Gravitational waves from a merged hyper-massive neutron star

November 14, 2018, Royal Astronomical Society
Graph showing data points from the LIGO gravitational wave observatory, plotting frequency against time. The GW170817 chirp in gravitational waves produced by the coalescence of two neutron stars is clearly visible as a sequence of dots in an ascending curve. After their merger, the gravitational-wave frequency declines for several seconds (see the line from 1843 to 1847 seconds), indicating a merged object with a decreasing spin rate. Van Putten and his team calculate that the chance of this being a false result is around 1 in 40,000. Credit: LIGO / M.H.P.M van Putten & M. Della Valle

For the first time astronomers have detected gravitational waves from a merged, hyper-massive neutron star. The scientists, Maurice van Putten of Sejong University in South Korea, and Massimo della Valle of the Osservatorio Astronomico de Capodimonte in Italy, publish their results in Monthly Notices of the Royal Astronomical Society: Letters.

Gravitational waves were predicted by Albert Einstein in his General Theory of Relativity in 1915. The waves are disturbances in space time generated by rapidly moving masses, which propagate out from the source. By the time the waves reach the Earth, they are incredibly weak and their detection requires extremely sensitive equipment. It took scientists until 2016 to announce the first observation of using the Laser Interferometer Gravitational Wave Observatory (LIGO) detector.

Since that seminal result, gravitational waves have been detected on a further six occasions. One of these, GW170817, resulted from the merger of two stellar remnants known as . These objects form after much more massive than the Sun explode as supernovae, leaving behind a core of material packed to extraordinary densities.

At the same time as the burst of gravitational waves from the merger, observatories detected emission in gamma rays, X-rays, ultraviolet, visible light, infrared and radio waves – an unprecedented observing campaign that confirmed the location and nature of the source.

A graph showing gamma-ray counts against time, whose initial peak is 1.7 seconds after the final coalescence of the two neutron stars. This short gamma-ray burst lasts for about three seconds during the period when the gravitational wave frequency declines, shown in Figure 1. Credit: A.M. Goldstein et al. / M.H.P.M. van Putten & M. Della Valle

The initial observations of GW170817 suggested that the two neutron stars merged into a black hole, an object with a gravitational field so powerful that not even light can travel quickly enough to escape its grasp. Van Putten and della Valle set out to check this, using a novel technique to analyse the data from LIGO and the Virgo sited in Italy.

Their detailed analysis shows the H1 and L1 detectors in LIGO, which are separated by more than 3,000 kilometres, simultaneously picked up a descending 'chirp' lasting around 5 seconds. Significantly, this chirp started between the end of the initial burst of gravitational waves and a subsequent burst of gamma rays. Its low frequency (less than 1 KHz, reducing to 49 Hz) suggests the merged object spun down to instead become a larger neutron star, rather than a black hole.

There are other objects like this, with their total mass matching known neutron star binary pairs. But van Putten and della Valle have now confirmed their origin.

Van Putten comments: "We're still very much in the pioneering era of gravitational wave astronomy. So it pays to look at data in detail. For us this really paid off, and we've been able to confirm that two stars merged to form a larger one."

Gravitational wave astronomy, and eking out the data from every detection, will take another step forward next year, when the Japanese Kamioka Gravitational Wave Detector (KAGRA) comes online.

Explore further: Gravitational waves provide dose of reality about extra dimensions

More information: Maurice H P M van Putten et al. Observational evidence for extended emission to GW170817, Monthly Notices of the Royal Astronomical Society: Letters (2018). DOI: 10.1093/mnrasl/sly166

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Nik_2213
5 / 5 (10) Nov 14, 2018
Does this mean it may be spinning very, very rapidly to stave off collapse ??
RNP
5 / 5 (11) Nov 14, 2018
@Nik_2113
The abstract to the paper linked above says that they believe a "magnetar" was formed. This is a HIGHLY magnetized neutron star that is formed spinning rapidly but is quickly decelerated by the interaction of its magnetic field with its environment.

