New gravity law

[Timo Moilanen]

 

 

That doesn't make much sense. Please show your working in detail.

If you dont understand this short calc. I'm afraid I can't help from here

 

 

If your model produces only incorrect results, then the model is wrong.

 

That is nothing to do with this website or me.

 

Why would anyone use an equation that gives the wrong results in 100% of cases?

EXACTLY what was the right answers again and how is the right way of finding them

If you dont understand this short calc. I'm afraid I can't help from here

EXACTLY what was the right answers again and how is the right way of finding them

All measurements and most predictions from them will come out similar or very near , except ALL stellar MASSES change RADICALLY . So there really is only one minor adjustment

[swansont]

 

All measurements and most predictions from them will come out similar or very near , except ALL stellar MASSES change RADICALLY . So there really is only one minor adjustment

 

 

If it's so minor, why haven't you done it?

 

But what will turn out to be the case is that when you make one adjustment, something else will fail. You get the earth-sun part correct, and the earth-moon calculation will fail, or the other planetary orbits will fail. (The other possibility is that everything reduces to Newtonian gravity, but that's unlikely)

[Sensei]

You need to insert r=1 before reducing r away obviously F=-GmM/(4Rr^2)*{ 2R-(r^2-R^2)*[a.s.o.]} => F=-GmM/4R*{2R-0*[ ] } => F=-GmM /2

To get 0 from (r^2-R^2), when r=1, you need to have R=1 (or -1).

But if you see what you called a.s.o.

there in denominator:

(r+R)(r-R)

 

If r=1 and R=1, then r=R, and denominator:

(1+1)(1-1)

equals 0.

And you will end up dividing by 0!

[Strange]

EXACTLY what was the right answers again and how is the right way of finding them

 

 

You can look up the height of a geostationary orbit (and how it is calculated): https://en.wikipedia.org/wiki/Geostationary_orbit

 

Note that (1) it is calculated using Newton's law of gravitation and (2) we have launched geostationary satellites and so we know that calculation is correct. we also, therefore, know that your equation is wrong. You can't keep denying this.

 

 

 

All measurements and most predictions from them will come out similar or very near , except ALL stellar MASSES change RADICALLY . So there really is only one minor adjustment

 

Please provide some evidence for this. So far the only evidence you have provided proves you wrong.

[Timo Moilanen]

After sniffing the air here I'm happy to see that my theory holds all arguments so far it's only wrong

Now I have searched ( not very deep ) and did'n find the R1, R2 and r values that are essential for me There's more to dig

To get 0 from (r^2-R^2), when r=1, you need to have R=1 (or -1).

But if you see what you called a.s.o.

there in denominator:

(r+R)(r-R)

 

If r=1 and R=1, then r=R, and denominator:

(1+1)(1-1)

equals 0.

And you will end up dividing by 0!

Yes and before the parenthesis it's multiplied by 0 to be sure and that part of equation is gone ( =0) With R = 0,9999999 it make F=- GmM/ 2,00000001 but its not my equation so?

 

 

You can look up the height of a geostationary orbit (and how it is calculated): https://en.wikipedia.org/wiki/Geostationary_orbit

 

Note that (1) it is calculated using Newton's law of gravitation and (2) we have launched geostationary satellites and so we know that calculation is correct. we also, therefore, know that your equation is wrong. You can't keep denying this.

 

 

Please provide some evidence for this. So far the only evidence you have provided proves you wrong.

The math is straight forward but as I expect you to mean the mass , so that is undone for any theorem

[Strange]

The math is straight forward but as I expect you to mean the mass , so that is undone for any theorem

 

 

I didn't mention mass, so why do you think I meant that.

 

When your results is compared with reality It is WRONG. Therefore your equation is wrong. Why can't you accept that?

Edited January 17, 2017 by Strange

[Mordred]

You know there is a very easy way to show this theory as completely wrong. Start with any scalar field of uniform mass distribution.

 

the use the formula [latex]F=\nabla\theta[/latex]

 

where [latex]\nabla[/latex]is the gradient

[latex]\theta[/latex] some scalar potential doesn't matter what.

 

the reason for the minus sign is to denote a conservative force

 

then from Stokes theorem we have for gravitational and electrostatic forces

 

[latex]V=C\frac{\hat{r}}{r^2}=C\frac{\hat{r}}{r^3}[/latex] to the immediate right of the first equal sign is gravitational the immediate right of the second equal sign is electrostatic.

