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Contraction effects at relativistic velocities

  1. Sep 22, 2004 #1
    Just an issue I want to be absolutely clear about.

    Scenario,

    WE propell a rod of iron with a dimensions of 100 meters *1 meters through space with the 100 meter length perpedicular to the direction of velocity.

    So the rod is traveling width edge forward. ( perpedicular to direction)

    the question is :

    Does the perpendicular length of our rod contract?

    ( if the velocity length is 1 meter and our width is 100 meters as per direction)

    And also

    Is there a table that is available to give contraction results vs veocity somewhere on the Net?

    say contraction vs Earth meter at velocities like 0.1c, 0.2c, 0.3c etc etc

    any help would be appreciated
     
  2. jcsd
  3. Sep 22, 2004 #2
    another version of the same question

    If we have two ships both traveling parallel to each other at relativistic velocities is the distance of separation subject to contraction?
     
  4. Sep 22, 2004 #3

    Ba

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    I don't think so not with both ships at the same line perpendicular to motion, because they are not moving at all relative to each other and the horizontal distance will not shrink no matter how fast they go.
     
  5. Sep 22, 2004 #4

    Doc Al

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    length contraction formula

    Assuming you mean the length perpendicular to the motion (the "100 meter" length), then no. Length contraction only occurs in the direction of motion.
    No need for a table, the formula is simple. Let [tex]L_0[/tex] be the length of an object in its own (rest) frame; then L will be the length of the object measured from a frame in which the object is moving at speed v parallel to its length:
    [tex]L = L_0 \sqrt{1 - v^2/c^2}[/tex]
     
  6. Sep 22, 2004 #5

    selfAdjoint

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    Dearly Missed

    This is relative to the frame which sees the rod moving with velocity v.
     
  7. Sep 22, 2004 #6

    Doc Al

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    Thanks, selfAdjoint. I think my wording was a bit convoluted! :yuck:
     
  8. Oct 4, 2004 #7
    ok lets say there is a ship 100ft long moving at .90c. a planet perpendicular has guns 101ft apart at rest. the guns line of fire are parrallel to each other. the planet plans it out so that the ship will be in between the bullets when the ship flies past the planet. would the bullets hit the ship or not?
     
  9. Oct 4, 2004 #8

    Janus

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    I'm going to assume that you mean for the guns to fire simultaneously according to the planet.

    The answer is that the bullets will miss the ship, according to both the ship and the planet. From the ships view, the guns do not fire simultaneously.
     
  10. Oct 5, 2004 #9
    no i mean simultaneously according to the ship
     
  11. Oct 5, 2004 #10

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    If they are fired simultaneously according to the ship, then the planet can't plan it out such that the ship is between the bullets when the ship flies past the planet. In that case, according to the planet, the guns do not fire simultaneously. For instance, if from the ship frame, the bullets hit the ship while being fired simultaneously, then from the planet frame, the guns will fire in a staggered order timed such that the bullets each hit the ship
     
  12. Oct 6, 2004 #11
    why is it not simultaneous from both frames? if it is simultaneous from the planet then the ship sees that farthest gun shooting off first. right?
     
    Last edited: Oct 6, 2004
  13. Oct 7, 2004 #12
    With regards to your first question - No. For the reason why please see the bottom section of - http://www.geocities.com/physics_world/sr/lorentz_contraction.htm

    As per the question above - Assume that the rockets are identical and that they fire their engines at a constant rate as measured in an instantaneous rest frame. Let them start from rest each lying on the x-axis, one in front of the other. Then let an observer in the inertial frame measure the distance between them. The inertial observer will detect no change in distance between the two rockets. Not let either of the observers who are at rest in the rockets measure the distance. Each observer will measure the distance increasing. This is known as Bell's Spaceship Paradox. See - http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

    The reason has to do with gravitational time dilation and Lorentz contraction.

    The observer in the trailing rocket ship can think of himself as being at rest in a uniform gravitational field. He's firing his engines so as to neither rise nor fall. This observer sees the the other rocket ship higher in the gravitational field. However, due to gravitational time dilation, the bottom observer will see the rocket as firing his engines at a rate which is faster than his and therefore he gains height.

    Pete
     
  14. Oct 8, 2004 #13
    there is no rocket trailing the other. there are two guns, parallel to each other, and firing perpendicular to the line of movement of the ship.

    l=bullets path
    -=ships path
    s=space

    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    -----------lShipl------------
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
     
    Last edited: Oct 8, 2004
  15. Oct 8, 2004 #14

    Doc Al

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    From the viewpoint of the ship, the gun in the rear fires first.
     
  16. Oct 9, 2004 #15

    Janus

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  17. Oct 9, 2004 #16
    the animation did not work for me.

    ssssssssss^ssss^sssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    -----------lShipl------------>
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssslsssslssssssssssss
    ssssssssssAssssBsssssssssss

    B would shoot first right?
     
  18. Oct 10, 2004 #17
    i got the animation to work. the ship that is moving in both animations looks like it is trying to out run the light. how can they see the explosion before they see the light of the flash. they could not possibly see the explosion until they see the light from the explosion.
     
  19. Oct 10, 2004 #18

    Janus

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    It's not that observer two sees the explosions happen before he sees the flashes, it is when observer 2 determines the explosions had to occur in order for him to see the flashes at the same time. Example: If a light flash originates 1 light sec from me, and I see that flash when my clock reads 12:00:01, I know that the flash started when my clock read 12:00:00

    And if the flash originated 2 light secs from me and I see the flash at 12:00:01, then it originated when my clock read 11:59:59, Thus, if I see two flashes at the same time and one originated further from me, then the further explosion occured before the closer.

    Also:
    Assume you have observers at the actuators in the same frame as observer two. They have clocks which are synchronized with each other and a clock carried by observer 2.

    The blue observer stops his clock when he is opposite the other blue marker (We can assume that he passes so closely to that marker that we don't have to consider light signal delay.)

    The green observer does the same when he passes the green marker.

    Thus each observer has a clock that records when the explosion from his marker occured. If you were to bring the clocks together you will find that that the green observer's clock will read less than the blue observer's, and thus according to synchonized clocks sharing the same frame as observer 2, the explosions did not occur at the same time.
     
  20. Oct 12, 2004 #19
    would time dilation come into effect?
     
  21. Oct 12, 2004 #20

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    Time dilation needs to be considered when we are comparing the clock rates between frames. Such as the rate that observer 1 would determine for frame 2's clocks, or the rate that frame 1's clocks would have as measured by observer 2. In this instance, we are only concerned with when the events occured in each frame according to its own clocks.
     
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