The more detailed Wikipedia description of a magnetar is here; https://en.wikipe...Magnetar
dirk_bruere
4 / 5 (4) Nov 14, 2018
Can we PLEASE have reports with relevent numbers not omitted
jonesdave
4 / 5 (16) Nov 14, 2018
Can we PLEASE have reports with relevent numbers not omitted


Free access paper is here;

https://academic..../5090425

Nik_2213
5 / 5 (7) Nov 14, 2018
@JD: Thank you, I'd searched for an open access / arxiv version, missed this.
@RNP : Having a nice, young magnetar to study as it spins down is very, very good.
Da Schneib
3.4 / 5 (5) Nov 14, 2018
@RNP, this implies that both neutron stars were of low enough mass that their combination, minus the energy loss, was insufficient to form an event horizon. I'm a little surprised by that. Can you give a source for more information?

I was rather skeptical of this article.
wduckss
2.3 / 5 (3) Nov 15, 2018
The super-massive black hole diameter is ~ 0.001 - 400 AU. The diameter of neutron stars is ~ 20 km.
Original Interpretation Gravitational waves are tied exclusively to the extensively large black holes. Now the authors migrate again contrary to my criticism and degrade Albert Einstein. How it all (deliberately) quickly forgets and continues to steal. http://phys.org/n...nal.html
https://www.acade...agiarism
RNP
5 / 5 (10) Nov 15, 2018
@Da Schneib
There is a paper on the maximum mass of neutron stars here; https://arxiv.org...0314.pdf

I think the thing to remember is that a large fraction of the mass of the merging objects is ejected from the merger producing all the heavy elements observed.
rrwillsj
5 / 5 (3) Nov 15, 2018
The way I'm reading this (and yes, I could be wrong interpreting the paper)

The way I'm reading this is that the gravitational attraction of the two original neutron stars caused the merger?

Triggering a reaction of explosive novae?

Resulting in the spectacular spectrum of radiation detected?

While ejecting enormous quantities of assorted elemental matter?

Which in turn was swept away into open Space. Propelled(?) carried(?) gathered(?) all the above(?) by the gravitational waves from this event?

There-a-bye, producing a revenant magnestar?

So, first came the Gravity, then came all the rest of the energies?
antialias_physorg
5 / 5 (8) Nov 15, 2018
Can we PLEASE have reports with relevent numbers not omitted

We can...we just need to click the provided link at the bottom of the article (as with most other articles on physorg it leads either to the full paper or at the very least to the abstract)
rrwillsj
4.7 / 5 (3) Nov 15, 2018
Shoot, wasn't fast enough. Correcting There-a-bye, producing a revenant magnestar?

To read There-a-bye, leaving behind a revenant magnestar?

Oh yeah. Considering the combined masses of such a giant? Wouldn't the spin be a product of the collision? Because what outside medium is there for EM to react against to cause spin?
While the effect of the masses of the two bodies merging is similar to an ice-skater pulling their arms in to increase their spin?
DarkHorse66
5 / 5 (7) Nov 15, 2018
@rrwillsj
Not "There-a-bye". The correct term is "thereby" :)
Cheers, DH66
Da Schneib
4 / 5 (8) Nov 15, 2018
@RNP, thanks. This about confirms my skepticism, since the minimum mass of a NS is 1.4M☉. If the current article is correct, then it must be as you say and on the order of 0.8M☉ must be ejected during the merger. I remain a bit skeptical but the paper this article is about does present a good line of evidence. I'll reserve judgement for now and wait for more.
Da Schneib
4.2 / 5 (10) Nov 15, 2018
Can we PLEASE have reports with relevent numbers not omitted

We can...we just need to click the provided link at the bottom of the article (as with most other articles on physorg it leads either to the full paper or at the very least to the abstract)
Another good trick is to take the title of the paper or some representative portion of it and google for that plus arxiv. I often find preprints that way.
RNP
5 / 5 (9) Nov 15, 2018
Hi all
You can search the arxiv site giving open access to papers in a multitude of disciplines by author, title, abstract words, or any other detail here;

https://arxiv.org...astro-ph

Make sure you set the archive to "All" if you are not sure which category the paper you are looking for is appropriate (astro, physics, gc-qc, etc) .
jonesdave
3.5 / 5 (11) Nov 15, 2018
Hi all
You can search the arxiv site giving open access to papers in a multitude of disciplines by author, title, abstract words, or any other detail here;

https://arxiv.org...astro-ph

Make sure you set the archive to "All" if you are not sure which category the paper you are looking for is appropriate (astro, physics, gc-qc, etc) .


