# Where is most "gravity", inside or out?

Discussion in 'Pseudoscience' started by nebel, Feb 29, 2016.

1. ### nebelValued Senior Member

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perhaps not. how could you have relativistic effects of gravity where here is none of it? ? The center line should not be shown to go down ( back in time*) but back up to a point at the normal level, a point, where no time dilation from the general value could be recorded.
It is the illustration that needs amending to satisfy the question where really is the most gravity measurable, in or out,
* from the "ALMA" thread in alternate theories.

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3. ### DaveC426913Valued Senior Member

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There is lots of it.

The net acceleration felt is zero - which is why the line is flat. But the gravitational potential is very high, which is why it isn't back up at the x-axis.

You are being attracted to every atom around you, it just happens that there are exactly as many on your left as on your right, so there is no net force.

Think of a slightly more conventional scenario.

There is a point between every two bodies in the universe where their gravitational pulls happen to balance. The Earth-Moon system has such a point. You could go there in a spaceship.

(Actually, it's not a point at all. It's a hyperbolic disc-like surface, extending to infinity. But let's keep it simple.)

By your reckoning, you could park your ship at that point and you would experience zero time dilation, even though you are in both the Earth's g-field and the Moon's g-field.

Except this is not true.

Last edited: Mar 16, 2018

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5. ### Janus58Valued Senior Member

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No, the steepness of the line represents strength of gravity at that point. Think of it as the amount of force per unit mass you would need to hold something stationary at that point. The depth is gravitational potential or specific gravitational energy. This is the potential energy per unit mass for an object there Moving a mass from a point lower on the graph to one higher would require an expenditure of energy.

While the force of gravity at the center of Sirius is zero, you are still at a lower potential than at the surface, in that you would have to expend energy to lift a mass "up-slope" from where the Sirius graph intersects the black line to where it intersects either green line.

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7. ### nebelValued Senior Member

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You are not internally stressed by half of the universe pulling in one direction, the other half in the other, The gravity disappears when balanced by an opposing tug.
There are such areas of balanced forces. lagrangians. There are also baricenters.
Gravity is a surface phenomenon, strongest there, going in or out makes it less. in a short fall to the center through the radius, getting ever smaller to zero at the near center, only coming to zero at infinity if you go out. A lot more of it in all directions.

Last edited: Mar 16, 2018
8. ### nebelValued Senior Member

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good point. falling in there, you would of course have gained a lot of energy too. The idea of the PO was that there is way more slope to push up on, Sisyphus like, on the outside in all directions to infinity, then in the small radii to the center, to reach that zero gravity floating feeling.
In both cases the hardest struggle, the steepest slope would be near the surface, and the surface is a lot closer to the center and surrounded, enclosing diminishing, less space in the interior than out to infinity.

Last edited: Mar 16, 2018
9. ### DaveC426913Valued Senior Member

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Correct. The net force cancels out.

No it doesn't. What disappears is the net effect.

You are still deep in a gravitational field.

10. ### nebelValued Senior Member

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You are in two opposing gravitational fields, but they do not add up, but subtract, cancel, not unlike a interfering wave?. so if there is no net effect of gravity, why should there be a relativistic effect? The question was about gravity inside versus outside, if the inside effects cancel out, but the outside field does not, does that not still leave more gravitational effects in the vastly greater outside, that an expanding body might move into?

11. ### DaveC426913Valued Senior Member

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Sorry, you'll have to take it up with Einstein.

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12. ### Q-reeusValued Senior Member

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By narrowly defining gravity as synonymous with 'gravitational force field', you trap yourself in an apparent dilemma. Gravity has more than one, context dependent, meaning in physics and standard dictionary definitions are inadequate. Many physicists reserve it's use for tidal gravity aka the second derivative of the metric, while elsewhere it can have the common use of 'g-field' aka first spatial derivative of metric. At any rate 'gravity' as a general phenomenon encompasses more than either of those two. In particular gravitational time dilation is a function of the metric itself - it's time component. Which is a function of the Newtonian potential φ and not it's gradient.
Redshift of light as a function of gravitational potential φ, and not the negative of it's gradient, -∇φ aka the Newtonian 'g-field', is an experimental fact e.g.:
https://arxiv.org/abs/1207.0177
See last para under 'Conclusion', where they write:
"...we may state that this value is well consistent with the expected value of +633 m s^−1 (due to the gravitational potential on the solar surface) as predicted from the general relativity theory. This means that we could confirm the solar gravitational redshift in a quantitatively satisfactory level within the precision of our analysis."

If you don't understand or accept that redshift and clock-rate (inverse of time-dilation) are equivalent concepts, then no further discussion will help.

13. ### nebelValued Senior Member

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Thank you for the clarification, link. The PO question referred to gravity in the broadest of terms, asserting, or questioning whether,- however one defines gravity, there is more of it on the outside of an entity than from the surface down to the center. In the case of tidal gravity, reversing the spaghettification experienced during entry into a gravity well with quicker restoration of normal girth on the way to the zero gravity center. I never questioned the validity of relativistic effects at all.
sorry for vulgarising the spirit of your instructive link and comment.

14. ### Q-reeusValued Senior Member

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You're welcome.
The one gravitational quantity that there is more of heading down from surface to the center - for a spherical solid body say, is the negative of gravitational potential φ. Hence it matters very much even there 'how one defines gravity'. For a thin shell, φ remains a constant everywhere inside the shell. Consistent with Newtonian g = -∇φ being everywhere zero there.
Will depend on the details of the body in question. If a constant density sphere, to a good approximation tidal gravity is zero everywhere below the surface. As it is in the case of a thin shell, despite g being non-zero at finite 0 < r < R (R the surface radius) in the former case, and g being zero for all r < R in the latter case.
No problem. Just hope any insight gained from this 'sticks'!

