Undergrad Relating orthogonal accelerations in special relativity

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The discussion focuses on relating orthogonal accelerations in special relativity between an inertial frame S and the instantaneous inertial frame S' of a particle. The acceleration components are defined as (a_x, a_y) in S and (a_x', a_y') in S'. Morin's argument suggests that the y-component acceleration can be expressed as a_y' = γ^2 a_y, but concerns arise regarding the differentiation of the γ factor. The reasoning is validated through the application of the chain rule, noting that the structure of Lorentz transformations indicates that the particle's velocity is aligned with the x-axis, leading to dy/dt = 0 in all frames. This highlights a potential shortcut in Morin's approach to the problem.
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Reasoning for the validity of relation between orthogonal accelerations in special relativity
We want to relate acceleration in two frames, an inertial frame S, and the instantaneous inertial reference frame of the particle on which it is being accelerated, S', which is moving in the ##x## direction at the moment. Let the acceleration in S be ##(a_x,a_y)## and in S' be ##(a_x',a_y')##. We want a relationship between them.

Now, here Morin argues that when consider the ##y## component we can write ##dy=dy'## and that ##dt'=dt/\gamma##, thus

$$a_y'=d^2y'/dt'^2=d^2y/(dt/\gamma)^2=\gamma^2a_y\tag1$$

But this seems wrong since we are taking the derivative of a factor of ##\gamma## here. If we go into a bit more detail then,


$$a_y=\frac{d^2y}{dt^2}=\frac d{dt}\bigg(\frac{dy'}{\gamma dt'}\bigg)=\frac1{\gamma^2}a_y'+\underbrace{\frac{dy'}{dt'}}_{v'}\frac d{dt}\bigg(\frac1{\gamma}\bigg)$$

but since the particle is at rest with respect to itself ##(v'=0)##, the second term goes to zero and we get ##(1)##. Is this reasoning correct?
 
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I think you are applying the chain rule here, and I agree that's the formally correct approach. However note that the structure of the Lorentz transforms, which are aligned parallel to the x axis, tells you that the velocity of the particle is parallel to x (otherwise you couldn't use these transforms to reach its rest frame). Hence ##dy/dt=0## in any frame, not just the rest frame.

I think that's why Morin is taking a shortcut here.
 
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