# How do 4-vectors change under an "accelerated" Lorentz transformation?

I assume that an observer moving with velocity $$\mathbf{v} = v\mathbf{n} = \mathbf{v}(t)$$ (with respect to another observer) has coordinates

where $$x^{\mu}$$ are the coordinates for the observer who is at rest.

Then, I expect

to be the inverse transform (even though I don't know how to actually invert the first relation, considering that $$\mathbf{v}$$ is arbitrarily dependent on $$t$$).

So far so good. The problem arises when trying to compute the transformation for the time component of a 4-vector. Try the partial derivatives, for size $$\begin{array}{rcl} \partial_{ct} & = & \tfrac{\partial ct'}{\partial ct}\partial_{ct'} + \tfrac{\partial x^{i'}}{\partial t}\partial_{i'} \\ & & \\ & = & \gamma^4\tfrac{v\partial_{ct}v}{c^2}ct'\partial_{ct'} + \gamma^4\tfrac{v\partial_{ct}v}{c^2}\tfrac{\mathbf{v}\cdot\mathbf{r}'}{c}\partial_{ct'} + \gamma\partial_{ct'} + \gamma^2\tfrac{v\partial_{ct}v}{c^2}\tfrac{\mathbf{v}\cdot\mathbf{r}'}{v^2}v^i\partial_{i'} \\ & & \\ & + & \left(\gamma-1\right)\tfrac{\partial_{ct}\mathbf{v}\cdot\mathbf{r}'}{v^2}v^i\partial_{i'} + \left(\gamma -1\right)^2\tfrac{\mathbf{v}\cdot\mathbf{r}'}{v^2}\tfrac{\mathbf{v}\cdot\partial_{ct}\mathbf{v}}{v^2}v^i\partial_{i'} + \left(\gamma^2-\gamma\right)\tfrac{\mathbf{v}\cdot\partial_{ct}\mathbf{v}}{v^2c}ct'v^i\partial_{i'} \\ & & \\ & + & \left(\gamma^2-\gamma\right)\tfrac{\mathbf{v}\cdot\mathbf{r}'}{v^2}\partial_{ct}v^i\partial_{i'} + \left(\gamma^2-\gamma\right)\tfrac{1}{c}ct'\partial_{ct}v^i\partial_{i'} \\ & & \\ & - & \left(\gamma^2-\gamma\right)\tfrac{\mathbf{v}\cdot\mathbf{r}'}{v^3}\partial_{ct}vv^i\partial_{i'} - \left(\gamma^2-\gamma\right)\tfrac{1}{vc}ct'\partial_{ct}vv^i\partial_{i'} \\ & & \\ & - & \gamma^2\tfrac{\partial_{ct}v^i}{c}ct'\partial_{i'} - \gamma^2\tfrac{\partial_{ct}v^i}{c}\tfrac{\mathbf{v}\cdot\mathbf{r}'}{c}\partial_{i'} - \gamma\tfrac{v^i}{c}\partial_{i'} \\ & & \\ \partial_i & = & \tfrac{\partial ct'}{\partial x^i}\partial_{ct'} + \tfrac{\partial x^{j'}}{\partial x^i}\partial_{j'} = -\gamma\tfrac{v^i}{c}\partial_{ct'} +\partial_{i'} + \left(\gamma-1\right)\tfrac{v^iv^j}{v^2}\partial_{j'}. \end{array}$$

How would you go about writing $$\partial_{ct}v$$ and $$\partial_{ct}\mathbf{v}$$ in terms of $$\partial_{ct'}v$$ and $$\partial_{ct'}\mathbf{v}$$? Is that required at all? To me, it seems like a circular problem where I'm trying to compute $$\partial_{ct}$$ in terms of $$\partial_{ct'}$$ and $$\partial_{i'}$$, only to find out that I need such a relation to begin with.