Chapter II

Absorbed Energy

148. Let us consider an oscillator which has just completed an emission and which has, accordingly, lost all its energy of vibration. If we reckon the time $t$ from this instant then for $t = 0$ we have $f = 0$ and $\frac{d f}{d t} = 0$ , and the vibration takes place according to equation (233). Let us write $E_{z}$ as in (149) in the form of a Fourier’s series:

(235)

$E_{z} = \sum_{n = 1}^{n = \infty} [A_{n} cos \frac{2 π n t}{T} + B_{n} sin \frac{2 π n t}{T}],$

where $T$ may be chosen very large, so that for all times $t$ considered $t < T$ . Since we assume the radiation to be stationary, the constant coefficients $A_{n}$ and $B_{n}$ depend on the ordinal numbers $n$ in a wholly irregular way, according to the hypothesis of natural radiation (Sec. 117). The partial vibration with the ordinal number $n$ has the frequency $ν$ , where

(236)

$ω = 2 π ν = \frac{2 π n}{T},$

while for the frequency $ν_{0}$ of the natural period of the oscillator

$ω_{0} = 2 π ν_{0} = \sqrt{\frac{K}{L}} .$

Taking the initial condition into account, we now obtain as the solution of the differential equation (233) the expression

(237)

$f = \sum_{1}^{\infty} [a_{n} (cos ω t - cos ω_{0} t) + b_{n} (sin ω t - \frac{ω}{ω_{0}} sin ω_{0} t)],$

where

(238)

$a_{n} = \frac{A_{n}}{L (ω_{0}^{2} - ω^{2})}, b_{n} = \frac{B_{n}}{L (ω_{0}^{2} - ω^{2})} .$

This represents the vibration of the oscillator up to the instant when the next emission occurs.

The coefficients $a_{n}$ and $b_{n}$ attain their largest values when $ω$ is nearly equal to $ω_{0}$ . (The case $ω = ω_{0}$ may be excluded by assuming at the outset that $ν_{0} T$ is not an integer.)

149. Let us now calculate the total energy which is absorbed by the oscillator in the time from $t = 0$ to $t = τ$ , where

(239)

$ω_{0} τ is large .$

According to equation (234), it is given by the integral

(240)

$\int_{0}^{τ} E_{z} \frac{d f}{d t} d t,$

the value of which may be obtained from the known expression for $E_{z}$ (235) and from

(241)

$\frac{d f}{d t} = \sum_{1}^{\infty} [a_{n} (- ω sin ω t + ω_{0} sin ω_{0} t) + b_{n} (ω cos ω t - ω cos ω_{0} t)] .$

By multiplying out, substituting for $a_{n}$ and $b_{n}$ their values from (238), and leaving off all terms resulting from the multiplication of two constants $A_{n}$ and $B_{n}$ , this gives for the absorbed energy the following value:

$\begin{matrix} \frac{1}{L} \int_{0}^{τ} d t \sum_{1}^{\infty} [\frac{A_{n}^{2}}{ω_{0}^{2} - ω^{2}} cos ω t (- ω sin ω t + ω_{0} sin ω_{0} t) \\ + \frac{B_{n}^{2}}{ω_{0}^{2} - ω^{2}} sin ω t (ω cos ω t - ω cos ω_{0} t)] . \end{matrix}$

In this expression the integration with respect to $t$ may be performed term by term. Substituting the limits $τ$ and $0$ it gives

$\begin{array}{l} \frac{1}{L} \sum_{1}^{\infty} \frac{A_{n}^{2}}{ω_{0}^{2} - ω^{2}} [- \frac{{sin}^{2} ω τ}{2} + ω_{0} (\frac{{sin}^{2} \frac{ω_{0} + ω}{2} τ}{ω_{0} + ω} + \frac{{sin}^{2} \frac{ω_{0} - ω}{2} τ}{ω_{0} - ω})] \\ + & \frac{1}{L} \sum_{1}^{\infty} \frac{B_{n}^{2}}{ω_{0}^{2} - ω^{2}} [\frac{{sin}^{2} ω τ}{2} - ω (\frac{{sin}^{2} \frac{ω_{0} + ω}{2} τ}{ω_{0} + ω} - \frac{{sin}^{2} \frac{ω_{0} - ω}{2} τ}{ω_{0} - ω})] . \end{array}$

