Why two mass attracts each other?

[hansda]

You explained nothing. You said that the attraction of the other planets on mercury is repulsive from the sun pov so you conclude that the attractipn from the sun is repulsive.

It is explained with the links, i provided there in post #717.

Attraction from the Sun is not repulsive. The effect of "additional precession on planet Mercury" due to the Sun is repulsive from the Sun point of view.

 

[Billy T]

inanimate love.

 

[1100f]

It is explained with the links, i provided there in post #717.

Attraction from the Sun is not repulsive. The effect of "additional precession on planet Mercury" due to the Sun is repulsive from the Sun point of view.

Look at the link that YOU provided. The precession is due to a force that is toward the sun. This is an attractive force. Look at the extra potential that CAUSES the precession. It is ATTRACTIVE.
Read the link that you provide.

 

[rpenner]

hansda -- your claim is both wrong and stupid.

If $$f(x) = \cosh (x) \, + \, 7 \, \cos (x)$$ then it can be decomposed into $$g(x) = 4 \, \cosh (x) \, + \, 4 \, \cos (x) + 7$$ and $$h(x) = -3 \, \cosh (x) \, + \, 3 \, \cos (x) - 7 $$. Now for all real x, $$h(x) \lt 0 \lt f(x) \lt g(x)$$ but it is stupid to say the ability to decompose a function into positive and non-positive components has anything to do with answering the question of whether the function itself is always positive.

http://www.wolframalpha.com/input/?...osh(x)+++7+cos(x)+,+4+cosh(x)+++4+cos(x)+++7}

Your claim that gravity should be split into Newtonian parts and non-Newtonian parts to point out a property that the non-Newtonian parts have is stupid when the phenomenon of gravity is the sum of both of them, and the sum is obviously attractive because Mercury orbits about the Sun. At no point is Mercury's net acceleration away from the sun. This is blindingly obvious.

Secondly, you are wrong not only in your overall claim, but also in the details when you claim the non-Newtonian parts are repulsive.

For a body in orbit with angular momentum L the Newtonian radial potential is:
$$V(r) = -\frac{GMm}{r} + \frac{ (M+m) L^2 }{ 2 M m r^2 }$$
which has both the attraction of gravity and the centrifugal force you get when looking at just the radial contribution, while the general relativity bit is approximately:
$$V(r) = -\frac{GMm}{r} \left[ 1 + \frac{ (M+m)^2 L^2 }{ c^2 M^2 m^2 r^2 } + \dots \right] + \frac{ (M+m) L^2 }{ 2 M m r^2 } $$
which basically says gravity gets stronger when the orbital speed of the of the planet is higher. So the precession of Mercury is not because gravity fails to be an exact 1/r^2 force by being weaker than Newton, but by being stronger than Newton. Either would lead to precession, but the details of the direction of precession matter to your claim.

 

[brucep]

hansda -- your claim is both wrong and stupid.

If $$f(x) = \cosh (x) \, + \, 7 \, \cos (x)$$ then it can be decomposed into $$g(x) = 4 \, \cosh (x) \, + \, 4 \, \cos (x) + 7$$ and $$h(x) = -3 \, \cosh (x) \, + \, 3 \, \cos (x) - 7 $$. Now for all real x, $$h(x) \lt 0 \lt f(x) \lt g(x)$$ but it is stupid to say the ability to decompose a function into positive and non-positive components has anything to do with answering the question of whether the function itself is always positive.

http://www.wolframalpha.com/input/?...osh(x)+++7+cos(x)+,+4+cosh(x)+++4+cos(x)+++7}

Your claim that gravity should be split into Newtonian parts and non-Newtonian parts to point out a property that the non-Newtonian parts have is stupid when the phenomenon of gravity is the sum of both of them, and the sum is obviously attractive because Mercury orbits about the Sun. At no point is Mercury's net acceleration away from the sun. This is blindingly obvious.

Secondly, you are wrong not only in your overall claim, but also in the details when you claim the non-Newtonian parts are repulsive.

For a body in orbit with angular momentum L the Newtonian radial potential is:
$$V(r) = -\frac{GMm}{r} + \frac{ (M+m) L^2 }{ 2 M m r^2 }$$
which has both the attraction of gravity and the centrifugal force you get when looking at just the radial contribution, while the general relativity bit is approximately:
$$V(r) = -\frac{GMm}{r} \left[ 1 + \frac{ (M+m)^2 L^2 }{ c^2 M^2 m^2 r^2 } + \dots \right] + \frac{ (M+m) L^2 }{ 2 M m r^2 } $$
which basically says gravity gets stronger when the orbital speed of the of the planet is higher. So the precession of Mercury is not because gravity fails to be an exact 1/r^2 force by being weaker than Newton, but by being stronger than Newton. Either would lead to precession, but the details of the direction of precession matter to your claim.

