Galilean Relativity

What is a Reference Frame§

Reference§

Imagine you ( $A$ ) and your friend ( $B$ ) are standing on a football field, and he measures everything relative to him, and you measure everything relative to you. That's the whole point of having a reference frame, as all inertial frames are empirically equivalent and thus must have a reference.

Displacement and Distance§

Suppose the displacement between you and him measured from you and relative to you is

$$\vec{r}_{AB}=(3,4)$$

This means, equivalently, measured from him, the displacement is

$$\vec{r}_{BA}=(-3, -4)$$

We will see later how the notation is defined and the invariance for displacement and distance.

When the ball is at $A$§

When the football is under your feet, you measure

$$\vec{r}_O=(0, 0)$$

Where $O$ is the ball.

But in his view, the ball is at

$$\vec{r}'_O = (-3, -4)$$

Where the $'$ indicates this is from $B$'s frame.

When the ball is at $B$§

When the football is under his feet, you measure

$$\vec{r}_O=(3, 4)$$

But in his view, the ball is at

$$\vec{r}'_O = (0, 0)$$

What is the point?§

The whole point of using a reference frame is just to have a way to measure object displacement and position relative to a known reference; in this case, it is you ( $A$ ) and him ( $B$ ). It is totally unambiguous when we look at the context that $\vec{r}_O = (0, 0)$ and $\vec{r}'_O = (-3, -4)$ are describing the same location even though they are different in the coordinate value. And they are empirically equivalent.

Invariant§

Quick definition§

Any measurement that is reference frame independent.

Displacement and Distance§

Either in frame $A$ or frame $B$, the displacement from $A$ to $B$ is defined to be

$$\vec{r}_{AB} = \vec{r}_B - \vec{r}_A$$

Within either frame, it will be measured to be

$$\vec{r}_{AB} = (3, 4)$$

And the distance is $d(A,B) = \sqrt{3^2 + 4^2} = 5$, measured from either frame.

The same goes for $\vec{r}_{BA} = \vec{r}_A - \vec{r}_B$.

Time§

Another invariant will be time intervals. While you may freely disagree on what time it is right now—it might be night for me while it is early in the morning for you—1 second passed on my side will be 1 second on your side.

Postulates on space and time§

Feel free to disagree (which is exactly what Einstein did to discover special relativity), but the invariance of displacement and time intervals are postulates of Galilean relativity and are motivated by common sense in everyday life. Note that, from now on, for any invariant values, I will freely omit $'$ regardless of the frame measured from, as it doesn't matter.

Moving Reference Frames§

In $B$§

Now imagine in $B$'s frame that he has the ball and kicks it toward you ($A$). The ball now has a speed of $v'_O$ towards you, while you run toward the ball with speed $v'_A$. Well, then, according to $B$, the ball's coordinate relative to him over time will be

$$\vec{r}'_O = \vec{v}'_O\cdot t$$

We already know that velocity is not invariant by Galileo's Ship, so let's derive the velocity of the ball in your ($A$'s) frame instead of just assuming it is $v'_O$ which is measured from $B$'s frame.

In $A$§

From $B$'s frame, the displacement to you ($A$) is $\vec{r}_{BA}$ which was originally $(-3, -4)$, but now you are moving, so

$$\vec{r}_{BA}=(-3, -4) + \vec{v}'_{A}\cdot t = \vec{r}'_A - \vec{r}'_B$$

Since $B$ will measure himself always at $(0, 0)$, thus

$$\vec{r}'_A = \vec{r}_{BA} = (-3, -4) + \vec{v}'_{A} \cdot t$$

Then, we know that displacement should be invariant, thus measuring the displacement from you ($A$) to the ball $O$ will be

$$\begin{align*} \vec{r}_{AO} &= \vec{r}'_O - \vec{r}'_A \\ &= \vec{v}'_O\cdot t - (-3, -4) - \vec{v}'_A \cdot t \\ &= -(-3, -4) + ( \vec{v}'_O - \vec{v}'_A )\cdot t \\ &= \vec{r}_O - \vec{r}_A \end{align*}$$

Again, $\vec{r}_A$ is you ($A$) measuring the displacement of yourself to yourself, which should be $(0, 0)$. Thus

$$\vec{r}_O = (3, 4) + \vec{v}_O\cdot t$$

Where

$$\vec{v}_O = \vec{v}'_O - \vec{v}'_A$$

Just for a sanity check, imagine the ball is kicked toward you, but then you run away from it with the same velocity, thus $\vec{v}_O = 0$. Then the ball should stay at a constant distance from you at all times; this justifies the equation above as we get

$$\vec{r}_O = (3, 4) = \text{const.} \qquad \vec{v}_O = 0$$

Rule for linearity of velocity§

This is one of the standard rules in Galilean relativity: velocities add or subtract linearly. As we derived above, if an object has a velocity in one frame, its velocity in another moving frame is simply the vector difference:

$$\vec{v}_O = \vec{v}'_O - \vec{v}'_A$$