Imagine the S’-frame moves at constant velocity (see the frames above this line):
![\[ \mathbf{v}=(v,0,0)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-4a4ecb6a326e86f280874cdf286aec78_l3.png "Rendered by QuickLaTeX.com")
relative to the S-frame. In the S’-frame an object moves with acceleration
![\[ \mathbf{a}'=(a'_x,a'_y,a'_z)=\left(\dfrac{du'_x}{dt'},\dfrac{du'_y}{dt'},\dfrac{du'_z}{dt'}\right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-5168515746be50a7045bc5331de1eb5c_l3.png "Rendered by QuickLaTeX.com")
QUESTION: What is the acceleration in the S-frame?
Of course it has to be something like this
![\[ \mathbf{a}=(a_x,a_y,a_z)=\left(\dfrac{du_x}{dt},\dfrac{du_y}{dt},\dfrac{du_z}{dt}\right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-8bcc55a773840bd852b407456acd462b_l3.png "Rendered by QuickLaTeX.com")
In order to get the relationship between both accelerations (and frames) we have to use the addition law of velocities from the previous post. Without loss of generality, we will use the rule (I) and we leave the general case as an exercise for the eager reader. We obtain:
![\[ a_x=\dfrac{du_x}{dt}=\dfrac{\dfrac{du_x}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\dfrac{u'_x+v}{1+\dfrac{u'_xv}{c^2}}\right]}{\dfrac{1}{c}\dfrac{d}{dt'}\left[\gamma(ct'+\beta x')\right]}=\dfrac{\dfrac{du'_x}{dt'}\left(1+\dfrac{u'_xv}{c^2}\right)-(u'_x+v)\dfrac{v}{c^2}\dfrac{du'_x}{dt'}}{\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(\dfrac{dt'}{dt'}+\dfrac{1}{c}\beta \dfrac{dx'}{dt'}\right)}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-218ae9a579296c37fb7d60372db8d724_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\[ a_x=\dfrac{\left(1-\dfrac{v^2}{c^2}\right)a'_x}{\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(1+\dfrac{u'_x v}{c^2}\right)}=\dfrac{1}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}a'_x\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-ac19aa1a0da48550b6d310a98d2ce6a5_l3.png "Rendered by QuickLaTeX.com")
Thus we get the first transformed component transformation for the acceleration:
![\[ \boxed{a_x=\dfrac{a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-ee81fcf38a4a2894d9eeec280aebd4da_l3.png "Rendered by QuickLaTeX.com")
For the transverse components, say 
( an analogue symmetrical calculation provides 
), we calculate
![\[ a_y=\dfrac{du_y}{dt}=\dfrac{\dfrac{du_y}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\dfrac{u'_y}{\gamma \left(1+\dfrac{u'_x v}{c^2}\right)}\right]}{\dfrac{1}{c}\dfrac{d}{dt'}\left[(ct'+\beta x')\right]}=\dfrac{\dfrac{du'_y}{dt'}\gamma \left(1+\dfrac{u'_x v}{c^2}\right)-u'_y \gamma \dfrac{v}{c^2}\dfrac{du'_x}{dt'}}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(\dfrac{dt'}{dt'}+\dfrac{1}{c}\beta\dfrac{dx'}{dt'}\right)}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-3f0bf0ffd96e78381492d3b530c84f39_l3.png "Rendered by QuickLaTeX.com")
Some basic algebra manipulations allow us to get:
![\[ a_y=\dfrac{\gamma \left(1+\dfrac{u'_x v}{c^2}\right)a'_y-\gamma\dfrac{u'_y v}{c^2}a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^2\left(1+\dfrac{u'_x v}{c^2}\right)}=\dfrac{a'_y}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_y v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-b70917a621b2c876774d225b21dc9aeb_l3.png "Rendered by QuickLaTeX.com")
Thus, the complete transformation of accelerations between frames from S’ to S(and from S to S’) are given by the following tables:
![\[ \boxed{\mbox{Transf.of acc.in SR:} S'\rightarrow S \begin{cases}a_x=\dfrac{a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}\\ \; \\ a_y=\dfrac{a'_y}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_y v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}\\ \; \\ a_z=\dfrac{a'_z}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_z v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}\end{cases}}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-1a236fdcdbf55dbb34defb9daaa9c149_l3.png "Rendered by QuickLaTeX.com")
![\[ \boxed{\mbox{Transf.of acc.in SR:} S\rightarrow S' \begin{cases}a'_x=\dfrac{a_x}{\gamma^3\left(1-\dfrac{u_x v}{c^2}\right)^3}\\ \; \\ a'_y=\dfrac{a_y}{\gamma^2 \left(1-\dfrac{u_x v}{c^2}\right)^2}+\dfrac{\dfrac{u_y v}{c^2}a_x}{\gamma^2\left(1-\dfrac{u_x v}{c^2}\right)^3}\\ \; \\ a'_z=\dfrac{a_z}{\gamma^2 \left(1-\dfrac{u_x v}{c^2}\right)^2}+\dfrac{\dfrac{u_z v}{c^2}a_x}{\gamma^2\left(1-\dfrac{u_x v}{c^2}\right)^3}\end{cases}}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-b6b7828b3f06e1af9d1adcd055ade776_l3.png "Rendered by QuickLaTeX.com")
where to obtain the second table we used the usual trick to map primed variables to unprimed variables and to map 
into 
.
One important comment is necessary in order to understand the above transformations. One could argue that the mathematical description of accelerated motions is beyond the scope of the special theory of relativity since it is a theory of inertial reference frames. THAT IDEA IS WRONG! Special Relativity (SR) IS based rather on the concept that every reference frame used for the calculations IS an inertial reference frame. It has NOTHING TO DO with accelerations between frames, and as we have showed you here, they can be calculated in the framework of SR!
In conclusion: the description of accelerated motions relative to inertial reference frames makes sense in SR and IS NOT subject to any limitation (unless, of course, you change/modify/extend the basic ideas and/or postulates of SR).
