# Gravity Question, new idea push or pull?

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Hello,

It has always been said that gravity pull's but does it really, what if it pushes just like water. If dark matter/energy is trying to fill the places in which it cant. Push a ball of in water or just your hand you can feel it yourself.

Then again what is the actual diffrence between pushing and pulling.

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It's not a new idea. The difference between pushing and pulling is in how you set up your equations. Pulling is a much, much simpler system.

For example, if it's a push, what is the source of the push (if it's a pull, in Newtonian terms it's mass), and how does something far, far away "know" to push on a mass if you have changed configuration (say, a newly launched satellite). You run into many such issues that are hard, if not impossible, to address.

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It's not a new idea. The difference between pushing and pulling is in how you set up your equations. Pulling is a much, much simpler system.

For example, if it's a push, what is the source of the push (if it's a pull, in Newtonian terms it's mass), and how does something far, far away "know" to push on a mass if you have changed configuration (say, a newly launched satellite). You run into many such issues that are hard, if not impossible, to address.

If I dig a deep, deep hole in the Earth, and lower myself into it hanging by my hands from a chain 1000 miles long, as my gravitational center gets closer to the Earth's gravitational center, am I going to be shorter because gravity is pushing down harder on me, or am I going to be taller because I'm being pulled by gravity? Will I experience a compression stress, or a tensile stress?

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If I dig a deep, deep hole in the Earth, and lower myself into it hanging by my hands from a chain 1000 miles long, as my gravitational center gets closer to the Earth's gravitational center, am I going to be shorter because gravity is pushing down harder on me, or am I going to be taller because I'm being pulled by gravity? Will I experience a compression stress, or a tensile stress?

Yeah but how can you get taller, also i dont think you get smaller unless you would throw a big rock into the hole youve made.

It's not a new idea. The difference between pushing and pulling is in how you set up your equations. Pulling is a much, much simpler system.For example, if it's a push, what is the source of the push (if it's a pull, in Newtonian terms it's mass), and how does something far, far away "know" to push on a mass if you have changed configuration (say, a newly launched satellite). You run into many such issues that are hard, if not impossible, to address.

Yeah but just like atoms, we have never seen one so what if you cant see the source but only its effects. Also what if we see black holes wrong since mass on earth keeps being created its not like it has stopped or anything so what if the unkown pushing force creates matter, that would realise black holes since you keep pomping things into it but it never filles.

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Yeah but just like atoms, we have never seen one so what if you cant see the source but only its effects. Also what if we see black holes wrong since mass on earth keeps being created its not like it has stopped or anything so what if the unkown pushing force creates matter, that would realise black holes since you keep pomping things into it but it never filles.

Conceptually, There's a problem: what goes in the equation used to predict the push? It's got to be the mass that's inside the sphere in question, because that's how it behaves, but the model has the source outside of the sphere.

You can't say it's unknown, because how do you then predict any values? You can't have mass creation, because that violates conservation of energy.

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Yeah but how can you get taller, also i dont think you get smaller unless you would throw a big rock into the hole youve made.

Because gravity changes the closer you get to the center of the Earth, by the formula 1/R2 (R being your distance from Earth's center).

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If I dig a deep, deep hole in the Earth, and lower myself into it hanging by my hands from a chain 1000 miles long, as my gravitational center gets closer to the Earth's gravitational center, am I going to be shorter because gravity is pushing down harder on me, or am I going to be taller because I'm being pulled by gravity? Will I experience a compression stress, or a tensile stress?

The naive answer (assuming constant density)is that gravity drops linearly (1/r^2 from distance, but you have an r^3 mass dependence*), but owing to the nonuniform density, it's more complicated

One would have a really hard time explaining that as a push from some external source.

*only mass inside the sphere contributes for a uniform distribution

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As an indication of how not new this is:

Le Sage's theory of gravitation is a kinetic theory of gravity originally proposed by Nicolas Fatio de Duillier in 1690 and later by Georges-Louis Le Sage in 1748. The theory proposed a mechanical explanation for Newton's gravitational force in terms of streams of tiny unseen particles (which Le Sage called ultra-mundane corpuscles) impacting all material objects from all directions. According to this model, any two material bodies partially shield each other from the impinging corpuscles, resulting in a net imbalance in the pressure exerted by the impact of corpuscles on the bodies, tending to drive the bodies together.

https://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation

That article also clearly explains several reasons why the idea cannot work.

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It is both, pull in one direction, push in another. Theoretically If the earths gravitational field was to be switched off for one hemisphere of the planet the whole planet would move in the direction of the hemisphere where the gravity has been removed. When the gravity of the planet is active as it is in the environment and constant for all point on a surface vector field there is no motion of the planet as gravity direction of action is opposed equally in every conceivable direction. However in the first stated example where the planet move that which gravity pulls off is pushed away from the planet when in the second example where gravity cancels itself out with respect to motion of the entire planet.

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It is both, pull in one direction, push in another.

No it isn't.

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Could you elaborate in order for me to see if we are opposing in view upon the concept of gravity's action upon objects. I cannot comment upon your current answer.

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Gravity does not push. Simple enough for you?

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A object falls to earth from space, this is most closely linked to the act of pulling a object. If the object could fall through the earth then as it did this the action of gravity upon the object would be closer to that of a pushing motion.

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A object falls to earth from space, this is most closely linked to the act of pulling a object. If the object could fall through the earth then as it did this the action of gravity upon the object would be closer to that of a pushing motion.

