different between gravitational force and gravitational field

Hi,

If there is a mass exactly between two stationary equal masses (ie big stars or blackholes) the gravitational force will cancel out and there will be no movement in the system (big stars artificially held in position). This means the gravitational forces cancel out for the mass in the middle, but if the two stars are removed, and just the mass in the middle is left, will the gravitational field still be the same? Do the two stars cause the gravitational field to become stronger even though there is no movement in the mass in the middle?

cheers, Jamie

Reply to
Jamie M
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To rephrase in a simpler way, is this true:

g force can be null, but if 2 strong forces are cancelling out at that point, the field intensity is still high?

cheers, Jamie

Reply to
Jamie M

Look up 'gradient'.

Jeroen Belleman

Reply to
jeroen Belleman

If the object has any extent, the null will only be in its middle, and its outsides will be pulled in two directions. If the two masses disappear, the refractory people walking on the surface of the star will suddenly feel heavier, as there will be less cancelation of the local star's gravity.

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John
Reply to
quiasmox

It seems like the field could only be known by its effects - force and movement - and if there isn't either, how could there be a field?

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John
Reply to
quiasmox

Hi,

There may be a force just not visible from our reference frame, ie maybe the situation described above with a null g force but high field intensity can explain dark energy, ie a group of mass creates a gravitational field that causes an overall anti gravity force.

cheers, Jamie

Reply to
Jamie M

Hi,

That's a good question, another question is if a mass creates a spacetime curvature, does the center point between the two masses spacetime curvature, where there is a point of no curvature, still have the same gravitational field "density" as an space with no masses?

Just because two forces cancel out doesn't mean there isn't still a force that would act in another dimension if it could. I think the overall expansion of the universe is an available dimension for the non-zero gravitational field density to expand into, which would cause an accelerated expansion rate.

cheers, Jamie

Reply to
Jamie M

Hi,

I think general relativity using three spatial dimensions, seems to require curvilinear coordinates:

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These allow the curvature of space time from the effects of gravity to be shown in 3dimensions, but also it may be missing some information compared to using a 4th dimension to describe gravitational field intensity, and 3 spatial dimensions with a fixed cartesian grid with no curvature.

This page shows a pair of masses and the resultant space time curvature:

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If you move them together it makes sense that the curvatures will combine, and form a flat non curved point between the masses, but the "height" of the surface will still be lower.

Here are some videos showing situations where the point between masses with a flat curvature has a high gravitational field intensity:

binary black hole event horizon combination:

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35seconds into this video, blackholes combining, shows a spot between the black holes where there could be a flat spot but the "height" is lower than surrounding space outside the black holes:
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Basically the curved spacetime with a flat spot between two masses has to have a greater gravitational field density! It is potential energy.

cheers, Jamie

Reply to
Jamie M

I'm pretty sure that space-time curvature right at the middle will be level (I wonder what this means if it's close to, or within, the event horizons of a pair of black holes?). But you're ignoring tides.

But as has been mentioned, there's still tidal forces. Ignoring any relativistic effects (because I don't know relativity -- does this mean I'm not relevant?) the acceleration due to gravity goes as distance squared. This, in turn, means that on either side of the null there'll be a pull toward the nearer stars.

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Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

Hi,

If you combine two blackholes, the event horizon forms a tube in space between them:

binary black hole event horizon combination:

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The outer ring height of two merging funnels is lower at the intersection where they are merging, than the outer ring of the funnels that aren't merging.

This shows that the flat space time between two masses can be at a different "height" than the flat space time in empty space with no masses, and even though there is no gravitational force, the gravitational field intensity is proportional to the "height" of the flat space time.

So gravity should be in 4 dimensions and spacetime curvature not used, unless it can account for gravitational field strength in flat space time between two masses.

cheers, Jamie

Reply to
Jamie M

Agreed. He is also going to need a very large number of light rigid rods to hold his various point mass objects in position.

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Regards, 
Martin Brown
Reply to
Martin Brown

I think of this as a descriptive error: the forces do NOT "cancel out", they are "in balance".

