# GR vs. Newton's calculation

1. Nov 7, 2006

### Sheyr

Can someone help me to calculate the difference between accelerations due to Newton’s law of gravitation and Einstein’s general relativity?

Let’s assume the gravitating mass M equals to 1kg and the distance r equals to 1m than the acceleration due to Newton is:

a = G * M / r^2 = 6,6742 m/s^2

What is the acceleration, in such a case, calculated using math of GR?

2. Nov 7, 2006

### lalbatros

I can calculate the motion in any coordinate system, from the invariant least action principle. This lead to differential equations in the choosen coordinate system and this is clear for me.

You may help me to communicate if you could give me the definition of the acceleration.

Thanks,

Michel

3. Nov 7, 2006

### Ich

It is
a = G*M/r² * 1/sqrt(1-2*G*M/(c²*r))

4. Nov 7, 2006

### lightarrow

You mean:

a = G * M / r^2 = 6,6742*10^-11 m/s^2

5. Nov 8, 2006

### Sheyr

Yes, I do :) Thank you Lightarrow

6. Nov 8, 2006

### pervect

Staff Emeritus
For specificity we should add that that's the acceleration one will measure with a local accelerometer. Since clocks tick at different rates due to gravitational time dilation, and rulers also experience gravitational effects, its important to specify how the acceleration is measured. So this is the answer that one would get from a local acclerometer (as opposed to one at infinity, for instance).

7. Nov 19, 2006

### Chris Hillman

Gravitational acceleration according to Newton versus Einstein?

Hi, Sheyr,

Since the setting for Newtonian gravitation (flat euclidean space with universal time) is quite different from the setting (curved Lorentzian manifold) for gtr, it should be easy to appreciate why it is not as easy to directly compare the predictions as you might wish.

One general principle we can state informally is this: in gtr, all fields carry energy, and all forms of mass-energy gravitate, so we should expect that the gravitational field itself gravitates, and that the central attraction of an isolated massive object should therefore be stronger in gtr than in Newtonian gravitation.

Here is a specific way to see that in a specific sense this expectation is fulfilled. Consider the Schwarzschild vacuum solution of the EFE, which models a spherically symmetric gravitational field, produced by some non-rotating isolated massive object, e.g. a simple model of a (nonrotating) star, or a (nonrotating) black hole. Consider the acceleration required for an observer in a spacecraft who is using his rocket engine to "hover" motionless over the object at some location r=r0.

Here, r is the Schwarzschild radial coordinate, which is closely analogous to radial coordinate of flat spacetime (in a polar spherical coordinate chart) in some ways (surface area of nested spheres of radius r=r0, expansion scalar of radially outgoing null congruence) and not analogous in others (Schwarzschild radial coordinate differences r1-r2 where r1 > r2 do NOT correspond to "radial distances" in the Schwarzschild geometry).

Nonetheless we can compute the magnitude of this acceleration (which is purely radial and OUTWARD pointing, as you should expect since, intuitively, it is opposing the INWARD attraction of the massive object which is the source of the ambient gravitational field) and compare this value with our naive Newtonian expectation. The answer turns out to be (in relativistic units in which G = c= 1)

m/r0^2/sqrt(1-2m/r0)

which is larger than the Newtonian value m/r0^2.

On the other hand, if we consider an observer in a rocket ship who fires his rocket engine radially inward with just the right thrust to produce an acceleration of m/r0^2, at each r=r0, then he should slowly fall toward the massive central object. This is indeed the case, in fact the tangent vector to his world line, Y, can be written (in the Schwarzschild chart)

Y = (1-m/r)/(1-2m/r) @/@t - m/r @/@r

But again, remember that our radial coordinate is not perfectly identifiable with the radial coordinate used in a polar spherical chart in Newtonian graviation, so you need to be cautious in interpreting such computations. In particular, many students are misled by incautiously interpreting a similar computation for the Reissner-Nordstrom electrovacuum solution. I'll leave that computation and its interpretation as an exercise.

Chris Hillman