

On a side note, our definition of time is completely invalid in this case, as "earth time" is not the same as "universe time", and if there is no earth, then any measurement of time is invalid.

Could you elaborate your reasoning of this?

 

[Mentallic]

I am guessing that you haven't taken calculus 2. Essentially an integral is a way of adding up little bits of an equation evaluated at different points.

In this case, it'd be a matter of finding the time it takes to travel some small distance x based on the radius (distance between the two neutrons), and integrating that over the entire distance.

On a side note, our definition of time is completely invalid in this case, as "earth time" is not the same as "universe time", and if there is no earth, then any measurement of time is invalid.

I've studied integration, I couldn't wrap my head around the required equation.

Why is Earth required to measure time? So if we lived on Mars then the measurement of time would be invalid? Nonsense...

 

[Mentallic]

To calculate it properly...
T = 5 10^{60}s \approx 10^{53} y

That doesn't seem right, since I already calculated the time to be exactly that, but with the assumption that the gravitational attraction between them increases as they get closer. I'm expecting it to be orders of magnitude faster than that value.

 

[DaveC426913]

On a side note, our definition of time is completely invalid in this case, as "earth time" is not the same as "universe time", and if there is no earth, then any measurement of time is invalid.

Yeah, that is kind of strange.

So, do you claim that these two neutrons do not have a decay rate if there is no Earth? The lack of an Earth somehere in the universe means that these neutrons behave by different laws of physics?

 

[rcgldr]

Prior threads that show the math for this (without using energy conservation method). Credit to Arildno for figuring how to solve one of the integrals:

point masses:
https://www.physicsforums.com/showthread.php?t=360987

masses of fixed size:
https://www.physicsforums.com/showthread.php?t=306442[/QUOTE]

 

[Janus]

That doesn't seem right, since I already calculated the time to be exactly that, but with the assumption that the gravitational attraction between them increases as they get closer. I'm expecting it to be orders of magnitude faster than that value.

I got 6.9e52 years.

Here's how:

Two objects sitting at rest at a given distance from each can be treated like two objects in orbit around each other but in extremely eccentric orbits, with a semi-major axis equal to half the distance between them with each at the periapsis of its orbit.

The period of the orbits of such bodies is found by:

P^2 = 4 \pi^2\frac{a^3}{G(M_1+M_2)}

Since falling in from periapsis to collision represents 1/2 of an orbit,

The time will be half that of the period of a whole orbit.

 

[DaveC426913]

I got 6.9e52 years.

Here's how:

Two objects sitting at rest at a given distance from each can be treated like two objects in orbit around each other but in extremely eccentric orbits, with a semi-major axis equal to half the distance between them with each at the periapsis of its orbit.

The period of the orbits of such bodies is found by:

P^2 = 4 \pi^2\frac{a^3}{G(M_1+M_2)}

Since falling in from periapsis to collision represents 1/2 of an orbit,

The time will be half that of the period of a whole orbit.

And what is their speed at perigee?

 

[betel]

I recheck my calc. I got the integral wrong but this will only change the numerical factor.
T= \sqrt{r_0}^3\frac{1}{\sqrt{Gm}}\frac{\pi}{\sqrt{2}}=1.9e53y
My formula is the same as Janus' but I get a different numerical result.

 

[Mentallic]

Is T in years or seconds?

 

[betel]

This is in years