Err, did I do this length contraction problem right?

  • #1
schattenjaeger
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it's a typical homework problem I suppose, a meter stick traveling at v=.8c at an angle of 60 degrees to v as seen in the stick's reference frame. I solved it out and got .6m, which is incidentaly the same answer I got for if it were traveling parallel to v.

is that, you know, how that works?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
860
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The stick only contracts along the direction it is moving (call that the x-direction). Since it as tilted at 60°, it's length along the x direction is .5m, while it's length in the y direction is [itex]\frac{\sqrt{3}}{2}m[/itex]. So it is the .5m that gets multiplied by a factor of .6, while the y component is constant. To find the total relativistic length, just use pythagorus.
 
  • #3
schattenjaeger
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okok, I did THAT, and got .3m, and didn't think that seemed right for the total length of the stick, so I stuck it back in the 30-60-90 triangle and got .6 for the length of the stick

err, does that make sense?

oh wait, I see, I almost did it right, the .3 then is the x component, and I use the same y component, which I didn't do


edit: So it's .916515m?
 
  • #4
LeonhardEuler
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Yeah, that's what I get.
 
  • #5
schattenjaeger
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Sweet, ok, another question

How many joules per kilogram of rest mass does it take to accelerate an object from rest to .9c?

E=gamma(mc^2)-mc^2, solve for E/m, and it should be some pretty gigantic number? I'm just checking, I got somethingx10^17
 
  • #6
LeonhardEuler
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Yes, I think that's right.
 
  • #7
schattenjaeger
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Hmm, is it? That's the formula for kinetic energy, so I found how much kinetic energy per kilogram something has at .9c, but is that also the energy required to accelerate it to that speed? Or could it just be the total energy gamma(mc^2)?
 

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