I always find the quickest way is to input the name of the paper into Google Scholar. If there is a free access version available, that will usually be the one linked on Scholar. Of course there is always sci-hub.tw, ahem, cough, didn't hear it from me. ;)
jonesdave
3.7 / 5 (9) Nov 15, 2018
I always find the quickest way is to input the name of the paper into Google Scholar. If there is a free access version available, that will usually be the one linked on Scholar. Of course there is always sci-hub.tw, ahem, cough, didn't hear it from me. ;)


Another useful site is libgen.io/ Particularly good for older stuff as well as books that have been scanned and OCR'd, etc. I was discussing some plasma cosmology woo by Anthony Peratt the other day, and I sure as hell wasn't paying for the book that it was in!

Disclaimer; this sh!t is illegal. Copyright and suchlike.
Nik_2213
5 / 5 (5) Nov 15, 2018
Have we cases where a fast-spinning magnetar on edge of mass instability 'bleeds off' speed, slows beyond critical threshold, collapses ? Or would such be too improbable to catch ?
jonesdave
3.4 / 5 (10) Nov 15, 2018
Have we cases where a fast-spinning magnetar on edge of mass instability 'bleeds off' speed, slows beyond critical threshold, collapses ? Or would such be too improbable to catch ?


The latter, I would guess.
Catching the first GWs from a NS merger, along with the EM signature was pay dirt for me. Unthinkable not so long ago. Bloody brilliant. Screwed the wooists, and stuffed up a fair bit of MOND as well. Happy days!
torbjorn_b_g_larsson
4.4 / 5 (7) Nov 15, 2018
I like this. Putting a massive spin on things. :-)

@RNP, this implies that both neutron stars were of low enough mass that their combination, minus the energy loss, was insufficient to form an event horizon. I'm a little surprised by that.


AFAIK the region where the merger could produce a neutron star and/or a black hole was unclear. In any case AFAIU we need more of these observations to probe and model.


stuffed up a fair bit of MOND


Yes, that 1.7 s delay - which is due to the X-ray emission coming delayed from a shell - with yet non-dispersive, universal speed limit waves for both EM (obviously) and gravity killed MOND in practice. I mean, it was already worse than LCDM on all scales including the idealized spirals it was ad hoc-ed for, and there are some epicyclic complex variants left at a guess (there always are), but it started out *ugly* and now is *pig ugly*.

GR FTW.
Da Schneib
3 / 5 (4) Nov 15, 2018
@torbjorn over here we'd say "butt ugly," but carry on!
TechnoCreed
5 / 5 (6) Nov 15, 2018
For more supplementary informations, let me suggest the wiki link https://en.wikipe...GW170817

I know that Wikipedia is not the best source of information but all the studies on this kilonova are linked at the bottom of the page.

Is it just me who fell this way or are there really many people here that do not know that this event happened on august 17 2017?
Ojorf
3.4 / 5 (5) Nov 16, 2018
Which in turn was swept away into open Space. Propelled(?) carried(?) gathered(?) all the above(?) by the gravitational waves from this event?


Pushed away by the radiation, not gravitational waves I think.
Ojorf
3.4 / 5 (5) Nov 16, 2018
^^^
While busy there, Firefox just crashed talking windows down with it, it just rebooted.

Wanted to add "magnetic coupling" flings the stuff away.
Ojorf
3 / 5 (4) Nov 16, 2018
And now I get a double post...
antialias_physorg
5 / 5 (6) Nov 16, 2018
Have we cases where a fast-spinning magnetar on edge of mass instability 'bleeds off' speed, slows beyond critical threshold, collapses ? Or would such be too improbable to catch

That would be an incredibly fine line to walk.

Thing is, if you want to have any effect like this that keeps it from gravitational collapse I think the surface must rotate close to the speed of light. These stars are tiny. The gravity gradient would be humongous if it were on the verge of collapse into a black hole (it's staggering as it is with neutron stars).

While neutron stars/magnetars can rotate incredibly quickly they 'only' come up to a few percent of that.
antialias_physorg
4.7 / 5 (7) Nov 16, 2018
Another good trick is to take the title of the paper or some representative portion of it and google for that plus arxiv.