Einstein orbits naturally precess and Newtonian orbits don't naturally precess. The rate of radial oscillation and the rate of angular velocity for Newtonian orbits are both M/r^3. No natural precession. They're different values for GR and don't have 1/1 correlation that Newton predicts. All Einstein orbits precess. No forces are counter balanced just the natural path of the object through spacetime. The Kerr metric describes how to calculate the total mass of a rotating object. The mass does increase because the total angular momentum is summed to all mass, associated with non rotation, as an equivalent. IE if an object spins to the limit c then it's total mass is twice what it was at rest. The contribution from spin is linear. My question is why do you think we seem to frame stuff in terms of Newton so much? Probably because in the more difficult calculations it becomes easier when using something like the 'Parameterized post-Newtonian formalism'. Seems less complicated when using the metric of GR for the easy stuff like discovery of natural paths. I'm probably the only person who studied GR before Newton. My initial involvement in SR was working on problems associated with the study of GR in a GR textbook. Doing weak field approximations. I have a different perspective. You do a great job revealing the other perspective. I think it's important to understand what GR predicts. It doesn't predict any forces being counterbalanced over an orbit.

 

[Tach]

Einstein orbits naturally precess and Newtonian orbits don't naturally precess.

It isn't clear what you mean by "Newtonian orbits don't naturally precess" but it is a fact that , in his works , Newton studied the orbits precession .

 

[rpenner]

brucep is talking about the Two Body problem in Universal Gravitation.

 

[Tach]

brucep is talking about the Two Body problem in Universal Gravitation.

This is precisely what Newton studied, the orbital precession in the case of a 2-body setup.

 

[Markus Hanke]

I think you haven't yet got the point. See my earlier post #717. I explained there as to why "perihelion precession of Mercury" should be considered as "repulsive" from "the Sun" point of view.

Having studied both GR and the perihelion precession calculations in some detail, I think I got the point pretty well. The additional precession ( compared to Newton ) is a result of the geometry of Mercury's orbit, plain and simple.

 

[brucep]

It isn't clear what you mean by "Newtonian orbits don't naturally precess" but it is a fact that , in his works , Newton studied the orbits precession .

It means for two bodies with no external perturbations Newtonian orbits will have 0 precession. Einstein orbits always naturally precess. This is what GR predicts.

Derive the natural precession rate of Einstein orbits. All Einstein orbits naturally precess.

Start with the Schwarzschild metric, in geometric units, setting theta at 0.

dTau^2 = (1-2M/r)dt^2 - dr^2/(1-2M/r) - r^2(dphi)^2

Substituting constants of geodesic motion E/m and L/m for dt and dphi

dt = [(E/m)/(1-2M/r)]dTau

dphi = [(L/m)/r^2]dTau

The solution relates squared values for radial motion (dr/dTau)^2, energy per unit mass (E/m)^2, and the effective potential per unit mass
(V/m)^2 = (1-2M/r)(1+[(L/m)^2/r^2]).

(dr/dTau)^2 = +/- (E/m)^2 - (1-2M/r)(1+[(L/m)^2/r^2])

Taking some license for the weak field and multiplying through by 1/2 after multiplying out the squared effective potential

1/2(dr/dTau)^2 = 1/2(E/m)^2 - [1/2 - M/r + (L/m)^2/2r^2 - M(L/m)^2/r^3]

setting (V/m)^2 = U/m

U/m = 1/2 - M/r + (L/m)^2/2r^2 - M(L/m)^2/r^3

1st derivative

d(U/m)/dr = M/r^2 - (L/m)^2/r^3 + 3M(L/m)^2/r^4

2nd derivative d'2(U/m)/dr'2 = rate of radial oscillation = w^2_r

w^2_r = M(r-6M)/r^3(r-3M)

Without writing down details the rate of angular velocity becomes

w^2_phi ~ (dphi/dTau)^2 = M/r^2(r-3M)


The difference.

w^2_phi - w^2_r = 6M^2/r^3(r-3M)

We can find a factor * M/r^3 which closely approximates
6M^2/r^3(r-3M)

That factor is 6M/r

(6M/r)(M/r^3) = 6M^2/r^4

The last step is further weak field approximation

(6M/r)^1/2 ~ 1/2(6M/r) = 3M/r

So a very close approximation for the rate of orbital precession, in the weak field is 3M/r. You can plug in numbers and get an answer that matches observation.