In order to separate the terms of different order of magnitude, this expression is to be transformed in such a way that the difference $ω_{0} - ω$ will appear in all terms of the sum. This gives

$\begin{matrix} \frac{1}{L} \sum_{1}^{\infty} \frac{A_{n}^{2}}{ω_{0}^{2} - ω^{2}} [\frac{ω_{0} - ω}{2 (ω_{0} + ω)} {sin}^{2} ω τ + \frac{ω_{0}}{ω_{0} + ω} sin \frac{ω_{0} - ω}{2} τ \cdot sin \frac{ω_{0} + 3 ω}{2} τ \\ + \frac{ω_{0}}{ω_{0} - ω} {sin}^{2} \frac{ω_{0} - ω}{2} τ] \\ + \frac{1}{L} \sum_{1}^{\infty} \frac{B_{n}^{2}}{ω_{0}^{2} - ω^{2}} [\frac{ω_{0} - ω}{2 (ω_{0} + ω)} {sin}^{2} ω τ - \frac{ω}{ω_{0} + ω} sin \frac{ω_{0} - ω}{2} τ \cdot sin \frac{ω_{0} + 3 ω}{2} τ \\ + \frac{ω}{ω_{0} - ω} {sin}^{2} \frac{ω_{0} - ω}{2} τ] . \end{matrix}$

The summation with respect to the ordinal numbers $n$ of the Fourier’s series may now be performed. Since the fundamental period $T$ of the series is extremely large, there corresponds to the difference of two consecutive ordinal numbers, $Δ n = 1$ only a very small difference of the corresponding values of $ω$ , $d ω$ , namely, according to (236),

(242)

$Δ n = 1 = T d ν = \frac{T d ω}{2 π},$

and the summation with respect to $n$ becomes an integration with respect to $ω$ .

The last summation with respect to $A_{n}$ may be rearranged as the sum of three series, whose orders of magnitude we shall first compare. So long as only the order is under discussion we may disregard the variability of the $A_{n}^{2}$ and need only compare the three integrals

$\begin{matrix} \int_{0}^{\infty} d ω \frac{{sin}^{2} ω τ}{2 {(ω_{0} + ω)}^{2}} = J_{1}, \\ \int_{0}^{\infty} d ω \frac{ω_{0}}{{(ω_{0} + ω)}^{2} (ω_{0} - ω)} sin \frac{ω_{0} - ω}{2} τ \cdot sin \frac{ω_{0} + 3 ω}{2} τ = J_{2}, \\ and \\ \int_{0}^{\infty} d ω \frac{ω_{0}}{(ω_{0} + ω) {(ω_{0} - ω)}^{2}} {sin}^{2} \frac{ω_{0} - ω}{2} τ = J_{3} . \end{matrix}$

The evaluation of these integrals is greatly simplified by the fact that, according to (239), $ω_{0} τ$ and therefore also $ω τ$ are large numbers, at least for all values of $ω$ which have to be considered. Hence it is possible to replace the expression ${sin}^{2} ω τ$ in the integral $J_{1}$ by its mean value $\frac{1}{2}$ and thus we obtain:

$J_{1} = \frac{1}{4 ω_{0}} .$

It is readily seen that, on account of the last factor, we obtain

$J_{2} = 0$

for the second integral.

In order finally to calculate the third integral $J_{3}$ we shall lay off in the series of values of $ω$ on both sides of $ω_{0}$ an interval extending from $ω_{1}$ ( $< ω_{0}$ ) to $ω_{2}$ ( $> ω_{0}$ ) such that

(243)

$\frac{ω_{0} - ω_{1}}{ω_{0}} and \frac{ω_{2} - ω_{0}}{ω_{0}} are small,$

and simultaneously

(244)

$(ω_{0} - ω_{1}) τ and (ω_{2} - ω_{0}) τ are large.$

This can always be done, since $ω_{0} τ$ is large. If we now break up the integral $J_{3}$ into three parts, as follows:

