Length Contraction: Solving for L

[Herricane]

Homework Statement

A rod of length L_0 moves with a speed v along the horizontal direction. The rod makes an angle of θ_0 with respect to the x'-axis.

a. Show that the length of the rod as measured by a stationary observer is given by L = L_o [1 - (v^2 / c^2) cos^2 (θ_o) ]^(.5)

Homework Equations

L = L_p/ gamma

The Attempt at a Solution

I have a few questions:

Is L proper L_o (the S frame?)
When I am trying to find the length of the rod as measured by a stationary observer do I refer to the graph to the right?

Horizontal length is all I need to worry about, correct?

L = L_o / gamma

x' = L cos θ_o

x' = x ( 1 - v^2/c^2 )^(-1/2)

L cos θ_o = x ( 1 - v^2/c^2 )^(-1/2)

L = x / [ (1 - v^2/c^2 )^(1/2) cos θ_o ] where x = L_o

L = L_o / [ (1 - v^2/c^2 )^(1/2) cos θ_o ]

Am I on the right track? I can't seem to make it look like:
L = L_o [1 - (v^2 / c^2) cos^2 (θ_o) ]^(.5)

 

[Mindscrape]

Wait, huh, is the picture you attached one given by the problem or something you made? The picture is really different from how I read the problem.

The way you're approaching the problem seems fine. But if you're going to find the x coordinates, you need to find two x coordinates because length is x2-x1. Looks like you're doing a strange mixing of length contraction and x coordinates. Either approach is okay, but you'll probably confuse yourself going between them (confused me at least).

 

[Herricane]

Yes I think that is my problem. I don't know how to draw the diagram. Can you explain me what you understood? I have read it several times and I just don't get it

 

[Mindscrape]

So you have an S' that moving relative to the S frame; their x-axis is collinear. In S' we have a proper length L_o, and the length makes an angle with respect to the x' axis. So the part that will be contracted is L_o*cosØ'. L_o*sinØ' will stay the same in both frames. S even sees a different angle than S' does.

 

[Herricane]

Thank you, I figured out the problem. I was making it way too complicated.