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In his paper "Uniformly accelerated reference frames in Special Relativity", E. Desloge describes the acceleration of an accelerated observer through what he refers to as a direct application of the Lorentz transformation and the use of two inertial reference frames. The inertial reference frames Desloge references are $K$ and $K'$ where $K'$ moves at a velocity $V$ with respect to $K$. He also defines the form of the Lorentz transformation he will be using to be: \begin{align} x' &= \frac{\left( x - V t\right)}{\sqrt{1-{V^2}}} \\ t' &= \frac{\left( t - V x\right)}{\sqrt{1-{V^2}}} \end{align} This is then applied to the below example. (I am paraphrasing for brevity). If $X(t)$ is a function which describes the world line of an accelerated observer $X$ as observed in the reference frame $K$, then through direct application of the Lorentz transformation Desloge obtains the equation: \begin{equation} \frac{d^2 X'}{dt'^2} = \frac{\ddot{X} {\left( 1 - V^2\right)}^{3/2}}{(1 - V \dot{X})^3} \end{equation} Where $\frac{d^2 X'}{dt'^2}$ is the acceleration of $X$ as viewed in the reference frame of $K'$.
I have worked on this for a few days now and I cannot derive the final equation from the previous two. Desloge gives very little other information on where this final equation comes from. Any help or points in the right direction would be greatly appreciated.

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  • $\begingroup$ Equations in current form doesn't makes sense, because parameter $V$ has couple of different dimensions even in same equation. I. E. dimensional analysis fails or I misunderstood something. $\endgroup$ – Agnius Vasiliauskas Jul 13 at 6:27
  • $\begingroup$ @AgniusVasiliauskas These equations are taken directly from the text. $V$ is the velocity between the inertial reference frame $K$ and $K’$. Because it doesn’t relate to $X$, $V$ I believe $V$ is treated as a constant. Desloge uses these two reference frames to define $\frac{d^2 X’}{dt’^2}$ so that later if $K$ is equal to $K’$ we can define the acceleration in $K$. $\endgroup$ – gfsch Jul 13 at 10:44
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What have you tried?
It might help to see your attempt to see where you got stuck.

Try using the chain rule:

$$ \begin{align} \color{red}{\frac{dx'}{dt'}} =\frac{dx'}{dt}\frac{dt}{dt'} &=\frac{\left(\displaystyle\frac{dx'}{dt}\right)}{\left(\displaystyle\frac{dt'}{dt}\right)}\\ &=\frac{ \displaystyle\frac{d}{dt}\left( \frac{x-Vt}{(1-V^2)^{1/2}} \right) }{\displaystyle \frac{d}{dt}\left( \frac{t-Vx}{(1-V^2)^{1/2}} \right) } \end{align} $$ Then, $$ \begin{align} \frac{d^2 x'}{dt'^2}=\frac{d}{dt'}\left( \color{red}{\frac{dx'}{dt'}} \right) =\frac{d\left( {\frac{dx'}{dt'}} \right)}{dt}\frac{dt}{dt'} &=\frac{\displaystyle\left(\frac{d\left( \frac{dx'}{dt'} \right)}{dt}\right)}{\left(\displaystyle\frac{dt'}{dt}\right)} \\ &=\frac{\displaystyle\frac{d}{dt}\left( \color{red}{\frac{dx'}{dt'}}\right)} {\displaystyle \frac{d}{dt}\left( \frac{t-Vx}{(1-V^2)^{1/2}} \right)} \end{align} $$

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  • $\begingroup$ Found my error. Thanks $\endgroup$ – gfsch Jul 23 at 17:15
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There are actually several ways one can derive the expression for the coordinates moving with constant acceleration, if you consider a bit more directly the geometry of curves in Minkowski space, but that's a deviation from your question. The formula you are interested in can be derived directly from the chain rule in calculus. It is just that the derivation is tedious.

I am going to start with the most general version of this formula. AFter that will reduce the result to the case of Lorenz transformations.