Another way is to google the mentioned researchers name and their affiliated institute. They mostly have the relevant numbers (if not a link to the paper itself) on their university websites. If I really want the paper (and it's only available paywalled elsewhere) I just ask them via the email-address posted there.

Of course, "google scholar" will often output the link to the paper directly.
https://scholar.google.com/
JaxPavan
4 / 5 (4) Nov 16, 2018
I would expect much of the matter would be ejected.

I don't know precisely what happens to the gravitational fields when two spinning neutron stars collide, but it stands to reason that it is zero in the middle, with outer perturbations as they actually collide.

It also stands to reason that wherever that gravitational field falls below what is necessary to compress neutrons into degenerate matter, then there would be a big boom, and they would be blasting apart as they came together.
rrwillsj
4.5 / 5 (2) Nov 17, 2018
I never understood the reasoning, that there can be zero-gravity at the center of a mass? As I comprehend it, gravitational attraction is always inward. A sphere of force. 100%

That the Earth's gravity as measured, is lumpy because large portions of the entire planet are more massive than the rest of the material making up the globe.

Just goes to show what a stochastic process of random collisions resulted in the proto-Earth, five billion years ago.
antialias_physorg
5 / 5 (5) Nov 17, 2018
I never understood the reasoning, that there can be zero-gravity at the center of a mass? As I comprehend it, gravitational attraction is always inward.

How much inward from the center can you go?

But to put it more precisely. The force that affects a test-mass is the sum over all the forces that cat on it. So if you pu a test mass (e.g. yourself) a the center of the Earth then there will be as much mass tugging gravitationally in one direction as there will be in the other..so they all cancel out.

Now don't confuse zero gravity with zero pressure. The pressure at the center of the Earth is immense, because all the stuff ABOVE the center (in every direction) *does* feel a net force (by the above sum-over-all-other-forces) downward...and that adds up to a huge amount of force *from all sides* that would be put on anyone unlucky enough to be standing at the center of the earth.
DarkHorse66
5 / 5 (6) Nov 17, 2018
This might help explain it in more detail:
https://www.quora...of-earth
https://www.youtu...Qjq675A8
Followed by:
https://www.youtu...8F7ZTrhI
Further example: where you might be close to the centre, but not at the actual centre (g still greater than 0, but less than surface gravity):
Since any gravitational force coming from all that mass above your head does cancel out, only the gravitational force being generated by whatever is below your feet (ie between your feet and the centre) actually counts.
Hope that this does help at least a little :)

Best Regards, DH66
JaxPavan
5 / 5 (4) Nov 17, 2018
@antialias

Right, so at the center of each neutron star there is no gravity but enough pressure to maintain the degenerate matter. At the radius there is enough gravity, with much less pressure, if any.

What I was trying to say is:

The big boom happens when two neutron stars approach each other and parts of the surface of each star get pulled in opposite directions that cancel out the gravity necessary to maintain the degenerate matter, but before they pile into each other to generate the pressure. Then if the stars are spinning to present additional degenerate matter at that "critical" radius. . .
granville583762
5 / 5 (3) Nov 17, 2018
Why is gravity zero at the centre of mass
This might help explain it in more detail:
https://www.quora...of-earth
Further example: where you might be close to the centre, but not at the actual centre (g still greater than 0, but less than surface gravity):
Since any gravitational force coming from all that mass above your head does cancel out, only the gravitational force being generated by whatever is below your feet (ie between your feet and the centre) actually counts.
Hope that this does help at least a little :)Best Regards, DH66

Simply, every particle attracts ever other particle
JaxPavan
3.7 / 5 (3) Nov 17, 2018
the zero gravity center of mass of the two neutron starts is in between them as they come together.
jonesdave
3.5 / 5 (8) Nov 17, 2018
I'm struggling to see where this conversation is going.
DarkHorse66
5 / 5 (6) Nov 17, 2018
the zero gravity center of mass of the two neutron starts is in between them as they come together.

No!
Each neutron star will have its own zero gravity center of mass, ON THE INSIDE, at their respective cores. Once the two are combined, THEN the new star will have one zero gravity center of mass.

Simply, every particle attracts ever other particle

Not true. That statement has nothing to do with zero gravity centers of mass.
The conversation was about an entire planet (or wrt Jax's question, about neutron stars), not individual particles.
At particle level it is about charges, not gravity. Particles of the same charge repel each other. Just ask two electrons...

