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I Velocity 4-Vector for Light Ray

  1. Dec 10, 2016 #1
    I hope I'm not violating Forum protocol, again.
    I tried posting this question in the Homework section but it got locked for violating homework protocol.

    My understanding for the relativistic transformation of a velocity u to u' is given by
    $$
    \begin{bmatrix}
    \gamma_{u'} \\
    \gamma_{u'} u'_x \\
    \gamma_{u'} u'_y \\
    \gamma_{u'} u'_z
    \end{bmatrix}
    =
    \begin{bmatrix}

    \gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\
    -\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\

    -\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\

    -\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2}

    \end{bmatrix}
    \begin{bmatrix}
    \gamma_{u} \\
    \gamma_{u} u_x \\
    \gamma_{u} u_y \\
    \gamma_{u} u_z
    \end{bmatrix}
    $$
    Where v is the velocity of reference frame S' with respect to reference frame S and
    $$
    \beta_x = v_x/c \\
    \beta_y = v_y/c \\
    \beta_y = v_y/c \\
    \beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2 \\
    \gamma = \frac{1}{\sqrt{1-\beta^2}} \\
    \gamma_u = \frac{1}{\sqrt{1-\beta_u^2}} \\
    \gamma_{u'} = \frac{1}{\sqrt{1-\beta_{u'}^2}}
    $$
    This seems to work for any object whose velocity is less that c. But, I got the impression from a previous post that I could transform the velocity of light rays the same way I transformed the velocity of massive objects.

    If I try to do that using the definitions I have above it means calculating $$\gamma_u$$ for a light ray. That's undefined. I have a 3-vector version of the velocity addition formula that doesn't require such a calculation and it works fine. Did I get the 4-vector structure wrong?
     
  2. jcsd
  3. Dec 10, 2016 #2

    vanhees71

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    Science Advisor
    2016 Award

    For a "light ray" there is no four-velocity in the usual sense, because it's light-like. You can take a photon (which is a quantum of the quantized electromagnetic field) momentum as a substitute. The corresponding four-momentum is
    $$(p^{\mu})=\begin{pmatrix} |\vec{p}| \\ \vec{p} \end{pmatrix}.$$
    It transforms as a four vector. The mass of a photon is ##p_{\mu} p^{\mu}=0##, i.e., it's a massless quantum. Of course this doesn't change under Lorentz transformations, thus a photon has a lightlike momentum in any frame, and the three-velocity is
    $$\vec{v}=\frac{c\vec{p}}{|\vec{p}|} \; \Rightarrow \; |\vec{v}|=c,$$
    i.e., the photon travels with the speed of light in any frame.
     
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