Recent content by azazello

I see, that makes a lot of sense. So in the primed frame, ##\gamma = 1## since the particle is initially at rest. Thus ##A^\prime = (0,0,\alpha,0)##, and transforming this gives ##A=(0,0,\alpha,0)##, which means that you can read off the three-acceleration as ##a_y = \alpha/\gamma^2##. (Obvious...

Sorry, but I still don't quite understand. Taking the three-acceleration in the primed frame as ##\alpha=a_y^\prime##, the 4-acceleration in the primed frame is $$A^\prime = (0,0,\gamma^2 \alpha, 0).$$ Applying a Lorentz transformation in the ##x##-direction then gives in the unprimed frame...

Thanks for your reply. I completely agree with you, but do you know where in my reasoning I went wrong? I guess a better solution is to write down the 4-acceleration in the primed frame and do a Lorentz transformation, but I'm curious as to what went wrong in the above.

I have been getting back to studying physics after a long break and decided to go through the problems in Rindler. But there is something I don't quite understand in this problem. To first answer the second part, Exercise II(12), I wrote $$\frac{du_2}{dt} = \frac{du_2}{du_2^\prime}...