Showing these two definitions of proper acceleration are equivalent

Question

This answer by Andrew Steane claims that a proper acceleration vector can be expressed in two different but equivalent ways:

$$\boldsymbol{a}_0=\frac{d\boldsymbol{v}}{dt}\qquad\boldsymbol{a}_0=\frac{d\boldsymbol{v}}{d\tau}$$

where $\boldsymbol{v}$ is coordinate velocity vector, $t$ is coordinate time, and $\tau$ is proper time. I am not sure how to show that they are equivalent.

First, I want to understand why $dt=d\tau$ holds in a momentarily comoving reference frame (MCRF). I think it can be shown using invariance of spacetime interval $ds^2=ds_\text{MCRF}^2$ where $ds$ is interval measured in the frame of the accelerating object. So, (assuming Minkowski metric)

$$-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2$$

In MCRF, I think $dx=dy=dz=0$ is true, right? How do I explain it? Assuming it is true, we get $dt=d\tau$.

Back to the main question, I found a nice derivation of proper acceleration vector by the same user:

That is, $$ \left( \begin{array}{c} a^0 \\ a^1 \\ a^2 \\ a^3 \end{array} \right) = \frac{d}{d\tau} \left( \begin{array}{c} \gamma c \\ \gamma {\bf v} \end{array} \right) = \gamma \frac{d}{dt} \left( \begin{array}{c} \gamma c \\ \gamma {\bf v} \end{array} \right) = \left( \begin{array}{c} \gamma \frac{d\gamma}{dt} c \\ \gamma \frac{d\gamma}{dt} {\bf v} + \gamma \frac{d\bf v}{dt}\end{array} \right). $$ In the instantaneous rest frame one has $\gamma = 1$ and $\dot{\gamma} = 0$ so in this frame one finds $$ a^\mu = \left( \begin{array}{c} 0 \\ \frac{d\bf v}{dt} \end{array} \right) = \left( \begin{array}{c} 0 \\ {\bf a}_0 \end{array} \right) $$

I wonder how do I show that $\frac{d\boldsymbol{v}}{dt}=\frac{d\boldsymbol{v}}{d\tau}$. I think I can do something like this:

$$\frac{d\boldsymbol{v}}{dt}=\frac{d\boldsymbol{v}}{d\tau}\frac{d\tau}{dt}=\frac{1}{\gamma}\frac{d\boldsymbol{v}}{d\tau}=\frac{d\boldsymbol{v}}{d\tau}$$

The last equality is justified because $\gamma=1$. Is that right?

Answer 1

The way you pose the question can be very misleading, because an expression like: $$ \mathbf{a} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} $$ may in general mean the time rate of change of the velocity of some other object which in general need not be ourselves (or in other words: need not be our own frame). As such, it will in general not be a true four acceleration, since the latter is always formed by differentiating the four velocity with respect to proper time $\tau$.

In the answer you linked to, Andrew very specifically qualified this equation by writing (emphasis is mine):

where $t$ is coordinate time and $\bf v$ is the velocity, both evaluated in that inertial frame in which the object is momentarily at rest at the event where we want to define the acceleration.

It is always more general then to write:

$$ \mathbf{a} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}\tau} $$

Because it is immediately understood that $\tau$ is the time as measured by the body whose four velocity is given by $\mathbf{v}$ and hence it forms a correct four acceleration vector. Now since, by the chain rule:

$$ \mathbf{a} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}\tau} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\tau} $$

Indeed there is an equality in the special case where $\frac{\mathrm{d}t}{\mathrm{d}\tau}=1$. To see why that is true in a body's own rest frame, we note that in general a four velocity is given by: $$ \mathbf{v} = \gamma(c,\vec{v}) = \frac{\mathrm{d}t}{\mathrm{d}\tau}(c,\vec{v}) $$

(As an aside, prove that $\gamma=\dfrac{\mathrm{d}t}{\mathrm{d}\tau}$ if you're not sure why that's true!)

Comparing this with the components of four velocity in the rest frame of the body: $$ \mathbf{v}_{\text{rest}} = (c,\vec{0}) $$

We see that: $$ \frac{\mathrm{d}t}{\mathrm{d}\tau} = 1 $$ $$ \vec{v} = \vec{0} $$

Where the first condition implies $t = \tau + k$ with $k$ being an irrelevant constant here fixed by initial conditions. All clocks in our frame run at the same rate. The second condition means that rather expectedly, we are not moving with respect to ourselves.

This however, needs to be understood with care when we're in an accelerating frame. The fact that $\vec{v}=0$ momentarily does not imply that the acceleration $\dot{\vec{v}}=\vec{a}$ also vanishes. Indeed, since we are using a family of MCRF's to describe the motion of an accelerating frame, in general $\vec{a}\neq 0$ while in each MCRF we also have $\vec{v}=\vec{0}$.

Finally, note that in fact you did write correctly that:

I think it can be shown using invariance of spacetime interval $ds^2=ds_\text{MCRF}^2$ where $ds$ is interval measured in the frame of the accelerating object. So, (assuming Minkowski metric) $$-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2$$ In MCRF, I think $dx=dy=dz=0$ is true, right? How do I explain it? Assuming it is true, we get $dt=d\tau$.

$\mathrm{d}x=\mathrm{d}y=\mathrm{d}z=0$ for the same simple reason that a frame is never moving with respect to itself. It can be beneficial for you to divide the above equation you wrote for the spacetime interval, by $\mathrm{d}\tau^2$ and try to derive the general form of the four velocity vector, it will also help you understand the relation that implies that $\vec{v}=0$ for a frame with respect to itself.