Best regards, DH66
jonesdave
3.5 / 5 (8) Nov 17, 2018
the zero gravity center of mass of the two neutron starts is in between them as they come together.


Errrmmm, no. Just no. Didn't do physics at uni, did you? The total mass of one star has a gravity, as does the other. Gravity is attractive, last time I looked.

Da Schneib
4.2 / 5 (5) Nov 17, 2018
@rrwillsj
I never understood the reasoning, that there can be zero-gravity at the center of a mass? As I comprehend it, gravitational attraction is always inward. A sphere of force.
This was first discovered by Newton, and is called the "shell theorem." You can look that up on Wikipedia, or find it in Newton's masterwork, the Principia Mathematica.

Without invoking all the math, the general idea is that at the center of a solid sphere, you are pulled equally in all directions. For every infinitesimal piece of the sphere pulling you up, there's another one pulling you down; for every one pulling right, one pulling left. Thus the total force is zero. Move away from the center, and there is more mass in the direction you are moving away from; thus, you feel increasing force from that side the further away you move.

Hopefully that will stimulate your imagination to understand why we are saying there is zero gravity at the center of the Earth.
Da Schneib
4.2 / 5 (5) Nov 17, 2018
@rrwillsj, also, one must keep in mind the superposition principle. This states that in the presence of equal and opposite forces, the total force is zero. The principle is supported by various mathematical theorems applicable to various forces and in various situations, but it is always true that equal and opposite forces cancel to zero. The basic idea is that if you have two forces acting on a body, then the total force on the body is the sum of the two forces taken separately. I can amplify a bit more if you like; just ask.
Da Schneib
4.2 / 5 (5) Nov 17, 2018
I don't know precisely what happens to the gravitational fields when two spinning neutron stars collide, but it stands to reason that it is zero in the middle, with outer perturbations as they actually collide
This seems intuitive but it ignores an important point. More in a moment
It also stands to reason that wherever that gravitational field falls below what is necessary to compress neutrons into degenerate matter, then there would be a big boom, and they would be blasting apart as they came together.
This is incorrect because you have neglected pressure. It is not gravity that holds a neutron star in its equation of state; it is pressure. Just because you can cancel gravity at some point by superposing equal and opposite gravity doesn't mean there's no pressure at that point. A great example is the center of the Earth; zero gravity, enormous pressure.
Da Schneib
4.2 / 5 (5) Nov 17, 2018
Right, so at the center of each neutron star there is no gravity but enough pressure to maintain the degenerate matter. At the radius there is enough gravity, with much less pressure, if any.
The "if any" is rather disingenuous considering the pressure of the atmosphere on Earth is obvious.

The big boom happens when two neutron stars approach each other and parts of the surface of each star get pulled in opposite directions that cancel out the gravity necessary to maintain the degenerate matter, but before they pile into each other to generate the pressure. Then if the stars are spinning to present additional degenerate matter at that "critical" radius. . .
While this is well reasoned, the problem is that the actual coalescence happens very quickly, in a matter of seconds, and there is insufficient time due to the long lifetime (relative to the coalescence time) of the neutron.

[contd]
Da Schneib
4.2 / 5 (5) Nov 17, 2018
[contd]
The lifetime of neutrons is on the order of tens of seconds; the coalescence time is very much shorter than this. Therefore the neutrons cannot decay fast enough to liberate many protons and electrons during the coalescence.

However, according to my conversation with @RNP above, these disruption effects do, in fact, occur; a sizeable amount of the material of both neutron stars may be ejected for quite likely just the reasons you have speculated on. So I gave you a 5 for that post.

Your second mistake is the same as your other one: you have confused pressure with gravity. Your fir4st mistake was assuming the neutrons would "pop" into protons and electrons immediately upon the degeneracy pressure being released. This is not so; neutrons have a lifetime, and do not immediately convert into protons and electrons. (Oh, and incidentally I left out the antielectron neutrinos, but for this discussion they are immaterial.)
Da Schneib
4.2 / 5 (5) Nov 17, 2018
Now, see how that works, @Jax? If you don't troll, I don't troll you. I will never outright lie or try to mislead you about physics. But if you troll, I troll back. Think about that a little while. It is my policy.