Assume you have a time-line curve $\big(\, t \,, \, x(t) \,\big)$ and at least twice continuously differentiable change of coordinates (curvilinear in general): \begin{align} &x' = x'(x, t)\\ &t' = t'(x, t) \end{align} your curve becomes a curve that can be parametrized as $\big(\, t' \,, \, x'(t') \,\big)$ where \begin{align} &x' = x'\big(x(t),\, t\big)\\ &t' = t'\big(x(t),\, t\big) \end{align} For example, in the case of Lorenz transformations \begin{align} &x' = \gamma \,\big(x \, - \, V\,t\big)\\ &t' =\gamma \,\big(-\, V\, x \, + \, t\big) &\text{ where } \, \gamma = \frac{1}{\sqrt{1-V^2}} \end{align} By the chain rule \begin{align} \frac{dx'}{dt'} \, &= \, \frac{dt}{dt'}\, \frac{dx'}{dt} \, = \, \frac{1}{\left( \frac{dt'}{dt}\right)} \, \frac{dx'}{dt} \end{align} Let us calculate both factors \begin{align} &\frac{dx'}{dt} \, = \, \frac{\partial x'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial x'}{\partial t}\\ &\frac{dt'}{dt} \, = \, \frac{\partial t'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial t'}{\partial t} \end{align} Thus, the first derivative becomes \begin{align} \frac{dx'}{dt'} \, = \, \frac{\,\frac{\partial x'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial x'\,}{\partial t}}{ \frac{\partial t'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial t'}{\partial t}} \end{align} For example, in the case of Lorenz, \begin{align} &\frac{dx'}{dt} \, =\, \gamma \,\left(\frac{dx}{dt} \, - \, V\right)\\ &\frac{dt'}{dt} \,=\, \gamma \,\left(-\, V\, \frac{dx}{dt} \, + \, 1\right) &\text{ where } \, \gamma = \frac{1}{\sqrt{1-V^2}} \end{align} $$ \frac{dx'}{dt'} \, = \, \frac{\, \frac{dx}{dt} \, - \, V}{1\, -\, V\, \frac{dx}{dt} \,} \, = \, \frac{\, \frac{dx}{dt} \, - \, V \,}{1 \, - \, V\, \frac{dx}{dt}} $$ Now, for the second derivative \begin{align} \frac{d^2x'}{dt'^2} \, =& \, \frac{d}{dt'}\left( \frac{dx'}{dt'}\right) \, = \, \frac{d}{dt'}\left( \frac{1}{\left(\frac{dt'}{dt}\right)}\, \frac{dx'}{dt} \right) \\ =& \, \frac{dt}{dt'}\,\frac{d}{dt}\left( \frac{1}{\left(\frac{dt'}{dt}\right)}\, \frac{dx'}{dt} \right) \, = \, \frac{dt}{dt'}\,\left[\, \frac{d}{dt}\left( \left(\frac{dt'}{dt}\right)^{-1} \,\right) \, \frac{dx'}{dt} \, + \, \left(\frac{dt'}{dt}\right)^{-1} \, \frac{d^2x'}{dt^2} \, \right] \\ =& \, \frac{dt}{dt'}\,\left[\, - \, \left(\frac{dt'}{dt}\right)^{-2} \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2}\, + \, \left(\frac{dt'}{dt}\right)^{-1} \, \frac{d^2x'}{dt^2} \, \right] \\ =& \, \left(\frac{dt'}{dt}\right)^{-1}\,\left[\, - \, \left(\frac{dt'}{dt}\right)^{-2} \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2}\, + \, \left(\frac{dt'}{dt}\right)^{-1} \, \frac{d^2x'}{dt^2} \, \right] \\ =& \, - \, \left(\frac{dt'}{dt}\right)^{-3} \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2} \, + \, \left(\frac{dt'}{dt}\right)^{-2} \, \frac{d^2x'}{dt^2} \\ =& \, \left(\frac{dt'}{dt}\right)^{-2} \, \frac{d^2x'}{dt^2} \, - \, \left(\frac{dt'}{dt}\right)^{-3} \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2}\\ =& \, \left(\frac{dt'}{dt}\right)^{-3} \,\left[\, \frac{dt'}{dt}\,\frac{d^2x'}{dt^2} \, - \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2} \,\right]\\ \end{align} where the formulas for the second derivatives can be derived fro mthe chain rule as follows \begin{align} \frac{d^2x'}{dt^2} \, =& \, \frac{d}{dt} \left(\frac{dx'}{dt}\right) \, = \, \frac{d}{dt} \left(\frac{\partial x'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial x'}{\partial t}\right) \\ =& \, \frac{\partial x'}{\partial x}\, \frac{d^2x}{dt^2} \, + \, \frac{\partial^2 x'}{\partial x^2}\,\left(\frac{dx}{dt}\right)^2\, + \, 2\, \frac{\partial^2 x'}{\partial x \partial t}\, \frac{dx}{dt} \, + \, \frac{\partial^2 