Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/223

and the work spent in producing the motion is equivalent to the kinetic energy. Hence

$T_{p{\dot {q}}}={\frac {1}{2}}(p_{1}{\dot {q}}_{1}+p_{2}{\dot {q}}_{2}+\And c.),$

(13)

where $T_{p{\dot {q}}}$ denotes the kinetic energy expressed in terms of the momenta and velocities. The variables $q_{1},q_{2},$ &c. do not enter into this expression.

The kinetic energy is therefore half the sum of the products of the momenta into their corresponding velocities.

When the kinetic energy is expressed in this way we shall denote it by the symbol $T_{p{\dot {q}}}$. It is a function of the momenta and velocities only, and does not involve the variables themselves.

563.] There is a third method of expressing the kinetic energy, which is generally, indeed, regarded as the fundamental one. By solving the equations (3) we may express the momenta in terms of the velocities, and then, introducing these values in (13), we shall have an expression for T involving only the velocities and the variables. When T is expressed in this form we shall indicate it by the symbol $T_{\dot {q}}$. This is the form in which the kinetic energy is expressed in the equations of Lagrange.

564.] It is manifest that, since $T_{p},T_{\dot {q}}$, and $T_{p{\dot {q}}}$ are three different expressions for the same thing,

$T_{p}+T_{\dot {q}}-2T_{p{\dot {q}}}=0,$

or

$T_{p}+T_{\dot {q}}-p_{1}{\dot {q}}_{1}-p_{2}{\dot {q}}_{2}-\And c.=0.$

(14)

Hence, if all the quantities $p,q,$ and ${\dot {q}}$ vary,

${\begin{aligned}&\left({\frac {dT_{p}}{dp_{1}}}-{\dot {q}}_{1}\right)\delta p_{1}+\left({\frac {dT_{p}}{dp_{2}}}-{\dot {q}}_{2}\right)\delta p_{2}+\And c.\\+&\left({\frac {dT_{\dot {q}}}{d{\dot {q}}_{1}}}-p_{1}\right)\delta {\dot {q}}_{1}+\left({\frac {dT_{\dot {q}}}{d{\dot {q}}_{2}}}-p_{2}\right)\delta {\dot {q}}_{2}+\And c.\\+&\left({\frac {dT_{p}}{dq_{1}}}+{\frac {dT_{\dot {q}}}{dq_{1}}}\right)\delta q_{1}+\left({\frac {dT_{p}}{dq_{2}}}+{\frac {dT_{\dot {q}}}{dq_{2}}}\right)\delta q_{2}+\And c.=0.\end{aligned}}$

(15)

The variations δp are not independent of the variations δq and $\delta {\dot {q}}$, so that we cannot at once assert that the coefficient of each variation in this equation is zero. But we know, from equations (3), that

${\frac {dT_{p}}{dp_{1}}}-{\dot {q}}_{1}=0,\quad \And c.,$

(16)

so that the terms involving the variations δp vanish of themselves.

The remaining variations $\delta {\dot {q}}$ and δq are now all independent, so that we find, by equating to zero the coefficients of $\delta {\dot {q}}_{1}$, &c,

$p_{1}={\frac {dT_{\dot {q}}}{d{\dot {q}}_{1}}},\quad p_{2}={\frac {dT_{\dot {q}}}{d{\dot {q}}_{2}}},\And c.;$

(17)