Reply to
Wond

Nah, he'll just tell his spherical cows to stay still. ;)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

This is really a question about tides. The Moon and Sun both cause a tidal bulge in the Earth, which interestingly is nearly symmetrical--there's an almost equal and opposite bulge on the side away from the other body.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

Yes, it will be zero.

No, they give that volume a greater gravitational potential than it would have if the stars weren't there.

A gravitational field is the way the force due to the field points at var ious places over a given volume, depending on the differences in potential across that volume. If there are no differences there are no forces. The lo cal field is flat but that tells you nothing about the "absolute" potential anywhere in that volume- you can only compare potential between points. IO W you need access to "infinity", where the rubber sheet is glued to the fra me.

It's like referring voltages to ground. Just because there's no current t hrough a circuit node, you can't assume that node is at zero volts.

Mark L. Fergerson

Reply to
Alien8752

across that volume. If there are no differences there are no forces. The local field is flat but that tells you nothing about the "absolute" potential

anywhere in that volume- you can only compare potential between points. IOW you need access to "infinity", where the rubber sheet is glued to the frame.

Hi,

Yes the gravitational "bias voltage" is unknown in spacetime, but the relative bias can be calculated knowing the masses and knowing the rate of spacetime expansion into surrounding space. Ie if the surrounding spacetime has a small gravitational bias, and a subset spacetime has a larger spacetime bias, ie due to matter's integrated curvature of spacetime in that area, then the area of spacetime with larger bias will expand into the area of smaller spacetime bias.

This can be observed in our own universe as dark energy, ie a massive supernova explosion, is an area of high spacetime bias relative (an additive term to the universes overall integrated spacetime bias). So the supernova will expand faster due to the expansion of the spacetime in that area. The question is can a spacetime expansion of a supernova be discriminated against an antigravity acceleration force on the matter being thrown out by the supernova? If so then that is some evidence this idea could be correct, that dark energy is a force resulting from the integration of the curvature in a given spacetime = gravitational bias (voltage).

This can also be applied to galactic filaments of galaxies, if the cylinder is modeled as a homogenous cylinder of point masses, then the integrated curvature of gravity will create a cylindrical bias force as well, and the spacetime expansion rate should grow the cylinders length and diameter and create the large spacetime voids around the filaments. If these filaments of galaxies are spread throughout the universe in a connected structure, they all are under compressive force as they are each pushing axially to expand spacetime.

cheers, Jamie

Reply to
Jamie M

There's three concepts here: force, field, and potential. The forces can balance, and the field is null, in this kind of three-body arrangement. The force (and the field) has tidal effects only, as far as the mass in the middle is concerned. But the potential of that mass in the middle is very strongly negative; it takes a LOT of energy to pull it out of that region and off to infinity, and that defines a potential (which is a scalar field, unlike gravity which is a vector field). From knowledge of the potential, one can

Reply to
whit3rd

There's three concepts here: force, field, and potential. The forces can balance, and the field is nulled locally, in this kind of three-body arrangement. The force (and the field) has tidal effects only, as far as the mass in the middle is concerned.

But the potential of that mass in the middle is very strongly negative; it takes a LOT of energy per unit mass to pull it out of that region and off to infinity, and that defines a potential (which is a scalar field, unlike gravity which is a vector field). From knowledge of the potential, one can mathematically predict the field. And from the field and knowledge of the mass of the 'object in the middle', one can calculate the force.

Thus, full information is

Reply to
whit3rd

Hi,

Yes the gravitational potential energy is the amount of spacetime curvature even in flat space time, which indicates that flat spacetime can be at different potential energy levels. I think this is a physical attribute of spacetime, otherwise the curvature in spacetime would be discontinuous (ie two flat sections can't be at the same physical potential and connected by a curve still). If the flat sections can be at different potentials then the curvilinear coordinate system in 3D is not a complete explanation of varying potential energy levels in flat spacetimes.

cheers, Jamie

Reply to
Jamie M

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