C is a constant for gravity [latex]C=-Gm_1M_2[/latex] for electrostatic we use the coulombs law of electrostatics. [latex]C=q_1q_2 4\pi\epsilon_o[/latex]

 

V=force

 

taking [latex]F_g=-\frac{Gm_1m_2\hat{r}}{r^2}=\frac{k\hat{r^2}}{r^2}[/latex]

 

starting from infinity [latex]\theta_g r-\theta_G\infty=-\int^\infty_r F_g*dr[/latex]

 

The above requires applications of Stokes theorem I am skipping over the cartesian coordinate expansions. However the simple consequence of these formulas is that the force is radial from a Centre of effective mass. <<<THIS IS OBVIOUS AS PLANETS ARE ROUND>>> due to conservation of energy.

 

Just as water droplets form round shapes. the work done by the coulombs force and conservation of energy itself shows that the force is radially inward.

 

if gravity was not radially outward from a center of mass then stars and planets would not be in the conserved shape they are in.

 

Neither would water droplets which the same fomulas above can be used for.

 

 

Well, it would be very easy for you to confirm this. You can calculate the orbital speed of a planet round the sun and compare it with what we measure.

 

 

Key wording Orbit, if the force was not radial outward from a Center of mass objects would have a different orbit.

 

So lets sit down and list OBSERVATION EVIDENCE YOUR THEORY IS WRONG

 

1) Shape of planets and Stars

2) path of Orbits

3) shape of galaxies

4) shape of water droplets

5) shape of any falling liquid with some cohesive force

6) pendulum tests to show the Earth isn't Flat

 

 

That isn't enough to convince you then you best step up your math because your formulas thus far are garbage.

 

However lets show gravity of a Planet and Star under GR shall we...

 

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
[latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex]
Which we can choose to be at the origin
[latex]\overrightarrow{x}=\overrightarrow{0}[/latex]
Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex]
[latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex]
Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex]
According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element.
[latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex]
Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere
[latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere.
The general relativistic form is.
[latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex]
By comparing the last two equations we can find the static mass distribution in spherical coordinates.
[latex](r,\theta\varphi)[/latex]
[latex]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}[/latex]
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
[latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex]
[latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex]
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)
[latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex]
Provided ds^2 is invariant
[latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation
[latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex]
With the line element invariance
[latex]\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[/latex]
The inverse of the metric tensor transforms as
[latex]\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[/latex]
In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as
[latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=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 [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex]
Specified by a non accelerated motion. Described as
[latex]\frac{d^2\xi^\mu}{ds^2}[/latex]
Where the line element ds=cdt such that
[latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex]
Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation.
[latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex]
This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation.
[latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex]
Where the new line element is given by
[latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex]
and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex]
Denote the metric tensor and the affine Levi-Civita connection respectively

Edited January 18, 2017 by Mordred

[Timo Moilanen]

You know there is a very easy way to show this theory as completely wrong. Start with any scalar field of uniform mass distribution.

 

the use the formula [latex]F=\nabla\theta[/latex]

 

where [latex]\nabla[/latex]is the gradient

[latex]\theta[/latex] some scalar potential doesn't matter what.

 

the reason for the minus sign is to denote a conservative force

 

then from Stokes theorem we have for gravitational and electrostatic forces

 

[latex]V=C\frac{\hat{r}}{r^2}=C\frac{\hat{r}}{r^3}[/latex] to the immediate right of the first equal sign is gravitational the immediate right of the second equal sign is electrostatic.

C is a constant for gravity [latex]C=-Gm_1M_2[/latex] for electrostatic we use the coulombs law of electrostatics. [latex]C=q_1q_2 4\pi\epsilon_o[/latex]

 

V=force

 

taking [latex]F_g=-\frac{Gm_1m_2\hat{r}}{r^2}=\frac{k\hat{r^2}}{r^2}[/latex]

 

starting from infinity [latex]\theta_g r-\theta_G\infty=-\int^\infty_r F_g*dr[/latex]

 

The above requires applications of Stokes theorem I am skipping over the cartesian coordinate expansions. However the simple consequence of these formulas is that the force is radial from a Centre of effective mass. <<<THIS IS OBVIOUS AS PLANETS ARE ROUND>>> due to conservation of energy.

 

Just as water droplets form round shapes. the work done by the coulombs force and conservation of energy itself shows that the force is radially inward.

 

if gravity was not radially outward from a center of mass then stars and planets would not be in the conserved shape they are in.

 

Neither would water droplets which the same fomulas above can be used for.

