Modern Physics - Extension of the Galilean Transformation?

[PFStudent]

Homework Statement

Conventionally, the Galilean Transformation relates two reference frames that begin at the same location and time with one reference frame moving at a constant velocity {\vec{v}} along a positive {x}-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: {K(x,y,z,t)} and {K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})} is the following,
{x^{\prime}} = {{x}-{vt}}
{y^{\prime}} = {y}
{z^{\prime}} = {z}
{t^{\prime}} = {t}

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

Homework Equations

Knowledge of Transformations.

The Attempt at a Solution

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

Taking reference frame: {K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})}; as the reference frame moving at constant velocity {\vec{v}} with respect to reference frame {K(x,y,z,t)} along a common {\vec{r}} direction we note that velocity {\vec{v}} now has vector components: {{\vec{v}} = {{{{v}_{x}}{\hat{i}}}+{{{v}_{y}}{\hat{j}}}+{{v}_{z}}{\hat{k}}}}}. It follows then that the Galilean Transformation equations must also reflect the displacements along the axes: {x}, {y}, and {z}; consequently the new Galilean Transformation becomes,
{x^{\prime}} = {{x}-{{{v}_{x}}{t}}}
{y^{\prime}} = {{y}-{{{v}_{y}}{t}}}
{z^{\prime}} = {{z}-{{{v}_{z}}{t}}}
{t^{\prime}} = {t}

Is that correct?

Thanks,

-PFStudent

 

[gabbagabbahey]

Taking reference frame: {K^{\prime}({x^{\prime}},{y^{\prime}},{z^{\prime}},{t^{\prime}})}; as the reference frame moving at constant velocity {\vec{v}} with respect to reference frame {K(x,y,z,t)} along a common {\vec{r}} direction we note that velocity {\vec{v}} now has vector components: {{\vec{v}} = {{{{v}_{x}}{\hat{i}}}+{{{v}_{y}}{\hat{j}}}+{{v}_{z}}{\hat{k}}}}}. It follows then that the Galilean Transformation equations must also reflect the displacements along the axes: {x}, {y}, and {z}; consequently the new Galilean Transformation becomes,
{x^{\prime}} = {{x}-{{{v}_{x}}{t}}}
{y^{\prime}} = {{y}-{{{v}_{y}}{t}}}
{z^{\prime}} = {{z}-{{{v}_{z}}{t}}}
{t^{\prime}} = {t}

Is that correct?

Sure, but you can do better even better than that...you can express \{v_x,v_y,v_z\} in terms of the speed v and the usual spherical polar and azimuthal angles \theta and \phi and then express those angles in terms of \{x,y,z\} if you like.

 

[PFStudent]

Hey,

Sure, but you can do better even better than that...you can express \{v_x,v_y,v_z\} in terms of the speed v and the usual spherical polar and azimuthal angles \theta and \phi and then express those angles in terms of \{x,y,z\} if you like.

Ok, we can build on this and establish that the Galilean Transformation (GT) for: 1, 2, and 3; dimensions is the following,

GT for 1-D
<br /> {{x}^{\prime}} = {{x}-{vt}}<br />

<br /> {{y}^{\prime}} = {y}<br />

<br /> {{z}^{\prime}} = {z}<br />

<br /> {{t}^{\prime}} = {t}<br />

GT for 2-D (Converted via using Polar Coordinates (r, \theta)*)
<br /> {{x}^{\prime}} = {{x}-{{v}{\left({\frac{x}{\sqrt{{x^2}+{y^2}}}}\right)}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{v}{\left({\frac{y}{\sqrt{{x^2}+{y^2}}}}\right)}{t}}}<br />
<br /> {{z}^{\prime}} = {z}<br />

<br /> {{t}^{\prime}} = {t}<br />

GT for 3-D (Converted via using Polar Coordinates (r, \theta, \varphi)**)
<br /> {{x}^{\prime}} = {{x}-{{v}{\left({\frac{x}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{v}{\left({\frac{y}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}<br />
<br /> {{z}^{\prime}} = {{z}-{{v}{\left({\frac{z}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}<br />
<br /> {{t}^{\prime}} = {t}<br />
-----------------------------------------------------------------------------------
<br /> {\theta} \equiv {{0}{\,}{\,}{\text{to}}{\,}{\,}{2{\pi}}{\,}{\,}{\text{Radians}}}^{*}<br />

<br /> {\varphi} \equiv {{0}{\,}{\,}{\text{to}}{\,}{\,}{\pi}{\,}{\,}{\text{Radians}}}^{**}<br />

Is that right?

