# 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. Mar 18, 2017 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 Mar 19, 2017 at 1:17

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.