Intuition as to why a constant four-acceleration must be zero In a tutorial, it was derived that a constant proper acceleration must be zero:
$$ u \cdot a = 0
$$
Taking the time derivative of both sides:
$$ a \cdot a + u \cdot \dot{a} = a^2 = 0$$
And so $a^\mu = (|\underline{a}|, \underline{a})$
Working in the instantaneous rest frame, where $u^\mu = (c, \underline{0})$
$$u \cdot a = c a_0 = 0$$
And so $|\underline{a}| = 0$ and $\underline{a} = \underline{0}$ since acceleration is real. In this frame $a^\mu = (0, 0, 0, 0)^T$
We can then transform from the rest frame to any intertial frame by pre-multiplying $a^\mu$ with the correct Lorentz transformation matrix. The resulting $a^\mu$ will be 0 and so a constant proper four-acceleration must be zero.
This seems like a surprising result, is there any good intuition behind it ?
Answer to RobPhy's comment: Can you show that the acceleration vector is perpendicular to the velocity vector when the circular orbit has constant speed?
$$\dfrac{d}{dt}|\underline{v}|^2 = 2 |\underline{v}| \dfrac{d}{dt}|\underline{v}|\\
\dfrac{d}{dt}|\underline{v}|^2 = \dfrac{d}{dt} \underline{v}\cdot \underline{v} = 2 \underline{a} \cdot \underline{v}\\
\dfrac{d}{dt}|\underline{v}| = \dfrac{1}{|\underline{v}|}\underline{a} \cdot \underline{v}\\
0 =\dfrac{1}{|\underline{v}|}\underline{a} \cdot \underline{v}
$$
therefore constant speed means the acceleration vector is perpendicular to the velocity vector
 A: Imagine a particle in Euclidean space that moves smoothly on the unit sphere. The time derivative of its position can't be constant over any nonzero time interval, since that would imply the point had moved in a straight line for that time, and there is no straight line in the sphere.
The same thing happens with four-velocity. It's a vector representing a spacetime direction only. The magnitude is set to 1 by convention, which means the vector always lies on the Minkowskian analog of a unit sphere centered at the origin. This "sphere" is negatively rather than positively curved, but like the Euclidean sphere it contains no straight lines, so the four-velocity can't move with constant "velocity" in the Minkowskian vector space for any nonzero length of time.
This is also why four-acceleration is perpendicular to four-velocity: the motion of the four-velocity has to be tangent to the surface it's confined to, and the tangent is perpendicular to the radius.
A: You are showing that constant four-acceleration must be zero, not that constant proper acceleration must be zero. This is obvious because if there were to be a non-zero constant four-acceleration, the three-velocity of the particle would exceed the speed of light. Imagine a constant four-acceleration given by $(A^0,\mathbf{A})$. The evolution of the three-velocity of the particle along its trajectory would follow the equation
\begin{align}
\frac{d}{d\tau}\mathbf{u}=\mathbf{A}\implies\mathbf{u}=\mathbf{u}_0+\mathbf{A}\tau
\end{align}
Thus, clearly, for a non-zero $\mathbf{A}$, this would result in superluminal travel.
However, constant proper-acceleration need not be zero because a constant proper-acceleration does not imply constant four-acceleration. A constant proper acceleration only implies a constancy of the length of the four-acceleration vector, i.e., $d_\tau(A^\mu A_\mu) = 0$, not $d_\tau A^\mu=0$. The motion of a particle under constant proper-acceleration is a very interesting and widely studied topic, such motion is known as hyperbolic motion.
A: The best intuitive explanation I can find is that constant accelerations have no place in special relativity, on account of a constant rate of speed increase would eventually surpass the speed of light (no matter how small that increase be). The concept that substitutes constant acceleration in SR is that of hyperbolic motion, which corresponds to constant proper acceleration.
A: After I made my comment, I deleted it because I wondered if I misunderstood the question and I didn't have any time to think about it... but I see the OP got the comment and updated the post. I have a little time now.
I'll go in sequence... putting the zeroes on the left.
I'll add vector-signs and hat-signs, but I'll only work in (1+1)-Minkowski spacetime with  $(+,-)$-signature.

