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I How to correctly use Lorentz transformations?

  1. Jul 7, 2016 #1
    Ok so... It's been a while since I first saw this problematic scenario and I want to know how to deal with it. The question arises in the context of special relativity. Suppose 2 objects moving at the same speed. The floor is the rest frame 'A' and the front object is the moving frame 'B'. The contradiction occurs when I want to find the coordinates of the middle object just as A observes that B is at x=d.
    Assuming any distance 's' (seen by B) between the moving objects, the coordinates should be:

    From A's viewpoint: ( d-s/γ , d/v )
    From B's viewpoint: ( -s , d/γv )
    Transforming A, you get: ( -s , d/γv + vs/c2 )
    Transforming B, you get: ( d-γs , d/v - γvs/c2 )

    Why the transformation of A's viewpoint doesn't give me the same result I got for B's viewpoint? (and viceversa). How to interpret this?
     
    Last edited: Jul 7, 2016
  2. jcsd
  3. Jul 7, 2016 #2

    Ibix

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    Objects are not frames. I think you mean that the floor is at rest in frame A, and the balls are at rest in frame B. One of the balls is at the origin in frame B.

    What is the "middle object"? You've defined two balls and a floor. None of these can sensibly be called the "middle" object. From your context I think you mean the ball that is not at the origin in frame B.

    This is the source of your problem. What do you mean by "just as"? What you mean is "where is the second ball at the same time as the first ball is at some point". The problem is that simultaneity is frame dependent, so frame A and frame B do not agree on what the italicised part of that sentence means. Therefore they do not agree on what event you are attempting to assign coordinates to. Unsurprisingly, you then get different answers for the time and, hence, the place that the second ball is at.

    Choose a frame whose simultaneity you want to use. Then you will be able to get a consistent answer.
     
  4. Jul 7, 2016 #3
    You're right. Inittially the (0,0) of A and B are the same, but after some time, B reaches x=d in A's frame.

    By "middle object" I mean the second ball.
    In my mind, I was asumming s/γ < d so... if B already reached x=d in A's frame, it was intuitive for me to call the second ball that way.

    Yes, that's what I mean so I assumed 'If A observes that the first ball coordinates are (d,t), B should observe (0,t/γ)', then I just have to substract the distance between the balls (as seen by each observer) keeping the time coordinates constant, i.e: A observes that the second ball is in (d-s/γ,t) - over gamma because the distance is proper from B - but B observes it is in (-s,t/γ).

    Taking into account that issue about simultaneity that you commented, would you say that (probably) keeping the time coordinates constant to compute the substraction is not the right procedure? - I'm still very confused and the contradiction is still in my mind.
    Certainly my teacher did that solving one exercise in class.
     
    Last edited: Jul 7, 2016
  5. Jul 8, 2016 #4

    Ibix

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    It's not generally correct. It will work sometimes. I'd personally avoid doing it with beginners because they are likely to make exactly the mistake you made - trying it in a situation where it doesn't work.

    In general, you can't take the difference between x coordinates and expect to get the distance between objects unless you measured the x coordinates at the same time. Imagine a beetle walking along a ruler. If you want its length you need to get the ruler readings at its nose and its tail at the same time, otherwise you get its length plus or minus its velocity times the delay between measurements.

    In relativity, two frames don't agree on what simultaneous is. So if I correctly measure the length of the beetle, a moving frame will not agree that I measured the positions simultaneously. So just transforming my measurements of x abd taking the difference won't give the correct answer.

    The one exception to this is that if the beetle is at rest in the moving frame (as the balls are at rest in frame B in your scenario). In that case it doesn't matter whether measurements are simultaneous or not since the objects aren't moving.

    So - when you start in your frame A and transform ##(x_1,t)## and ##(x_2,t)## into frame B and take the distance between the balls to be ##x_1'-x_2'## you get away with it. The fact that measurements weren't made simultaneously doesn't matter because the positions don't change. But when you take coordinates ##(x_3',t')## and ##(x_4',t')## in frame B and transform into frame A then it all goes wrong because ##x_3## and ##x_4## were not measured simultaneously and the positions of the balls change with time.
     
  6. Jul 8, 2016 #5
    Sure thing, one can't measure distances that way. If one wants to know the spatial separation between 2 events in A they should be simultaneous but, clearly, if they were simultaneous in B they won't be simultaneous in A.

    What I usually do to avoid this problem is the following:
    ( d , d/v ) represent the coordinates of the first ball (as A observes them).
    ( d-γs , d/v - γvs/c2 ) represent the coordinates of the second ball (Obtained via Lorentz transformations).

    In order to make simultaneous both events I modify d/v - γvs/c2 in an ammount of +γvs/c2 and the spatial coordinate d-γs in an ammount of (Velocity x Delay): +γv2s/c2.

    That said, d-γs+γv2s/c2 = d-s/γ.
    Now I'm able to compare that coordinate with the spatial coordinate of the first ball. That gives me the distance between the two balls (as A observes them), namely, this distance is: d-(d-s/γ) = s/γ (as one would expect).
     
    Last edited: Jul 8, 2016
  7. Jul 8, 2016 #6
    In the other hand, the computation of the distance between the balls (observed by B) is very straight-forward:
    ( 0 , d/γv ) represent the coordinates of the first ball (as B observes them).
    ( -s , d/γv + vs/c2 ) represent the coordinates of the second ball (Obtained via Lorentz transformations).

    You just have to compute 0-(-s) = s, since you don't have to worry about delays in the measurements because, as you said, both balls are at rest in B's frame.
    This result is exactly what one would expect.

    Now, the question persists: If both analysis are giving me the expected results what is the correct procedure?
     
  8. Jul 8, 2016 #7
    That's because those A's viewpoint and B's viewpoint are referring to two distinct events, they are not different viewpoints of the same event. The fact that two events are distinct is invariant, every frame will agree. You cannot transform one and expect it to coincide with the other in some frame.
     
  9. Jul 8, 2016 #8

    Mister T

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    Here in the US we have, for the most part, two flavors of introductory college-level physics. One is designed for pre-meds and the like and almost always has no calculus in it. The other is for engineering, physics, and chemistry majors. The textbooks used in both usually devote a chapter to relativity. Almost without exception they deal with cases where the object of interest is at rest in one of the two reference frames. It is only in the one section of the chapter devoted to the "relativistic addition of velocities" that they make a departure from that practice.

    This practice often leaves both students and instructors confused, I think.
     
  10. Jul 8, 2016 #9

    Ibix

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    The correct procedure for what? I don't think your route is the cleanest one I've seen, but it works. You're looking at two different things here, as @Vitro says, so having two different answers is what you would expect.

    Have you come across a Minkowski diagram before? They're just displacement-time graphs, with the convention that time goes up the page and that scales are chosen so that something moving at the speed of light has a track with a slope of ±1. But you can draw both the ##x## and ##t## axes and the ##x'## and ##t'## axes, which can help to see what's going on.

    Here's a Minkowski diagram showing your setup.
    Minkowski1.png
    The black axes are the ##x,t## coordinates used in frame A. The green axes are the ##x',t'## axes used in frame B. The red line is the so-called worldline of the first ball, moving to the right at 0.6c. The blue line is the worldline of the second ball, following it. I've marked ##x=d## and a dotted line leading up to the place and time where the front ball is at ##d##. Then I've drawn a black dotted line to where the back ball is at the same time according to frame A (marked as A). I've also drawn a green dotted line to where the back ball is at the same time according to frame B (marked as B).

    You can see that these are not the same place and time. You are calculating the coordinates of the two different events.
     
    Last edited: Jul 9, 2016
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