# Confused about the Equivalence Principle and Inertial Reference Frames

1. Apr 30, 2013

### JPaquim

Hey everyone,

I started reading up on GR a couple of days ago, and I'm somewhat stuck on the concept of a free-falling IRF. I understand that an observer on a free-falling small spaceship would experience the laws of physics in a rather simple form, eliminating the need for a force of gravity in the model, and thus would call it an inertial frame of reference, as in all proper accelerations being zero.

What I don't get is how this can be reconciled with the traditional relation between IRFs, that if you have another reference frame moving with constant translational speed with respect to the first, then the second will be an IRF as well.

Let's imagine a celestial body with an associated uniform gravitational field at sufficiently large distances. Let's also impose that it should not be rotating nor translating around some other object, in order for an observer on its surface to be qualified as an IRF in the traditional sense. If you define a second IRF to be the one associated with a spaceship in free fall, then clearly the observer on the surface cannot be considered to also be an IRF. The two observers' law of physics differ in the inclusion or exclusion of a force associated with gravity. So which one is more fundamental?

I'm sorry if I'm not wording this correctly, but my main problem is that it seems that there are two conflicting definitions of what an IRF should be, based on whether you want to include the force of gravity in your model of physics or not.

Cheers

2. Apr 30, 2013

### Staff: Mentor

In the GR description, neither one is experiencing a force of gravity, but the ground-bound one is not in an inertial reference frame. He's in an accelerated non-inertial frame because he's experiencing a force from the ground under his feet pushing him upwards; he knows this because he's standing on a spring scale that's reading some non-zero value. If the ground under his feet were to suddenly disappear (trapdoor opens?), he would find himself in free-fall, the scale would read zero, and he'd be in an inertial reference frame (at least until his free-fall trajectory intersected the surface again).

3. May 1, 2013

### JPaquim

Ok, so you agree that the IRF of GR does not coincide with the same concept in classical mechanics, right?

4. May 1, 2013

### Staff: Mentor

That is correct. In GR a local IRF is one where an accelerometer at rest reads 0. In Newtonian mechanics an IRF is one where an accelerometer at rest reads -g where g is the local gravitational field as determined by Newton's law of gravitation. They coincide only where g=0 (and then only to the degree that Newton's law of gravitation is a suitable approximation for GR).

5. May 1, 2013

### JPaquim

Ok, I understand now. But why call them both IRF's, if they're clearly different concepts?

6. May 1, 2013

### Staff: Mentor

They both have the feature that the laws of physics take a simple form in the respective inertial reference frames. They also both have the definitive feature that objects not subject to external forces travel in straight lines.

7. May 2, 2013

### A.T.

The concept is the same: In both cases an IRF is a RF where no inertial forces appear, or where the sum of interaction forces determines coordinate acceleration of objects via F=ma.

The difference is that the force of gravity is an interaction force in Newtons model, and an inertial force in GR. This redefinition of gravity has consequences on what is considered an inertial frame.

8. May 2, 2013

### JPaquim

Ok, Thank you very much, that explains why the same designation is used.

Cheers

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