I am trying to understand the following example:
Relativistic Television
A non-flat screen, older-style television display works by
accelerating electrons over a short distance to relativistic speed,
and then using electromagnetic fields to control where the electron
beam strikes a fluorescent layer at the front of the tube. Suppose the
electrons travel at $6.00\times{10}^{7}$ m/s through a distance of
$0.200$ m from the start of the beam to the screen. (a) What is the time of travel of an electron in the rest frame of the television set? (b)
What is the electron’s time of travel in its own rest frame?
The electron beam in a cathode ray tube television display.
Strategy for (a): Calculate the time from $vt=d$. Even though the speed is relativistic, the calculation is entirely in one frame of reference, and relativity is therefore not involved.
My question is why isn't relativity involved in measuring the time it takes for the electron to travel across the distance of $0.2$ m?
Best Answer
The answer is in the question. The calculation is entirely in one frame of reference. You only Lorentz transform when switching frames.
You may be confused because many introductory texts treat mass thus:
" $ m:=m_0$, $E^2=m^2c^4 + p^2c^2$ "
And then introduce a pseudo-Lorentz transformation for mass in the unprimed frame,
" $m_{relativistic} = \gamma m = \gamma m_0$ "
This is false - or at least, misleading, awkward, and useless. $m$ is the mass as measured in the frame doing the measuring, period.
$m \ne m_0$ for $v \ne 0$
The frame transformation for mass is $m' = m/\gamma$. An object's rest mass $m_0$, that is, its mass in its own frame, is the primed frame of whatever unprimed frame measured $m$.
Mass and its consequences (force, momentum, etc) are no exception, and you never Lorentz transform unless you're switching frames.