Special relativity: Lorentz equations

  • Thread starter aznkid310
  • Start date
  • #1
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Homework Statement



A pole vaulter holds a 16ft pole. A barn has doors at both ends, 10 ft apart. The pole-vaulter on the outside of the barn begins running toward one of the open barn doors, holding the pole in the level direction he's running. When passing through the barn, the pole fits entirely in the barn at once. According to the stationary observer in the barn, which occurs first, the front end of the pole leaving the barn first or the back end entering, and what is the time interval between these events?



Homework Equations



I realize that the 'proper length' Lo = 16ft, and L = 10 ft. From there, I can get the velocity v and the factor y_v [measure of the departure of relativistic expectations].
I can then use the lorentz transformations, but I am having trouble finding the values

The Attempt at a Solution



L = (y_v)*Lo

10 = sqrt[1-(v^2/c^2)]*Lo

Solving for v: v = 0.78c

Thus, y_v = sqrt[1-(0.78^2)] = 0.626

Since we want to know t_2 - t_1, i used:

t_2 - t_1 = (y_v)*[(v/c^2)*(x'_2 - x'_1) + (t'_2 - t'_1)]

I dont know how to find the x primes and t primes (distances and times according to pole vaulter).
 

Answers and Replies

  • #2
2,268
7
10/16=0.625

if the coordinates in one frame are (x,t)
then the coordinates in another frame are (gamma*(x-vt), gamma*(t-vx))

but only if c=1. otherwise there is another term which I dont know because I never bother with it.
 
  • #3
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the origin of the 2 systems cooncide at t=0
 
  • #4
109
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are you saying that i can use the time dilation formula: t = (y_v) * t'?
 
  • #5
2,268
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no. time and distance are both involved. there is a lass of simultaneity. just use the formula I gave you.

btw, the stationary observer sees the pole fit into the barn. its the pole vaulter that sees it differently. so the question is worded wrongly.

from the stationary observers point of view:
back end of pole enters barn (0,0)
front end of pole leaves barn (10,0)
 

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