x'}{\partial t^2} \\ &\\ \frac{d^2t'}{dt^2} \, =& \, \frac{d}{dt} \left(\frac{dt'}{dt}\right) \, = \, \frac{d}{dt} \left(\frac{\partial t'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial t'}{\partial t}\right) \\ =& \, \frac{\partial t'}{\partial x}\, \frac{d^2x}{dt^2} \, + \, \frac{\partial^2 t'}{\partial x^2}\,\left(\frac{dx}{dt}\right)^2\, + \, 2\, \frac{\partial^2 t'}{\partial x \partial t}\, \frac{dx}{dt} \, + \, \frac{\partial^2 t'}{\partial t^2} \end{align} So to summarize, $$\frac{d^2x'}{dt'^2}\, = \, \left(\frac{dt'}{dt}\right)^{-3} \,\left[\, \frac{dt'}{dt}\,\frac{d^2x'}{dt^2} \, - \, \frac{dx'}{dt} \, \frac{d^2t'}{dt^2} \,\right]\\$$ where \begin{align} &\frac{dx'}{dt} \, = \, \frac{\partial x'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial x'}{\partial t}\\ &\frac{dt'}{dt} \, = \, \frac{\partial t'}{\partial x}\, \frac{dx}{dt} \, + \, \frac{\partial t'}{\partial t}\\ \frac{d^2x'}{dt^2} \, =& \,\frac{\partial x'}{\partial x}\, \frac{d^2x}{dt^2} \, + \, \frac{\partial^2 x'}{\partial x^2}\,\left(\frac{dx}{dt}\right)^2\, + \, 2\, \frac{\partial^2 x'}{\partial x \partial t}\, \frac{dx}{dt} \, + \, \frac{\partial^2 x'}{\partial t^2} \\ \frac{d^2t'}{dt^2}\, =& \, \frac{\partial t'}{\partial x}\, \frac{d^2x}{dt^2} \, + \, \frac{\partial^2 t'}{\partial x^2}\,\left(\frac{dx}{dt}\right)^2\, + \, 2\, \frac{\partial^2 t'}{\partial x \partial t}\, \frac{dx}{dt} \, + \, \frac{\partial^2 t'}{\partial t^2} \end{align} In the case of Lorenz transformations, \begin{align} &\frac{dx'}{dt} \, =\, \gamma \,\left(\frac{dx}{dt} \, - \, V\right)\\ &\frac{dt'}{dt} \,=\, \gamma \,\left(-\, V\, \frac{dx}{dt} \, + \, 1\right)\\ &\frac{d^2x'}{dt^2} \, =\, \gamma \,\frac{d^2x}{dt^2} \\ &\frac{d^2t'}{dt^2} \,=\, -\,\gamma\, V\, \frac{d^2x}{dt^2}\\ &\text{ where } \, \gamma = \frac{1}{\sqrt{1-V^2}} \end{align} Thus $$\frac{d^2x'}{dt'^2}\, = \, \left(\,-\,\gamma\, V\, \frac{dx}{dt} \, + \, \gamma\,\right)^{-3} \,\left[\, \left(-\, \gamma\, V\, \frac{dx}{dt} \, + \, \gamma\right) \, \gamma \,\frac{d^2x}{dt^2} \, + \, \left(\gamma\,\frac{dx}{dt} \, - \, \gamma\,V\right)\, \gamma\, V\, \frac{d^2x}{dt^2} \,\right]$$ $$= \, \left(\,-\,\gamma\, V\, \frac{dx}{dt} \, + \, \gamma\,\right)^{-3} \,\left[\, \left(1\, -\, V\, \frac{dx}{dt} \, \right) \, + \, \left(\frac{dx}{dt} \, - \,V\right)\, V\, \right] \, \gamma^2 \,\frac{d^2x}{dt^2} $$ $$= \, \gamma^{-3}\,\left(\,1\, - \,V\, \frac{dx}{dt} \, \right)^{-3} \,\left[\, 1\, -\, V\, \frac{dx}{dt} \, + \, V\,\frac{dx}{dt} \, - \,V^2\, \right] \, \gamma^2 \,\frac{d^2x}{dt^2}$$ $$= \, \gamma^{-1}\,\left(\,1\, - \,V\, \frac{dx}{dt} \, \right)^{-3} \,\big(\, 1\, - \,V^2\, \big) \,\frac{d^2x}{dt^2}$$ $$= \, \frac{\,\gamma^{-1} \,\big(\, 1\, - \,V^2\, \big) \,}{\left(\,1\, - \,V\, \frac{dx}{dt} \, \right)^{3}}\,\,\frac{d^2x}{dt^2}$$ $$= \, \frac{\,\big(\, 1\, - \,V^2\, \big)^{\frac{1}{2}} \,\big(\, 1\, - \,V^2\, \big) \,}{\left(\,1\, - \,V\, \frac{dx}{dt} \, \right)^{3}}\,\,\frac{d^2x}{dt^2}$$ And here is the formula

$$\frac{d^2x'}{dt'^2}\, = \, \frac{\,\big(\, 1\, - \,V^2\, \big)^{\frac{3}{2}} \,}{\left(\,1\, - \,V\, \frac{dx}{dt} \, \right)^{3}}\,\,\frac{d^2x}{dt^2}$$

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  • $\begingroup$ Just before you begin the second derivative calculation, your expression for $dx’/dt’$ has a typo. The denominator has a sign error. (This expression for velocity should look like relative velocity formula.) [The final expression for the acceleration agrees with what was sought.] $\endgroup$ – robphy Jul 24 at 20:06
  • $\begingroup$ @robphy Thank you for pointing it out. I fixed it. $\endgroup$ – Futurologist Jul 24 at 21:29

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