Modern Physics - Extension of the Galilean Transformation?

1. Jan 31, 2010

PFStudent

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Knowledge of Transformations.

3. 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

Last edited: Jan 31, 2010
2. Jan 31, 2010

gabbagabbahey

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.

3. Feb 10, 2010

PFStudent

Hey,

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
$${{x}^{\prime}} = {{x}-{vt}}$$

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

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

$${{t}^{\prime}} = {t}$$

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

$${{t}^{\prime}} = {t}$$

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

$${\varphi} \equiv {{0}{\,}{\,}{\text{to}}{\,}{\,}{\pi}{\,}{\,}{\text{Radians}}}^{**}$$

Is that right?

Thanks,

-PFStudent

Last edited: Feb 11, 2010
4. Feb 10, 2010

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

5. Feb 10, 2010

PFStudent

Hey,

Well, I assume this part is right,
$${{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}$$

$${{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}$$

$${{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}}$$

$${{t}^{\prime}} = {t}$$
so where do I go from here?

Thanks,

-PFStudent

Last edited: Feb 10, 2010
6. Feb 10, 2010

willem2

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

7. Feb 11, 2010

PFStudent

Hey,
Well, since we're beginning at,
$${{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}$$
$${{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}$$
$${{z}^{\prime}} = {{z}-{{{v}_{z}}{t}}}$$
$${{t}^{\prime}} = {t}$$

We can let,
$${\vec{v}} = {{\vec{v}}_{\rho}}$$

$${{\vec{v}}_{r}} = {{\vec{v}}_{x}}+{{\vec{v}}_{y}}$$
$${{\vec{v}}_{\rho}} = {{{\vec{v}}_{r}}+{{\vec{v}}_{z}}}$$

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

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

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

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

Thanks,

-PFStudent