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

1. Feb 10, 2010

### PFStudent

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.
$${{\sum_{}^{}}{\vec{F}}} = {m{\vec{a}}}{\,}{\,}{\text{[N.II.L.]}}$$

GT for 1-D
$${{x}^{\prime}} = {{x}-{vt}}$$

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

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

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

GT for 2-D
$${{x}^{\prime}} = {{x}-{{v}{\left({\frac{x}{\sqrt{{x^2}+{y^2}}}}\right)}{t}}}$$
$${{y}^{\prime}} = {{y}-{{v}{\left({\frac{y}{\sqrt{{x^2}+{y^2}}}}\right)}{t}}}$$
$${{z}^{\prime}} = {z}$$

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

GT for 3-D
$${{x}^{\prime}} = {{x}-{{v}{\left({\frac{x}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}$$
$${{y}^{\prime}} = {{y}-{{v}{\left({\frac{y}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}$$
$${{z}^{\prime}} = {{z}-{{v}{\left({\frac{z}{\sqrt{{x^2}+{y^2}+{z^2}}}}\right)}{t}}}$$
$${{t}^{\prime}} = {t}$$

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. Feb 10, 2010

### Gokul43201

Staff Emeritus
How would you proceed in the 1-D case?

3. Feb 10, 2010

### PFStudent

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

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

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

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

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

I can proceed as follows,
$${{x}^{\prime}} = {{x}-{vt}}$$

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

$${{\frac{d{{x}^{\prime}}}{dt}}} = {{\frac{dx}{dt}}-{v}}$$

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

$${\frac{{{d}^{2}}{{x}^{\prime}}}{d{{t}^{2}}}} = {{\frac{{{d}^{2}}{x}}{d{{t}^{2}}}}-{0}}$$

$${\left({{a}^{\prime}}\right)} = {\left({a}\right)}$$

$${{a}^{\prime}} = {a}$$

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

Thanks,

-PFStudent

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

### PFStudent

Hey,

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

Thanks,

-PFStudent