D O C . 2 9 0 T H E O R Y O F R A D I O M E T E R F O R C E S 2 9 1
Let each molecule that hits the plate from the negative x side have the velocity
and leave the plate again at the velocity u in the negative x-direction. Let of
such collisions occur per unit area and unit time. , u, and are the correspond-
ing quantities for the other side of the plate. Here it is assumed that the velocities
of these molecules leaving the plate after collision are the same on both sides. Then
the condition of pressure equality is
(6)
Furthermore, the heat flow on both sides of the plate must be equally large,
which is expressed by the equation
. (7)
By dividing both these equations, one initially obtains
(8)
If this is inserted into (6), one gets, by replacing by v and this by :
(9)[4]
If there is a hole in the plate of surface area σ, which is small compared to the
free path length, then obviously more molecules must go through in the
direction of decreasing x per unit time than is measured in the opposite direction
, that is, the intensity of a molecular flow pushing (back) through the hole,
the apparent velocity of flow of which is given by the equation
(10)
From (9) and (10) follows the equation
(10a)[5]
which forms the counterpart to equation (5).
§3. Cause of Certain Radiometer Forces in Denser Gases. These results essen-
tially fall within the scope of the laws of the Knudsen gas, in which the effective
body dimensions are small compared to the free path length within the gas. Even
so, they also offer the key to certain radiometer phenomena in denser gases.
un νn
up νp
p
m
--- - νn( u un) + νp( u up) + = =
2f
m
---- - νn( un 2 u2) – νp( u2 up 2) – = =
2f
m
---- - un u – u up – = =
νn νp +
2
---------------- -
nu
6
----- -
νp νn –
1nf
6
------ -
p
- =
νp νn)σ – (
[p. 5]
νp νn –
νp νn – nv –=
v
1
6p
-- -
f
-–=--