- #1

azazello

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- Homework Statement
- Problem 11, chapter IV of Rindler's "Introduction to Special Relativity": Use the fact that ##A## as given in (23.5) is a four-vector to give an alternative solution to the first part of Exercise II(12).

That exercise reads: In ##S^\prime## a particle is momentarily at rest and has acceleration ##\alpha## in the ##y^\prime##-direction. What is the magnitude and direction of its acceleration in ##S##?

- Relevant Equations
- The expression for the four-vector acceleration (23.5) is $$A = \gamma \left(\frac{d\gamma}{dt}c, \frac{d\gamma}{dt}\mathbf{u} + \gamma\mathbf{a} \right).$$

Time dilation: ##\frac{dt}{dt^\prime} = \gamma.##

Velocity transformations: $$u_1 = \frac{u_1^\prime +v}{1 + u_1^\prime v/c^2},\, u_2 = \frac{u_2^\prime}{\gamma(1+u_1^\prime v/c^2)},\, u_3 = \frac{u_3^\prime}{\gamma(1+u_1^\prime v/c^2)}.$$

It should also be noted that the frame ##S^\prime## is moving in the positive ##x##-direction with speed ##v## as seen from the frame ##S##.

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 $$\frac{du_2}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt^\prime}\frac{dt^\prime}{dt}.$$ The first and the last factors are both ##\frac{1}{\gamma}## as can be seen from the time-dilation equation and the velocity transformation with ##u_1^\prime = 0##. The middle factor is the proper acceleration ##\alpha##. Thus, I get ##\alpha/\gamma^2##. For the ##x##- and ##z##-components, the acceleration is still zero since ##\frac{d u_1^\prime}{dt^\prime} = \frac{d u_3^\prime}{dt^\prime} =0##.

In order to do this with the four-vector, note that in the lab frame ##S##, we have the velocity ##\mathbf{u} = (v, 0, 0)##. Starting at rest, we get (I think) ##\frac{d\gamma}{dt} = 0##. Using the expression for the four-acceleration, I get $$A = (0,0,\gamma^2 \alpha, 0),$$ which is different from the previous result by a factor of ##\gamma^4##. Where did I go wrong?

To first answer the second part, Exercise II(12), I wrote $$\frac{du_2}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt^\prime}\frac{dt^\prime}{dt}.$$ The first and the last factors are both ##\frac{1}{\gamma}## as can be seen from the time-dilation equation and the velocity transformation with ##u_1^\prime = 0##. The middle factor is the proper acceleration ##\alpha##. Thus, I get ##\alpha/\gamma^2##. For the ##x##- and ##z##-components, the acceleration is still zero since ##\frac{d u_1^\prime}{dt^\prime} = \frac{d u_3^\prime}{dt^\prime} =0##.

In order to do this with the four-vector, note that in the lab frame ##S##, we have the velocity ##\mathbf{u} = (v, 0, 0)##. Starting at rest, we get (I think) ##\frac{d\gamma}{dt} = 0##. Using the expression for the four-acceleration, I get $$A = (0,0,\gamma^2 \alpha, 0),$$ which is different from the previous result by a factor of ##\gamma^4##. Where did I go wrong?