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15. ### nebelValued Senior Member

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I used this situation you described, to argue in a fringe/alternate theories thread I started ("alma" look back time), that the past time inside an expanding shell model of the universe is void of all. (if I am permitted to cross comment here, plug a related theme).

PS: When asking the gravity OP question here, I was not thinking of empty shells, of which there are not many around, sadly, because they would be really entertaining,- but thought of naturally occurring entities filled with ever denser matter toward the center. Of course, empty shells would have less gravity on the inside, but apparently solid bodies have too.

Last edited: Mar 17, 2018
16. ### Q-reeusValued Senior Member

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Sorry I can't pass that statement well. The standard FLRW model of our universe assumes an expanding spatially homogeneous matter density. Which is somewhat different to the case of an expanding matter shell 'universe'.
In the expanding spherical shell scenario, and assuming a constant rate of expansion of shell mean radius R, light sources separated by a constant spatial distance will experience a constant relative redshift independent of time - according to GR solution for such an expanding spherical shell. There also there being a linear relation between separation distances and relative redshift.
There is a serious issue within GR relating to the spatial metric within the expanding (or for that matter static) shell - but I won't go into that here.
Pass.

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17. ### Janus58Valued Senior Member

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This is a common misconception for layman about Relativity; That gravitational time dilation is due to the difference "felt" locally by a clock. This is not the case. Time dilation is due to the difference in potential, or as mentioned above the specific energy per unit mass to move the clock between the points where we are considering its tick rate.
Another way to look at this is that the gravitational time dilation as viewed from a far removed point has the same magnitude as the time dilation for velocity alone for an object moving at escape velocity for the position in the gravity field of the clock.

Escape velocity is derived from the setting the sum of the specific kinetic energy and specific gravitational potential energy to 0. For points at or above the surface of mass M it looks like this.

$0 = \frac{v^2}{2}- \frac{GM}{r}$

Where r is the distance from the center of mass M.

Solving for v gives

$v = \sqrt{\frac{2GM}{r}}$

If we substitute this for v in the standard SR equation for time dilation due to relative velocity, we get

$T = \frac{t}{\sqrt{1-\frac{2GM}{rc^2}}}$

which is the standard gravitational time dilation equation.

If you want to find the time dilation for the center of a uniformly dense sphere*, the term for gravitational potential at the center is

$-\frac{3GM}{2R}$

where R is the radius of the sphere.
Solving for escape velocity and substituting in to the SR time dilation equation like we did above gives us a gravitational time dilation at the center of

$T = \frac{t}{\sqrt{1-\frac{3GM}{Rc^2}}}$

At the surface r=R, so the time dilation will be

$T = \frac{t}{\sqrt{1-\frac{2GM}{Rc^2}}}$

Time dilation at the center is more even though gravity at the surface is greater.

*Large celestial object are not going to be of uniform density, as interior pressures cause a rise in density as you move towards the center. This will result in some divergence from this equation. This divergence, however, does not affect the general gist of the argument.

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18. ### Q-reeusValued Senior Member

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I will have to clarify a few statements made earlier. In #111 where I wrote:
"Will depend on the details of the body in question. If a constant density sphere, to a good approximation tidal gravity is zero everywhere below the surface. As it is in the case of a thin shell, despite g being non-zero at finite 0 < r < R (R the surface radius) in the former case, and g being zero for all r < R in the latter case."

That in red was assuming measurements were made within a tiny bubble cavity inside the sphere. Strictly though, tidal gravity for a truly solid sphere of constant density reverses sign below the surface, and is non-zero.

In #113, where I wrote: "In the expanding spherical shell scenario, and assuming a constant rate of expansion of shell mean radius R,..."

that should have read "constant rate of accelerated expansion of shell...."
Which is unrealistic as it implies some kind of 'dark energy' filling the shell interior, voiding the initial assumption of an empty interior. Realistically, there will be decelerated expansion owing to mutual gravitational attraction between all parts of the shell. Then interior redshift steadily decreases with time, but is still directly proportional to assumed constant separation distances between sources and receivers. Expansion (or contraction) of the shell does not result in interior spatial separations being 'dragged along' with the shell - unless of course there is direct mechanical contact. Unlike in our universe where at large scales, Hubble flow results in steadily increasing separations.

19. ### RainbowSingularityValued Senior Member

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Gravity is a Radiant energy ?

20. ### nebelValued Senior Member

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Gravity as a tension of spacetime, or as a field, falls off into the infinite distance from the surface of a body with the inverse square ratio. t shares that feature with radiant energy. Gravity is not energy, can be a source of potential energy though. A planet with low solar radiation falling on it, has also a low gravitational attraction from the sun, hence a lower orbital velocity to counter/balance that. or?

21. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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What is that supposed to mean?

22. ### nebelValued Senior Member

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Nebel:A planet with low solar radiation falling on it, has also a low gravitational attraction from the sun, hence a lower orbital velocity to counter/balance that. or?

Origin:What is that supposed to mean?

Origin: It was just meant to highlight the similarity in strength between gravitational pull and available radiation from a far away source, both governed by the inverse square laws. so,
Both gravity and sun light would be only ~ 1/1500 of our's, at the median 9th planet orbital distance. . (showing my age and giving Pluto's confirmed status as the outermost Bode/ Titius outpost).

23. ### DaveC426913Valued Senior Member

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You are confused.

While it is true that solar flux falls off at the same rate over distance as gravity, that has nothing to do with the brightness of the star.

A star that puts 15,000 times more light than our sun would still - after applying your 1/1500th factor - bathe its Earth in 10x as much solar radiation.