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Relativity with stationary rod

  • #1
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An observer in S' sees a stationary rod lying in the x'-y' plane and making an angle [itex]\phi'[/itex] with the x' axis. Show that
[itex]\tan{\phi'}=\frac{1}{\gamma} \tan{\phi}[/itex]

where [itex]\phi[/itex] is the angle the rod makes with the x axis according to an observer in S.

so far I have [itex]\tan{\phi'} = \frac{y'}{x'}=\frac{y}{\gamma(x-vt)}=\frac{1}{\gamma} \frac{y}{x-vt}[/itex]

if i could just get rid of that vt, i'd be there but i can't seem to get it to go away...
 

Answers and Replies

  • #2
dx
Homework Helper
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I assume the rod is moving in the x direction. In the S frame, let the length in the x direction be Lx and it's length in the y direction be Ly. In the S' frame, the length in the y direction will be the same i.e. L'y = Ly, and the length in the x direction will be shorter by a factor of √(1 - v²), i.e. L'x = √(1 - v²) Lx. So we get

tan θ = Ly / Lx

tan θ' = L'y / L'x = Ly/Lx√(1 - v²)
 
  • #3
dx
Homework Helper
Gold Member
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Your mistake was that you took tan θ to be y/x. If the coordinates of the tips of the rod are (x1, y1) and (x2, y2), then tan θ will be ∆y/∆x where ∆x = x2 - x1 and ∆y = y2 - y1. You should apply the Lorentz transform to both (x1, y1) and (x2, y2) to obtain ∆x' and ∆y'. tan θ' will then be ∆y'/∆x'.
 

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