Physics Asked by gfsch on August 4, 2020
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 tright)}{sqrt{1-{V^2}}}
t’ &= frac{left( t – V xright)}{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^2right)}^{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.
What have you tried?
It might help to see your attempt to see where you got stuck.
Try using the chain rule:
$$ begin{align} color{red}{frac{dx'}{dt'}} =frac{dx'}{dt}frac{dt}{dt'} &=frac{left(displaystylefrac{dx'}{dt}right)}{left(displaystylefrac{dt'}{dt}right)} &=frac{ displaystylefrac{d}{dt}left( frac{x-Vt}{(1-V^2)^{1/2}} right) }{displaystyle frac{d}{dt}left( frac{t-Vx}{(1-V^2)^{1/2}} right) } end{align} $$ Then, $$ begin{align} frac{d^2 x'}{dt'^2}=frac{d}{dt'}left( color{red}{frac{dx'}{dt'}} right) =frac{dleft( {frac{dx'}{dt'}} right)}{dt}frac{dt}{dt'} &=frac{displaystyleleft(frac{dleft( frac{dx'}{dt'} right)}{dt}right)}{left(displaystylefrac{dt'}{dt}right)} &=frac{displaystylefrac{d}{dt}left( color{red}{frac{dx'}{dt'}}right)} {displaystyle frac{d}{dt}left( frac{t-Vx}{(1-V^2)^{1/2}} right)} end{align} $$
Correct answer by robphy on August 4, 2020
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),, tbig) &t' = t'big(x(t),, tbig) end{align} For example, in the case of Lorenz transformations begin{align} &x' = gamma ,big(x , - , V,tbig) &t' =gamma ,big(-, V, x , + , tbig) &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} , - , Vright) &frac{dt'}{dt} ,=, gamma ,left(-, V, frac{dx}{dt} , + , 1right) &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} , - , Vright) &frac{dt'}{dt} ,=, gamma ,left(-, V, frac{dx}{dt} , + , 1right) &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 $$frac{d^2x'}{dt'^2}, = , left(,-,gamma, V, frac{dx}{dt} , + , gamma,right)^{-3} ,left[, left(-, gamma, V, frac{dx}{dt} , + , gammaright) , gamma ,frac{d^2x}{dt^2} , + , left(gamma,frac{dx}{dt} , - , gamma,Vright), gamma, V, frac{d^2x}{dt^2} ,right]$$ $$= , left(,-,gamma, V, frac{dx}{dt} , + , gamma,right)^{-3} ,left[, left(1, -, V, frac{dx}{dt} , right) , + , left(frac{dx}{dt} , - ,Vright), V, right] , gamma^2 ,frac{d^2x}{dt^2} $$ $$= , gamma^{-3},left(,1, - ,V, frac{dx}{dt} , right)^{-3} ,left[, 1, -, V, frac{dx}{dt} , + , V,frac{dx}{dt} , - ,V^2, right] , gamma^2 ,frac{d^2x}{dt^2}$$ $$= , gamma^{-1},left(,1, - ,V, frac{dx}{dt} , right)^{-3} ,big(, 1, - ,V^2, big) ,frac{d^2x}{dt^2}$$ $$= , frac{,gamma^{-1} ,big(, 1, - ,V^2, big) ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$ $$= , frac{,big(, 1, - ,V^2, big)^{frac{1}{2}} ,big(, 1, - ,V^2, big) ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$ And here is the formula
$$frac{d^2x'}{dt'^2}, = , frac{,big(, 1, - ,V^2, big)^{frac{3}{2}} ,}{left(,1, - ,V, frac{dx}{dt} , right)^{3}},,frac{d^2x}{dt^2}$$
Answered by Futurologist on August 4, 2020
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