3M_Sun = 4431m
r_mean Mercury = 5.8x10^10 meters
415.1539069 times Mercury orbits the Sun in 100 Earth years
360 degrees per year
3600 arcseconds per degree
etc...

 

[ash64449]

You are correct that it would expand even without cosmological constant; however, what we observe is that it expands at an accelerating rate, and that necessitates the presence of the constant.

OK. Confused on what exactly does cosmological constant. I thought Einstein used cosmological constant to make the universe appear static. But you say that the presence of constant makes the universe accelerate it's expansion?? So what is happening here???

Ok. if cosmological constant can make universe accelerate,does that mean dark energy has some relationship between it?

 

[ash64449]

Gravity is mostly attractive but it can be repulsive also.

Consider that i answered why gravity was mainly attractive. I am not going to be involved in that discussion that gravity is repulsive.

 

[Markus Hanke]

OK. Confused on what exactly does cosmological constant. I thought Einstein used cosmological constant to make the universe appear static. But you say that the presence of constant makes the universe accelerate it's expansion?? So what is happening here???

Ok. if cosmological constant can make universe accelerate,does that mean dark energy has some relationship between it?

The situation is as follows - in the absence of a cosmological constant, we get a universe which expands at a constant rate. Einstein wasn't comfortable with that, so he added a negative cosmological constant into his EFEs, which has the effect of counteracting the expansion. Its numerical value was chosen such that the universe overall becomes stationary.
Nowadays, we know from observation that the universe appears to be expanding at an accelerating rate - to achieve this, a positive cosmological constant is needed in the EFEs. Currently its value stands somewhere at the order of 10^-122 ( in dimensionless Planck units ), so it is very small and thus has no measurable local effects.

 

[ash64449]

The situation is as follows - in the absence of a cosmological constant, we get a universe which expands at a constant rate. Einstein wasn't comfortable with that, so he added a negative cosmological constant into his EFEs, which has the effect of counteracting the expansion. Its numerical value was chosen such that the universe overall becomes stationary.
Nowadays, we know from observation that the universe appears to be expanding at an accelerating rate - to achieve this, a positive cosmological constant is needed in the EFEs. Currently its value stands somewhere at the order of 10^-122 ( in dimensionless Planck units ), so it is very small and thus has no measurable local effects.

So why dark energy to accelerate expansion when we got cosmological constant?

 

[eram]

So why dark energy to accelerate expansion when we got cosmological constant?

In a hand-wavy fashion, we could say that the C.C. represents dark energy.

 

[Tach]

It means for two bodies with no external perturbations Newtonian orbits will have 0 precession.

This is incorrect, look here.

 

[eram]

It means for two bodies with no external perturbations Newtonian orbits will have 0 precession. Einstein orbits always naturally precess. This is what GR predicts.

Derive the natural precession rate of Einstein orbits. All Einstein orbits naturally precess.

Start with the Schwarzschild metric ..... per year
3600 arcseconds per degree
etc...

Okay, I've put all that stuff in TeX.

Derive the natural precession rate of Einstein orbits. All Einstein orbits naturally precess.

Start with the Schwarzschild metric, in geometric units, setting $$\theta$$ at $$0$$.

$${d\tau}^2=\left(1-\frac{2M}{r}\right){dt}^2-\frac{dr^2}{1-\frac{2M}{r}}-r^2{d\phi}^2 $$


Substituting constants of geodesic motion: $$\dfrac{E}{m}$$ and $$\dfrac{L}{m}$$ for $$dt$$ and $$d\phi$$

$$dt=\left(\frac{\frac{E}{m}}{1-\frac{2M}{r}}\right)d\tau$$

$$d\phi =\left( \frac{\frac{L}{m}}{r^2}\right)d\tau$$


The solution relates squared values for radial motion: $${\left(\frac{dr}{d\tau}\right)}^2$$, energy per unit mass: $${\left(\frac{E}{m}\right)}^2$$, and the effective potential per unit mass: $${\left(\frac{V}{m}\right)}^2=\left(1-\frac{2M}{r}\right)\left[1+\frac{{\left(\frac{L}{m}\right)}^2}{r^2}\right]$$.

$${\left(\frac{dr}{d\tau}\right)}^2=\pm{\left(\frac{E}{m}\right)}^2-\left(1-\frac{2M}{r}\right)\left[1+\frac{{\left(\frac{L}{m}\right)}^2}{r^2}\right]$$


