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Lorentz Invariance of Plane Wavefront

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data

    For a plane, monochromatic wave, define the width of a wavefront to be the distance between two points on a given wavefront at a given instant in time in some reference frame. Show that this width is the same in all frames using 4-vectors and
    in-variants.


    2. Relevant equations
    • $$ \vec{X} = (ct, \vec{x}) $$
    • $$ \vec{K} = (\omega/c, \vec{k})$$
    • $$ \vec{U} = (\gamma_u c, \gamma_u \vec{u})$$

    3. The attempt at a solution

    I have tried in vein to create an appropriate invariant quantity to show that this 'width' is invariant. I know that this question can also be done by writing equations for the movement of the two ends of the width, and taking a Lorentz transform, but I am also struggling to set this up. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 7, 2016 #2

    TSny

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    I assume the wavefront is for an electromagnetic wave.

    Here’s a diagram as a hint.

    upload_2016-10-7_9-5-34.png

    It shows the wavefront as viewed in a particular inertial frame at two different times. It indicates two events, one occurring at the red point of the wavefront at time ##t## and one occurring at the green point of the same wavefront at time ##t+\Delta t##.

    For these events, what would be a natural four-vector to consider?
     
  4. Oct 7, 2016 #3
    Yeah, sorry for not being specific; it is indeed a plane wavefront corresponding to an EM wave.
    I might be getting myself in a pickle here, but wouldn't ##\vec{U}## be a good option to consider, seeing as the direction of the rays remains unchanged? Or ##\vec{K}## due to the nature of the wavefront as a plane.
     
  5. Oct 7, 2016 #4

    TSny

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    The problem is concerned with the spatial distance, ##w##, between the two points on the wavefront. Can you think of a 4-vector associated with the two events ##E_1## and ##E_2## in which spatial distance would be involved?
     
  6. Oct 7, 2016 #5
    Well, you could define two four vectors
    ##\begin{align}
    \vec{X}_1 & = (ct, \vec{x}_1) \\
    \vec{X}_2 & = (c(t+\Delta t), \vec{x}_2)
    \end{align}##

    such that their difference gives
    ##\begin{align}
    \vec{R}= \vec{X}_2 - \vec{X}_1 = (c\Delta t, \Delta \vec{x})
    \end{align}##

    where we have defined ## \Delta \vec{x} = \vec{x}_2 - \vec{x}_1 = w ##. Am I at all on the right track?
     
  7. Oct 7, 2016 #6

    TSny

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    Definitely on the right track. However, for the two events shown, ## \Delta \vec{x} \neq \vec{w} ## if ##\vec{w}## is the spatial vector between the red and green points at the same time ##t##.

    But ##\vec{w}## can be thought of as part of ## \Delta \vec{x}## .
     
  8. Oct 7, 2016 #7
    Sorry yes, putting ##\Delta \vec{x} = w## was a grave transgression; I might have the solution. I will update this post if I achieve it!
     
  9. Oct 8, 2016 #8
    Update: I believe I have now found a solution, which I will post below.

    Define the four vectors
    ##\begin{align}
    \vec{X}_1 & = (ct_1,\vec{r}_1)^T \\
    \vec{X}_2 & = (ct_2,\vec{r}_2)^T
    \end{align}##
    that correspond to the displacements of the ends of the interval ##\vec{w}## in the wave-front, so that at any instant, if measured simultaneously, the distance between them gives
    ## \begin{align}
    \vec{X}_r \cdot \vec{X}_r = \left( \vec{X}_2 - \vec{X}_1 \right)^2 = w^2
    \end{align} ##
    at some time ##t = t_0##. Consider these vectors at some later time ##t_0 + \delta t##. We know that the extra displacement will be ## c \delta t \vec{\widehat{k}} ##, where ##\vec{\widehat{k}}## is the unit vector of the direction of the wave-vector of the wave-front. Introducing this into ##\vec{X}_1## and ##\vec{X}_2##, it clear that ##\vec{X}_r## does not change. This means that for a given frame, the relationship
    ## \begin{align}
    \vec{X}_r \cdot \vec{X}_r = w^2
    \end{align} ##
    remains true for all times. However, the left-hand side of this equation is frame invariant, meaning that the right-hand side is also invariant. Thus, the width (and thus area, as we can arbitrarily choose the end-points of ##\vec{X}_1## and ##\vec{X}_2## as long as they remain on the plane of the wavefront) is the same across all frames.

    Does this look good?
     
  10. Oct 8, 2016 #9

    TSny

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    Note that you are taking the two events ##E_1## and ##E_2## to occur at the same time in some particular inertial frame. These two events will not necessarily occur at the same time in another inertial frame.

    All frames agree on the space-time interval ##\vec{X}_r \cdot \vec{X}_r## between the two events. However, would all frames necessarily interpret this space-time interval as representing the "width of the wavefront"?

    Is there a nice interpretation of the space-time interval between the events ##E_1## and ##E_2## when they occur at different times (as shown in the figure in post #2)?
     
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