$$\require{cancel}$$

# 15.20: Acceleration

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Figure XV.33 shows two references frames, $$\Sigma$$ and $$\Sigma'$$, the latter moving at speed $$v$$ with respect to the former. A particle is moving with acceleration $$\bf{a'}$$ in $$\Sigma'$$. (“ in $$\Sigma'$$ ” = “referred to the reference frame $$\Sigma'$$ ”.) The velocity is not necessarily, of course, in the same direction as the acceleration, and we’ll suppose that its velocity in $$\Sigma'$$ is $$\bf{u'}$$. The acceleration and velocity components in $$\Sigma'$$ are $$a'_{x'}, a'_{y'}, u'_{x'}, u'_{y'}$$.

What is the acceleration of the particle in $$\Sigma$$? We shall start with the $$x$$-component.

The $$x$$-component of its acceleration in $$\Sigma$$ is given by

$a_{x}=\frac{du_{x}}{dt}, \label{15.20.1}$

where

$u_{x}=\frac{u_{x}'+v}{1+\frac{u_{x}'v}{c^{2}}} \label{15.16.2}$

and

$t=\gamma\left(t'+\frac{vx'}{c^{2}}\right) \label{15.5.19}\tag{15.5.19}$

Equations 15.16.2 and 15.5.19 give us

$du_{x}=\frac{du_{x}}{du'_{x}}du'_{x}=\frac{du'_{x}}{\gamma^{2}(1+\frac{u'_{x}v}{c^{2}})^{2}} \label{15.20.2}$

and

$dt\ =\frac{\partial t}{\partial t'}dt'+\frac{\partial t}{\partial x'}dx'=\gamma dt'\ +\ \frac{\gamma v}{c^{2}}dx' \label{15.20.3}$

On substitution of these into Equation $$\ref{15.20.1}$$ and a very little algebra, we obtain

$a_{x}=\frac{a'}{\gamma^{3}(1+\frac{u'_{x}v}{c^{2}})^{3}} \label{15.20.4}$

The $$y$$-component of its acceleration in $$\Sigma$$ is given by

$a_{y}=\frac{du_{y}}{dt}, \label{15.20.5}$

We have already worked out the denominator $$dt$$(Equation $$\ref{15.20.3}$$). We know that

$u_{y}=\frac{u'_{y'}}{\gamma(1+\frac{u'_{x'}v}{c^{2}})} \label{15.16.3}\tag{15.16.3}$

from which

$du_{y}=\frac{\partial u_{y}}{\partial u'_{x'}}+\frac{\partial u_{y}}{\partial u'_{y'}}\partial u'_{y'}=\frac{1}{\gamma}\left(-\frac{vu'_{y'}}{c^{2}\left(1+\frac{vu'_{x'}}{c^{2}}\right)^{2}}du'_{x'}+\frac{1}{1+\frac{vu'_{x'}}{c^{2}}}du'_{y'}\right). \label{15.20.6}$

Divide Equation $$\ref{15.20.6}$$ by Equation $$\ref{15.20.3}$$ to obtain

$a_{y}=\frac{1}{\gamma^{2}}\left(-\frac{vu'_{y'}}{c^{2}\left(1+\frac{vu'_{x'}}{c^{2}}\right)^{2}}a'_{x'}+\frac{1}{1+\frac{vu'_{x'}}{c^{2}}}a'_{y'}\right). \label{15.20.7}$