# Poincare transform

1. Aug 24, 2007

### phatm

After a lot of searching I cannot for the life of me find the transformation for a Momentarily Co-moving Reference Frame. Essentially what I'm looking for is the transformations from inertial frame Sigma to inertial frame Sigma[']. Sigma['] is moving at speed v relative to Sigma not along any axis in Sigma. The origins of the inertial reference frames do not coincide at t=t'=0, so I cannot use the general Lorentz transforms. Does anyone have any idea where to find the elements of the Poincare transform. Any help is greatly appreciated.
Andrew.

2. Aug 24, 2007

### meopemuk

You can find a general Lorentz transformation (boost velocity has an arbitrary direction) in https://www.physicsforums.com/showthread.php?t=179810. For a general boost coupled with space-time translation you should just subtract the components of the translation vector from the boost-transformed coordinates (t', x',y',z'). In the most general case (arbitrary boost + arbitrary rotation + translation by the 4-vector $(t_0, x_0,y_0,z_0)$) the coordinate transformation is given by

$$\left[ \begin{array}{c} t' \\ x' \\ y' \\ z' \end{array} \right] = \Lambda \left[ \begin{array}{c} t \\ x \\ y \\ z \end{array} \right] - \left[ \begin{array}{c} t_0 \\ x_0 \\ y_0 \\ z_0 \end{array} \right]$$

where $\Lambda$ is a pseudoorthogonal 4x4 matrix.

Eugene.

Last edited: Aug 24, 2007
3. Aug 25, 2007

### phatm

Thanks for the reply Eugene. So say we had a observer frame $$\Sigma$$ and a MCRF $$\Sigma'$$. At time time t=t'=0 $$\Sigma'$$ is moving with velocity $$\vec{\beta}$$ in some arbritary direction in $$\Sigma$$ at say $$x_0=y_0=z_0=1$$ in $$\Sigma$$, with no rotation of $$\Sigma'$$ in relation to $$\Sigma$$. Then by what you stated, $$\Lambda$$ is just the general Lorentz boost and the whole equation would be

$$\left[ \begin{array}{c} t' \\ x' \\ y' \\ z' \end{array} \right] = \begin{pmatrix} \gamma & -\gamma \beta_1 & -\gamma \beta_2 & -\gamma \beta_3 \\ -\gamma \beta_1 & 1 + \frac{(\gamma - 1)\beta_1^2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} \\ -\gamma \beta_2 & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & 1+\frac{(\gamma - 1)\beta_2^2}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} \\ -\gamma \beta_3 & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} & 1 + \frac{(\gamma - 1)\beta_3^2}{\beta^2} \end{pmatrix}\left[ \begin{array}{c} t \\ x \\ y \\ z \end{array} \right] - \left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1\end{array} \right]$$
I'm not sure if I have that correct. Also, do you know what the $$\Lambda$$ matrix would be for rotations. Any references to texts with some simple explicit examples would be really really appreciated. Thank you very much for the reply,
Andrew.

4. Aug 30, 2007

### meopemuk

Your formula looks right.

You can derivations of arbitrary boost and rotation matrices in sections 1.3.2 and D.5 of http://www.arxiv.org/abs/physics/0504062

Eugene.

5. Aug 31, 2007

### phatm

Thanks for the reply Eugene. The link you gave me was very useful. Just one final question. The inverse the above transform, would that be
$$\left[ \begin{array}{c} t \\ x \\ y \\ z \end{array} \right] = \begin{pmatrix} \gamma & \gamma \beta_1 & \gamma \beta_2 & \gamma \beta_3 \\ \gamma \beta_1 & 1 + \frac{(\gamma - 1)\beta_1^2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} \\ \gamma \beta_2 & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & 1+\frac{(\gamma - 1)\beta_2^2}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} \\ \gamma \beta_3 & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} & 1 + \frac{(\gamma - 1)\beta_3^2}{\beta^2} \end{pmatrix}\left[ \begin{array}{c} t' \\ x' \\ y' \\ z' \end{array} \right] + \left[ \begin{array}{c} t'_{0} \\ x'_{0} \\ y'_{0} \\ z'_{0} \end{array} \right]$$
?. I've been unable to find it so far. Thanks for the replies,
Andrew.

6. Aug 31, 2007

### meopemuk

No, this is not correct formula for the inverse Poincare transformation. You can find the correct expression in eq. (I.11) of the reference I gave you. Note that the inverse boost matrix $\Lambda^{-1}$ can be obtained from $\Lambda$ simply by changing the sign of the boost velocity.

Eugene.

7. Aug 17, 2011

### Albertgauss

Why is the 0,1,1,1 4-vector subtracted? How can I convince myself that makes sense?

8. Aug 17, 2011

### Albertgauss

One other question on the shift of the origin by 0,-1,-1,-1. Why is the x0=1 not length-contracted as viewed in S'? If x0 is 1 in frame S, and S' is moving, shouldn't the x0 be length contracted somehow also? Just trying to understand is all.