Relativistic factor between coordinate acceleration and proper acceleration I did a recent question about relativistic kinematics here: Generalizing a relativistic kinematics formula for spatial-acceleration dependence.
I have a confusion. In the textbooks I've seen, they put the relationship between proper acceleration and coordinate acceleration as
$$ \alpha = \gamma^3 \frac{dv}{dt} $$
However, when I try to do the derivation myself, I get a factor of $\gamma^4$ instead. I'm not sure where the error is.
My derivation is like this:
$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$
where we used $\frac{dt}{d \tau} = \gamma $
Now, 
$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2} $$
$$\frac{d \gamma}{d \tau} = \gamma \frac{d \gamma}{dt} = \gamma (\frac{ \gamma^3 }{c^2} \frac{dx}{dt} \frac{d^2 x}{dt^2})$$
$$\frac{d^2 x}{d \tau^2} =  \gamma^4 \frac{v^2}{c^2} \frac{d^2 x}{dt^2} + \gamma^2 \frac{d^2 x}{dt^2} = \gamma^2 \frac{d^2 x}{dt^2} (\gamma^2 \frac{v^2}{c^2} + 1) = \gamma^4 \frac{d^2 x}{dt^2}$$
where on the last step, $v= \frac{dx}{dt}$ .
I don't understand where is the mistake.
 A: From Wiki:

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.

