2 different formulas for proper acceleration derived from the same equation

Question

I was looking at this answer which eventually stated this about the proper acceleration (when $\vec v$ is parallel to $\vec a$): $$ \vec \alpha =\gamma^3 \vec a = \gamma^3 \dfrac{d\vec v}{dt}=\dfrac{du}{dt}=\dfrac{d}{dt}\dfrac{d\vec x}{d \tau} $$ Now, I'm programming a relativistic integrator and to do this I used the first and last part of this equation and switched the derivatives around: $$ \vec \alpha= \dfrac{d}{d\tau} \dfrac{d\vec x}{dt} = \dfrac{d\vec v}{d\tau} $$ Now this seems to work as I can recreate the hyperbolic curve obtained from Rindler coordinates if I choose $\alpha$ to be constant. (I have numerically compared this and it works). So to clarfiy I would do this integration step: $dv = \alpha d\tau$, where $dv$ is the change of velocity in the coordinate frame.

Now this all seems fine until we look back at the first equation and use $dt = \gamma d\tau$: $$ \vec \alpha = \gamma^3 \dfrac{d\vec v}{dt}=\gamma^2 \dfrac{d\vec v}{d\tau} \neq \dfrac{d\vec v}{d\tau} $$ This is not equal to the previously acquired formula. What's going on here? Am I not allowed to just convert $d\tau$ to $dt$?

Answer 1

and switched the derivatives around

This is your error. If you write the whole thing in terms of derivatives of $t$ you should have $$ \frac{d}{dt}\frac{d\vec x}{d \tau} = \frac{d}{dt} \left( \gamma \frac{d\vec x}{d t} \right) = \gamma \frac{d^2 \vec{x}}{dt^2} + \frac{d\gamma}{dt} \frac{d\vec{x}}{dt} $$ but $$ \dfrac{d}{d\tau} \dfrac{d\vec x}{dt} = \gamma \frac{d}{dt} \left( \frac{d\vec x}{d t} \right) = \gamma \frac{d^2 \vec{x}}{dt^2} $$ and these two expressions will not be equal in general.

Answer 2

No you cannot switch the derivatives in general. This is not like Schwartz' theorem for partial derivatives. In general: $$ \begin{align} \frac{d}{dx}\frac{d}{dy} &= \frac{d}{dx}\frac{dx}{dy}\frac{d}{dx} \\ &= \frac{d^2x}{dxdy}\frac{d}{dx}+\frac{dx}{dy}\frac{d^2}{dx^2} \\ &= \frac{d^2x}{dxdy}\frac{d}{dx}+\frac{d}{dy}\frac{d}{dx} \end{align} $$ so you have the first extra term which spoils your "identity".

Hope this helps.