How Does Length Contraction Affect Measurements in Different Reference Frames?

[richard7893]

Homework Statement

A meter stick at rest in S' is tilted at an angle of 30 degrees to the x'axis. If S'
moves a speed beta = 0.8 with respect to S.
a)How long is the the meter stick as measured in S ?
b)What angle does it make with the x-axis?

Homework Equations

l=l naught/gamma

The Attempt at a Solution

I have part a) as 0.72m. only parallel motion of meter stick will contract according to frame S. to find length in S frame I took 1m*cos(30)/gamma to see how much meter stick contracted relative to motion and it was 0.52. and took the answer to this and used pythagorean to get .72m( sqrt((.52)^2+sin^2(30)) , which is the total length observed in S. this has to mean that the answer in part b is 30 degrees because we assumed sin 30 degrees when I used the pythagorean in part a). Am I correct?

 

[jdwood983]

I have part a) as 0.72m. only parallel motion of meter stick will contract according to frame S. to find length in S frame I took 1m*cos(30)/gamma to see how much meter stick contracted relative to motion and it was 0.52. and took the answer to this and used pythagorean to get .72m( sqrt((.52)^2+sin^2(30)) , which is the total length observed in S. this has to mean that the answer in part b is 30 degrees because we assumed sin 30 degrees when I used the pythagorean in part a). Am I correct?

Doesn't seem to be correct. If you are contracting the x-direction while keeping the y-direction constant, doesn't this imply that

<br /> \theta_2=\tan^{-1}\left[\frac{y}{x}\right]\neq\tan^{-1}\left[\frac{y&#039;}{x&#039;}\right]<br />

where x&#039; is the relativistic frame.

 

[jdwood983]

whoops, misread your post and made some notes that are irrelevant now.

You have already identified S_x. You now need to find the angle \theta_2 that this contracted meter stick appears to the stationary observer. (Hint: you can use the equation in the previous post)

 

[richard7893]

I think what you wrote makes sense. When we use the pythagorean thm in part a) to find the total lenth observed is S we assume the "y component" of length remains unchenged from the moving frame to stationary frame, this is why we use sin30 in the pythagorean. however the angle from which S' sees the meterstick is different because the "x component" of length is changed in S' and changes the angle. Was this your logic in coming up with the arctan(y/x') equation? ANd do you mean theta in your equation not theta prime because we are looking for the angle in the stationary S frame? theta prime = arctan (y'/x') (which was given in question: 30 degrees) Is that what you meant to put? Theta prime is angle in moving frame S'. I am assuming you meant to put Theta = arctan (y'/x) where y'=y

 

[jdwood983]

I think what you wrote makes sense. When we use the pythagorean thm in part a) to find the total lenth observed is S we assume the "y component" of length remains unchenged from the moving frame to stationary frame, this is why we use sin30 in the pythagorean. however the angle from which S' sees the meterstick is different because the "x component" of length is changed in S' and changes the angle. Was this your logic in coming up with the arctan(y/x') equation?

Yes, this is the logic used here. You should always remember SOH CAH TOA to remember which components of the triangle are needed to find the angles. In this case, you know the y-component and the x'-component and can easily find the angle \theta.

ANd do you mean theta in your equation not theta prime because we are looking for the angle in the stationary S frame? theta prime = arctan (y'/x') (which was given in question: 30 degrees) Is that what you meant to put? Theta prime is angle in moving frame S'. I am assuming you meant to put Theta = arctan (y'/x) where y'=y

This is true, it should not be \theta&#039;. If anything, it should be \theta_2 to indicate it is not the same angle to begin with. The equation should read (and I edited the previous posts to make it read this)

<br /> \theta_2=\tan^{-1}\left[\frac{y}{x}\right]<br />

 

[gabbagabbahey]

This is true, it should not be \theta&#039;. If anything, it should be \theta_2 to indicate it is not the same angle to begin with. The equation should read (and I edited the previous posts to make it read this)

<br /> \theta_2=\tan^{-1}\left[\frac{y}{x&#039;}\right]<br />

I think you are confusing yourself here; \theta=\tan^{-1}\left[\frac{y}{x}\right] and \theta&#039;=\tan^{-1}\left[\frac{y&#039;}{x&#039;}\right]...the problem statement tells you that the stick is at rest in S&#039; and makes an angle of 30 degrees w.r.t. the x&#039;-axis...that means \theta&#039;=30^\circ and you are looking for \theta.

 

[jdwood983]

I think you are confusing yourself here; \theta=\tan^{-1}\left[\frac{y}{x}\right] and \theta&#039;=\tan^{-1}\left[\frac{y&#039;}{x&#039;}\right]...the problem statement tells you that the stick is at rest in S&#039; and makes an angle of 30 degrees w.r.t. the x&#039;-axis...that means \theta&#039;=30^\circ and you are looking for \theta.

Whoops...that's twice I've made prime & no-prime mistakes. Thanks for catching that one too.