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?

  • $\begingroup$ Related question and directly applicable answer: physics.stackexchange.com/questions/66839/… $\endgroup$ Jun 15, 2013 at 13:08
  • $\begingroup$ @AlfredCentauri That question is one of sources of my confusion. Also when I start from $u = v\gamma$ then differentiate it with respect to $t$, then assuming $du/dt = \alpha$ and $dv/dt = a$ I can obtain the $\alpha = a\gamma^3$ formula. I always overlooked the mistake that I assumed $\alpha = du/dt$. But it still yields the correct formula... $\endgroup$
    – Calmarius
    Jun 18, 2013 at 7:36
  • $\begingroup$ But $d(\gamma v)/dt$ is the proper acceleration $\alpha$. $d(\gamma v)/d\tau$ is not the proper acceleration but is, instead, the spatial part of the four-acceleration. These are not the same thing. $\endgroup$ Jun 18, 2013 at 11:05

4 Answers 4


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.

  • 4
    $\begingroup$ +1. I agree that the term "proper velocity" for $\gamma_u u$ is awful since, unlike proper time and proper acceleration, it is not invariant but rather explicitly frame dependent. $\endgroup$ Jun 17, 2013 at 11:14

Because it's a mistake. You can see the formulae in this section


of that article that they surely meant the proper time. I have already changed the word "coordinate" to "proper" in that sentence on Wikipedia. Proper acceleration, like any proper quantity, is expected to be independent of the choice of the inertial system and the definition involving the coordinate time is demonstrably dependent on the boosts.

Well, even the definition with the proper time in the denominator depends on the reference frame – but only through the rotations of the spacelike vector $A^\mu$, not its length.

Alternatively, the sentence could have been corrected by adding the comment that the derivative has to be computed in the instantaneous rest frame of the accelerating object. With this qualification, one could have kept the coordinate time.

  • $\begingroup$ But the acceleration measured by an accelerometer is invariant isn't it? And, as you point out, in the instantaneous rest frame, the coordinate time equals the proper time thus... $\endgroup$ Jun 15, 2013 at 11:01
  • $\begingroup$ That's the point. The reading on an accelerometer is invariant - but the derivative of the velocity with respect to coordinate time (in a general inertial system) is not. $\endgroup$ Jun 16, 2013 at 6:23
  • $\begingroup$ Oh, so the Wikipedia was wrong. It's clear now. $\endgroup$
    – Calmarius
    Jun 17, 2013 at 7:29
  • 2
    $\begingroup$ @Calmarius, the Wiki article isn't wrong as I and Pulsar show below. $\endgroup$ Jun 17, 2013 at 10:44
  • $\begingroup$ @Luboš Have a look at the posts from Alfred and me. The confusion has to do with nomenclature. $\endgroup$
    – Pulsar
    Jun 17, 2013 at 10:55

From the Wiki article Four-acceleration:

In special relativity, four-acceleration is a four-vector and is defined as the change in four-velocity over the particle's proper time:

$$\mathbf{A} =\frac{d\mathbf{U}}{d\tau}=\left(\gamma_u\dot\gamma_u c,\gamma_u^2\mathbf a+\gamma_u\dot\gamma_u\mathbf u\right) =\left(\gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c},\gamma_u^2\mathbf{a}+\gamma_u^4\frac{\left(\mathbf{a}\cdot\mathbf{u}\right)}{c^2}\mathbf{u}\right)$$ where

$$\mathbf a = {d\mathbf u \over dt}$$


$$\dot\gamma_u = \frac{\mathbf{a \cdot u}}{c^2} \gamma_u^3 = \frac{\mathbf{a \cdot u}}{c^2} \frac{1}{\left(1-\frac{u^2}{c^2}\right)^{3/2}}$$

and $\gamma_u$ is the Lorentz factor for the speed $u$. A dot above a variable indicates a derivative with respect to the coordinate time in a given reference frame, not the proper time $\tau$.

In an instantaneously co-moving inertial reference frame $\mathbf u = 0$, $\gamma_u = 1, $ and $\dot\gamma_u = 0$, i.e. in such a reference frame : $\mathbf{A} =\left(0, \mathbf a\right)$

Geometrically, four-acceleration is a curvature vector of a world line.

Therefore, the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration that a moving particle "feels" moving along a world line.

Just as proper time is an invariant, the magnitude of the four-acceleration is an invariant and, as you can see, equal to a, the magnitude of the coordinate time derivative of u.

Also, from the Wiki article Proper acceleration:

In relativity theory, proper acceleration is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is thus acceleration relative to a free-fall, or inertial, observer who is momentarily at rest relative to the object being measured.

Given this notion of proper acceleration, the acceleration relative to an inertial frame relatively non-moving at that instant, it is easy to see why the derivative is with respect to coordinate time.


The answer to this question is evidently in dispute so, in the following, I will make the case that the magnitude of the coordinate time derivative of proper velocity is an invariant and, rather than add to my previous answer, I will argue for the case here.

Consider the usual standard configuration of two coordinate systems, (t', x'), (t, x), with uniform relative velocity $v$, where the velocity of the particle in the primed and un-primed frames is $u'$ and $u$ respectively.

From the relativistic velocity addition formula, we have:

$u' = \dfrac{u - v}{1- \frac{uv}{c^2}}$

Using $t' = \gamma_v(t - \frac{vx}{c^2})$, it can be shown that:

$\dfrac{du'}{dt'} = \dfrac{du}{dt} \dfrac{1}{\gamma^3_v(1-\frac{uv}{c^2})^3} $

This is a general result but, in the case that $u'$ is instantaneously zero, $u =v$ and we have:

$\dfrac{du'}{dt'} = \gamma^3_u \dfrac{du}{dt}$

But, $\frac{du'}{dt'}$, at that instant, is the invariant proper acceleration $\alpha$ since, at that instant, the primed frame is the MCRF (Momentarily Co-moving Reference Frame).

Thus, we have the general result that:

$\alpha = \gamma^3_u \dfrac{du}{dt}$.

But, it is also the case that the derivative, with respect to coordinate time, of proper velocity $\gamma_u u$ is:

$\dfrac{d(\gamma_u u)}{dt} = \gamma^3_u \dfrac{du}{dt}$

thus establishing the result that:

$\alpha = \dfrac{d(\gamma_u u)}{dt}$

  • 1
    $\begingroup$ +1 My main issue was thinking proper velocity was measured entirely in the proper frame of the particle, and therefore's zero - it isn't! So a wikipedia link to proper velocity and an explanation of what it is will probably help the OP. $\endgroup$ Jun 17, 2013 at 2:29
  • $\begingroup$ So basically what you are saying is that $du/dt$ and $du/d\tau$ is equal because in a momentarily co-moving frame both $t$ and $\tau$ approaches zero? This would explain why I got the correct formula by the way I mentioned at the comment of the OP. $\endgroup$
    – Calmarius
    Jun 18, 2013 at 7:39
  • $\begingroup$ @Calmarius, I'm not sure how you get that out of my answer. First, be aware that in my answer above, $u'$ and $u$ are coordinate velocities. Second, the derivative of proper velocity (a three-vector) with respect to coordinate time is proper acceleration. Third, the derivative of four-velocity (a four-vector) with respect to proper time is four-acceleration. We "feel" proper acceleration, not four-acceleration. Four-acceleration is covariant whilst proper acceleration is invariant. $\endgroup$ Jun 18, 2013 at 10:40

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