Garth said:
Certainly! An illustrative example to think about.

I claim that under that acceleration an on-board torque free gyroscope will precess relative to the fixed stars.
My understanding of this case is, if we let the spaceship have a relative 3-velocity initially also in the direction of Sirius parallel to the 3-acceleration vector:
\frac{d\mathbf{S}}{d\tau} = (\mathbf{u} \wedge \mathbf{a})\cdot d\mathbf{S}
The gyroscope's intrinsic angular momentum 4-vector is always orthogonal to the 4-velocity.
It will precess in 3-space orthogonally to the plane defined by the gyro's axis and the ship-Sirius direction 3-vector by an amount (while speeds are non-relativistic):
\frac{a \sin(\theta)}{c} radians/sec
where \theta is the angle between the direction of the gyro's spin axis and the ship-Sirius direction 3-vector and a is the magnitude of the acceleration.
Garth
Let's introduce a coordinate system, with Sirius lying in the 'x' direction.
y
^
^
^
-------> x (Sirius)
and define vectors using their components (t,x,y,z), in that order
Your claim, as I understand it, is that a gyroscope with a spin axis in the 'y' direction precesses.
Now for a spin axis in the 'y' direction, in a comoving frame we should haveS = \left( 0,0, L, 0 \right)
Let's use your formula (a^u) S to find the initial rate of change of S in a comoving frame.
Then
u = \left( 1,0,0,0 \right)
a = \left( 0, a, 0, 0 \right)
see MTW 6.4 pg 166 for the components of a, which we can see is orthogonal to u as it should be. Then
a \wedge u = \left( \begin{array} {cccc} <br />
0&-a&0&0\\<br />
a&0&0&0\\<br />
0&0&0&0\\<br />
0&0&0&0 \end{array} \right)<br />
We can see that \left( a \wedge u \right) \cdot S = 0
[add]
Why can we see this? Because (a ^ u) only has components (tx and xt). This is because a and u only have t and x components, hence a^u can only have t and x components. Because the wedge product is anti-symmetric, the tt and xx components are 0, only the tx and xt components exist, and they must be negatives of each other.
But the vector that we multiply by has only a y component. The only way to get a nonzero result is from y components in the wedge product. But all such components are zero.
Note that we also expect (0,0,L,0) to boost in the x direction to (0,0,L,0), i.e. we don't expect a boost in the x direction to change the y component of angular momentum given that it transforms as a 4-vector. Thus we don't really expect acceleration in the x direction (which is a series of small boosts) to change the y component of momentum. Which is what we just calculated in more detail.
[end add]
i.e. there is no change in S, a gyroscope with its spin axis initially in the 'y' direction does not precess.