Find Poincare Transform for Momentarily Co-Moving Reference Frame

In summary: 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{(\
  • #1
phatm
3
0
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.
Thanks in advance,
Andrew.
 
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  • #2
phatm said:
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.
Thanks in advance,
Andrew.

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 [itex] (t_0, x_0,y_0,z_0) [/itex]) the coordinate transformation is given by

[tex]
\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]
[/tex]

where [itex] \Lambda [/itex] is a pseudoorthogonal 4x4 matrix.

Eugene.
 
Last edited:
  • #3
meopemuk said:
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 [itex] (t_0, x_0,y_0,z_0) [/itex]) the coordinate transformation is given by

[tex]
\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]
[/tex]

where [itex] \Lambda [/itex] is a pseudoorthogonal 4x4 matrix.

Eugene.

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

[tex] \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][/tex]
I'm not sure if I have that correct. Also, do you know what the [tex]\Lambda[/tex] 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
phatm said:
Then by what you stated, [tex]\Lambda[/tex] is just the general Lorentz boost and the whole equation would be

[tex] \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][/tex]
I'm not sure if I have that correct.

Your formula looks right.

phatm said:
Also, do you know what the [tex]\Lambda[/tex] matrix would be for rotations.

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
meopemuk said:
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.
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
[tex] \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][/tex]
?. I've been unable to find it so far. Thanks for the replies,
Andrew.
 
  • #6
phatm said:
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
[tex] \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][/tex]
?. I've been unable to find it so far. Thanks for the replies,
Andrew.

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 [itex]\Lambda^{-1}[/itex] can be obtained from [itex]\Lambda [/itex] simply by changing the sign of the boost velocity.

Eugene.
 
  • #7
Why is the 0,1,1,1 4-vector subtracted? How can I convince myself that makes sense?
 
  • #8
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.
 

1. What is a Poincare Transform for a Momentarily Co-Moving Reference Frame?

A Poincare Transform for a Momentarily Co-Moving Reference Frame is a mathematical representation of the relationship between two reference frames that are moving at a constant velocity relative to each other. It describes how physical quantities, such as position and time, are transformed between the two frames.

2. How is a Poincare Transform calculated?

A Poincare Transform is calculated using the Lorentz transformation equations, which take into account the relative velocity and time dilation between the two reference frames. The equations involve the speed of light, the velocity of the two frames, and the difference in time between the two frames.

3. What is the significance of a Poincare Transform in physics?

The Poincare Transform is significant in physics because it allows us to understand and predict how physical quantities behave in different reference frames. It is a fundamental concept in the theory of relativity and is essential in many fields, including astrophysics, cosmology, and particle physics.

4. Can a Poincare Transform be applied to non-inertial reference frames?

Yes, a Poincare Transform can be applied to non-inertial reference frames, but it becomes more complex. In this case, the equations must take into account additional factors such as acceleration and non-uniform motion.

5. How does a Poincare Transform relate to Einstein's theory of relativity?

The Poincare Transform is a mathematical expression of Einstein's theory of relativity, which states that the laws of physics are the same in all inertial reference frames. It is based on the principle of the constancy of the speed of light and explains how physical quantities such as space and time are affected by the relative motion between two frames.

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