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Error in special relativity albert einstein assumed

  1. Mar 18, 2010 #1
    Error in special relativity
    albert einstein assumed (in the time dilation) a box-car moving with velocity V with a mirror in its ceiling and observer 1 sitting in the box car holding flash light vertically so when he sends a flash pulse , this pulse will go vertically , hit the mirror , and return back to the observer 1 so delta time = 2d/c.
    while according to observer 2 who is outside the box car what he have seen is that the flash light starts from observer 1 old position , and ends in observer 1 new position . (where difference in the distance is (delta time)*V) so according to observer 1 the light pulse has moved a longer distance than 2d.
    where delta time can be calculated afterward using pythogeran therom.
    and here lies the error
    that pulse went vertically , hit the mirror , and returned back , it didnt return back to observer 1 but it returned back to its same starting position as y vector of velocity of this light pulse is equal to c , while the x-vector is equal to zero(not affected by V of the box-car, and it has to be un-affected because if x-vector of the light pulse velocity has a value then the total V of the light will be more than c and its a fact for observer 1 and 2 that light velocity is c)
    so since x-vector is zero the light didnt change its x-coordinate so when light pulse returned back to its position (which was observer 1 oldpostion) the distance between the light pulse and observer 1 after delta time is (delta time * V) since the box car moved.
    while according to observer 2 the light pulse starts from observer 1 old position , and ends in observer 1 old position so simply delta time according to observer 2 the light pulse moved 2d
    so to both observer the light pulse moved the same distance. with the same velocity so equal delta time.
    This not an argue , i just wantto know whats wrong in the above...so i am not gonna argue just asking to understand.
     
    Last edited: Mar 18, 2010
  2. jcsd
  3. Mar 18, 2010 #2

    Doc Al

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    Re: relativity

    Why do you think the light doesn't return back to observer 1? Pretend observer 2 isn't even there. Someone in the box car is just reflecting the light off the (very tall) ceiling; to him, the light just moves vertically. Everything works just like you'd think, regardless of the box car's motion.

    Now add observer 2; Clearly observer 2 sees the light move sideways, since the boxcar is moving.
     
  4. Mar 18, 2010 #3

    Fredrik

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    Re: relativity

    I think you need to keep in mind that no matter whose point of view you're using to describe what happens, the description must include the same events. In this scenario there are three events: 1. The flashlight emits light. 2. The light gets reflected. 3. The reflected light hits the flashlight.

    When you describe these three events from observer 1's point of view, the light is going straight up and then straight down, and when you describe the same three events from observer 2's point of view, it isn't.

    The "up" component of the velocity of light in observer 2's coordinates is less than c. If it had been c, the speed of light would have been greater than c.
     
  5. Mar 18, 2010 #4
    Re: relativity

    as i said , the light being reflected by the moving mirror(due to the car motion) doesnt affect the position of the x-cordinate of the light it will just reflect it perpendicular .
    and the velocity of light is equal to c according to observer 1 and 2 since the direction of the light in both cases is downward BTW i am not talking in the case where light particles moving with the box-car (as it shouldnt) and that what i stated above that why do u think light particles move with the boxing car and when it get reflected back it comes to a new position(delta time * V) , untrue because from where did the light particles gain horizontal velocity to move relative to the boxing car?
     
  6. Mar 18, 2010 #5

    JesseM

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    Re: relativity

    Light always has the same speed regardless of the motion of the emitter, but the velocity (which includes direction) does depend on the motion of the emitter. This follows from the first postulate of relativity, which says the laws of physics work the same way in all frames. So, if you have a laser emitter standing vertically on the floor of a train car, and when the train is at rest relative to the ground the beam goes straight up and hits a point on the ceiling directly above the laser, then the beam still has to hit the same point on the ceiling if the train car is moving at constant velocity relative to the ground. If you're in a windowless train car that's moving inertially (not accelerating), any experiment you do inside the car has to give the same results regardless of how the train car is moving relative to outside landmarks; that's just an application of the first postulate.
     
  7. Mar 18, 2010 #6

    Matterwave

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    Re: relativity

    If you assume this, then there IS a preferred reference frame wherein there is absolute rest. This frame can be found as the frame in which a light shone straight up, will come straight down. So, if I were on the ground, i could tell that I was preferred, because if I were in the boxcar, the light would look like it's drifting. This is directly contrary to the postulate of relativity.
     
  8. Mar 18, 2010 #7

    Doc Al

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    Re: relativity

    From the view of observer 1, the mirror doesn't move.

    From the view of observer 1, the light just bounces vertically up and down. Of course the light moves with the boxcar.

    The light gained horizontal velocity due the motion of the car. Since the speed of light is constant from any frame, that implies that the vertical component of the light's velocity is less than c from observer 2's frame. A beam of light that is vertical in one frame, appears slanted from a horizontally moving frame.
     
  9. Mar 18, 2010 #8
    Re: relativity

    well, that cant be because if the train moving with constant velocity , and light or laser moving vertically with velocity c so to hit the same point as if the train wasnt moving it has to have also velocity in the x direction (let it V) so total velocity of the light will be sqrt(v^2 + c^2) so greater than c . and thats impossible
    the light cannot move vertically with speed c and move horizontally relative to the train because its velocity then will be greater than c.
    @ dr al said "that implies that the vertical component of the light's velocity is less than c from observer 2's frame. A beam of light that is vertical in one frame, appears slanted from a horizontally moving frame."
    yes observer 2 will see the vertical component of light is less than c, BUT THAT IF THE LIGHT IN FIRST PLACE WAS REFLECTED BACK IN THE NEW POSITION(V*DELTA TIME) but that what i am talking about that light not be reflected to new position but same position( as stated above light cannot gain velocity in x direction while moving vertically with speed of c)
     
    Last edited: Mar 18, 2010
  10. Mar 18, 2010 #9

    JesseM

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    Re: relativity

    But that's the point, in the frame where the light is moving diagonally, it moves slower than c in the vertical direction, which is how you can derive the time dilation equation via the idea of a light clock. In the frame where the train is moving, the vertical speed v1 will be less than c, and the horizontal speed v2 (which is just the velocity of the train) is also less than c, such that sqrt(v12 + v22) = c.
     
