5 6 D O C U M E N T 3 0 J A N U A R Y 1 9 2 2 wanders through the bisulf.-of-carb. tube. Meanwhile the wave-fan wanders at the relat. velocity V − U from the tail of the group to the head of the group, while new waves are constantly forming at the tail and just as many are dying away at the head and three observers running along at the head, at the middle, and at the tail of the group persistently get to see virtually vertical standing wave planes, even if the group travels for kms through the bisulfide-of-carbon (thanks to Gibbs’s theorem!). So now, Einstein, I’m pretty sure of myself.[6] 1.) During passage through the slit (or during a fraction of this time), the canal-ray particle transmits a fan-shaped group of waves into the apparatus, therefore ultimately into the bisulfide- of-carbon tube. The inclination of the foremost and hindmost planes upon entry into the tube amount to + , or −ε , resp. 2.) Follow (with Gibbs) the wandering of this group through the bisulf.-of-carbon.—It wanders with “group”-velocity U(λ) toward the right.– 3.) Meanwhile the individ. waves are moving relative to the group at the veloc. V(λ) − U(λ) from left (tail) to right (head of the group) and at the tail end new waves are con- stantly being born, while at the head end just as many are dying. 4.) If we follow an individual wave plane, it rotates (in heaven’s name) as much rapidly as you calculate it (on this point you, Rayleigh, and Gibbs agree)—and if we !!could!! follow an individual wave as it wanders through the whole tube, it would ultimately rotate by some angle the tube only has to be long enough. 5.) But we can’t do so—because the wave exists only as long as it lies within the group. 6.) So we just need to check which wave inclinations can occur within the group.—This is easily done with the help of Gibbs’s theorem. 7.) For Gibbs proves: If we let a system of coordinates move along in parallel to itself at the group’s velocity U(λ) toward the right, then—since the waves are mov- ing at the greater veloc. V(λ), new waves are constantly consecutively occupying the 0 point of the running coordinate axes. But all these consecutive waves have exactly the same inclination[7] [The proof goes such that he demonstrates: In the time required for the (n + 1)th wave behind the nth to reach the 0 point, it has rotated just as much as it had been askew to its predecessor.] (at which inclination, of course, depends on whether the 0 point of the running coordinate axes is moving at the front, in the middle, or at the rear of the group). ε (“wave veloc.”)