Modern Physics - Extention of the Lorentz Transformation?

[PFStudent]

Homework Statement

Conventionally, the Lorentz Transformation relates two reference frames that begin at the same location and time with one reference frame moving at a constant velocity [tex]{\vec{v}}[/tex] along a positive [itex]{x}[/itex]-axis (which is common to both reference frames) with respect to the other reference frame. It follows that the transformation relating the two reference frames: [tex]{K(x,y,z,t)}[/tex] and [tex]{K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})}[/tex] is the following,
[tex]{x^{\prime}} = {{\gamma}{\left({x-vt}\right)}}[/tex]
[tex]{y^{\prime}} = {y}[/tex]
[tex]{z^{\prime}} = {z}[/tex]
[tex]{t^{\prime}} = {t}[/tex]

Consider the following, what would the Lorentz Transformation equations be if one reference frame was moving with a constant velocity [tex]{\vec{v}}[/tex] along a radial direction [itex]{\vec{r}}[/itex] (which is common to both reference frames) with respect to the other reference frame? Given reference frames: [tex]{K(x,y,z,t)}[/tex] and [tex]{K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})}[/tex]; find this Lorentz Transformation.

Homework Equations

Knowledge of Transformations.
Einstein's Two Postulates on Relativity (The Principle of Relativity and The Constancy of the Speed of Light).
[tex]
{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}} = {{{c}^{2}}{{t}^{2}}}
[/tex]
[tex]
{{{{x}^{\prime}}^{2}}+{{{y}^{\prime}}^{2}}+{{{z}^{\prime}}^{2}}} = {{{c}^{2}}{{{t}^{\prime}}^{2}}}
[/tex]

The Attempt at a Solution

Conventionally, in a Lorentz Transformation we are only concerned with the constant velocity [tex]{\vec{v}}[/tex] of one reference frame moving along a common [itex]{x}[/itex]-axis between both reference frames with respect to the other reference frame. Consequently, the vector components of [tex]{\vec{v}}[/tex] are:
[tex]{{\vec{v}} = {{v}_{x}}{\hat{i}}[/tex]

Taking reference frame: [tex]{K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})}[/tex]; as the reference frame moving at constant velocity [tex]{\vec{v}}[/tex] with respect to reference frame [tex]{K(x,y,z,t)}[/tex] along a common [itex]{\vec{r}}[/itex] direction we note that velocity [tex]{\vec{v}}[/tex] now has vector components: [tex]{{\vec{v}} = {{{{v}_{x}}{\hat{i}}}+{{{v}_{y}}{\hat{j}}}+{{v}_{z}}{\hat{k}}}}}[/tex]. It follows then that the Lorentz Transformation equations must also reflect the displacements along the axes: [itex]{x}[/itex], [itex]{y}[/itex], and [itex]{z}[/itex]; but mathematically how would I show this?

Thanks,

-PFStudent

 

[anathema]

Dear PFStudent,

Firstly, it is important to note that the Lorentz Transformation equations are derived from Einstein's two postulates of relativity and are used to relate two reference frames in relative motion. Therefore, the equations will remain the same regardless of the direction of motion.

In the case of one reference frame moving with a constant velocity {\vec{v}} along a radial direction {\vec{r}}, we can still use the Lorentz Transformation equations as they are, with some minor adjustments. Let us consider the reference frame {K(x,y,z,t)} as the stationary frame and {K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})} as the moving frame with velocity {\vec{v}} along {\vec{r}}.

Firstly, we need to define the axes for both frames. In the stationary frame {K}, the axes are {x}, {y}, and {z} while in the moving frame {K^{\prime}}, the axes are {x^{\prime}}, {y^{\prime}}, and {z^{\prime}}. It is important to note that {\vec{r}} is perpendicular to the {x}-axis in both frames.

Now, we can apply the Lorentz Transformation equations as follows:

{x^{\prime}} = {{\gamma}{\left({x-vt}\right)}}

In this case, {\vec{v}} is along {\vec{r}} and perpendicular to the {x}-axis. Therefore, we can write {\vec{v}} as {\vec{v} = {v_{r}}{\hat{r}}} where {\hat{r}} is the unit vector in the direction of {\vec{r}}. Substituting this in the above equation, we get:

{x^{\prime}} = {{\gamma}{\left({x-{v_{r}}{\hat{r}}\cdot{t}}\right)}} = {{\gamma}{\left({x-{v_{r}}t}\right)}}

Similarly, we can apply the Lorentz Transformation equations for {y} and {z} as they remain unchanged.

{y^{\prime}} = {y}

{z^{\prime}} = {z}

For {t^{\prime}}, we use the time dilation equation:

{t^{\prime}} = {{\gamma}{\left