Lorentz transformation of the Gravitational constant

[Dilema]

Since force is transformed via: F'_x= F_x ; F'_y= F_y/ ϒ; F'_z=F_z/ ϒ

(F' is the force related to the moving frame, F is the force on the rest frame and ϒ=1/√1-v²/c² ).I expect that G (Gravitational constant) will be transformed between moving and rest frame in order to satisfy force transformation since

F=-Gm₁m₂/r²Can you please refer me to a paper or textbook on that specific topic?It is posable to see that substituting the mass Lorentz transformation m=moϒ and Lorentz-FitzGerald's contraction - r_o/ϒ without applying a transformation on G, do not settle the force to get the force transformation like. Instead it gets F'=-Gm₁m₂ ϒ³/r². Namely F'=F ϒ³.Can you please refer me to a paper or textbook on that specific topic?

 

[Dale]

Newtonian gravity is not compatible with relativity.

 

[Ibix]

As Dale says, you can't tweak Newtonian gravity to make it work in relativity. The instantaneous action at a distance implicit in Newton's theory violates the "nothing travels faster than light" rule of relativity.

If you want to understand relativistic gravity you need to learn general relativity. Sean Carroll's lecture notes are a good free online place to start.

 

[PeterDonis]

force is transformed

In relativity, gravity is not a force, so this transformation law doesn't apply to gravity.

 

[Orodruin]

In relativity, gravity is not a force, so this transformation law doesn't apply to gravity.

He could however have made the same question about Coulomb's constant. The answer being that Coulomb's law is not consistent with relativity except for in the case of a stationary field source.

 

[PeterDonis]

He could however have made the same question about Coulomb's constant.

Yes, fair point.

 

[Orodruin]

Yes, fair point.

This makes me think of the alt-text of this XKCD: https://xkcd.com/1489/

 

[Dilema]

As Dale says, you can't tweak Newtonian gravity to make it work in relativity. The instantaneous action at a distance implicit in Newton's theory violates the "nothing travels faster than light" rule of relativity.

If you want to understand relativistic gravity you need to learn general relativity. Sean Carroll's lecture notes are a good free online place to start.

And yet the escape velocity from a black hole is given by the application of Newton law c=(GM/r)^0.5. r is Schwarzschild radius. Newton's law is good enough for small mass. Apparently, back hols turn to be small mass compared to galaxies.

 

[Ibix]

And yet the escape velocity from a black hole is given by the application of Newton law c=(GM/r)^0.5. r is Schwarzschild radius. Newton's law is good enough for small mass. Apparently, back hols turn to be small mass compared to galaxies.

The formula is the same, but $r$ is not the same. And escape velocity from the "surface" of a black hole is a meaningless calculation since you can't escape - that alone should tell you that you are not doing things right.

 

[Orodruin]

And yet the escape velocity from a black hole is given by the application of Newton law c=(GM/r)^0.5. r is Schwarzschild radius. Newton's law is good enough for small mass. Apparently, back hols turn to be small mass compared to galaxies.

This is a common misconception. Just because the numerical value of where you would have Newtonian escape velocity equal to c happens to be the same as the Schwarzschild radius does not hold any deeper meaning. The concept of escape velocity is meaningless here outside of overpopularized treatments.

 

[Dale]

And yet the escape velocity from a black hole is given by the application of Newton law c=(GM/r)^0.5. r is Schwarzschild radius. Newton's law is good enough for small mass. Apparently, back hols turn to be small mass compared to galaxies.

As others have said, this doesn't mean what you think it means. But more importantly, Newtonian gravity is more than just the escape velocity formula. You still "can't tweak Newtonian gravity to make it work in relativity" as @Ibix said.

 

[Ibix]

Just because the numerical value of where you would have Newtonian escape velocity equal to c happens to be the same as the Schwarzschild radius does not hold any deeper meaning.

In fact, if you construct a velocity $v$ from $G$, a mass $M$, and a distance $r$ then it's a trivial consequence of dimensional analysis that $v^2\propto GM/r$ with a dimensionless constant of proportionality. That the constant happens to be 2 in both cases is a coincidence, but not an especially large one.