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Passage from the book relativity i am not able to understand

  1. Aug 17, 2012 #1
    Below is a passage from a book "RELATIVITY THE SPECIAL AND GENERAL THEORY by ALBERT EINSTEIN" and i am not able to understand the trajectories of the stone from the train and from the ground.
    I stand at the window of a railway carriage which is travelling uniformly, and drop a stone on the embankment, without throwing it. Then, disregarding the influence of the air resistance, I see the stone descend in a straight line. A pedestrian who observes the misdeed from the footpath notices that the stone falls to earth in a parabolic curve. I now ask: Do the “positions” traversed by the stone lie “in reality” on a straight line or on a parabola?"
     
  2. jcsd
  3. Aug 17, 2012 #2

    Mentz114

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    Both. It depends on your point of view. Relative to the train the stone has no horizontal velocity but to the pedestrian it has some.
     
  4. Aug 17, 2012 #3

    tiny-tim

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    Hi RohitRmB! :smile:

    The train is moving horizontally with speed v.

    The stone is dropped, which means that it receives no impulse, so its initial velocity is also v horizontally (and 0 vertically).

    So it follows a parabola (as seen by the pedestrian), with vx = v, so x = vt, y = -1/2 gt2.

    Since you are on the train, your vx is the same as the stone, so you are always vertically above the stone.

    So in your frame (the train's frame) of reference, the horizontal component of velocity of the stone is always 0 …

    ie, you see the stone fallng vertically (in a straight line) :wink:
     
  5. Aug 17, 2012 #4

    HallsofIvy

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    Your assumption that one of the "parabola" or "straight" is "real" is in error. The path depends on the observer. (And this has nothing to do with Einstein's relativity. This is result of Galillean relativity.)
     
  6. Aug 18, 2012 #5
    if i were standing on a moving train and i drop a stone, neglecting the effect of air, would'nt i see it moving backward because when i leave the stone from my hands, even when i am still standing on the train, even when initial the stone was at rest relative to me?
     
  7. Aug 18, 2012 #6

    tiny-tim

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    Hi RohitRmB! :wink:
    no, it's because the stone was at rest relative to you

    that it starts with the same velocity that you had, ie v horizontally and 0 vertically

    since there are no horizontal forces on it, its horizontal component of velocity will always be 0, the same as yours :smile:
     
  8. Aug 18, 2012 #7
    But meanwhile it is falling, i am moving ahead with the train, whereas there is no horizontal component of velocity with the stone, so would'nt it appear to be going in opposite direction to me?
     
  9. Aug 18, 2012 #8

    tiny-tim

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    yes there is!!

    it starts with the same horizontal component that you had
     
  10. Aug 18, 2012 #9

    Mentz114

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    The first post tells you that there is a horizontal component in the pedestrian frame - pay attention !
     
  11. Aug 18, 2012 #10

    ghwellsjr

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    Next time you're riding in a car, why don't you try the experiment? Just drop something and see if it appears to you to fall straight down or go flying backwards. As long as you drop it within the car it will go straight down. If you drop it out the window, the air will force it backwards, but your quote said "disregarding the influence of the air resistance", which, of course, you can't do, but you have to do what Einstein said later on in his book in chapter 7:
    "We shall imagine the air above [our embankment] to have been removed".
     
  12. Aug 18, 2012 #11

    cepheid

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    Actually there is. After all, it had a horizontal component of velocity when it was moving along with the train, and it retains that after being dropped. For more details, try Googling Newton's First Law of Motion.

    Halls, bear in mind that RohitRmB wasn't necessarily assuming that. This question here:

    was in fact posed by Einstein, in the book.
     
  13. Aug 19, 2012 #12
    thanks everybody, i have understood and my problem is now solved.
    actually at first i was not able to imagine that when i leave the stone, the stone is now disconnected from my hand but it still moves forward along with me, so after leaving the stone from my hand as i move forward in the train the stone is also moving at the same speed with me, and the gravitational force acting on the stone is perpendicular to the direction of velocity so it will not affect its horizontal velocity, so i will see it falling straight down.
    where as for a observer on the ground, the stone has a positive horizontal velocity as well as vertical velocity downwards, so for him it follows a parabolic trajectory towards the ground.
     
  14. Aug 19, 2012 #13
    thanks everybody for your help,
    first i was not able to visualize
     
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