# Lorentz boost matrix in terms of four-velocity

1. Dec 11, 2013

### 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?

2. Dec 11, 2013

### 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.

Last edited: Dec 11, 2013
3. Dec 12, 2013

### robphy

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

4. Dec 12, 2013

### 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:
$$\hat{v} = (\beta \gamma)^{-1}(v - \gamma u)$$
$$\Lambda = I - 2 uu + (\gamma + 1)^{-1}(uu + vv + uv + vu)$$
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:
$$\Lambda \cdot u = v$$

5. Dec 12, 2013

### Staff: Mentor

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

6. Dec 12, 2013

### CarlosMarti12

7. Dec 13, 2013

### Bill_K

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