Taking some license for the weak field and multiplying through by $$\frac{1}{2}$$ after multiplying out the squared effective potential:
$$\frac{1}{2}{\left(\frac{dr}{d\tau}\right)}^2=\frac{1}{2}{\left(\frac{E}{m}\right)}^2-\left[\frac{1}{2}-\frac{M}{r}+\frac{{\left(\frac{L}{m}\right)}^{2}}{2r^2}-\frac{{\left(\frac{L}{m}\right)}^{2}}{r^3}M\right]$$

Setting $${\left(\frac{V}{m}\right)}^2=\frac{U}{m}$$

$$\frac{U}{m}=\frac{1}{2}-\frac{M}{r}+\frac{{\left(\frac{L}{m}\right)}^{2}}{2r^2}-\frac{{\left(\frac{L}{m}\right)}^{2}}{r^3}M$$



1st derivative:
$$\frac{d}{dr}\left(\frac{U}{m}\right)=\frac{M}{r^2}-\frac{{\left(\frac{L}{m}\right)}^{2}}{r^3}+3M\frac{{\left(\frac{L}{m}\right)}^{2}}{r^4}$$


2nd derivative:

$$\frac{d^2}{dr^2}\left(\frac{U}{m}\right)=$$ rate of radial oscillation $$={\omega_r}^2$$

$${\omega_r}^2=\frac{M(r-6M)}{r^3(r-3M)}$$



Without writing down details the rate of angular velocity becomes:
$${\omega_\phi}^2$$~$${\left(\frac{d\phi}{d\tau}\right)}^2=\frac{M}{r^2(r-3M)}$$


The difference:
$${\omega_\phi}^2-{\omega_r}^2=\frac{6M^2}{r^3(r-3M)}$$


We can find a factor which closely approximates $$\frac{6M^2}{r^3(r-3M)}$$ when multiplied by $$\frac{M}{r^3}$$.

That factor is $$\frac{6M}{r}$$.

$$\left(\frac{6M}{r}\right)\left(\frac{M}{r^3}\right) = \frac{6M^2}{r^4}$$


The last step is further weak field approximation:

$$\sqrt{\frac{6M}{r}}$$~$$\frac{1}{2}\left(\frac{6M}{r}\right)=\frac{3M}{r}$$

So a very close approximation for the rate of orbital precession in the weak field is $$\frac{3M}{r}$$. You can plug in numbers and get an answer that matches observation.


3M⊙ = 4431 m
r☿ = 5.8 x 10^10 m

Mercury orbits the Sun 415.1539069 times in 100 Earth years
360 degrees per year
3600 arcseconds per degree
etc...

 

[eram]

I'm probably the only person who studied GR before Newton. My initial involvement in SR was working on problems associated with the study of GR in a GR textbook. Doing weak field approximations. I have a different perspective.

Woah. I would think so. I can't imagine why anyone would pick up a GR textbook before Newton. It is insanely simple compared to GR. Just one formula, based on Kepler and Galileo's laws.

You definitely have a different perspective. Then again you've probably had the most non-conventional education.

 

[hansda]

Einstein orbits naturally precess and Newtonian orbits don't naturally precess. The rate of radial oscillation and the rate of angular velocity for Newtonian orbits are both M/r^3. No natural precession. They're different values for GR and don't have 1/1 correlation that Newton predicts. All Einstein orbits precess. No forces are counter balanced just the natural path of the object through spacetime. The Kerr metric describes how to calculate the total mass of a rotating object. The mass does increase because the total angular momentum is summed to all mass, associated with non rotation, as an equivalent. IE if an object spins to the limit c then it's total mass is twice what it was at rest. The contribution from spin is linear. My question is why do you think we seem to frame stuff in terms of Newton so much? Probably because in the more difficult calculations it becomes easier when using something like the 'Parameterized post-Newtonian formalism'. Seems less complicated when using the metric of GR for the easy stuff like discovery of natural paths. I'm probably the only person who studied GR before Newton. My initial involvement in SR was working on problems associated with the study of GR in a GR textbook. Doing weak field approximations. I have a different perspective. You do a great job revealing the other perspective. I think it's important to understand what GR predicts. It doesn't predict any forces being counterbalanced over an orbit.

Good post. Thanks.

 

[hansda]

Look at the link that YOU provided. The precession is due to a force that is toward the sun. This is an attractive force. Look at the extra potential that CAUSES the precession. It is ATTRACTIVE.
Read the link that you provide.

Here the additional 'attractive term' is used for calculating 'additional perihelion advance', as per GR correction to Newtonian calculation. This 'additional term' does not cause 'additional precession'. This 'additional attractive term' is added with the Sun's gravitational force. The Sun's gravitational force decides the orbit of a planet. The Sun's attractive force does not cause any precession.


It is not mentioned in the link that this 'additional term' causes 'additional precession'. May be it is your conclusion.