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Homework Help: Modern Physics - Extension of the Galilean Transformation?

  1. Jan 31, 2010 #1
    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 [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}} = {{x}-{vt}}[/tex]
    [tex]{y^{\prime}} = {y}[/tex]
    [tex]{z^{\prime}} = {z}[/tex]
    [tex]{t^{\prime}} = {t}[/tex]

    Consider the following, what would the Galilean 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 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 [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 Galilean Transformation equations must also reflect the displacements along the axes: [itex]{x}[/itex], [itex]{y}[/itex], and [itex]{z}[/itex]; consequently the new Galilean Transformation becomes,
    [tex]{x^{\prime}} = {{x}-{{{v}_{x}}{t}}}[/tex]
    [tex]{y^{\prime}} = {{y}-{{{v}_{y}}{t}}}[/tex]
    [tex]{z^{\prime}} = {{z}-{{{v}_{z}}{t}}}[/tex]
    [tex]{t^{\prime}} = {t}[/tex]

    Is that correct?

    Thanks,

    -PFStudent
     
    Last edited: Jan 31, 2010
  2. jcsd
  3. Jan 31, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Sure, but you can do better even better than that...you can express [itex]\{v_x,v_y,v_z\}[/itex] in terms of the speed [itex]v[/itex] and the usual spherical polar and azimuthal angles [itex]\theta[/itex] and [itex]\phi[/itex] and then express those angles in terms of [itex]\{x,y,z\}[/itex] if you like.
     
  4. Feb 10, 2010 #3
    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
    [tex]
    {{x}^{\prime}} = {{x}-{vt}}
    [/tex]

    [tex]
    {{y}^{\prime}} = {y}
    [/tex]

    [tex]
    {{z}^{\prime}} = {z}
    [/tex]

    [tex]
    {{t}^{\prime}} = {t}
    [/tex]

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

    [tex]
    {{t}^{\prime}} = {t}
    [/tex]

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

    [tex]
    {\varphi} \equiv {{0}{\,}{\,}{\text{to}}{\,}{\,}{\pi}{\,}{\,}{\text{Radians}}}^{**}
    [/tex]

    Is that right?

    Thanks,

    -PFStudent
     
    Last edited: Feb 11, 2010
  5. Feb 10, 2010 #4
    The 2d and 3d cases aren't right.

    (r, [itex]\theta[/itex], [itex]\phi[/itex]) 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
     
  6. Feb 10, 2010 #5
    Hey,

    Well, I assume this part is right,
    [tex]
    {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}
    [/tex]

    [tex]
    {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}
    [/tex]

    [tex]
    {{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}}
    [/tex]

    [tex]
    {{t}^{\prime}} = {t}
    [/tex]
    so where do I go from here?

    Thanks,

    -PFStudent
     
    Last edited: Feb 10, 2010
  7. Feb 10, 2010 #6
    I don't think there is anywhere else you want to go.
     
  8. Feb 11, 2010 #7
    Hey,
    Well, since we're beginning at,
    [tex]
    {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}}
    [/tex]
    [tex]
    {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}}
    [/tex]
    [tex]
    {{z}^{\prime}} = {{z}-{{{v}_{z}}{t}}}
    [/tex]
    [tex]
    {{t}^{\prime}} = {t}
    [/tex]

    We can let,
    [tex]
    {\vec{v}} = {{\vec{v}}_{\rho}}
    [/tex]

    [tex]
    {{\vec{v}}_{r}} = {{\vec{v}}_{x}}+{{\vec{v}}_{y}}
    [/tex]
    [tex]
    {{\vec{v}}_{\rho}} = {{{\vec{v}}_{r}}+{{\vec{v}}_{z}}}
    [/tex]

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

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

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

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

    Thanks,

    -PFStudent
     
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