As far as whether you are a YEC who is trolling more subtly, that will emerge in time and I will judge based on your actions. You probably shouldn't have joined up with the trolls in the first place if you actually wanted to learn something about astrophysics.

I challenged you elsewhere on exactly these grounds and you ignored it and kept on trolling. You can see what conclusions I drew from that. It's a bad policy on your part.
jonesdave
3.5 / 5 (8) Nov 17, 2018
Now, see how that works, @Jax? If you don't troll, I don't troll you. I will never outright lie or try to mislead you about physics. But if you troll, I troll back. Think about that a little while. It is my policy.

As far as whether you are a YEC who is trolling more subtly, that will emerge in time and I will judge based on your actions. You probably shouldn't have joined up with the trolls in the first place if you actually wanted to learn something about astrophysics.

I challenged you elsewhere on exactly these grounds and you ignored it and kept on trolling. You can see what conclusions I drew from that. It's a bad policy on your part.


Aha! A message for Jax Pillock: \o/

https://www.youtu...Ul7oT9sA

I fart in your general direction.
antialias_physorg
5 / 5 (5) Nov 18, 2018
Each neutron star will have its own zero gravity center of mass, ON THE INSIDE, at their respective cores.

There will also be a point between them where the attraction of one will cancel out the attraction of the other.

What happens when they touch is anyone's guess. At the immense speeds that these stars tend to rotate you get close to the "immovable wall meets insurmountable force"-type of scenario.

But as to stability: The 'inside' points are stable due to the pressure. When neutron stars get very close there is this 'zero gravity' area that touches the surface, but it's very small and the gradients are incredibly steep.

I haven't done the calcs but I'm willing to bet that the Roche Limit of neutron star-on-neutron star is super tiny.
Also the last moments before they touch happen very fast, so there's not really a lot of time for neutrons to decay due to lack of degeneracy pressure. (The outer layers of a neutron stars aren't neutrons, anyhow)

rrwillsj
5 / 5 (3) Nov 18, 2018
DH66, DS, a_p, jd, I want to thank every one of you for your explanations & the url's you provided.

However, {oh, that word!} I have to confess that I cannot accept the concept that, in close proximity, gravity cancels gravity. I am thinking that would violate both the Inverse Square Rule & Gravity as a Constant of Mass.

I concede your point that the mathematics of gravity physics allows for a zero-G effect at the center of a Mass. I certainly do not have the math chops to refute that claim.

It still remains an itch on my brain that those concepts are not correctly explaining reality, as I visualize it. But truthfully? How real is my sense of reality?

"Philosopher know thy self & be self-aware of your limitations!"

Please do not feel any obligation to further enlighten me. I have already consumed your time & knowledge more than I deserve.

This old dog ain't learning any new tricks! I'll just keep shufflin' along with my All Gravity, All the Time woowhooee.
DarkHorse66
5 / 5 (3) Nov 18, 2018
@rrwillsj:
There is no violation of the Inverse Square Rule. From https://en.wikipe..._theorem
A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass. This can be seen as follows: take a point within such a sphere, at a distance r from the centre of the sphere. Then you can ignore all the shells of greater radius, according to the shell theorem. So, the remaining mass m is proportional to r^3 (because it is based on volume), and the gravitational force exerted on it is proportional to m/r^2 (the inverse square law), so the overall gravitational effect is proportional to r^3/r^2 = r, so is linear in r.
(looks better on the wikipage) I am assuming that you know how to multiply these two.
Bear in mind that the linear relationship in r is a product of the interaction of two different things &is an end product.
...cont
DarkHorse66
5 / 5 (3) Nov 18, 2018
cont...
I cannot accept the concept that, in close proximity, gravity cancels gravity

