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

[PFStudent]

Homework Statement

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.

Homework Equations

Newton's Second Law.
<br /> {{\sum_{}^{}}{\vec{F}}} = {m{\vec{a}}}{\,}{\,}{\text{[N.II.L.]}}<br />

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
<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
<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 />

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

 

[Gokul43201]

How would you proceed in the 1-D case?

 

[PFStudent]

Hey,

How would you proceed in the 1-D case?

Well, in the 1-D case we are only considering motion along one axis, where {\vec{v}} is a constant, hence the following Galilean Transformation,
<br /> {{x}^{\prime}} = {{x}-{vt}}<br />

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

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

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

So, my guess is that by beginning with the following,
<br /> {{x}^{\prime}} = {{x}-{vt}}<br />

I can proceed as follows,
<br /> {{x}^{\prime}} = {{x}-{vt}}<br />

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

<br /> {{\frac{d{{x}^{\prime}}}{dt}}} = {{\frac{dx}{dt}}-{v}}<br />

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

<br /> {\frac{{{d}^{2}}{{x}^{\prime}}}{d{{t}^{2}}}} = {{\frac{{{d}^{2}}{x}}{d{{t}^{2}}}}-{0}}<br />

<br /> {\left({{a}^{\prime}}\right)} = {\left({a}\right)}<br />

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

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

Thanks,

-PFStudent

 

[PFStudent]

Hey,

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

Thanks,

-PFStudent