# Can the world line of a particle be successively differentiated to any order to always give a four-vector?

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.

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).