Lorentz Transformations in xy & xyz Directions | Pat

[patapat]

So I'm looking at some Lorentz transformation equations and it says
x'=$$\gamma$$(x-vt)
t'=$$\gamma$$(t-vx/c$$^{2}$$)
y'=y
z'=z

I'm assuming the values for y', y, z' and z only hold true when the inertial frames of S and S' are moving at a relative velocity in the x-direction. With this being said, what would the transformations be if the inertial frames were in an xy or xyz direction? Thanks in advance.

-Pat

 

[bernhard.rothenstein]

lorentz transformation plane motion

So I'm looking at some Lorentz transformation equations and it says
x'=$$\gamma$$(x-vt)
t'=$$\gamma$$(t-vx/c$$^{2}$$)
y'=y
z'=z

I'm assuming the values for y', y, z' and z only hold true when the inertial frames of S and S' are moving at a relative velocity in the x-direction. With this being said, what would the transformations be if the inertial frames were in an xy or xyz direction? Thanks in advance.

-Pat

As far as I know the situation is known as plane motion. Your question is answered in

Relativistic motion in a plane
Byron L. Coulter
Am. J. Phys. 48, 633 (1980) Full Text: [ PDF (486 kB) GZipped PS Order ]
I think it can be simplified.

 

[meopemuk]

So I'm looking at some Lorentz transformation equations and it says
x'=$$\gamma$$(x-vt)
t'=$$\gamma$$(t-vx/c$$^{2}$$)
y'=y
z'=z

I'm assuming the values for y', y, z' and z only hold true when the inertial frames of S and S' are moving at a relative velocity in the x-direction. With this being said, what would the transformations be if the inertial frames were in an xy or xyz direction? Thanks in advance.

-Pat

If $(\mathbf{r},t)$ are space-time coordinates of an event in the reference frame O, and the reference frame O' moves with velocity $\mathbf{v} = c \vec{\theta} \theta^{-1} \tanh \theta$ with respect to O, then space-time coordinates $(\mathbf{r}',t')$ of the same event in O' can be obtained by formulas

$$\mathbf{r}' = \mathbf{r} + \frac{\vec{\theta}}{\theta}(\mathbf{r} \cdot \frac{\vec{\theta}}{\theta}) (\cosh \theta - 1) - \frac{\vec{\theta}}{\theta} ct \sinh \theta$$

$$t' = t \cosh \theta - (\mathbf{r} \cdot \frac{\vec{\theta}}{\theta}) \frac{\sinh \theta}{c}$$

These formulas are derived by the same procedure as momentum-energy Lorentz transformations (see eq. (4.2) - (4.3) in http://www.arxiv.org/physics/0504062 )

Eugene.

 

[Meir Achuz]

$$t'=\gamma(t-{\vec r}\cdot{\vec v}/c^2)$$

 

[Meir Achuz]

The previous reply is three equations, I couldn't get Latex to do line by line.
r_\parallel and r_\perp are parallel and perp to v.

 

[CompuChip]

And if you know matrix multiplication,
$$B = \begin{pmatrix} \gamma & -\gamma \beta_1 & -\gamma \beta_2 & -\gamma \beta_3 \\<br /> -\gamma \beta_1 & 1 + \frac{(\gamma - 1)\beta_1^2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} \\<br /> -\gamma \beta_2 & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & 1+\frac{(\gamma - 1)\beta_2^2}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} \\<br /> -\gamma \beta_3 & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} & 1 + \frac{(\gamma - 1)\beta_3^2}{\beta^2}<br /> \end{pmatrix}$$
where $\beta = (\beta_1, \beta_2, \beta_3)$ is a unit vector in the direction of the relative velocity, and
$x' = B x$ for $x = (c t, x, y, z)$ and similar for the transformed system $x'$.

Source: Jackson, Classical Electrodynamics, chapter 11.7[/size]

 

[patapat]

I had a feeling there was no simple answer, thanks guys.