Are the components of 4-vectors the physically measured quantities?

I am very confused with the difference between components of four-acceleration and coordinate acceleration. If I was in an inertial frame observing an accelerated object I would say its four-acceleration is

$$a^{\mu} = \Big( \gamma^4 \frac{\mathbf{v} \cdot \mathbf{a}}{c},\gamma^2\mathbf{a}+\gamma^4 \frac{\mathbf{v} \cdot \mathbf{a}}{c^2} \mathbf{v}\Big).$$

I am confused about what the spatial components actually tell me. Do these tell me the physical acceleration that I would measure? Or is $\mathbf{a} = \frac{d^2\mathbf{r}}{dt^2}$ what I would measure?

I came across this confusion while trying to show that the acceleration observed by an intertial observer is $a = a'/\gamma^3$ where $a'$ is the acceleration in the instantaneous rest frame of the accelerated object. This relied on using $\mathbf{a}$ as the physically measured acceleration in each frame.

Similarly, the 4-velocity has the form $U^\mu = (\gamma(v) c, \gamma(v) \mathbf{v})$. Is $\mathbf{v}$ what an inertial observer would measure or is it actually $\gamma(v) \mathbf{v}$?

In general, are the components of four-vectors the physically measured quantities? Or do we 'build' the component of four-vectors from coordinate measurements such as $\mathbf{r}, \frac{d\mathbf{r}}{dt}, \frac{d^2\mathbf{r}}{dt^2}$, etc., which are the physical measurements?

Of course, it depends on how you do the measuring! Suppose there's an accelerating object with a clock attached to it, and you measure the change in its four-velocity at two times.

• If you divide by the time elapsed on your clock, you'll be measuring $du^\mu / dt$.
• If you divide by the time elapsed on the object's clock, you'll be measuring $a^\mu = du^\mu / d\tau$.

The same goes for measuring the four-velocity $u^\mu = dx^\mu / d\tau$. If you instead divide by $dt$, you get the coordinate velocity $dx^\mu / dt$. All of these are perfectly good physical quantities, and which one is important depends on what you're doing.

• If you want to know the path of the object in your frame, you want everything in terms of coordinates. In particular, you don't even want $du^\mu/dt$ since $u^\mu = dx^\mu / d\tau$, you want $d^2 x^\mu /dt^2$.
• If you want to know the force the object is experiencing, you want $a^\mu$ because $F^\mu = m a^\mu$.

Physically, if you have a lab setup which measures acceleration using a ruler and stopwatch, it's measuring coordinate acceleration $d^2x^\mu/dt^2$. On the other hand, if you mount an accelerometer to the object, it will be measuring $a^\mu$ in the object's frame.