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1. Jul 11, 2016

### Figaro

From the weak EP, inertial mass is equivalent to the gravitational mass. The weak EP implies the universality of gravitation. Some implications of the EP are:
1) Gravity is inescapable - there is no such thing as gravitationally neutral object with respect to which we can measure the acceleration due to gravity so acceleration due to gravity is not something that can be reliably defined.
2) Gravity is not a force, a force is something that leads to acceleration, and the definition of zero acceleration is moving freely in the presence of whatever gravitational field happens to be around.

1) If acceleration due to gravity is not something that can be reliably defined, what about 9.8m/s^2?

2) I know that gravity is not a force but how do we resolve the usual way of calculating the "force due to gravity" and "acceleration due to gravity" as in some general physics books? I know that these are due to the equivalence of inertial mass and gravitational mass but can someone expound on it?

Reference: Spacetime and Geometry by Carroll

2. Jul 11, 2016

### vanhees71

In the (1+3)-formulation of the motion of particles in a gravitational field you have clearly the usual notion of "force", "acceleration", etc., and there's also a non-relativistic limit, valid for weak gravitational fields and slow motion (speeds much less the the speed of light). In the non-relativistic limit you are back at Newton's law, and the value $g=9.81 \; \mathrm{m/s}$ is determined by Newton's Law,
$$g=\frac{G M_{\text{Earth}}}{r_{\text{Earth}}^2}.$$
There's nothing mysterious about this. As any good theory also GR provides the correct limit to the successful predecessor theory of gravity, i.e., Newton's gravitational law.

In my opinion, it is misleading to teach students that "gravity is no force". Of course the motion of particles, when described in 4D space time is determined by the geodesics of space-time, but the emphasis is that it is on space-time not space! Space is, of course, an observer (i.e., frame) dependent concept as is the concept of force, acceleration, etc. However, measurements and observations are made by observers (a tautology in fact!) and thus require the introuction of a reference frame.

3. Jul 11, 2016

### A.T.

Gravity doesn’t cause proper acceleration (that an accelerometer measures), but one aspect of it can be quantified as coordinate acceleration (dv/dt, e.g. 9.8m/s^2).

These are different models which give approximately the same results in most everyday applications. Even in GR gravity can locally be interpreted as an "inertial froce", or more geometrically like in the animation below.

4. Jul 11, 2016

### Staff: Mentor

That number tells us how the distance between the surface of the earth and a nearby object changes. If we were in a sealed windowless box we would be unable to measure it - there would be no problem determining that a dropped object accelerated towards the floor at 9.8m/s^2, but no way of knowing how much of that is "gravitational acceleration" caused by the nearby earth and how much is due to the box being accelerated upwards.
Classical physics does treat gravity as a force, and classical physics is still widely used because it works well enough in most problems. You can calculate the behavior of an object near the surface of the earth either by assuming that the earth exerts a gravitational force on the object or that the mass of the earth curves spacetime in such a way that the natural free-falling path of the object will bring it ever closer to the surface of the earth. The second way produces more exact results, but it is a lot more work and we often don't need that umpteenth decimal point of accuracy. There's an essay by Isaac Asimov that's worth reading: http://chem.tufts.edu/answersinscience/relativityofwrong.htm

5. Jul 11, 2016

### MeJennifer

The difference between an object moving by force and by gravity is that the prior undergoes proper acceleration while the latter does not.

6. Jul 11, 2016

### FactChecker

Nice video. Epstein also wrote a book, "Relativity Visualized" that I think is very enjoyable reading.

7. Jul 11, 2016

### Staff: Mentor

You are assuming that introducing a "reference frame" requires introducing a particular split of spacetime into space and time. That is only true if you interpret "reference frame" to mean "coordinate chart"--and then only if you pick a particular kind of coordinate chart, roughly speaking, one that has one timelike and three spacelike coordinates everywhere. (And this choice can be very non-intuitive--for example, in Schwarzschild spacetime, if you are dealing with a black hole, the only such choice known is Kruskal coordinates, whose "space" and "time" don't correspond to anything physically intuitive.)

But there is another interpretation of "reference frame" which actually models measurements and observations better and does not require introducing a particular split of spacetime into space and time, at least not globally. This interpretation takes "reference frame" to mean "frame field"--roughly speaking, an assignment of an orthonormal tetrad to every event in spacetime, subject to some constraints like continuity. The reason this models measurements and observations better than a coordinate chart is that an orthonormal tetrad, by definition, specifies a "unit clock", "unit ruler", and "three perpendicular spatial directions", i.e., it realizes our intuitive sense of a "reference frame" at a given event, whereas a coordinate chart does not.

Such an orthonormal tetrad, at a given event, implies a particular local split of spacetime into space and time at that event--roughly speaking, the split that is "natural" to an observer whose 4-velocity at that event is the timelike vector in the tetrad (but even this is not unique, since there will be multiple such observers with differing proper accelerations). But there is no requirement that all of the local splits must fit together into a single global split, and in fact in many cases they won't, even in flat spacetime (consider, for example, the frame field of Langevin observers).

8. Jul 12, 2016

### vanhees71

Sure, I've tacitly assumed that observers are local observers. A physical theory does not only consist of the mathematical framework but also needs to make contact with observations, and I don't see, how you can make sense of GR if not defining such local reference frames as you describe in the previous posting. Of course, there are infinitely many local reference frames, and the experimental setup (defining the "observer") determines which one is convenient to describe the outcomes.

Concerning the question here, I find it ridiculus to answer it by saying that there are no gravitational forces and no acceleration. It depends on the physical setup, whether this is true. Of course, if you define everything covariant and work with invariant quantities, there's no proper acceleration and in this sense no force, but here the question was specific to how the $g=9.81 \mathrm{m/s}$ has to be interpreted, and for me the clear answer is that it refers to an observer who is at rest relative to the Earth and thus is not in a locally free falling reference frame and thus feels the force of the Earth counterbalancing the gravitational force. An object, initially at rest with respect to such an observer won't stay at rest but fall down to Earth. Thus there's a gravitational force acting on it from the point of view of this observer, and the acceleration is independent of the mass of the object in accordance with the (weak) equivalence principle. It is also clear that you get the value $g=9.81 \mathrm{m/s}$ from GR, e.g., by assuming approximately the Earth to be a spherical object and plug in the corresponding Schwarzschild solution to calculate the gravitational force with respect to the observer at rest relative to the Earth. Since in this case the non-relativistic limit for sure is valid, you are lead back to Newton's law of gravity and get the said value of the acceleration.

It's clear that one must be careful when interpreting frame dependent quantities, and physically it's always save to work with covariant quantities. Nevertheless one should not say that GR is counterintuitively claiming that gravity is no force.