Lorentz boost matrix in terms of four-velocity

[CarlosMarti12]

As I understand it, the value of a 4-vector x in another reference frame (x') with the same orientation can be derived using the Lorentz boost matrix \bf{\lambda} by x'=\lambda x. 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 \beta and \gamma, which are defined as
$$\beta=\frac{v}{c}$$
$$\gamma=\frac{1}{\sqrt{1-\beta\cdot\beta}}$$
where v is the 3-velocity and c is the speed of light. My question is this: How can the components of \lambda be written in terms of the 4-velocity U alone?

I know that U_0=\gamma c and U_i=\gamma v_i=\gamma c\beta_i for i\in\{1,2,3\}, but I'm having trouble deriving the components for \lambda using the matrices based on \beta and \gamma. An example of one of these matrices can be found at Wikipedia. How can I rewrite this matrix in terms of U alone?

 

[dauto]

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.

 

[robphy]

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

 

[Bill_K]

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

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

 

[Chestermiller]

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

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

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

 

[CarlosMarti12]

Thanks for the answers everyone! Unfortunately I don't have access to http://scitation.aip.org/content/aapt/journal/ajp/50/9/10.1119/1.12748

Could the matrix be written in the form I+\frac{UU^T}{U_0+c}, including time components?

 

[Bill_K]

Answer:
\Lambda = I - 2 uu + (\gamma + 1)^{-1}(uu + vv + uv + vu)

Could the matrix be written in the form I+\frac{UU^T}{U_0+c}, including time components?

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