Page:Amazing Stories Volume 21 Number 06.djvu/157

We have

3.4

$F=f(x)=f(y)=f(z)$

4. Let m₁ be any material particle, (2.1), and Vₜᵤ its velocity at a given instant parallel to the direction, u; (u = x, y, z). Then the momentum of m₁ is

4.1

$M_{1}=m_{1}V_{iu}$

If m₁, m₂, . . ., mₙ are n such particles called the set Sₙ, (22), then at the given instant

4.2

$M_{u}=\sum _{i=l}^{n}m_{1}V_{iu}$

and Mᵤ is the momentum of Sₙ in the direction u; (u = x, y, z).

Then there is a mean value for Vᵢᵤ which will be denoted by Vₒᵤ, for which

4.3

$M_{u}=V_{ou}\sum _{i=l}^{n}m_{o}V_{ou}$

where

4.4

$M_{o}=\sum _{i=l}^{n}m_{i}$

5. The energy of mᵢ due to its velocity Vᵢᵤ is

5.1

$E_{t}u=1/2m_{i}V_{iu}^{2}$

and the total energy of Sₙ in the direction u, (u = x, y, z), is

5.2

$E_{u}=1/2\sum _{i=l}^{n}m_{i}V_{iu}^{2}$

Then there is a mean value Qᵤ for Vᵢᵤ such that

5.3(see 4.4)

$E_{u}=1/2m_{o}Q_{u}^{2}$

Then

5.4

$Q_{u}={\sqrt {\frac {2E_{u}}{m_{o}}}}$

is the velocity Sₙ would have if its internal energy were translated into linear velocity along u.

6. Let

6.1

$E_{u}=1/2m_{o}V_{ou}^{2}+I_{u}$

From 6.1 it is evident that (Eᵤ — Iᵤ) is the energy Sₙ would have due to its velocity Vₒᵤ is each m₁ had the same velocity Vₒᵤ; (S.2). There- fore Iᵤ must be the component of Eᵤ, (the total energy of Sₙ), that is internal energy. From 6.1 and 5.3 we have

6.2

$I_{u}=1/2(Q_{u}-V_{ou}^{2})m_{o}$

This expression may be written as

6.3

${\begin{aligned}I_{u}&=1/2\sum _{i=l}^{n}m_{i}V_{iu}^{2}-1/2\sum _{i=l}^{n}m_{i}V_{ou}^{2}\\&=1/2\sum _{i=l}^{n}m_{i}(V_{iu}^{2}-V_{ou}^{2})\end{aligned}}$

Now let I′ᵤ be the u-component of the internal energy of Sₙ relative to Vₒᵤ taken at rest. Then

6.4

$I'_{u}=1/2\sum _{i=l}^{n}m_{i}(V_{iu}-V_{ou})^{2}$

On squaring, separating, and collecting terms, we get

6.5

$I'_{u}=1/2\sum _{i=l}^{n}m_{i}(V_{iu}^{2}+V_{ou}^{2})-V_{ou}\sum _{i=l}^{n}m_{i}V_{iu}$

Subtracting 6.5 from 6.3 we get, (4.3);

6.6

$I'_{u}-I_{u}=0$

Whence,

6.7

$I_{u}=1/2\sum _{i=l}^{n}m_{i}(V_{iu}-V_{ou})^{2}$

Since this is equivalent to 6.3 it follows that the total energy of Sₙ is equal to the sum of its internal energy and its energy of velocity.

7. Consider the total energy of Sₙ as being that of an unknown mass, mᵤ, along u, of velocity Vₒᵤ. Then

7.1

$E_{u}=1/2m_{u}V_{ou}^{2},V_{ou}\neq 0$

Then we may consider mᵤ as being a sum of the mass mₒ, (4.4), and an increment Dmᵤ. Thus

7.2

$m_{u}=m_{o}+Dm_{u}$

From 6.1 and 7.1 we have

7.3

$m_{u}V_{ou}^{2}=m_{o}V_{ou}^{2}+2I_{u}$

whence, from 7.2 we get

7.4

$V_{ou}^{2}Dm_{u}=2I_{u}$

From 7.1 we have

7.5

$V_{ou}^{2}=2E_{u}/m_{u}$

Substituting in 7.4 and transposing,

7.6

$Dm_{u}=m_{u}I_{u}/E_{u}$

whence, from 7.2 we get

7.7

$Dm_{u}={\frac {m_{o}r}{l-r}}$

where

7.8

$r=I_{u}/E_{u}$

Then, from 7.2 we get

7.9

$m_{u}={\frac {m_{o}}{l-r}}$

8. From the law of conservation of energy it is evident that if all the energy E of Sₙ were given out to any system, then Sₙ would come to rest relative to that system and every m₁ would also come to rest in that system.

Conversely, if Sₙ were to gain back its original velocity and internal energy from that system it would have to do so by diverting part of the acquired energy to internal energy and use the remainder for acceleration to its former velocity.

Hence, it is evident that r in 7.8 is in reality this diversion fraction. It follows that

8.1

$r'=l-r={\frac {E_{u}-I_{u}}{E_{u}}}$

is the fraction of the energy, or the work done on Sₙ, that will produce linear velocity. Hence,