This, quite literally, has to do with the the fact that the push/pull (read force)of gravity is DIRECTIONAL &therefore we use arrows(vectors)to represent that direction. Rewatch https://www.youtu...Qjq675A8 When you reach the bits where you see the arrows(vectors), try the following: imagine you are standing directly in front of the tip of an arrow. That arrow is a now person who is pushing you hard &is making you move in the direction they are pushing you(in the case, backwards). At the same time there is another arrow pointing to your back.This is another person who is pushing you back in the opposite direction to the first.If person2 is pushing less hard than the 1st, you will still be moving backwards, just more slowly (the 2nd person is partially overcoming the effects of the pushing of the first).If person2 is pushing harder than the 1st, they are not only fully counteracting...cont
DarkHorse66
5 / 5 (3) Nov 18, 2018
cont...
the 'push' from the 1st, but are forcing you to move forward as well.If person2 is pushing equally hard as the 1st, you will find that you are moving neither forwards, nor backwards. Because the two 'pushes' are from opposite directions, they have cancelled out each other's effect, You are in equilibrium, ie because of the opposing dirctions (maths) they will have opposing signs that represent their directions of movements/force. Eg, x is +ve='forward & is y -ve=backwards. If If x & y are equal in magnitude (size https://en.wikipe...ematics) , but in opposing directions, then their magnitude alone means that x=y. But since the forces come from opposite directions, x will = +x, but y will = -x. So when you add up the two vectors (yes, that simple here), you get: x + (-x) = 0. The 0 means that both are pushing at equal strength against each other & cancelling out, &you are going nowhere. That is equilibrium. This model can be extended...cont
DarkHorse66
5 / 5 (3) Nov 18, 2018
cont...
to all directions coming at you. Eg, another two people are pushing on you from the left & right at the same time as each other, & at the same time as the first. Here too, those two new guys will cancel each other out. The maths become a little more complicated (summation), but as soon as you can cancel a pair of forces out, by making them=0, you can drop them from the overall equation. So, ultimately, the NET effect of all these people(=grav forces) pushing on you EQUALLY from all directions is there is no dominant direction for you to be pushed in, so grav net=0 at centre. Btw it is the strength of that 'push' that creates the the massive pressure that you would experience FROM ALL SIDES.
Gravity as a Constant of Mass.
Do you mean "the law of conservation of mass"? https://www.quora...ant-Mass
What you need to be mindful of is that gravity is product of mass, NOT the other way around! The strength of gravity is directly proportional...cont
DarkHorse66
5 / 5 (3) Nov 18, 2018
cont...
to the density of whatever mass is generating it. That is the very reason a neutron star generates such high gravity. The mass of even a teaspoon of matter is so high because it has been crushed to such a high density.
I hope that all this makes the concepts a LITTLE clearer. :)

Best Regards, DH66
DarkHorse66
5 / 5 (3) Nov 18, 2018
"Philosopher know thy self & be self-aware of your limitations!"

At least you do have that awareness. That makes you more open to leaning stuff, even if it is only bitesize bits at a time. KUDOS
Please do not feel any obligation to further enlighten me. I have already consumed your time & knowledge more than I deserve.

What obligation??? It becomes a pleasure when someone genuinely wants to know. That is when a person deserves the time and effort the most.
This old dog ain't learning any new tricks!

Not true. We learn all the time (if we are open to it), even when we are not aware of doing so. Besides, if you were truly not learning, what are you doing on this site. And asking open questions?
I'll just keep shufflin' along with my All Gravity, All the Time woowhooee.

Don't.

Keep asking, and plugging away. You might just 'get it' sometime(s).

Best Regards, DH66 :D

DarkHorse66
5 / 5 (3) Nov 18, 2018
@rrwillsj:
add-on. And a warning to others.

The link that I pasted in, in my second post won't work. I made the mistake of doing a copy'n'paste from an earlier post. When one does that, this stupid editor seems to cut out the symbols that it has previously truncated into dots and when you then click on it, that is all that google sees. The correct procedure is to go to the post where you posted the link originally from the wanted page. Open that link and re-copy'n'paste from the inside of the wanted page.
So here is a (proper) repost of the link:
https://www.youtu...Qjq675A8

Best Regards, DH66
Here is
Da Schneib
3.7 / 5 (3) Nov 18, 2018
@DH, well said:
What obligation??? It becomes a pleasure when someone genuinely wants to know. That is when a person deserves the time and effort the most.
Same goes for me, @rrwillsj.
rrwillsj
5 / 5 (2) Nov 18, 2018
Oh DarkHorse! You are one of those strict, classical pedagogues belaboring us lazy students with your cane of education.

I see what I think you intend with your Dr. Domuch pushme/pullme example. I agree with equal forces applied that I would be rendered inert between them. But what I am dubious about is the contradiction that opposing direction of force are canceled.

I can agree that smarter people than me accept the supposition that the center of a Mass must be at zero-gee. That does not seem reasonable to me.