 

 

Key wording Orbit, if the force was not radial outward from a Center of mass objects would have a different orbit.

 

So lets sit down and list OBSERVATION EVIDENCE YOUR THEORY IS WRONG

 

1) Shape of planets and Stars

2) path of Orbits

3) shape of galaxies

4) shape of water droplets

5) shape of any falling liquid with some cohesive force

6) pendulum tests to show the Earth isn't Flat

 

 

That isn't enough to convince you then you best step up your math because your formulas thus far are garbage.

 

However lets show gravity of a Planet and Star under GR shall we...

 

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

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

Which we can choose to be at the origin

[latex]\overrightarrow{x}=\overrightarrow{0}[/latex]

Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex]

[latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex]

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

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

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

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

[latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere.

The general relativistic form is.

[latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex]

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

[latex](r,\theta\varphi)[/latex]

[latex]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}[/latex]

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

[latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex]

[latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex]

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)

[latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex]

Provided ds^2 is invariant

[latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation

[latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex]

With the line element invariance

[latex]\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[/latex]

The inverse of the metric tensor transforms as

[latex]\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[/latex]

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

[latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=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 [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex]

Specified by a non accelerated motion. Described as

[latex]\frac{d^2\xi^\mu}{ds^2}[/latex]

Where the line element ds=cdt such that

[latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex]

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

[latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex]

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

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

Where the new line element is given by

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

and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex]

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

As you begin (physics course) " for any uniform massdistribution " and this is very true for any "force" that is exactly proportional to d M/r^2 , and we probably can assume I do not think gravity is one of them , but should be calculated from cg or whatever midpoint gravity can be called and is situated dr= 2R^2/5r closer to observation height than sph. midpoint so dM/(r-2R^2/5r)^2

and that I'm going to explore further and it give almost impossible to " swallow " different perspective and after all maybe classic mechanics also is right ( Newton , Euler a.s.o)

As you begin (physics course) " for any uniform massdistribution " and this is very true for any "force" that is exactly proportional to d M/r^2 , and we probably can assume I do not think gravity is one of them , but should be calculated from cg or whatever midpoint gravity can be called and is situated dr= 2R^2/5r closer to observation height than sph. midpoint so dM/(r-2R^2/5r)^2

and that I'm going to explore further and it give almost impossible to " swallow " different perspective and after all maybe classic mechanics also is right ( Newton , Euler a.s.o)

And I would be very interested in opinions of my integrals since they don't contradict any "modern belief " and should there for be harder to "declare" wrong .

As you begin (physics course) " for any uniform massdistribution " and this is very true for any "force" that is exactly proportional to d M/r^2 , and we probably can assume I do not think gravity is one of them , but should be calculated from cg or whatever midpoint gravity can be called and is situated dr= 2R^2/5r closer to observation height than sph. midpoint so dM/(r-2R^2/5r)^2

and that I'm going to explore further and it give almost impossible to " swallow " different perspective and after all maybe classic mechanics also is right ( Newton , Euler a.s.o)

And I would be very interested in opinions of my integrals since they don't contradict any "modern belief " and should there for be harder to "declare" wrong .

Newton would have loved the first (two parted) one

[imatfaal]

!

Moderator Note

 

OK - I have had enough.

 

You are just soapboxing and ignoring arguments - you do not get to merely assert your theory's correctness.

 

Last chance to engage (ie provide an argument which starts with agreed foundation and builds to your hypothesis - and NOT a mere claim that your ideas will be shown to be right in the fullness of time) with member's counter-arguments before the thread is locked. Do not respond to this moderation within the thread

 

[Phi for All]

Personally, I think choosing to quote yourself in a discussion is a very good way to continue making the same mistakes, over and over again. Just sayin'.

[Timo Moilanen]

s

Sorry again . I'm too upset coz I published new formula and got banned from that site (Ursa). I wrote wrong and correct now g =Ti*M*(1/r^2 -3R^2/(14r^4)) and Ti =7,42*10^-11 Nm2/kg2 . I know my writing is bad but hardly a reason for banning me from a cite , so I suppose my work is stolen now . yours Timo Moilanen

[hypervalent_iodine]

!

Moderator Note

We don't care about what happened in another forum. We only care that you follow the rules here. On that note, it doesn't seem to me that you've improved on anything following imatfaal's moderator comment. If you aren't willing to facilitate discussion in your own thread, then we have no desire to allow your thread to continue.

I am closing this pending review.