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Starting with the world line of a particle given by $x^{\mu}$, this can be successively differentiated with the particle's proper time $\tau$ to give the four-velocity from the four-position, four-acceleration from the four-velocity. I'd expect to be able to do this to any order creating another four-vector, but I'm not sure if I'm missing some mathematical subtlety which doesn't guarantee this.

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In special relativity in the usual coordinate system, the answer is yes, any number of derivatives with respect to proper time gives you a four-vector with respect to Lorentz transformations.

But in general, the answer is no. If $x^\mu(\tau)$ are the coordinates of the worldline as a function of its proper time (assuming a timelike worldline), then $$ \frac{dx^\mu}{d\tau} \tag{1} $$ is always a four-vector, but $$ \frac{d^2 x^\mu}{d\tau^2} \tag{2} $$ is not always a four-vector (or any other tensor). In flat spacetime with metric $$ d\tau^2=\eta_{\mu\nu}dx^\mu\,dx^\nu \tag{3} $$ with coordinate-independent coefficients $\eta_{\mu\nu}$, it is a four-vector with respect to coordinate transformations that preserve the form (3). But in general it is not a four-vector. In general, to get a four-vector corresponding to the worldline's "acceleration," the quantity (3) should be replaced by $$ \frac{dx^\nu}{d\tau}\nabla_\nu\frac{dx^\mu}{d\tau} \tag{4} $$ where $\nabla$ is the covariant derivative. Explicitly, \begin{align} \frac{dx^\alpha}{d\tau}\nabla_\alpha\frac{dx^\mu}{d\tau} &= \frac{dx^\alpha}{d\tau}\left( \partial_\alpha\frac{dx^\mu}{d\tau} +\Gamma^\mu_{\alpha\beta} \frac{dx^\beta}{d\tau} \right) \\ &= \frac{d^2 x^\mu}{d\tau^2} +\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} \tag{5} \end{align} where the coefficients $\Gamma$ are the Christoffel symbols. The quantity (4) or (5) is a four-vector in any spacetime and in any coordinate system. When the metric has the form (3) with coordinate-independent $\eta_{\mu\nu}$ (which implies both flat spacetime and a special coordinate system), then $\Gamma=0$ and (4) reduces to (2).

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Yes. When you differentiate a four-vector with respect to a scalar, you get another four-vector.

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    $\begingroup$ This is assuming that the four-trajectory of the particle is actually differentiable to whatever order you want in the first place, which may not be guaranteed. $\endgroup$ – probably_someone Feb 13 at 1:00

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