Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/222

and the complete variation of T_p is

${\begin{aligned}\delta T_{p}&=\sum \left({\frac {dT_{p}}{dp}}\delta p+{\frac {dT_{p}}{dq}}\delta q\right),\\&=\sum \left(\left({\frac {dp}{dt}}+{\frac {dT_{p}}{dq}}\right)dq\right).\end{aligned}}$

(7)

But the increment of the kinetic energy arises from the work done by the impressed forces, or

$\delta T_{p}=\sum {(F\delta q)}.\,$

(8)

In these two expressions the variations δq are all independent of each other, so that we are entitled to equate the coefficients of each of them in the two expressions (7) and (8). We thus obtain

$F_{r}={\frac {dp_{r}}{dt}}+{\frac {dT_{p}}{dq_{r}}},$

(9)

where the momentum p_r and the force F_r belong to the variable q_r.

There are as many equations of this form as there are variables. These equations were given by Hamilton. They shew that the force corresponding to any variable is the sum of two parts. The first part is the rate of increase of the momentum of that variable with respect to the time. The second part is the rate of increase of the kinetic energy per unit of increment of the variable, the other variables and all the momenta being constant.

The Kinetic Energy expressed in Terms of the Momenta and Velocities.

562.] Let $p_{1}$, $p_{2}$, &c. be the momenta, and ${\dot {q}}_{1}$, ${\dot {q}}_{1}$, &c. the velocities at a given instant, and let ${\text{p}}_{1}$, ${\text{p}}_{2}$, &c., ${\dot {\text{q}}}_{1}$, ${\dot {\text{q}}}_{2}$, &c. be another system of momenta and velocities, such that

${\text{p}}_{1}=np_{1},\quad {\dot {\text{q}}}_{1}=n{\dot {q}}_{1},\And c.$

(10)

It is manifest that the systems ${\text{p}},{\dot {\text{q}}}$ will be consistent with each other if the systems, $p,{\dot {q}}$ are so.

Now let n vary by δn. The work done by the force F₁ is

$F_{1}\delta q_{1}={\dot {\text{q}}}_{1}\delta p_{1}={\dot {q}}_{1}p_{1}n\delta n.$

(11)

Let n increase from 0 to 1, then the system is brought from a state of rest into the state of motion $({\dot {q}}p$ and the whole work expended in producing this motion is

$({\dot {q}}_{1}p_{1}+{\dot {q}}_{2}p_{2}+\And c.)\int _{0}^{1}{n\,dn}.$

(12)

But

$\int _{0}^{1}{n\,dn}={\frac {1}{2}},$