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Velocity 4-vector

  1. Nov 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Given
    Space Time Coordinate of object <t, x, y, z> = <0, 0, 0, 0>
    Velocity of object as Vector <betaX, betaY, betaZ> = <.866, 0 ,0>
    Velocity of target reference frame as Vector <betaX, betaY, betaZ> = <-.866, 0 ,0>

    Transform velocity of object to the target reference frame using the 4-vector technique.

    2. Relevant equations

    3. The attempt at a solution
    Following directions from a source on the internet........

    Velocity as a 4-vector (tau, vx, vy, vz)
    = <1, betaX, betaY, betaZ>
    = <1, 0.866, 0.0, 0.0>

    magnitude of 4-vector
    = Sqrt( 1 + betaX * betaX + betaY * betaY + betaZ * betaZ )
    = 1.3228590249909473

    Velocity as a 4-vector normalized (tau, x, y, z)
    = <1/magnitude, betaX/magnitude, betaY/magnitude, betaZ/magnitude>
    = <0.7559384493044097, 0.6546426970976188, 0.0, 0.0>

    Transform the 4-vector where L is the Lorentz Transform populated for the target velocity:
    v' = Lv

    resulting in v'(tau, x, y, z)
    = <2.6454852575216505, 2.6183403845739543, 0.0, 0.0>

    ??? I should be getting a vx value somewhere around .98c. Totally lost.
     
  2. jcsd
  3. Nov 22, 2016 #2

    PeroK

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    Science Advisor
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    If you think you're totally lost, I'm lost trying to figure out what you're doing. First, I suggest stating what the 4-vector technique is. Is that the Lorentz transformation of 4-vectors? Perhaps you could state the relevant transformation equations?
     
  4. Nov 22, 2016 #3

    PeroK

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    To give a start and also help you with some latex. If a particle has a velocity ##u## in the x-direction with gamma factor ##\gamma_u##, then its four-velocity is:

    ##(u_{\mu}) = (\gamma_u, \gamma_u u, 0,0)##
     
  5. Nov 23, 2016 #4
    That helps a lot. I am going to spend some time working the problem (and my latex).
     
  6. Nov 24, 2016 #5
    I think I got it!
    It took a couple days to learn latex but PeroK's hint and seeing the formulas clearly made me realize what I was doing wrong.

    Here's the latex by the way (still working on format)

    The Lorentz Transformation is defined as

    $$\lambda=
    \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}
    $$

    Where v is the velocity of S with respect to S' and
    $$\beta_x = v_x/c$$
    $$\beta_y = v_y/c$$
    $$\beta_y = v_y/c$$
    $$\beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2$$
    $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$

    The velocity 4-vector in reference frame S' is defined as
    $$u'_\mu=
    \begin{bmatrix}
    \gamma_{u'} \\
    \gamma_{u'} u'_x \\
    \gamma_{u'} u'_y \\
    \gamma_{u'} u'_z
    \end{bmatrix}
    $$
    Where u' is the 3-vector velocity of an object with respect to the S' reference frame and gamma is the gamma value using u'.

    The velocity 4-vector in reference frame S is defined as
    $$u_\mu=
    \begin{bmatrix}
    \gamma_{u} \\
    \gamma_{u} u_x \\
    \gamma_{u} u_y \\
    \gamma_{u} u_z
    \end{bmatrix}
    $$
    Where u is the 3-vector velocity of an object with respect to the S reference frame and gamma is the gamma value using u.

    Thanks PeroK !
     
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