Can you quote the source (the "tutorial")?

Maybe I am misunderstanding the situation.

Presumably $\hat u$ is a 4-velocity and
$
\def\DT{ \frac{d}{d\tau} }
\def\DTU{ \frac{d\hat u}{d\tau} }
\def\DTA{ \frac{d\tilde a}{d\tau} }
%
\tilde a=\DT\hat u
$
is the 4-acceleration.

So,
$$0=\hat u \cdot \tilde a,$$
which follows from $0=\DT(1)
=\DT\left( \hat u \cdot \hat u\right)=2\DTU\cdot \hat u=2\tilde a \cdot \hat u$ (analogous to the perpendicularity of the acceleration-vector and the velocity-vector for uniform circular motion (hence, my comment)).
Note that $0=\hat u \cdot \tilde a$ means that since $\tilde u$ is timelike, it must be that a non-zero $\tilde a$ is spacelike .
Taking the derivative yields:
\begin{align}
0&=\DT\left(\hat u \cdot \tilde a\right)\\
&=\DTU \cdot \tilde a +\hat u \cdot \DTA\\
&=\ \ \tilde a\cdot\tilde a \ \ +\hat u \cdot \DTA
\end{align}
So, we have
$$\tilde a\cdot \tilde a = -\hat u \cdot\DTA $$
Note, however, that $\displaystyle\left( -\hat u \cdot\DTA \right) \neq 0$.

In fact,  it must be negative (in my signature convention) since a nonzero $\tilde a$ is spacelike.
Thus, [assuming I understood the situation correctly]
 what you wrote $\qquad$ "And so $a^\mu = (|\underline{a}|, \underline{a})$" $\qquad$ seems incorrect....

since it seems to suggest that $\tilde a$ is lightlike---but it's not---$\tilde a$  is  spacelike.

So, the rest of the proof that asserts $\tilde a=0$ fails.

For completeness....
If we define $g^2$ as
$$-g^2 \equiv \tilde a \cdot \tilde a,$$
then it follows that
$$\tilde a = g\hat a,$$
and
$$\DTA= g^2\hat u + f\hat a,$$
where
$\hat a$ is the spacelike unit-vector orthogonal to $\hat u$, and
where $f$ has to be determined. (Check that $\quad -g^2 = -\displaystyle\hat u \cdot \DTA \quad $.)
To find $f$, observe
$$\hat a \cdot \DTA = \hat a\cdot \left( g^2\hat u + f\hat a\right) = 0 - f =-f$$
and
\begin{align}
\DT(\tilde a \cdot \tilde a)
&=2\tilde a \cdot \DTA\\
\DT( -g^2)
&=
2(g\hat a)\cdot \DTA\\
-2g\frac{dg}{d\tau}
&=2g \left( \hat a \cdot \DTA \right)
\end{align}
Thus,
\begin{align}
-2g\frac{dg}{d\tau}
&=2g \left( \quad -f \quad\right)
\end{align}
or
\begin{align}
f &=\frac{dg}{d\tau}
\end{align}
Thus,
$$\DTA= g^2\hat u + \frac{dg}{d\tau}\hat a,$$
From this, "uniform [proper] acceleration" ($\frac{dg}{d\tau}=0$)
can be characterized by
the condition
$$\DTA \stackrel{const\ g}{=} g^2\hat u $$
which I learned from
J. Dwayne Hamilton, The uniformly accelerated reference frame,
American Journal of Physics 46, 83 (1978); https://doi.org/10.1119/1.11169 .

(Use $0=\DT\left(\hat u \cdot \tilde a\right)$ to get $\tilde a \cdot \tilde a= -g^2$.)
