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B Lorentz Transformation direction of motion

  1. Apr 17, 2017 #1
    Hi I was looking at the Lorentz transformation and I see that it moves in the x axis if vt is positive.
    How can I re-arrange the lorentz transformations in a way that will cause the moving frame of reference to get closer to me. I was trying with x'=gamma(x-vt) but I don't know what x is equal to in this case.

    I'm doing all this because I think when you move further away from the observer at rest the clock from the moving frame gets slower and slower. so the opposite must be true.

    But yes, what does x denote in x'=gamma(x-vt) and how can I reverse the motion of the moving frame so it comes towards me? I was thinking I would just put a minus sign on the v so it becomes x'=gamma(x+vt);
     
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  3. Apr 17, 2017 #2

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    The moving frame extends forever in all directions. The part in the direction of motion is going away from you but the part in the opposite direction is coming at you. If there is a very long train going past you, the front of the train is moving away from you but the rear of the train is coming toward you.
    That is not how it works. All moving clocks in the moving frame are ticking slower, regardless of where they are. For clocks on a long train, all those clocks seem to you as though they are going slow, regardless of if they are in the front going away or in the rear coming toward you. People on the train see no problems with their clocks. They think that all your clocks are running slow. The reason that both reference frames can think that the other's are running slow is that the two reference frames can not agree on how to synchronize clocks that are separated in the direction of motion.
     
  4. Apr 18, 2017 #3
    I just want to figure out incoming movement. Moving towards the observer.
    How would I plot that in a computer program, or what would I use to do that.

    I know that the Lorentz boost talks about moving frames of reference at speed v opposed to one that is standing still. so I assumed that that movement is along the x axis and that it's a movement outward(away from observer).

    Where x'=gamma(x-vt);
    and t'=gamma(t-vx/c^2)

    like the time dilation equation derived from the Lorentz transformations t'=t(sqrt(1-(v^2/c^2))
    With that equation ^ the difference between t' and t grows bigger as time goes by. or you need to explain to me why the time dilation function makes the difference between t' and t grow bigger as time passes.

    So I was thinking what would happen if the train instead of going away comes closer to the observer.
     
  5. Apr 18, 2017 #4

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    The Lorentz transformation just explains the disagreement between what the moving frame measures and what the stationary frame measures. If the stationary frame measures an object coming toward it at a certain velocity and position, then that is all you need and can plot that. It doesn't matter if the object is approaching or departing. Remember that anything approaching one point in the stationary frame is leaving another point of the stationary frame. There is no disagreement between the two points of the stationary frame.
     
  6. Apr 18, 2017 #5
    How would you set v and x up so that it would move towards you?

    I do understand that the Lorentz transformations are some sort of coordinate transformation (linear coordinate transformation). But it's happening at velocity v. Which suggests that v is the speed the other frame of reference is traveling at. I'm curious how to you figure direction in those equations. do you set v to be negative for incoming or x to be negative.
     
  7. Apr 18, 2017 #6

    PeroK

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    How would you do that in classical phyics?
     
  8. Apr 18, 2017 #7

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    A positive velocity is a velocity in the direction where the x coordinate is increasing. That is the same whether the current x position is positive or negative.
     
  9. Apr 18, 2017 #8

    Ibix

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    What the Lorentz transforms do is relate coordinates on maps of spacetime drawn by people in inertial motion relative to one another. They don't directly relate observations. It's analogous to how a rotation relates coordinates on a map with north pointing up the page to one with north pointing off at some other angle. Neither tells you directly what you would see because there's no you on the map.
     
  10. Apr 18, 2017 #9
    Show us how you figure out outgoing movement.
     
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