Rotating cylinder on x'-axis in S' frame. Find twist per unit length in S frame

[crissyb1988]

Homework Statement

A cylinder rotating uniformly about the x' axis of S' will seem twisted when observed instantaneously in S, where it not only rotates but also travels forward. If the angular speed of the cylinder in S' is ω, prove that in S the twist per unit length is yωv/c(squared). Here S and S' are inertial frames of reference in the standard configuration with respect to one another. y= gamma factor

Homework Equations

twist per unit length = yωv/c(squared)
Lorentz equations
Inverse Lorentz equations

The Attempt at a Solution

By twist per unit length ii think it means dθ/dx where the x-axis lines up with the axis of the cylinder?.

We can write the angular speed as
ω= dθ'/dt',
and then transposing we get
dθ'=ωdt'
because theta is in the z-y plane we can say that dθ'=dθ ?

So subbing in dt'=y(dt-vdx/c2) we get

dθ= ωy(dt-vdx/c2)

divide thru by dx we get

dθ/dx= ωy( dt/dx - v/c2)

dθ/dx = ωy/v - ωyv/c2

The answer should be dθ/dx = ωyv/c2 . BTW I don't think it matters about the negative sign but why am i left with ωy/v ?

Would really appreciate some hints :)

 

[bloby]

It's the comparison, at fixed time t in S of elements of the cylinder separated by dx AND that correspond to elements in S' with same θ' for a fixed t'. The thing is (x,t) and (x+dx,t) correspond to two different times t₁' & t₂' in S'. Elements of the cylinder with θ' at t₁' are at θ'+ω(t₂'-t₁') at t₂'.