# Is gravity an inertial force?

1. Jul 2, 2010

### dulrich

I've often thought that one way to restate the equivalence principle is that the force of gravity is a manifestation of using a frame that is not in free-fall. In other words, it's an inertial force akin to the Coriolis and centrifugal forces in a rotating frame of reference.

However, I just read the following quote from http://plato.stanford.edu/entries/newton-principia/#NewLawMot
For some reason, I've never seen the third law being invoked to distinguish real and "apparent" forces.

Now, the deflection of objects in a gravitational field is due to the curvature of spacetime, so I should think of it as inertial. But spacetime wouldn't be curved except a second object were doing the curving. It truly is an interaction between the two via the intermediary of the gravitational field. Hence, it is not really inertial in the same sense as the Coriolis force.

So I guess gravity stands alone: it's the only inertial force that is also "real".

Does this make any sense?

2. Jul 2, 2010

### bcrowell

Staff Emeritus
It's an excellent way to make the distinction, in Newtonian mechanics. In Newtonian mechanics, real forces are made by one object acting on another object. If a "force" isn't made by one object on another object, then you can't even define Newton's third law.

A general comment on your OP is that you seem to be mixing statements about GR and Newtonian mechanics, which isn't necessarily valid.

In Newtonian mechanics, gravity is not an inertial force, it's a real force.

In GR, the gravitational field is indeed treated differently from other fields (it's the only one treated geometrically), but force isn't even a really useful concept.

3. Jul 2, 2010

### haushofer

But in Newtonian mechanics I can do the coordinate transformation

$$x^i \rightarrow x^i + \frac{1}{2}a^i t^2$$

where a is a uniform acceleration (and this transfo is ofcourse not an element of the Galilei group). My "geodesic equation" x-dubbeldot is zero becomes

$$\ddot{x}^i + a^i = 0$$

If I now use the equivalence principe (which is perfectly allowed in classical mechanics) I rewrite for the gravitational potential phi

$$\ddot{x}^i + \frac{\partial\phi}{\partial x^i} = 0$$

Can't I regard gravity as being an inertial force in this sense?

4. Jul 2, 2010

### bcrowell

Staff Emeritus
Suppose planets A and B are interacting gravitationally. You can do a transformation like this to set one of the planets' accelerations equal to zero, but you can't make both accelerations vanish, just one of them.

5. Jul 2, 2010

### Passionflower

I suppose I do not see the big dichotomy. Rewrite Newton's laws in a geometric form (Newton-Cartan) and it is "suddenly" force free?

6. Jul 3, 2010

### haushofer

That's true, that's something I haven't realized :)

7. Jul 6, 2010

### hendriko373

The difference between a real gravitational force and a fictious inertial force is that the Riemann tensor does not vanish in a region of space-time when there is a real force, as opposite to a fictious force. The reason is that when there is curvature it is impossible to find a GCT such that the metric tensor is Minkowskian everywhere. When considering a fictious inertial force such a trafo is possible and the observer concludes the force is not a real one. So you should be careful to consider gravitation as an intertial phenomenon. Curvature is an observer independent feature of physical reality, whereas intertial forces are by definition not. (The mathematical reason being that you cannot fix the second order derivative of the metric completely by changing coordinates).

8. Jul 6, 2010

### Ich

The EP could also be a weird coincidence, like in Newonian mechanics. Your statement is the core of GR, where the EP is interpreted that way.
Of course. Gravitational force being fictitious does not mean that there's no gravity.

9. Jul 6, 2010

### physics.alex

sorry pls correct me if I am wrong.

Didn't Einstein say that Gravity is nothing but a geometry curved only? It is NOT a force and we are living in curved space.

isn't it?
Alex

10. Jul 6, 2010

### haushofer

Yes. Due to the equivalence principle the notion of "acceleration due to gravity" is ill-defined.