Is acceleration due to gravity a 4-vector?

##a^{\mu} = \ddot{x}^{\mu} + \Gamma^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}## is a 4-vector but the two individual terms on the RHS are not 4-vectors. It's very easy to show that they both fail to transform as 4-vectors but that their sum does transform as a 4-vector.

 

##x^{\mu}## are not position vectors. The transformation ##x^{\mu} \rightarrow x^{\mu'}(x^{\nu})## will not be the same as the transformation matrix ##V^{\mu} \rightarrow V^{\mu'}## which is where your error is; in fact if we write the former transformation as ##x^{\mu'} = x^{\mu'}(x^{\nu})## then ##V^{\mu'} = \frac{\partial x^{\mu'}(x^{\nu})}{\partial x^{\mu}}V^{\mu}## so clearly we are talking about different transformations for the two.

 

To simplify the notation, I'm going to drop the indices. It should be clear how to put them back.

In general, if you converting from a coordinate [itex]x'[/itex] to a coordinate [itex]x[/itex], then it's a nonlinear, continuous, invertible function:

[itex]x = F(x')[/itex]

So,

[itex]\frac{dx}{d \tau} = \frac{\partial F}{\partial x'} \frac{dx'}{d\tau}[/itex]

So even if [itex]F(x')[/itex] is nonlinear, the relationship between velocities is linear:

[itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex]

where [itex]\Lambda = \frac{\partial F}{\partial x'}[/itex]

But note: the transformation between coordinates is not the same as the transformation between velocities:

[itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex] versus [itex]x = F(x')[/itex]

The only time that you can use the same matrix [itex]\Lambda[/itex] to transform coordinates or velocities is when:

[itex]F(x') = \Lambda x'[/itex]

which is only possible when [itex]\frac{\partial \Lambda}{\partial x'} = 0[/itex]