- #1
etotheipi
Two frames measure the position of a particle as a function of time: S in terms of x and t and S', moving at constant speed v, in terms of x' and t'. The acceleration as measured in frame S is $$ \frac{d^{2}x}{dt^{2}} $$ and that measured in frame S' is $$ \frac{d^{2}x'}{dt'^{2}} $$My question is how can we write the expression for the acceleration in frame S' in terms of that measured in frame S, noting the two coordinate transformations $$x' = \gamma(x-vt)$$ and $$t' = \gamma(t-vx)$$ I have had a go at the first derivative
\begin{align}
\frac{dx'}{dt'} &= \gamma \frac{dx}{dt'} - \gamma v \frac{dt}{dt'} \\ &= \gamma \frac{dx}{dt}\frac{dt}{dt'} - \gamma v \frac{dt}{dt'}
\end{align}
I tried deriving the time transformation with respect to t
\begin{align}
t' &= \gamma t - \gamma xv
\\ \frac{dt'}{dt} &= \gamma - \gamma v \frac{dx}{dt}
\end{align}
so that $$\frac{dt}{dt'} = \frac{1}{\gamma - \gamma v \frac{dx}{dt}}$$ but substituting this in doesn't really help and even then I'm not sure how to handle the second derivative. The aim would be to have some relationship like $$ \frac{d^{2}x}{dt^{2}} = f(\frac{d^{2}x'}{dt'^{2}})$$ where $$f$$ is some function. I was wondering if anyone could give me some tips or guidance since I don't really know if I'm going about this the right way. Thanks a bunch.
\begin{align}
\frac{dx'}{dt'} &= \gamma \frac{dx}{dt'} - \gamma v \frac{dt}{dt'} \\ &= \gamma \frac{dx}{dt}\frac{dt}{dt'} - \gamma v \frac{dt}{dt'}
\end{align}
I tried deriving the time transformation with respect to t
\begin{align}
t' &= \gamma t - \gamma xv
\\ \frac{dt'}{dt} &= \gamma - \gamma v \frac{dx}{dt}
\end{align}
so that $$\frac{dt}{dt'} = \frac{1}{\gamma - \gamma v \frac{dx}{dt}}$$ but substituting this in doesn't really help and even then I'm not sure how to handle the second derivative. The aim would be to have some relationship like $$ \frac{d^{2}x}{dt^{2}} = f(\frac{d^{2}x'}{dt'^{2}})$$ where $$f$$ is some function. I was wondering if anyone could give me some tips or guidance since I don't really know if I'm going about this the right way. Thanks a bunch.
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