  11. Mar 18, 2010 #10
    Re: relativity

    plz read the post above :) , as i said that the light cannot move diagonally in the first place as when it will reflect back it will reflect in the same position it started, so it will appear to observer1 as moving vertically not diagonally
     
  12. Mar 18, 2010 #11

    JesseM

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    Re: relativity

    In the frame of observer1 on the train it moves vertically, in the frame of observer2 on the ground who sees the train in motion it moves diagonally. What's the problem with this? It would also be true that if observer1 tossed a ball vertically in the air it would move diagonally in the frame of observer2, although in that case the speed would not be the same in both frames.

    Is your argument that if its vertical speed is less than than c in observer2's frame, then it must also be less than c in observer1's frame? If so then you're forgetting about time dilation, which says the time that observer1 measures for the light to go from flashlight to ceiling is less than the time in observer2's frame, because in observer2's frame the clock of observer1 is running slow. Please read the light clock link I gave above if this isn't clear.
     
  13. Mar 18, 2010 #12

    Matterwave

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    Re: relativity

    I think you are still positing a notion of absolute rest, or an aether. Both ideas are unnecessary in the context of special relativity.

    If there is no absolute rest, there is no "it must be bounced back vertically". This is because "vertical" is different for the two observers.
     
  14. Mar 18, 2010 #13

    Dale

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    Re: relativity

    Think about this logically. How can the light pulse logically both return back to observer 1 and not return back to observer 1? This is a clear contradiction, your logic is unsound. Suppose observer 1 is a photodiode which triggers a bomb. Does the bomb explode or not? It cannot possibly both explode and not explode.
     
  15. Mar 19, 2010 #14
    Re: relativity

    thats because the light will not be reflected back to the flash light but it will be reflected back to a postion = postion of the flashlight- (V*delta time)
     
  16. Mar 19, 2010 #15
    Re: relativity

    it seems you dont read my posts as i explained why should light start at pos (x,y) and return back to pos(x,y) while when the car and the observer 1 will be in pos(x+(t*V) ,y)
    and observer2 didnt even move so basically
    observer 2 and the light didn't move from their position and observer 1 will see light bouncing up and reflecting back but not in the same position.
    even if you assume that light reflected back from the mirror again to the flash light(which is impossible) observer 2 will not assume then that light moved diagonally , he will assume that light changed it's position as it was moving with train and observer1(which is again impossible ) and both will have the same delta time!
     
  17. Mar 19, 2010 #16
    Re: relativity

    jeesm , the train will move horizontally or even diagonally according to observer2 but not the light , as the light have nothing to do with moving train and mirror as it will just start from postion1 and back to it(as i said million times above) and will not return to a new position(which was assumed x +v*t) as light moved vertically and back again(no velocity in x direction) so:
    to observer1: light moved vertically upward and returned back not in the same position as observer 1 has moved a little bit(which equal to v*t) while observer 2 will see light bouncing up and down normally.
     
  18. Mar 19, 2010 #17

    Fredrik

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    Re: relativity

    I don't know why you think so, but it's wrong. You're supposed to describe the same sequence of events from two different points of view. If observer 1's description includes the three events I listed in my previous post, then so does observer 2's description.

    Again, both descriptions must include the same events, so if (and only if) this is impossible in observer 2's description, then it's impossible in observer 1's description.

    Now you seem to be describing a different sequence of events than you did at first. That's fine, as long as you make sure that both descriptions include the same events. So for example, you can consider the scenario where observer 2 (on the ground outside the train) sees light go straight up and straight down. This scenario includes the following events:

    1. Light is emitted from a flashlight on the floor near the front of the train car.
    2. Light is reflected by a mirror in the ceiling near the middle of the train car.
    3. Light reaches the floor near the rear of the train car.

    So in this scenario observer 1 would have to describe the light as going "diagonally".
     
  19. Mar 19, 2010 #18

    Dale

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    Re: relativity

    I read and understood your posts. The problem is that your posts are illogical (self contradictory). You have specifically proposed that in one frame the light pulse returns to the flashlight, and in another frame it does not. Remember that the two different frames are simply two different and equivalent ways of describing the same situation. It is simply not logically possible for something to both happen and not happen. Just think about the implications of what you have proposed. It should be clear that if the pulse returns to the flashlight in one frame it must do so in all frames.
     
  20. Mar 19, 2010 #19

    Ich

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    Re: relativity

    You didn't understand the whole point of the example. As there clearly must be an x-component to the light speed, it follows logically that the y-component must be smaller acordingly, so that the speed remains c. That's a valid derivation of time dilatation.

    If I'm allowed a personal remark: If you, as a layman, get to clearly nonsensical results by trying to follow a popular account of time dilatation, it would be more appropriate to title your post "An error in my understanding?".
     
  21. Mar 19, 2010 #20

    Matterwave

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    Re: relativity

    @thetrice: In your scenario, observer 2 is obviously favored. You remark again and again that the light will "come back to the same position". But "the same position" is only from observer 2's point of view! Special relativity says that if you are traveling with uniform velocity relative to another observer, you couldn't tell whether it was you moving towards them, or them moving towards you; both views are equally valid. If you take this postulate as fact, there is no reason to think that observer 2 is favored, and the light must work according to his perspective. The light must actually go back to the flashlight because it was emitted by observer 1's flashlight!
     
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