$J_{3} = \int_{0}^{\infty} = \int_{0}^{ω_{1}} + \int_{ω_{1}}^{ω_{2}} + \int_{ω_{2}}^{\infty},$

it is seen that in the first and third partial integral the expression ${sin}^{2} \frac{ω_{0} - ω}{2} τ$ may, because of the condition (244), be replaced by its mean value $\frac{1}{2}$ . Then the two partial integrals become:

(245)

$\int_{0}^{ω_{1}} \frac{ω_{0} d ω}{2 (ω_{0} + ω) {(ω_{0} - ω)}^{2}} and \int_{ω_{2}}^{\infty} \frac{ω_{0} d ω}{2 (ω_{0} + ω) {(ω_{0} - ω)}^{2}} .$

These are certainly smaller than the integrals:

$\int_{0}^{ω_{1}} \frac{d ω}{2 {(ω_{0} - ω)}^{2}} and \int_{ω_{2}}^{\infty} \frac{d ω}{2 {(ω_{0} - ω)}^{2}}$

which have the values

(246)

$\frac{1}{2} \frac{ω_{1}}{ω_{0} (ω_{0} - ω_{1})} and \frac{1}{2 (ω_{2} - ω_{0})}$

respectively. We must now consider the middle one of the three partial integrals:

$\int_{ω_{1}}^{ω_{2}} d ω \frac{ω_{0}}{(ω_{0} + ω) {(ω_{0} - ω)}^{2}} \cdot {sin}^{2} \frac{ω_{0} - ω}{2} τ .$

Because of condition (243) we may write instead of this:

$\int_{ω_{1}}^{ω_{2}} d ω \cdot \frac{{sin}^{2} \frac{ω_{0} - ω}{2} τ}{2 {(ω_{0} - ω)}^{2}}$

and by introducing the variable of integration $x$ , where

$x = \frac{ω - ω_{0}}{2} τ$

and taking account of condition (244) for the limits of the integral, we get:

$\frac{τ}{4} \int_{- \infty}^{+ \infty} \frac{{sin}^{2} x d x}{x^{2}} = \frac{τ}{4} π .$

This expression is of a higher order of magnitude than the expressions (246) and hence of still higher order than the partial integrals (245) and the integrals $J_{1}$ and $J_{2}$ given above. Thus for our calculation only those values of $ω$ will contribute an appreciable part which lie in the interval between $ω_{1}$ and $ω_{2}$ , and hence we may, because of (243), replace the separate coefficients $A_{n}^{2}$ and $B_{n}^{2}$ in the expression for the total absorbed energy by their mean values $A_{0}^{2}$ and $B_{0}^{2}$ in the neighborhood of $ω_{0}$ and thus, by taking account of (242), we shall finally obtain for the total value of the energy absorbed by the oscillator in the time $τ$ :

(247)

$\frac{1}{L} \frac{τ}{8} (A_{0}^{2} + B_{0}^{2}) T .$

If we now, as in (158), define $I$ , the “intensity of the vibration exciting the oscillator,” by spectral resolution of the mean value of the square of the exciting field-strength $E_{z}$ :

(248)

$\bar{E_{z}^{2}} = \int_{0}^{\infty} I_{ν} d ν$

we obtain from (235) and (242):

$\bar{E_{z}^{2}} = \frac{1}{2} \sum_{1}^{\infty} (A_{n}^{2} + B_{n}^{2}) = \frac{1}{2} \int_{0}^{\infty} (A_{n}^{2} + B_{n}^{2}) T d ν,$

and by comparison with (248):

$I = \frac{1}{2} (A_{0}^{2} + B_{0}^{2}) T .$

Accordingly from (247) the energy absorbed in the time $τ$ becomes:

$\frac{I}{4 L} τ,$

that is, in the time between two successive emissions, the energy $U$ of the oscillator increases uniformly with the time, according to the law

(249)

$\frac{d U}{d t} = \frac{I}{4 L} = a .$

Hence the energy absorbed by all $N$ oscillators in the time $d t$ is:

(250)

$\frac{N I}{4 L} d t = N a d t .$