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Homework Help: Modern Physics - Invariability of Newton's 2nd Law under a GT?

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the form of Newton's Second Law is invariant under:
    (a). a Galilean Transformation (GT) in 1-Dimension.
    (b). a Galilean Transformation (GT) in 2-Dimensions.
    (c). a Galilean Transformation (GT) in 3-Dimensions.

    2. Relevant equations
    Newton's Second Law.
    [tex]
    {{\sum_{}^{}}{\vec{F}}} = {m{\vec{a}}}{\,}{\,}{\text{[N.II.L.]}}
    [/tex]

    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
    [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
    [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]

    3. The attempt at a solution
    I'm not sure exactly where to begin here. Particularly, how to proceed in the: 2-D and 3-D; cases since I have extra variables to deal with in those GT equations.

    Thanks,

    -PFStudent
     
    Last edited: Feb 10, 2010
  2. jcsd
  3. Feb 10, 2010 #2

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How would you proceed in the 1-D case?
     
  4. Feb 10, 2010 #3
    Hey,
    Well, in the 1-D case we are only considering motion along one axis, where [itex]{\vec{v}}[/itex] is a constant, hence the following Galilean Transformation,
    [tex]
    {{x}^{\prime}} = {{x}-{vt}}
    [/tex]

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

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

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

    So, my guess is that by beginning with the following,
    [tex]
    {{x}^{\prime}} = {{x}-{vt}}
    [/tex]

    I can proceed as follows,
    [tex]
    {{x}^{\prime}} = {{x}-{vt}}
    [/tex]

    [tex]
    {{\frac{d}{dt}}{\Bigl[{{x}^{\prime}}\Bigr]}} = {{\frac{d}{dt}}{\Bigl[{{x}-{vt}}\Bigr]}}
    [/tex]

    [tex]
    {{\frac{d{{x}^{\prime}}}{dt}}} = {{\frac{dx}{dt}}-{v}}
    [/tex]

    [tex]
    {{\frac{d}{dt}}{\left[{\frac{d{{x}^{\prime}}}{dt}}\right]}} = {{\frac{d}{dt}}{\left[{{\frac{dx}{dt}}-{v}}\right]}}
    [/tex]

    [tex]
    {\frac{{{d}^{2}}{{x}^{\prime}}}{d{{t}^{2}}}} = {{\frac{{{d}^{2}}{x}}{d{{t}^{2}}}}-{0}}
    [/tex]

    [tex]
    {\left({{a}^{\prime}}\right)} = {\left({a}\right)}
    [/tex]

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

    So, I guess by showing that,
    [tex]
    {{a}^{\prime}} = {a}
    [/tex]
    this implies that N.II.L. is invariant under a (1-D) Galilean Transformation? If so, why?

    Thanks,

    -PFStudent
     
    Last edited: Feb 10, 2010
  5. Feb 11, 2010 #4
    Hey,

    Any help on this, still not sure if I'm on the right path here.

    Thanks,

    -PFStudent
     
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