Sunday, 14 June 2020

special relativity - Relativistic factor between coordinate acceleration and proper acceleration


I did a recent question about relativistic kinematics here: Generalizing a relativistic kinematics formula for spatial-acceleration dependence.


I have a confusion. In the textbooks I've seen, they put the relationship between proper acceleration and coordinate acceleration as



$$ \alpha = \gamma^3 \frac{dv}{dt} $$


However, when I try to do the derivation myself, I get a factor of $\gamma^4$ instead. I'm not sure where the error is.


My derivation is like this:


$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$


where we used $\frac{dt}{d \tau} = \gamma $


Now,


$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2} $$


$$\frac{d \gamma}{d \tau} = \gamma \frac{d \gamma}{dt} = \gamma (\frac{ \gamma^3 }{c^2} \frac{dx}{dt} \frac{d^2 x}{dt^2})$$


$$\frac{d^2 x}{d \tau^2} = \gamma^4 \frac{v^2}{c^2} \frac{d^2 x}{dt^2} + \gamma^2 \frac{d^2 x}{dt^2} = \gamma^2 \frac{d^2 x}{dt^2} (\gamma^2 \frac{v^2}{c^2} + 1) = \gamma^4 \frac{d^2 x}{dt^2}$$


where on the last step, $v= \frac{dx}{dt}$ .



I don't understand where is the mistake.




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