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!