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Lorentz boost matrix in terms of four-velocity

  1. Dec 11, 2013 #1
    As I understand it, the value of a 4-vector [itex]x[/itex] in another reference frame ([itex]x'[/itex]) with the same orientation can be derived using the Lorentz boost matrix [itex]\bf{\lambda}[/itex] by [itex]x'=\lambda x[/itex]. More explicitly,
    $$\begin{bmatrix}
    x'_0\\
    x'_1\\
    x'_2\\
    x'_3\\
    \end{bmatrix}
    =
    \begin{bmatrix}
    \lambda_{00}&\lambda_{01}&\lambda_{02}&\lambda_{03}\\
    \lambda_{10}&\lambda_{11}&\lambda_{12}&\lambda_{13}\\
    \lambda_{20}&\lambda_{21}&\lambda_{22}&\lambda_{23}\\
    \lambda_{30}&\lambda_{31}&\lambda_{32}&\lambda_{33}\\
    \end{bmatrix}
    \begin{bmatrix}
    x_0\\
    x_1\\
    x_2\\
    x_3\\
    \end{bmatrix}
    $$
    I have seen examples of these components written in terms of [itex]\beta[/itex] and [itex]\gamma[/itex], which are defined as
    $$\beta=\frac{v}{c}$$
    $$\gamma=\frac{1}{\sqrt{1-\beta\cdot\beta}}$$
    where [itex]v[/itex] is the 3-velocity and [itex]c[/itex] is the speed of light. My question is this: How can the components of [itex]\lambda[/itex] be written in terms of the 4-velocity [itex]U[/itex] alone?

    I know that [itex]U_0=\gamma c[/itex] and [itex]U_i=\gamma v_i=\gamma c\beta_i[/itex] for [itex]i\in\{1,2,3\}[/itex], but I'm having trouble deriving the components for [itex]\lambda[/itex] using the matrices based on [itex]\beta[/itex] and [itex]\gamma[/itex]. An example of one of these matrices can be found at Wikipedia. How can I rewrite this matrix in terms of [itex]U[/itex] alone?
     
  2. jcsd
  3. Dec 11, 2013 #2
    Keep in mind that the matrix λ is not a 4-tensor because one of its indexes is in one reference frame while the other index is in a different reference frame. Because of that Wikipedia's format is the correct one.
     
    Last edited: Dec 11, 2013
  4. Dec 12, 2013 #3

    robphy

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    Possibly interesting reading:

    Fahnline "A covariant four‐dimensional expression for Lorentz transformations"
    http://scitation.aip.org/content/aapt/journal/ajp/50/9/10.1119/1.12748

    Fahnline "Manifestly covariant, coordinate‐free dyadic expression for planar homogeneous Lorentz transformations"
    http://scitation.aip.org/content/aip/journal/jmp/24/5/10.1063/1.525833

    Celakoska "On Isometry Links between 4-Vectors of Velocity"
    http://www.emis.de/journals/NSJOM/Papers/38_3/NSJOM_38_3_165_172.pdf
     
  5. Dec 12, 2013 #4

    Bill_K

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    Expressing the Lorentz transformation in vector form is easy as pie. Suppose the initial 4-velocity is u, the final 4-velocity is v. There are many Lorentz transformations that take u into v, but one that's preferred is a simple boost in the u - v plane. This transformation can be written down in a completely covariant form, depending only on u and v, in four steps:

    1) The usual Lorentz boost along the z-axis is
    [tex]\Lambda = \left(\begin{array}{cc}\gamma&\beta \gamma\\\beta \gamma&\gamma\end{array}\right)[/tex]
    2) The Lorentz boost in an arbitrary direction (quoted in several previous threads) can be written in 3-d form:
    [tex]\Lambda = \left(\begin{array}{cc}\gamma&\beta \gamma \hat{v}\\\beta \gamma \hat{v}&I + (\gamma - 1)\hat{v} \hat{v}\end{array}\right)[/tex]
    where [itex]\hat{v}[/itex] is the unit 3-vector in the direction of the boost.

    3) Write this [itex]\Lambda[/itex] in dyadic form:
    [tex]\Lambda = I + (\gamma - 1)(uu + \hat{v}\hat{v}) + \beta \gamma(u\hat{v} + \hat{v}u)[/tex]
    4) Replace the unit 3-vector [itex]\hat{v}[/itex] by the 4-vector v:
    [tex]\hat{v} = (\beta \gamma)^{-1}(v - \gamma u)[/tex]
    Answer:
    [tex]\Lambda = I - 2 uu + (\gamma + 1)^{-1}(uu + vv + uv + vu)[/tex]
    Here [itex]\gamma[/itex] encodes the relative velocity between u and v: [itex]\gamma = u \cdot v[/itex]. As a check, note that the transformation does take u into v:
    [tex]\Lambda \cdot u = v[/tex]
     
  6. Dec 12, 2013 #5
    Hi Bill_K.

    A couple of years ago, I derived a dyadic relationship for the transformation very similar to this one that I think might be of interest to you. But, I'm reluctant to present it on PF because of my rather novice status with respect to SR. I wanted to send you a private message, but apparently, this is not possible. I've written up my derivation in a Word Document. If you are interested, is there a way of sending the document to you. Please feel free to respond in a private message.

    Chet
     
  7. Dec 12, 2013 #6
  8. Dec 13, 2013 #7

    Bill_K

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    Carlos, What you have corresponds to the one term, ##(\gamma + 1)^{-1}vv##. As you can see, there's more to it than that!
     
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