Wikipedia says:
In relativity theory, proper acceleration[1] is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object.
and says:
In the standard inertial coordinates of special relativity, for unidirectional motion, proper acceleration is the rate of change of proper velocity with respect to coordinate time.
Why coordinate time?
As far as I know the moving observer itself measures the proper things. It's clock measures proper time, it's speedometer measures proper velocity.
So if the observer accelerates with $10m/s^2$ (and feels that proper acceleration), isn't the speed shown by the local speedometer changes $10m/s$ per second on the local clock?
Answer
The problem has to do with an inconsistent nomenclature between proper velocity and proper acceleration. Let me first explain what proper acceleration is, as defined in the wiki article and most relativity textbooks.
A relativistic object follows a path in four-dimensional spacetime $(t(\tau),\vec{x}(\tau))$. It's more intuitive to convert this four-dimensional motion into the more familiar three-dimensional motion $\vec{x}(t)$, within a certain inertial reference frame. How can we do this? Obviously, we can calculate the coordinate path $\vec{x}(t)$ if we know the coordinate acceleration $\vec{a}(t)$ of the object. The question then is how we can relate $\vec{a}(t)$ to the force that's acting on the object. In other words, suppose that the object itself feels an acceleration $\vec{a}{}'(t')$ in its own reference frame. What is the relation between $\vec{a}(t)$ and $\vec{a}{}'(t')$?
First of all, we can think of an accelerating object as an object that 'jumps' between inertial frames. So at any instant, we can say that the object is located in a particular inertial frame, where an inertial observer will see the object accelerating with $\vec{a}{}'(t')$. Indeed, suppose that the accelerating frame is described by coordinates $(t',\vec{x}{}')$. Now define a co-moving inertial frame that momentarily coincides with the accelerating frames, so that is is also described by $(t',\vec{x}{}')$. Now, let's call $\vec{a}{}'$ the acceleration of the object, measured by the inertial observer.
In an infinitesimal time interval $\text{d}t'$ he will see the object moving with a velocity $\text{d}\vec{v}{}' = \vec{a}{}'\text{d}t'$. In the accelerating frame, the situation is completely symmetrical: an accelerating observer will see the inertial observer moving away with the same velocity $-\text{d}\vec{v}{}'$ (obviously in the opposite direction), so he will measure the same acceleration $-\vec{a}{}' = -\text{d}\vec{v}{}'/\text{d}t'$. The difference is that the accelerating observer 'feels' the acceleration himself, so he concludes that he is in fact the one who's accelerating, due to some force, with acceleration $\vec{a}{}'$.
So now we'd like to find a way to transfer $\vec{a}{}'(t')$ to $\vec{a}(t)$. In other words, we're looking for a quantity that
- is equal to $\vec{a}{}'(t')$ in the accelerating frame (or the co-moving inertial frame),
- is easily transferred from one inertial frame to another. In other words, it can be written in the form of a Lorentz-invariant.
