Accelerating a gravitational field

Question

Newton's equation for gravity calculates gravitational force between two bodies $$F= GMm/r^2$$

My question is: How does the force change as these two bodies accelerate together through the universe and approach light speed?

You might argue there should be no change because both bodies are accelerating together and stay the same relative to each other. Consider a planet and a moonbase. They people on each site will experience the same time dilation according to special relativity. Special relativity allows us to adjust the kinematics equation for an object as it approaches the speed of light. The Lorentz factor concisely describes this

$\gamma=1/\sqrt{1 \ - \ v^2/c^2}$.

So, relativistic mass increases with velocity $M=m\gamma$ The force then increases as gamma squared (because the mass of the planet and the moon both increase) $F= G \frac{Mm}{ (1 \ - \ v^2/c^2) \ r^2}$. If I read that right, it says force goes to infinity as a planetary pair reach lightspeed together.

I found this: http://www.einsteins-theory-of-relativity-4engineers.com/support-files/velocity-effects-on-gravityY.pdf But it seems not exactly relevant as it discusses the measurement of gravity between two masses moving at relativistic speeds to each other. I am talking about the gravity within a solar system that is accelerating and reaching a velocity away from ours approaching light speed.

If gravitational force indeed increases then orbital mechanics of the solar system would be different than ours. Solar systems very far away (which happen to be close to light speed due to the Hubble expansion) would have a much greater gravitational milieu. That means that the planets orbit their sun a lot faster than expected at a given distance from their sun.

Please point out the error in my analysis gently. This discussion assumes the gravitational constant,G, does not change as objects accelerate and reach light speed.

Answer 1

Observers in different reference frames may disagree on what they measure, but the consequences of their observations must be consistent. That is, a moving ladder can't be length contracted in one reference frame to fit through a barn, but too long to fit through that barn in another reference frame. With the proper application of Lorentz transforms, you can always find a base truth that all reference frames will agree on.

Else, there's some reference frame where Earth is contracted down into being incredibly dense and we turn into a black hole. Sure, someone in that reference frame will see Earth as being so incredibly dense, but actually turning into a black hole is something that all reference frames (or more particularly Earth's rest frame) must agree on.

Do you see the problem with your proposal now? Your calculation is flawed because you're calculating a force based on your reference frame, but an observer in a different reference frame will disagree.

Not to mention that relativistic mass simply isn't a very good concept and has been generally abandoned by the physics community since it so easily leads to misunderstandings like this.

That's not even getting into how special relativity just fundamentally doesn't play nicely with Newtonian gravity. That's the fundamental reason why Einstein had to develop general relativity in the first place.

Answer 2

So, relativistic mass increases with velocity $M=m \gamma$ The force then increases as gamma squared (because the mass of the planet and the moon both increase) $F=\frac{GMm}{(1 − v^2/c^2) \ r^2}$ . If I read that right, it says force goes to infinity as a planetary pair reach lightspeed together.

Consider an observer on the surface of the Earth that sees an apple falling a distance of $h'$ in time $t'$ in the y direction with acceleration $a' = h'/t'^2$. An alien passing with a relative velocity of $v$ in the $x$ direction, will say the gravitational acceleration of the apple was $ a = h'/(t' \gamma)^2 = a'/\gamma^2$ due to time dilation. The alien will also consider the gravitational force acting on the apple to be \begin{align} F &= ma \\ &= (m'\gamma) (a'/\gamma^2) \\ &= m'a'/\gamma \\ &= F'/\gamma. \end{align}

This is in agreement with how transverse forces transform in relativity. As you can see there has been no increase in the gravitational force of the Earth from the point of view of the passing alien and no danger of the Earth becoming a black hole. Nothing changes if we consider the alien to be stationary and the Earth to be moving or the other way around. The observer on the surface of the Earth will always consider the acceleration of gravity to be $9.8 \ \mathrm{m/s^2}$ and the passing alien will always say it is $9.8/\gamma^2 \ \mathrm{m/s^2}$.

It turns out that the force parallel to the motion does not transform so there is no increase in gravity in that direction either.

The modern treatment is that we should only use invariant mass and not relativistic mass and this mostly due to the confusion that ensues when we consider relativistic gravitational interactions. However there is some validity in considering $m \gamma$ to be the inertial mass of the object. Inertial mass is a measure of how a mass resists acceleration due a force and is also used for the calculation of momentum in the momentum-energy relationship, where we use $m\gamma v$ for the momentum. When analysing relativistic collisions we use also $m\gamma$ in the calculations. It can also be noted that when I derived the relativistic force from the relativistic acceleration I had to use $m\gamma$. The concept of relativistic mass has its uses, but it should not be confused with gravitational mass increasing with relative velocity.