Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trying to reconcile Lorentz Transformation and Length Contraction

  1. Sep 15, 2013 #1
    Suppose I am in a stationary frame of reference S and there is a lamp post at a distance X from my origin in the positive X direction. Say you move at a velocity V along that axis and the distance of the lamp post in your frame of reference S' is X'. Then by Lorentz transformation equation

    X' = (X - Vt)/√(1-[V/c]^2)

    But as per length contraction, you should measure a lesser distance between you and the lamp post than what I measure between you and the lamp post. But the above Lorentz transformation equation says the opposite, that is, the distance in your frame of reference is greater than that in mine? Please point out where I am going wrong.
  2. jcsd
  3. Sep 15, 2013 #2


    User Avatar
    Science Advisor
    Gold Member

    You have to use a value of t so that t' comes out to be 0. You probably just set t=0 which will produce a non-zero value for t', correct?
  4. Sep 15, 2013 #3
    Ok George. Setting t'=0 gets the right answer.

    t' = (t-VX/(c^2))/√(1-[V/c]^2)
    so t = VX/(c^2) for t' to be zero

    Sub t in X' = (X - Vt)/√(1-[V/c]^2) gives the right answer

    X' = X √(1-[V/c]^2)

    But what does "Setting t' = 0" mean physically? I could not figure out
  5. Sep 15, 2013 #4


    User Avatar

    It means making a measurement that is simultaneous with the (x',t') = (0,0) event in my coordinates. You and the lamp post are moving in my frame, so I must make sure that I measure the distances to you and to the lamp post simultaneously. You are at distance x'=0 at my time t'=0, so I must measure the distance x'=X' to the lamp at time t'=0 too.
    Last edited: Sep 15, 2013
  6. Sep 15, 2013 #5
    That makes it clear. Thanks, dvf and George
  7. Sep 15, 2013 #6


    User Avatar
    Science Advisor

    This with lentgh contraction may seem very confusing, so one must be very careful and state problem precisely. In your example, we have two coordinates, 0 and X, in the system you consider as stationary. The spatial dístance between any events (0,t1) and (X,t2) is X, regardless of which times t1 and t2 we have.

    These two events will in the system you consider as moving have the coordinates (X'1,t'1) and (X'2,t'2) (can be calulated by the Lorentz transformation), but here it is clear that the spatial distance X'2-X'1 will depend upon the times t'1 and t'2. So which times shall we choose?
    The convention in SR is to choose t'1=t'2, that is, the times should be equal in the moving system. This is logical, since that is the system we are measuring in.

    In your example, we can take t1=0, t'1=0 and X'1=0. Then we must also take t'2=0. Therefore we must find t2 such that (X,t2) transforms to (X'2,0) and then calculate X'2.
  8. Sep 16, 2013 #7
    ok, to calculate the length of a "stationary" object in a moving frame

    1) Take the coordinates (x1, t1) of one endpoint of the object as measured in the stationary frame and using Lorentz transformation equations, calculate the equivalent coordinates (x1', t1') in the moving frame.

    2) Now take the coordinates (x2, t2) of the other endpoint of the object as measured in the stationary frame and say x2 is greater than x1. Dont even have a look at the value t2, just erase it and put t1+[v/c^2][x2-x1] if the moving frame is moving along positive x direction (or put t1-[v/c^2][x2-x1] if moving along negative x direction)
    With this new coordinates (x2, t1+[v/c^2][x2-x1]), calculate the equivalent coordinates (x2', t2') in moving frame using Lorentz transformation equations

    3) We eventually end up in t1' = t2' and the length of the object in the moving frame is x2' - x1'

    Will this method work under all conditions?
    Last edited: Sep 16, 2013
  9. Sep 16, 2013 #8


    User Avatar
    Science Advisor
    Gold Member

    I haven't checked the details of your work but I suspect it is correct.

    However, once you have done the work that you did in post #3, it is sufficient to extrapolate to all other inertial cases and simply observe that the Proper Length of an object in its rest frame becomes Length Contracted by the factor 1/gamma in any frame in which it is moving.

    I think the main issue with your details is that it appears to be restrictive to a single pair of endpoint events but the length of an inertial object is the same for all pairs of simultaneous endpoint events. Furthermore, for non-inertial objects, there are additional factors that may lead to erroneous results if you're not careful to apply the correct calculation during the transition intervals. Special Relativity does not provide any answers as to how the length of an object changes during acceleration so there is that added complication. However, if you specify the coordinates of the endpoints during acceleration in one inertial frame, then the Lorentz Transformation process will correctly give you the coordinates in any other inertial frame but there won't be a single equation to describe the changing Length Contraction during the acceleration interval.

    I think the best way to understand these issues is to come up with some specific examples and draw spacetime diagrams for different frames. The Length Contraction them becomes immediately visible.
  10. Sep 17, 2013 #9
    It is correct. You perceive the universe beyond your frame as contracted, as it passes at -v.
    Consider having a stick extending to the post at t=0, which is already contracted. That would be the length measured by S, but in your frame the stick is longer by 1/γ, since you can't detect your own length contraction.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook