Does the proper four-acceleration $A^{\mu} = (0,0)?$

Let the proper four-position vector $$x^{\mu}(\tau) = (0, \tau)$$. Differentiating this successively wrt $$\tau$$ I get the four-velocity $$u^{\mu}(\tau) = (0, 1)$$ and then the four-acceleration $$A^{\mu}(\tau) = (0, 0)$$. But the proper acceleration should be $$A^{\mu}(\tau) = (a, 0)$$ which seems to contradict the previous definition. Where is the error in the logic of the first paragraph?

• You are in the rest frame of the accelerating particle, which is not an inertial system. – lalala Feb 14 at 16:58
• Yes, your calculation works for an inertial frame, and in an inertial frame the proper acceleration is of course zero as you found. If you have an accelerating frame then you need to take a covariant derivative not just $d/d\tau$. – John Rennie Feb 14 at 17:41
• @JohnRennie isn't it the opposite way around? it's zero in the accelerating non-inertial frame and the OP needs to take a covariant derivative for the inertial frame. – John McVirgo Feb 14 at 19:10
• I presume your convention is to write $(space, time)$. In this case your world-line ($x^\mu=x^\mu\left(\tau\right)$) is not the world-line of an accelerating particle. That the error. The world-line of an accelerating particle has non-vanishing curvature, which is independent of reference frame. See the chapter on acceleration in SR in the MTW book. The world-line of a uniformly accelerating particle, in cartesian basis, would look something like $x^\mu\left(\tau\right)=\left(time, space\right)^\mu=\left(\cosh a\tau, \sinh a \tau\right)^\mu$ – Cryo Feb 14 at 22:21