# Is any reference frame truly inertial?

I'm aware of the potential duplicate question here. However, that question centres on the Newtonian argument of there being a force, and hence acceleration. However, my issue is with the expansion of the universe.

An inertial reference frame takes one point to be that which is experiencing 'proper' time and so on, with this frame not accelerating. Even with the lack of acceleration due to external forces, I don't see how this is reconciled with the fact that space itself is expanding. The scale factor $a(t)$ gives the relative expansion of the universe, which surely means any reference frame itself is expanding and changing, preventing it from being truly inertial. I'd expect this to be completely negligible on most scales, but am curious anyway. Is this correct, or am I fundamentally misunderstanding something?

• A free falling reference frame is (locally) inertial. – Count Iblis Mar 18 '17 at 22:15
• inertial reference frame is a model. It works OK for some experiments in predicting future and doesn't work for other. It seems to me that you are asking: "does it work in general?", to which, of course, answer is no – aaaaaa Mar 19 '17 at 1:17

## 1 Answer

"Inertial" loses its sense in general relativity. The analysis is much more complex, and "inertial" systems are simply done away with. We have the concept of "locally inertial", but this just means that we can always choose our coordinates such as at a point the metric will be the Minkowski metric, with null first derivatives: this is a general property, and you don't use it to distinguish frames.

What is interesting to do is understand if a moving particle that follows geodesics in the FRW metric. For simplicity, let us take an $1+1$ dimensional space, of metric $$g=-dt^2+a(t)^2dx^2.$$ This is a flat space similar to the FRW one. Let us suppose $a$ to be never $0$, to avoid singularities.

Suppose to have a massive particle that is moving on a geodesic, parametrized by $x^a(\tau)=(t(\tau),x(\tau))$. Furthermore, suppose an affine parametrization, such as $t'(\tau)^2-a(t)^2x'(\tau)^2=1$ for each value of $\tau$. Equations of motion for an affine parametrization can be found from the Lagrangian $$L=\frac12(-t'(\tau)^2+a(t(\tau))^2x'(\tau)^2).$$ To solve this problem, we can use the fact that $x$ is cyclic: motion equations will be $$x'(\tau)=\frac{C}{a(t)^2};\\ t'(\tau)=\sqrt{1+a(t)^2 x'(\tau)^2}=\sqrt{1+\frac{C^2}{a(t(\tau))^2}}.$$ We can choose initial conditions such as $x^a(0)=(0,0)$ and $x'(0)=v t'(0)$, that means $$t'(0)=\gamma, \quad x'(0)=v\gamma,$$ where $\gamma$ is the usual dilation factor, relative to $t$. We are now ready to solve the problem. You can try seeing what happens by using an explicit law for $a(t)$. Without specifying the form of $a$, we can note that, if $a$ is a monotonic function of $t$, the object that we can interpret as velocity $x'/t'$ effectively decreases with time.

This teaches us an important lesson: a particle that begins moving in the FRW universe with some velocity will lose its velocity. As velocity is not commonly used in GR, it is better to talk in terms of energy: a free particle in FRW experiences energy loss due to universe growing. This is the reason for the cosmological redshift, when you apply this method to light particles (that travel on null geodesics). So, to conclude: in this particular frame we can see the energy of the particle getting dissipated, so we can infer that it will be slower. Even if there is no force acting on it, according to GR.

EDIT: as requested, I include an addendum to improve the answer.

The core answer to your question is "There is no notion of inertial frame in GR". The rest of my answer provided an example of a free particle (from the GR point of view) that experiences a change of velocity. This was just to show that the definition "an inertial frame is a frame where free particles move on straight lines" really makes no sense: if by straight lines you intend segments, that's not true. If you intend geodesics, that's true in every frame. Simply put, "inertial frames" do not exist in GR.