You're instead finding the rate of change of proper velocity with respect to proper time and, so, are picking up an extra Lorentz factor.  See Four-acceleration.
A: Proper acceleration is not the same as the four-vector acceleration. Proper acceleration is defined as a Lorentz-invariant acceleration, that is, an acceleration that all inertial observers agree upon. Let's define a few concepts:
The proper velocity $\vec{u}$ is the derivative of the position with respect to proper time:
$$
\vec{u} = \frac{\text{d}\vec{x}}{\text{d}\tau}=\gamma\vec{v}.
$$
The proper velocity is the spacial component of the velocity four-vector $\boldsymbol{U}$
$$
\boldsymbol{U} = (c\gamma,\vec{u}),
$$
and the acceleration four-vector $\boldsymbol{A}$ is 
$$
\boldsymbol{A} = \frac{\text{d}\boldsymbol{U}}{\text{d}\tau} = \left(c\frac{\text{d}\gamma}{\text{d}\tau},\frac{\text{d}\vec{u}}{\text{d}\tau} \right).
$$
The relativistic scalar product $\boldsymbol{A}\bullet\boldsymbol{A} = A_0^2 - (\vec{A})^2$ is Lorentz-invariant. We get
$$
\boldsymbol{A}\bullet\boldsymbol{A} = c^2\left(\frac{\text{d}\gamma}{\text{d}\tau}\right)^2 - \left(\frac{\text{d}\vec{u}}{\text{d}\tau}\right)^2.
$$
Now,
$$
\frac{\text{d}\gamma}{\text{d}\tau} = \gamma^4\frac{\vec{v}\cdot\vec{a}}{c^2},
$$
where $\vec{a} = \text{d}\vec{v}/\text{d}t$ is the classical acceleration vector. On the other hand,
$$
\frac{\text{d}\vec{u}}{\text{d}\tau} = \frac{\text{d}\gamma}{\text{d}\tau}\vec{v} + \gamma\frac{\text{d}\vec{v}}{\text{d}\tau} = \gamma^4\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\vec{v} + \gamma^2\vec{a}.
$$
Let's call $v_\parallel$ the component $\vec{v}$ parallel to $\vec{a}$, and $v_\perp$ the perpendicular component. It follows that $\vec{v}\cdot\vec{a}=v_\parallel a$, so that
$$
\begin{align}
\boldsymbol{A}\bullet\boldsymbol{A} &=
 \gamma^8\frac{v_\parallel^2 a^2}{c^2} - \gamma^8\left(\frac{v_\parallel^2 a^2}{c^2}\right)\left(\frac{v^2}{c^2}\right) - 2\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^4a^2\\
&= \gamma^6\frac{v_\parallel^2 a^2}{c^2} - 2\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^4a^2\\
&= -\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^6a^2\left(1 - \frac{v_\parallel^2}{c^2}- \frac{v_\perp^2}{c^2}\right)\\
&= -\frac{\gamma^6}{\gamma_\perp^2} a^2,
\end{align}
$$
where
$$
\gamma_\perp = \frac{1}{\sqrt{1 - v_\perp^2/c^2}}.
$$
So $\boldsymbol{A}\bullet\boldsymbol{A}$ is always negative. If we now define the proper acceleration vector $\vec{\alpha}$
$$
\vec{\alpha} = \frac{\gamma^3}{\gamma_\perp} \vec{a} = \frac{\gamma^3}{\gamma_\perp}\frac{\text{d}\vec{v}}{\text{d}t},
$$
then $\alpha^2 = -\boldsymbol{A}\bullet\boldsymbol{A}$ is indeed Lorentz-invariant. This is why the proper acceleration is defined like this. 
If $v_\perp=0$, then the formula reduces to
$$
\vec{\alpha} =\gamma^3 \frac{\text{d}\vec{v}}{\text{d}t} = \frac{\text{d}\vec{u}}{\text{d}t}.
$$
So in this case, the proper acceleration is the derivative of the proper velocity with respect to coordinate time $t$, not the proper time. The derivative of the proper velocity with respect to proper time $\tau$ is in fact the spatial part of the four-vector acceleration
$$
\vec{A} = \frac{\text{d}\vec{u}}{\text{d}\tau} = \frac{\text{d}^2\vec{x}}{\text{d}\tau^2}.
$$
That's where the extra $\gamma$ comes from.
A: The proper acceleration in the formula you're trying to derive refers to what's measured in the inertial frame instantaneously co-moving with the particle, and not in the accelerated co-moving frame of the particle where the measured acceleration is zero.
This is how I would do the derivation using either a 3-vector or 4-vector approach:
 3-vector approach
Let frames $S$ and $S'$ move at a constant velocity, and $p$ be an accelerated point. The strategy is to find how the acceleration of $p$ transforms between $S$ and $S'$, and then to make the velocity of $S'$ equal to the instantaneous velocity of $p$ which then gives us the relationship between the proper acceleration and lab acceleration of $p$. Without loss of generality and to simplify things, $p$ accelerates along $x$, and $S'$ moves along the $x-\text{axis}$ at constant velocity $v$ as usual.
In $S'$, the coordinates of $p$ are $(x',y',z',t')$ which can be expressed in terms of the $S$ coordinates via the Lorentz transformations. Using these we can also find $\frac{dt'}{dt}$:
$$ x' =\gamma(x-vt),\quad t' = \gamma(t - vx/c^2),\quad \frac{dt'}{dt}= \gamma(1-\frac{vu_x}{c^2})$$ 
$p$ is accelerated along the x'-axis:
$$\begin{align*}
\frac{d^2x'}{dt'^2}
&=\frac{d}{dt'}\frac{dt}{dt'}\frac{d}{dt}\gamma(x - vt)
=\frac{dt}{dt'}\frac{d}{dt}\frac{u_x - v}{1-\frac{vu_x}{c^2}}
=\frac{1}{\gamma(1-\frac{vu_x)}{c^2}}
\cdot
\frac{1}{\gamma^2(1-\frac{vu_x}{c^2})^2}
\frac{d^2x}{dt^2}\\\
&=\frac{1}{\gamma^3(1-\frac{vu_x}{c^2})^3}\frac{d^2x}{dt^2}
\end{align*}$$
We now make $S'$ the co-moving and therefore proper frame of $p$ with $u_x$ the velocity of $p$ in $S$ which we now set equal to $v$, finally giving
$$\frac{d^2x'}{dt'^2} = \gamma^3\frac{d^2x}{dt^2}$$
 4-vector approach 
The four acceleration is given by
$$\left(
{\gamma_u}^4\frac{\mathbf{a\cdot u}}{c},
{\gamma_u}^2\mathbf{a}+
{\gamma_u}^4\frac{(\mathbf{a\cdot u)}}{c^2}\mathbf{u}
\right)$$
In the proper frame where $\mathbf{u=0}$, this immediately simplifies to $(0,\mathbf{a_p})$. We just Lorentz tranform the time part to establish the relationship between the lab and proper accelerations
$$
\begin{align*}
\gamma^4\frac{\mathbf{a_lv}}{c^2}&= 
\gamma\left(0 + \frac{\mathbf{va_p}}{c^2}\right)\\
\mathbf{a_p}&=\gamma^3\mathbf{a_l}
\end{align*}
$$
method in question
It's not clear to me what your strategy is, but it could be made to work based upon the 3-vector approach above:

$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$
where we used $\frac{dt}{d \tau} = \gamma $

Correct

Now, 
$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2} $$

This won't work because $\gamma$ is a function of the relative velocity $v$ between the two frames where the acceleration is being measured, and we keep this velocity constant:
$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \gamma^2 \frac{ d^2 x}{dt^2} $$
The term on the RHS isn't an acceleration for any frame because it's mixing the space coordinate from the proper frame with the time coordinate from the lab frame. You need to transform $x$ to $x'$ using the Lorentz transformation $x=\gamma(x' + vt)$ so that finally, keeping $\gamma$ and $v$ constant:
$$\frac{d^2 x}{d \tau^2} = \gamma^2 \frac {d^2x}{dt^2} = \gamma^2\frac {d^2}{dt^2}\gamma (x' + vt)= \gamma^3\frac{d^2x'}{dt^2}$$
