15.24: Kinetic Energy
- Page ID
- 8477
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)If a force \( \bf{F}\) acts on a particle moving with velocity \( \bf{u}\), the rate of doing work – i.e. the rate of increase of kinetic energy \( T\) is \( \dot{T}=\bf{F\cdot u}\). But \( \bf{F=\dot{p}}\) where \( \bf{p}=m\bf{u}=\gamma m_{0}u\).
(A point about notation may be in order here. I have been using the symbol \( \bf{v}\) and \( v\) for the velocity and speed of a frame \( \Sigma'\) relative to a frame \( \Sigma\), and my choice of axes without significant loss of generality has been such that \( \bf{v}\) has been directed parallel to the \( x\)-axis. I have been using the symbol \( \bf{u}\) for the velocity (speed = \( u\)) of a particle relative to the frame \( \Sigma\). Usually the symbol \( \gamma\) has meant \( \left(1-\frac{v^{2}}{c^{2}}\right)^{-\frac{1}{2}}\), but here I am using it to mean \( \left(1-\frac{u^{2}}{c^{2}}\right)^{-\frac{1}{2}}\). I hope that this does not cause too much confusion and that the context will make it clear. I toyed with the idea of using a different symbol, but I thought that this might make matters worse. Just be on your guard, anyway.)
We have, then
\[ \bf{F}=m_{0}(\dot{\gamma}u+\gamma\dot{u}) \label{15.24.1} \]
and therefore
\[ \dot{T}=m_{0}(\dot{\gamma}u^{2}+\gamma\bf{\dot{u}\cdot u}). \label{15.24.2} \]
Making use of Equations 15.23.5 and 15.23.6 we obtain
\[ \dot{T}=\dot{\gamma}m_{0}c^{2} \label{15.24.3} \]
Integrate with respect to time, with the condition that when \( \gamma\) = 1, \( T\) = 0, and we obtain the following expression for the kinetic energy:
\[ T=(\gamma-1)m_{0}c^{2}. \label{15.24.4} \]
Exercise. Expand \( \gamma\) by the binomial theorem as far as \( \frac{u^{2}}{c^{2}}\), and show that, to this order, \( T=\frac{1}{2}mu^{2}\).
I here introduce the dimensionless symbol
\[ K=\frac{T}{m_{0}c^{2}}=\gamma-1 \label{15.24.5} \]
to mean the kinetic energy in units of \( m_{0}c^{2}\). The second half of this was already given as Equation 15.3.5.