A first candidate would be the vector-part of the four-acceleration: $$ \vec{A} = \frac{\text{d}^2\vec{x}}{\text{d}\tau^2} = \frac{\text{d}\vec{U}}{\text{d}\tau}, $$ where $$ \vec{U} = \frac{\text{d}\vec{x}}{\text{d}\tau} = \gamma\vec{v}, $$ is (unfortunately) called the proper velocity. We find $$ \vec{A} = \frac{\text{d}\gamma}{\text{d}\tau}\vec{v} + \gamma\frac{\text{d}\vec{v}}{\text{d}\tau} = \gamma^4\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\vec{v} + \gamma^2\vec{a}, $$ where $\vec{v}$ and $\vec{a}$ are the coordinate velocity and acceleration of the object in the inertial rest-frame. $\vec{A}$ clearly meets our first criterion: in the co-moving inertial frame (with primed coordinates), $\vec{A}$ becomes $\vec{A}{}'$; we have $\vec{v}{}'=\vec{0}$, so $\gamma'=1$, and the coordinate acceleration is $\vec{a}{}'$, so that $\vec{A}{}'=\vec{a}{}'$. However, $\vec{A}{}'$ is not a Lorentz-invariant: while the proper time $\text{d}\tau$ is Lorentz -invariant, the coordinate displacement $\text{d}\vec{x}$ is not. So the relation between $\vec{A}$ and $\vec{A}{}'$ is not obvious. This was to be expected, since we're missing the time-component of the four-acceleration: $$ A^0 = \frac{\text{d}U^{\!0}}{\text{d}\tau} = c\frac{\text{d}\gamma}{\text{d}\tau} = \gamma^4\frac{\vec{v}\cdot\vec{a}}{c}. $$ And now we are able to construct a Lorentz-invariant. In this post, I showed that $$ \sqrt{-A_\mu A^\mu} = \sqrt{(\vec{A})^2 - (A^0)^2} = \frac{\gamma^3}{\gamma_\perp} a, $$ where $$ \gamma_\perp^{-1} = \sqrt{1 - v_\perp^2/c^2} $$ and $v_\perp$ is the component of $\vec{v}$ perpendicular to $\vec{a}$. So if we now define the vector $$ \vec{\alpha} = \frac{\gamma^3}{\gamma_\perp} \vec{a}, $$ we have the quantity that we sought: in the co-moving frame, $\vec{\alpha}{}' = \vec{a}{}'$, and $\alpha=||\vec{\alpha}||$ is a Lorentz-invariant, so that $\alpha=\alpha'$. In other words, if we know the value of $\vec{a}{}'=\vec{\alpha}{}'$ in the accelerating frame, we can immediately obtain the corresponding acceleration in the rest-frame: $\vec{a}= \vec{\alpha}\gamma_\perp/\gamma^3$. There is one complication though: only the norm of $\vec{\alpha}$ is Lorentz-invariant, but not its orientation. So in order for this to work, we have to assume that the accelerating frame is orientated in such a way that the direction of $\vec{a}{}'$ is the same as $\vec{a}$. In other words, the rest-frame observer also needs to know the orientation of $\vec{a}{}'$ in his rest-frame.
It's this $\alpha$ that is usually called the proper acceleration. In the particular case that the object is accelerating parallel to its velocity, we get $$ \vec{\alpha} = \gamma^3\vec{a} = \frac{\text{d}(\gamma\vec{v})}{\text{d}t} = \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}\vec{x}}{\text{d}\tau}\right). $$ The mixture of $\text{d}t$ and $\text{d}\tau$ arises from the fact that $\vec{\alpha}$ provides a link between two reference frames: the coordinate rest-frame and the 'proper' accelerating frame. That's all there is to it: the proper acceleration is a handy tool to calculate accelerations in different reference frames.
A well-known application is the relativistic rocket, where the passengers experience a constant proper acceleration of $1\;g$, parallel to their velocity. The corresponding acceleration in the rest-frame is then $a=g/\gamma^3$ (note that it decreases as $v$ increases), and the rocket has a so-called hyperbolic motion.
Unfortunately, the nomenclature is non consistent. It certainly makes sense to call $\alpha$ the proper acceleration: it is the acceleration felt by a non-inertial observer, and it is Lorentz-invariant, consistent with e.g. proper time $\tau$. However, for some unfathomable reason, people decided to call $\vec{U}=\text{d}\vec{x}/\text{d}\tau$ the proper velocity, which imo is a mistake. Just like $\vec{A}$, it is not a Lorentz-invariant. If one were consistent, then the Lorentz scalair $$\sqrt{U_\mu U^{\!\mu}} = \sqrt{\gamma^2c^2 - \gamma^2v^2} = c$$ should be called the proper velocity. Indeed, you could say that we are moving through spacetime at the speed of light.
The problem is made worse because, since $\vec{U}$ is called proper velocity, some authors (although a minority) actually call $\vec{A}$ the proper acceleration. Hence the confusion.
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