Of course it wouldn't. It would fall the rough ocean and the Earth due to the pull of gravity.

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If you could explain exactly why gravity is a pull only phenomenon in a satisfactory and convincing way then you will change my mind.

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If you could explain exactly why gravity is a pull only phenomenon in a satisfactory and convincing way then you will change my mind.

Burden of proof: you are the one making a crackpot claim so it is up to you to provide the the evidence and/or theory to support it.

But, to get the ball rolling: we only have evidence of gravity as an attractive force. Newtown's law of gravitation only describes it as an attractive force. Over to you.

Edited by Strange

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Over to you but not in mainstream physics.

Thats why have a Speculations forum. Do not EVER post answers in mainstream physics based upon your personal models/misconceptions.

The purpose of mainstream sections is to assist Students pass their physics tests. Not to push incorrect answers.

Under GR gravity is not a force, it is the result of spacetime curvature. There is no force involved on freefall under GR.

As such there is no antigravity as this is counter to geodesic motion.

Edited by Mordred

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If you could direct me to a forum where progress is acceptable then that would be much appreciated.

How is one to make progress with a utter reluctance to question that which came before him.

The evidence is obvious for some, the ball would continue in its current direction of motion beyond the increasing field strength into a area of decreasing field strength.

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here is your equations of gravity under GR.

The speculations forum is specifically where you want to approach personal modelling or explore other possibilities under the rules in Speculations. Not the mainstream sections.

Any physics forum will either lock a thread or take moderator action for posting outside the mainstream answers under mainstream physics sections.

In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass

$M=\int_Dd^3x\rho(\overrightarrow{x})$ concentrated at a point $\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})$

Which we can choose to be at the origin

$\overrightarrow{x}=\overrightarrow{0}$

Sources outside region D the following Newton potential at $\overrightarrow{x}$

$\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}$

Where $G_n=6.673*10^{-11}m^3/KG s^2$ and $r\equiv||\overrightarrow{x}||$

According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element.

$ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2$

Where $d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2$ denotes the volume element of a 2d sphere

$\theta\in(0,\pi)$ and $\varphi\in(0,\pi)$ are the two angles fully covering the sphere.

The general relativistic form is.

$ds^2=g_{\mu\nu}(x)dx^\mu x^\nu$

By comparing the last two equations we can find the static mass distribution in spherical coordinates.

$(r,\theta\varphi)$

$G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}$

Now that we have defined our static multi particle field.

Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance.

Ok so now the Principle of Equivalence.

You can google that term for more detail

but in the same format as above

$m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}$

$\overrightarrow{g}-\bigtriangledown\phi_N$

Denotes the gravitational field above.

Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism)

$x\mu\rightarrow\tilde{x}^\mu(x)$

Provided ds^2 is invariant

$ds^2=d\tilde{s}^2$ an infinitesimal coordinate transformation

$d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha$

With the line element invariance

$\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x$

The inverse of the metric tensor transforms as

$\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x$

In GR one introduces the notion of covariant vectors $A_\mu$ and contravariant $A^\mu$ which is related as $A_\mu=G_{\mu\nu} A^\nu$ conversely the inverse is $A^\mu=G^{\mu\nu} A_\nu$ the metric tensor can be defined as

$g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu$ where $\delta^\mu_nu$=diag(1,1,1,1) which denotes the Kronecker delta.

Finally we can start to look at geodesics.

Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories $\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)$

Specified by a non accelerated motion. Described as

$\frac{d^2\xi^\mu}{ds^2}$

Where the line element ds=cdt such that

$ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu$

Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation.

$d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)$

This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation.

$\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0$

Where the new line element is given by

$ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu$ and $g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}$

and $\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}$

Denote the metric tensor and the affine Levi-Civita connection respectively.

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If you could direct me to a forum where progress is acceptable then that would be much appreciated.

How is one to make progress with a utter reluctance to question that which came before him.

The evidence is obvious for some, the ball would continue in its current direction of motion beyond the increasing field strength into a area of decreasing field strength.

Most of the non-serious science sites will let you wildly speculate. We get students trying to pass classes, so your "progress" isn't appreciated. If you think you have some evidence to support your ideas, you can try our Speculations section.

Most people who come here want a rigorous approach to science. You can get what you're looking for just about anywhere.

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If you could direct me to a forum where progress is acceptable then that would be much appreciated.

How is one to make progress with a utter reluctance to question that which came before him.

The evidence is obvious for some, the ball would continue in its current direction of motion beyond the increasing field strength into a area of decreasing field strength.

This is incorrect to the principle of least action. Which correlates your potential vs kinetic energy. Here is the classical Feyman lecture on the subject.

http://www.feynmanlectures.caltech.edu/II_19.html

I highly recommend you study the Feyman website. The first volume will teach you the proper physics.

Edited by Mordred

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The push is going to be away from the earth into space matter. Explanation requires a deal of intellect.

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The push is not going to be anything.... because it is a pull - and attraction between masses. Seems pretty well covered above.

QUOTE: "Explanation requires a deal of intellect"

Yes - it requires intellect to explain it correctly... I suppose it requires intellect to come up with an idea that is totally wrong too - it shows you are thinking about it.

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!

Moderator Note

stupidnewton's hijack of this discussion to promote an alternative model has been split

http://www.scienceforums.net/topic/101647-push-gravity-hijack-from-gravity-push-or-pull/

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