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Special Relativity. Lorentz Boosts.

  1. Nov 16, 2013 #1
    The question is given a straight rod parallel to the x axis moving in the y direction at speed u, to find the angle between the rod and x'axis in R'. (the chosen method to solve being to write equations for the endpoints and transform them to R')

    I am looking at my book's solution and don't understand some parts.

    ( I think these are stemming from my physical intuition as to what is producing a non-zero angle in R' *, essentially how the y components are transformed , and how the principle of measuring the endpoints simultaneously should be tied in , and in which frame R or R' should measuring the endpoints be simultaneous. )

    Here's my method:

    So if i say the rod has length l, the endpoints in R are given by (x1,y1,z1)= (0,ut1,0) and (x2,y2,z2)= (l,ut2,0), and in R' are given by (x1',y2',z2')=(0',ut1,0) and(l',ut2,0).

    Where 0'= γ(0-vt1)=-vγt1 [1] and l'= γ(l-vt2) [2], so this length contraction part is fine.

    Okay and t1'=γ(t1-0)= γt1 and t2'=(t2-vl/c^2)

    y1'=ut1= [3]
    y2'=ut2 [4]

    So in terms of t':

    t2=t2'/γ +vl/c^2

    So using these in [1],[2],[3], and [4] :

    θ=(U/γ(t2'-t1')+uvl/c^2 )/(-v(t2'+t1')+l/γ)

    Now the book then gives the final answer as uvγ/c^2, which implies that t1'=t2'-(so I conclude the measurement is simultaneous in R')

    But my question is the rod is parallel to the x-axis in R, so θ=0, => Δy=0=u(t2-t1) s.t t2=t1 (measurement also simultaneous in R). But if I simply look at Δy' = u(t2-t1) also as frame is moving in the x direction Δy=Δy', so why doesn't this instantly imply θ'=0?

    Last Question
    I don't really understand why Δy=Δy'? * So the argument is that because motion is in the x direction, so I understand if the y co-ordinates were not time dependent. But Δy=uΔt , so I would have thought that Δy'=uΔt' , where Δt' is given by the normal Lorentz Boost - physically the reason being that time runs differently in both frames...?

    Many Thanks to anyone who can shed some light on this.
  2. jcsd
  3. Nov 17, 2013 #2
    You have not explained what R and R' are.
  4. Nov 17, 2013 #3


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    R and R' are simply the two inertial reference frames.

    To understand the Lorentz-transformation properties of the rod, it's better to work with four-vectors, e.g., the endpoints in R are described by the four vectors
    [tex]x_1=\begin{pmatrix}c t \\ 0 \\ u t \\ 0\end{pmatrix}, \quad x_2=\begin{pmatrix}c t \\ l \\ u t \\ 0 \end{pmatrix}.[/tex]
    These you transform to R' using a Lorentz boost in [itex]x[/itex] direction. An observer in R' again measures all distances and angles by reading off the coordinates of the end points at the same time t' in his frame of reference. This gives rise to both the length contraction and the non-zero angle wrt. to the [itex]x'[/itex] axis for the observer in R'.
  5. Nov 18, 2013 #4
    Okay so in frame R' the coordinates are read off at the same time t', but in R at t.
    So I think my confusion lies in the ,perhaps, notation used by the book for t.
    As motion is in x direction y remains unchanged, the book denotes these as ut=y1=y2=y1'=y2'.
    So simply by looking at θ'=Δy=Δx=y2'-y1'/x2'-x1', from the above would you not immediately deduce θ'=0.
    Looking at the rest of the solution I can see that when we express t in terms of t', a difference in t' for each end point arises as t' in the lorentz boost contains x in the formula, which is 0 and l respectively.
    So this is why I would go for t1' and t2' instead?

    Thanks in advance. :smile:
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