# I Force applied by a moving object

1. Mar 6, 2017

### Ibix

[Moderator's note: post moved from previous thread on same topic, and edited to remove content no longer relevant.]

I think you are trying to ask: what is the gravitational force between two objects under certain circumstances in general relativity. GR does not model gravity as a force, so your question does not have an answer.

Last edited by a moderator: Mar 6, 2017
2. Mar 6, 2017

### CBH

Let us consider to objects, m1 and m2 of 0 charge.
Scenario 1:
both objects are at rest, and have a distance r between them so that the gravitation force applied is:
F= Gmm/r^2. Let's call this quantity $[Neutons] Scenario 2: In the frame of reference in which m2 is at rest, m1 is moving in speed V (e.g.: 0.5c) towards m2. What is the gravitational force felt by m2? Let us assume that$ is very low, so that the speed and acceleration of m2 are negligible, at the moment I'm asking about.

Applying (probably wrong) methods of special relativity, we get that the distance should shorten by gamma, and the mass should increase by gamma (will the increase in effective mass effect the force??)
So according to calculations I'm told are wrong new F = gamma^3 * \$

I would love an explanation a first-year physics student can understand, but I'm mostly interested in the answer at this point. I'll take the general relativity course in third-year probably.
I wanna see if I can reach something close to the answer using methods that I know.

Extra question:
Same thing with Xoloumb's law, assuming m1,m2 are of negligible mass, one has charge q and the other charge -q.

3. Mar 6, 2017

### Staff: Mentor

[Moderator's note: post moved from previous thread on same topic, and quote updated to reference this thread.]

The implicit model of gravity you are assuming, where gravity is still a Newtonian force but you make "corrections" due to special relativity, for things like relativistic mass, length contraction, time dilation, etc., does not work. Einstein spent a couple of years finding that out, from 1905 to 1907, before he shifted tracks and started looking for a relativistic theory of gravity based on the equivalence principle (which ended up being GR, but took him until 1915 to figure out completely).

In the correct relativistic theory of gravity, GR, as Ibix has said, gravity is not a force. What's more, the "source" of gravity is not "mass" (particularly not relativistic mass), it is the stress-energy tensor. The SET is relativistically covariant, which means, roughly, that how an object acts as a source of gravity does not depend on your choice of coordinates.

4. Mar 7, 2017

### CBH

Could you please just answer, what is the acceleration of m2 at the moment I described? from the point of view I described? please? a = something (m/s^2). please. just calculate this and tell me the answer.

5. Mar 7, 2017

### Ibix

Zero. The object is in free fall, and one of Einstein's key realisations (the equivalence principle) was that free-falling observers feel no forces so feel no acceleration.

You can try to measure something like the rate of change of distance to the surface of the planet, but this involves defining "distance to the surface of the planet" and there isn't a unique way to do this in curved spacetime. This answer (called coordinate acceleration) is arbitrary.

You can think of the acceleration required to stay at constant altitude. But that doesn't mesh well with a fast-moving infaller.

The basic problem is that you are trying to import flat-space notions of acceleration with respect to a global inertial coordinate system. That has a natural interpretation as "acceleration" without further interpretation. But there is no global inertial coordinate system possible in curved spacetime, so "acceleration due to gravity" is a notion you need to leave at the door?

6. Mar 7, 2017

### CBH

2ndly, the sun is applying a force on me right now.
I could make an experiment and measure my acceleration.
Let us assume only the sun and I exist (and ignore stuff like the fact the sun is spinning), so I'd measure an acceleration of about 0.1 meter/second^2

Now, let us imagine a world in which the sun is moving towards me at a speed of 0.5 c?
How much will I measure my acceleration towards the sun?

I promise that once I'll take gen. rel. course I'll break my head really hard on trying to figure this stuff out myself.
Even though, my current and obviously ignorant belief is that curving space is just one way of describing reality and that it's not necessarily necessary. Maybe I'll find out I'm wrong once I properly study the subject.

Even if this question is hard to answer, the answer is simple. Either you can calculate this or you can't.

Last edited by a moderator: Mar 8, 2017
7. Mar 7, 2017

### Ibix

How? What will you measure?

8. Mar 7, 2017

### Staff: Mentor

Not according to GR. In GR gravity is not a force. That's why objects moving solely under gravity feel no force: because there isn't one.

There are two types of acceleration that you need to carefully distinguish.

What you are talking about is called "coordinate acceleration" in GR, and as its name implies, it only makes sense once you have chosen a system of coordinates. You appear to be choosing a system of coordinates, at least in your initial case, in which the Sun is at rest. In such coordinates, yes, the Earth will have a coordinate acceleration, which can be approximated by the Newtonian calculation you have done.

If you then switch to a system of coordinates in which the Sun is moving at 0.5 c, you have a problem: how to define the transformation between these two systems of coordinates. If there were no gravity present, you could use the Lorentz transformation of SR; but there is gravity present, so SR no longer applies globally. So there is no unique answer to your question because there is no unique way to define coordinates in which the Sun is moving at 0.5c, as there would be in SR.

GR also defines a different kind of acceleration, called "proper acceleration". This is the acceleration that you actually measure with an accelerometer, or feel as weight. The Earth has zero proper acceleration because it is in a free-fall orbit; more generally, any object moving solely under gravity has zero proper acceleration; it is weightless. This is true regardless of what coordinates you choose, and GR focuses on quantities that are invariant in this sense.