What is the most accurate gravitational force equation?

jaketodd

The Newtonian gravitational force formula is F=G([m1*m2]/D^2)

Using limits, it has the property that as D goes to zero, F goes to infinity.

Did Einstein fix this? If so, what's his equation for gravitational force? Please accommodate me by defining all terms so I can work with it.

Thanks,

Jake

mathman

Even Newton's law has a practical problem as D -> 0, namely the fact that the objects have a non-zero size.

Simple example - two balls of non-zero radius. D is the distance between the centers and has a minimum value of the sum of the radii.

Passionflower

No, point masses fail completely in GR.

Lee Wang

Do point masses actually exist though?

yossell

if point masses don't exist then what's the alleged problem for Newton?

D H

One big problem with Newtonian gravity is that it implicitly assumes an infinite transmission speed for the gravitational force. Another problem is that Newtonian gravity does not quite match reality; in a sense this is an even bigger problem than the first problem. For example, the discrepancy between what Newtonian mechanics said Mercury's should look like and what astronomers saw was an open problem at the end of the 19th century. General relativity solved both of these issues.

jaketodd

So what's Einstein's equation for gravitational force?

Jonathan Scott

In General Relativity, the normal way to describe motion in a gravitational field is not using forces but rather using the fact that free-falling objects follow a geodesic in curved space-time. The fact that space-time is curved means that one cannot use it directly as a coordinate system, but instead one has to describe the shape of space in terms of some other coordinate system (in the same way that to create a flat map of a large area of the earth, one has to assume some convention for projecting the actual area to the map).

If you want to map that back to Newtonian terms, the main difference (for "weak field" situations such as in the solar system) is that space is curved and "shrinks" slightly relative to a typical coordinate system which can be applied around a central mass. This adds a term to the force law which depends on the square of the speed.

In the simple case of a dominant central mass with lighter objects moving around it, such as the solar system, the usual practical convention is to use "isotropic" coordinates, in which the speed of light relative to the coordinate system varies slightly with potential but the scale factor at any point is the same for the x, y and z directions. Basically, rulers shrink by (1-Gm/rc²) and clocks run slow by the same factor, which means that relative to the coordinate system the speed of light is decreased by the square of this factor, approximately (1-2Gm/rc²).

In that case, the force law for a particle of total energy E moving with momentum p = Ev/c² in a Newtonian gravitational field g is as follows:

dp/dt = (E/c²) g (1 + v²/c²)

You can also divide both sides by the energy to get the following:

d(v/c²)/dt = (1/c²) g (1 + v²/c²)

Note that this is expressed in terms of coordinate values, which differ slightly from local values. In particular, c in the above expression is the coordinate speed of light (which depends on potential), not the local standard value.

This is effectively the same as a special relativity extension of the Newtonian force plus an extra v²/c² term due to the curvature (and scale factor) of space. This means that the rate of change of coordinate momentum for something moving at (or near) the speed of light is twice what Newtonian theory alone would predict.

Note that the above rate of change of momentum does not depend on the direction of travel; it applies regardless of whether the test particle is travelling tangentially, radially or anywhere in between. Also, the rate of change of momentum is directed exactly towards the source (at least in this simplified case where only the central source has a significant mass), which means that angular momentum is conserved too.

For something moving downwards at the speed of light c, it may seem odd that its momentum is increasing, but if you look at the momentum expression Ev/c² you will see that for speed of light travel, the momentum is E/c, where E is constant for free fall, but c decreases (slightly) as you get closer to the source. In contrast, the expression for the rate of change of velocity (that is, acceleration) is much more complicated and depends on the direction of travel.

Passionflower

Jonathan, you describe a perhaps possible and certainly tempting view on GR.

However how do you explain things past the event horizon? The speed of light, in this view, would have to become imaginary, or do I misunderstand completely what you are saying?

Jonathan Scott

The semi-Newtonian description applies only to weak gravitational situations (where Gm/rc² is much less than 1).

The same isotropic coordinate system (where the coordinate speed of light remains the same in all directions) can be used in combination with the Schwarzschild solution to describe strong gravity, although it is more conventional to use Schwarzschild coordinates. The physics is unaffected by the choice of coordinate system (in the same way as the layout of the Earth is unaffected by the projection used to map it).

Passionflower

That is of course very true, but the interesting thing that I noticed in your writing is that you gave a physical interpretation, e.g. the variable speed of light.

Again, it is very interesting view but I think problematic past the event horizon of for instance the Schwarzschild metric. I was simply interested if you had a perspective on this.

Jonathan Scott

Any description using the "coordinate speed of light" as a scalar function of potential only works in an isotropic coordinate system. In any other coordinate system, the speed of light varies with direction.

I've never even thought about whether isotropic coordinates can be continued past the event horizon; it seems likely to run into trouble, because "inwards" and "outwards" are physically different, but I don't have time to look at it at the moment. Perhaps you could look up the standard transformation between isotropic and Schwarzschild coordinates and try it out yourself.