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Physics LibreTexts

15.20: Acceleration

  • Page ID
    8473
  • [ "article:topic", "authorname:tatumj" ]

    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

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

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