Thanks,

-PFStudent

 

[willem2]

The 2d and 3d cases aren't right.

(r, \theta, \phi) are the components of the velocity of the moving frame. If you would convert them back to cartesian coordinates you would get v_x, v_y and v_z back, and the equations you already had at the end of your first post. I don't think gabbagabbahey's post makes any sense. v_x, v_y and v_z are constants that don't depend on x, y and z

 

[PFStudent]

Hey,

The 2d and 3d cases aren't right.

(r, \theta, \phi) are the components of the velocity of the moving frame. If you would convert them back to cartesian coordinates you would get v_x, v_y and v_z back, and the equations you already had at the end of your first post. I don't think gabbagabbahey's post makes any sense. v_x, v_y and v_z are constants that don't depend on x, y and z

Well, I assume this part is right,
<br /> {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}<br />

<br /> {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}<br />

<br /> {{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}}<br />

<br /> {{t}^{\prime}} = {t}<br />
so where do I go from here?

Thanks,

-PFStudent

 

[willem2]

Hey,


Well, I assume this part is right,
<br /> {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}<br />

<br /> {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}<br />

<br /> {{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}}<br />

<br /> {{t}^{\prime}} = {t}<br />
so where do I go from here?

I don't think there is anywhere else you want to go.

 

[PFStudent]

Hey,

I don't think there is anywhere else you want to go.

Well, since we're beginning at,
<br /> {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}<br />
<br /> {{z}^{\prime}} = {{z}-{{{v}_{z}}{t}}}<br />
<br /> {{t}^{\prime}} = {t}<br />

We can let,
<br /> {\vec{v}} = {{\vec{v}}_{\rho}}<br />

<br /> {{\vec{v}}_{r}} = {{\vec{v}}_{x}}+{{\vec{v}}_{y}}<br />
<br /> {{\vec{v}}_{\rho}} = {{{\vec{v}}_{r}}+{{\vec{v}}_{z}}} <br />

Allowing us to write the following,
<br /> {{x}^{\prime}} = {{x}-{{\left({{{v}_{r}}{\cos{\theta}}}\right)}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{\left({{{v}_{r}}{\sin{\theta}}}\right)}{t}}}<br />
<br /> {{z}^{\prime}} = {{z}-{{\left({{{v}_{\rho}}{\sin{\varphi}}}\right)}{t}}}<br />
<br /> {{t}^{\prime}} = {t}<br />

<br /> {{x}^{\prime}} = {{x}-{{\left({\left(v\right)}{\sin{\varphi}}\right)}{\left(\cos{\theta}\right)}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{\left({\left(v\right)}{\sin{\varphi}}\right)}{\left(\sin{\theta}\right)}{t}}}<br />
<br /> {{z}^{\prime}} = {{z}-{{\left({\left(v\right)}{\sin{\varphi}}\right)}{t}}}<br />
<br /> {{t}^{\prime}} = {t}<br />

<br /> {{x}^{\prime}} = {{x}-{{{v}{\left(\sin{\varphi}\right)}}{\left(\cos{\theta}\right)}{t}}}<br />
<br /> {{y}^{\prime}} = {{y}-{{{v}{\left(\sin{\varphi}\right)}}{\left(\sin{\theta}\right)}{t}}}<br />
<br /> {{z}^{\prime}} = {{z}-{{v}{\left({\sin{\varphi}}\right)}{t}}}<br />
<br /> {{t}^{\prime}} = {t}<br />

However, is the above incorrect since we are measuring the angles in the the reference frame {K} as opposed to {{K}^{\prime}}?

Thanks,

-PFStudent