Page:Die Kaufmannschen Messungen.djvu/3

${\mathsf {for}}\ 0<\xi <\xi '\ {\mathsf {is}}\ {\mathfrak {E}}_{1}=1$ ${\mathsf {for}}\ \xi '<\xi <{\frac {x_{1}}{2}}\ {\mathsf {is}}\ {\mathfrak {E}}_{1}=\varkappa -\lambda \xi$

2)

Then, because of the continuity of ${\mathfrak {E}}_{1}$:

$\varkappa -\lambda \xi '=1$ und $\varkappa -\lambda {\frac {x_{1}}{2}}=$0.

The value of the constant ξ' I assumed to be as large, so that that the value of the "electric field integral" is the same as Kaufmann's. Its value is:^[1]

$(x_{1}-x_{1})\cdot \left\{\int _{x_{0}}^{x_{1}}{\mathfrak {E}}_{1}dx-{\frac {1}{x_{1}-x_{0}}}\int _{x_{0}}^{x_{1}}dx\int _{x_{0}}^{x}{\mathfrak {E}}_{1}dx\right\}=1,565$.

Substituting the above values, it follows:

${\frac {1}{2}}(x_{2}-x_{1})\cdot \left({\frac {x_{1}}{2}}+\xi '\right)=1,565$

and from this:

ξ = 0,593 ϰ = 2,468 λ = 2,475.

To reduce the electric field strength to the absolute electrostatic unit: ${\mathfrak {E}}$, or to the absolute electromagnetic unit: ${\mathfrak {E}}_{m}$, one has:^[2]

${\mathfrak {E}}_{m}={\mathfrak {E}}_{1}\cdot {\frac {25\cdot 10^{10}}{0,1242}}={\mathfrak {E}}\cdot 3\cdot 10^{10}$

3)

Whether the simplistic assumptions made in relation to the electric and magnetic field are really sufficient for the relevant calculations, will be shown below.

§ 3. Magnetic deflection.

If we introduce into the equations of motion (§ 1) the momentum vector (quantity of motion)

${\frac {\partial H}{\partial q}}=p$

4)

and also the electromagnetic unit of the electric field strength (${\mathfrak {E}}_{m}$) and for the electrical elementary quantum (ε), then they are:

${\frac {d}{dt}}\left(p{\frac {\dot {x}}{q}}\right)=-\epsilon {\dot {z}}{\mathfrak {H}}$

5)

${\frac {d}{dt}}\left(p{\frac {\dot {y}}{q}}\right)=\epsilon {\mathfrak {E}}_{m}$

6)

${\frac {d}{dt}}\left(p{\frac {\dot {z}}{q}}\right)=\epsilon {\dot {x}}{\mathfrak {H}}$.

7)

Because ${\mathfrak {H}}$ is constant, (5) and (7) can be integrated with respect to time t. Dividing the two resulting equations, t, p and q are entirely eliminated, and a second integration yields the equation of the trajectory of projection on the xz-plane, a circle, which goes through the points x = 0, z = 0, x = x₁, z = 0 and x = x₂, $z={\bar {z}}$ and is determined by it. The current coordinates x, z of the points of this circle can be represented as functions of one variable parameter: the angle φ which is the tangent of the circle in the direction of motion on the x-axis, and it is positive when the motion is to the side of the positive z-axis:

$x=\varrho \sin \varphi +{\frac {x_{1}}{2}},\qquad z=-\varrho \cos \varphi +{\frac {x_{1}}{2}}ctg\ \varphi _{1}$.

8)

Where

$\varrho ={\frac {x_{1}}{2\sin \varphi _{1}}}$

9)

is the radius of the circle and φ₁ the value of φ for x = x₁. In these equations it is already expressed that for x = 0 and x = x₁, z = 0. If it is considered that x = x₂, $z={\bar {z}}$, then we have the values:

$tg\ \varphi _{1}={\frac {x_{1}{\bar {z}}}{(x_{2}-x_{1})x_{2}+{\bar {z}}^{2}}}$

10)

and

$\sin \varphi _{2}={\frac {2x_{2}-x_{1}}{x_{1}}}\sin \varphi _{1}$

11)

and also $\varrho$ from (9).

By inserting (8) into (5) or (7) we have:

${\frac {p}{q}}\cdot {\dot {\varphi }}=\epsilon {\mathfrak {H}}$.

12)

Now it is:

$q^{2}=\varrho ^{2}{\dot {\varphi }}^{2}+{\dot {y}}^{2}$,

for which we can put with sufficient approximation:^[3]

$q=\varrho {\dot {\varphi }}$.

13)

Therefore:

$p=\epsilon {\mathfrak {H}}\varrho =\epsilon {\mathfrak {H}}{\frac {x_{1}}{2\sin \varphi _{1}}}$.

14)

The momentum p of each electron is independent of time and, without entering into a special theory, can be calculated from the magnetic deflection ${\bar {z}}$.

Since p is independent of time t, the same follows for the velocity q and by (12) for the angular velocity ${\dot {\varphi }}$. So angle φ is linearly dependent on time t.

§ 4. Electrical deflection.

From (6) it follows:

${\frac {p}{q}}{\frac {d^{2}y}{d\varphi ^{2}}}\cdot {\dot {\varphi }}^{2}=\epsilon {\mathfrak {E}}_{m}$

and from this according to (12) and (13)

${\frac {d^{2}y}{d\varphi ^{2}}}={\frac {\varrho }{q}}{\frac {{\mathfrak {E}}_{m}}{\mathfrak {H}}}$

15)

1. l. c. p. 526 and p. 547.
2. l. c. p. 547
3. l. c. p. 527.