@dialectphilosophy, 1.3.1

…

As long as object 1 and object 2 have the same velocity-relative-to-the-ground as that of the car, $\displaystyle{v}$ , i.e.

$\displaystyle{v_1=v_2=v}$ ,

no matter what the value of $\displaystyle{v}$ is, the distance between object 1 and object 2 is always the same. In other words, you cannot deduce the value of the $\displaystyle{v}$ by observing the separation changes between any two objects/points within the car.

Even in another case where $\displaystyle{v_2 \ne v_1}$ ,

the separation between the 2 objects is

$\displaystyle{ \begin{aligned} &x_2(t) - x_1(t) \\ &= \Bigl( x_2(0) + v_2 t \Bigr) - \Bigl(x_1(0) + v_1 t \Bigr) \\ &= \Bigl( x_2(0) - x_1(0) \Bigr) + \Bigl( v_2 - v_1 \Bigr) t \\ \end{aligned}}$

In other words, the separation $\displaystyle{(x_2(t) - x_1(t))}$ depends only on the initial separation $\displaystyle{(x_2(0) - x_1(0))}$ and the velocity of object 2 relative to the object 1, $\displaystyle{(v_2 - v_1)}$ .

Let this relative velocity be $\displaystyle{v_{21}}$ :

$\displaystyle{ \begin{aligned} v_{21} &= v_2 - v_1 \\ \end{aligned}}$

Although the value of either $v_1$ or $v_2$ depends on the observer’s own velocity, $v_{21}$ does not. In other words, although either $v_1$ or $v_2$ would have a different value under Galilean transformation, $v_{21}$ would stay the same.

$\displaystyle{ \begin{aligned} v_{21}' &= v_2' - v_1' \\ &= (v_2 + V) - (v_1 + V) \\ &= v_2 - v_1 \\ &= v_{21} \\ \end{aligned}}$

The observer can directly see are positions $x_1(0), x_2(0), x_1(t),$ and $x_2(t)$ . With them and the time measured $t$ , he can deduce the value of $v_{21}$ .

He can deduce the relative velocity $v_2 - v_1$ by the separations $x_2(t) - x_1(t)$ and $x_2(0) - x_1(0)$ . However, he still cannot deduce $v_2$ nor $v_1$ unless he is able to look outside the car window. Thus, he cannot deduce the car speed $v$ just by observing the positions and velocities of the objects inside the car.

For simplicity, assume object 1 is actually a point of the car itself. So $v_1$ is actually the speed of the car, $v$ . Then the calculation

$\displaystyle{ \begin{aligned} &x_2(t) - x_1(t) \\ &= \Bigl( x_2(0) + v_2 t \Bigr) - \Bigl(x_1(0) + v_1 t \Bigr) \\ &= \Bigl( x_2(0) - x_1(0) \Bigr) + \Bigl( v_2 - v_1 \Bigr) t \\ \end{aligned}}$

becomes

…

— Me@2023-12-06 11:06:23 AM

Physics Town

continues

@dialectphilosophy, 1.3.1

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