Though I have no proof or any supporting evidence for my opinion. It's just a step too far for my imagination.

I do not accept the "shell" theorem. Ah geez, Sir Newton's ghost is going to boink me over my head with an apple!

In my inconsiderate opinion. Any homogenized Mass, whether nanograms or gazillions of tons will have an equal Gravitational attraction in all directions right down to the central point. Always attracting.

Da Schneib
3 / 5 (2) Nov 18, 2018
Superposition is a hard concept to get your head around. I learned it in EE.

In gravity, it might help you to remember that gravity is a deformation in spacetime. Thus, if you have this much gravity pulling in one direction, and the same amount pulling in the opposite direction, the two deformations cancel one another. This is what other posters mean when they say they're "opposite vectors."
DarkHorse66
5 / 5 (2) Nov 18, 2018
In my inconsiderate opinion. Any homogenized Mass, whether nanograms or gazillions of tons will have an equal Gravitational attraction in all directions right down to the central point. Always attracting.
Yes, that is what is happening; you should have been able to work that out from the video.Unfortunately you are forgetting what I also said a few posts beforehand; that
where you might be close to the centre,but not at the actual centre(g still greater than 0, but less than surface gravity): Since any gravitational force coming from all that mass above your head does cancel out, only the gravitational force being generated by whatever is below your feet (ie between your feet and the centre) actually counts.
As you said, it is gravitational ATTRACTION.If the stuff coming from above your head did count, it would be attracting you UPWARDS, towards the ceiling(that is a surface too);&that would mean that the direction of force vector would be pointing the other way!..cont
DarkHorse66
5 / 5 (3) Nov 18, 2018
cont...
Hence if you are at the centre, there is no mass beneath your feet to do any attracting in its own right.
You also need to consider the following: gravity might be attracting, but it is in ONE direction ONLY. Even for a sphere (where it does all terminate at a point (the centre)). In the case for a sphere, it is towards a central point. inwards &that is why you are not floating to the sky. It is all pushing you DOWN (technically towards the centre) & it does not matter where you are, on the surface of the sphere, &for two opposite points on the surface, those attracting grav forces coming from opposing directions have to meet somewhere; in this case at the centre of the earth. If you were to overshoot the centre, that attracting grav force coming from the opposing direction would begin to push you back down, just like on the 'other side'. Rewatch vid#2 https://www.youtu...8F7ZTrhI
...cont
DarkHorse66
4.7 / 5 (3) Nov 18, 2018
cont...
I am glad that you have managed to understand the push/pull example that I gave,&not see a contradiction.I did mean that example in a quite literal sense&as a direct demonstration of what is happening.
But what I am dubious about is the contradiction that opposing direction of force are canceled.
It is not the opposing directions of force that are being cancelled, but rather their effect. They are cancelling out each other's effect.Remember,they are coming from all points on the surface &have all just met in the middle/centre.One point that a lecturer repeatedly made"If you want to get a better handle on a problem, draw it"Draw a simple diagram with a circle&arrows that denote the direction of the attraction.You should be able to quickly see what I am talking about.When you look at the centre where all the arrows are meeting, ask yourself, "where do they go from there?" I think you will find that the answer will be"Nowhere.They all terminate at that central point"
DH66:)
DarkHorse66
5 / 5 (4) Nov 18, 2018
btw, just one thing I do take issue with:
Oh DarkHorse! You are one of those strict, classical pedagogues belaboring us lazy students with your cane of education.

I would rather not be described with statements like that; it does come across as a little mocking. Do you feel as though I am forcing this stuff onto you? :(
I sincerely hope not, that would be rather disappointing.. :((

Regards, DH66

rrwillsj
5 / 5 (2) Nov 19, 2018
My apologies DarkHoorse, it is my snarky sense of the satirical absurd compels me to laugh at myself. Humor is how I deal with my failures.

I do want to express my appreciation to you and DaSchneib for the effort you have both put into attempting to overcome my limitations of comprehension. I really enjoy this challenge of intellectual stimulation.

Personaly. I think Science is Fun!
"Knowledge is not a border. Knowledge is a frontier."

Hey! I think I just made that up... Thanks guys for the inspiration.

et al: In case it is not original to me, please let me know who I need to properly credit.

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