Exercises — Reference frames — Galilean transformations

(b) ${\vec{a}}^{\prime }=\vec{a}$. What we did: recalled the last line of the derivation chain. Why: velocity differs by the constant $\vec{V}$; differentiating a constant gives zero, so the shift disappears at the acceleration level. Options (a) and (c) mix up the velocity shift ($\vec{V}$) and the position shift ($\vec{V}t$) — those live one and two derivatives lower.

(i) In $S^{\prime }$ the passenger is at rest: $u^{\prime }=0$. (ii) Ground velocity $u=u^{\prime }+V=0+15=15\text{ m/s east}$. Why: "at rest on the train" means zero velocity in the primed frame; the ground observer still sees them carried along at the train's speed. This is the whole meaning of "relative to what".

$u=u^{\prime }+V=1.5+1.2=2.7\text{ m/s in }+x.$ Why plain addition works: both velocities point the same way ($+x$), so the vector equation collapses to scalar addition of like signs.

Take $+x$ as positive. Then $u^{\prime }=-1.5$, $V=+1.2$: $u=u^{\prime }+V=-1.5+1.2=-0.3\text{ m/s}.$ The negative sign means the person still drifts in $-x$ (the belt cannot keep up with them), at $0.3\text{ m/s}$. Why sign-tracking matters: opposite directions become opposite signs; the algebra then tells you the net direction automatically.

We are given $u=0$ (ground) and want $u^{\prime }$. Rearrange $u=u^{\prime }+V$: $u^{\prime }=u-V=0-20=-20\text{ m/s},\text{ i.e. }20\text{ m/s west relative to the train.}$ Why: to appear frozen to the ground, the ball must move backward on the train at exactly the train's speed — the two shifts cancel.

Set $+x$ = east, $+y$ = north. Then ${\vec{u}}^{\prime }=(0,8)$ and $\vec{V}=(6,0)$. $\vec{u}={\vec{u}}^{\prime }+\vec{V}=(0+6,8+0)=(6,8)\text{ m/s}.$ Speed (Pythagoras on the right triangle with legs $6$ and $8$ — see figure): $|\vec{u}|=\sqrt{6^2+8^2}=\sqrt{100}=10\text{ m/s}.$ Direction. The angle $\theta$ north-of-east satisfies $\tan \theta =\frac{\text{north leg}}{\text{east leg}}=\frac{8}{6}$, so $\theta ={\tan }^{-1}\left(\frac{8}{6}\right)\approx 53.13^{\circ }\text{ north of east}.$ Why $\tan$ and ${\tan }^{-1}$? On the right triangle the ground-velocity arrow is the hypotenuse; $\tan \theta$ = opposite/adjacent encodes how "steep" that arrow tilts toward north. ${\tan }^{-1}$ answers the reverse question "which angle has this steepness?" Both components are positive, so $\vec{u}$ sits in quadrant I and the naive ${\tan }^{-1}$ needs no correction.

${\vec{u}}^{\prime }=(-8,6)$, $\vec{V}=(6,0)$: $\vec{u}=(-8+6,6+0)=(-2,6)\text{ m/s}.$ Speed: $|\vec{u}|=\sqrt{(-2{)}^2+6^2}=\sqrt{40}=2\sqrt{10}\approx 6.32\text{ m/s}.$ Direction. The $x$-component is negative, the $y$-component positive → the arrow points into quadrant II (up-left). The naive calculator value ${\tan }^{-1}(6/-2)={\tan }^{-1}(-3)\approx -71.57^{\circ }$ lands in quadrant IV — wrong, because $\tan$ repeats every $180^{\circ }$ and cannot tell quadrant II from quadrant IV. Fix by adding $180^{\circ }$: $\theta =-71.57^{\circ }+180^{\circ }=108.43^{\circ }\text{ from }+x.$ Why the fix: $\tan$ collapses two opposite directions onto one number; the sign pair $(-,+)$ pins us to quadrant II, so we shift the principal value into that quadrant.

$+x$ = east. $u_A=+32$, $u_B=-18$. B's frame plays the role of $S^{\prime }$ with $\vec{V}=u_B$: $u_{A/B}=u_A-u_B=32-(-18)=50\text{ m/s (eastward, toward B).}$ Why a sum, not a difference: they move oppositely, so subtracting a negative adds — the closing speed is the sum of the two road-speeds.

${\vec{u}}_{A/B}={\vec{u}}_A-{\vec{u}}_B=(20-0,0-15)=(20,-15)\text{ m/s}.$ Speed: $\sqrt{20^2+15^2}=\sqrt{625}=25\text{ m/s}$. Direction: $x>0$, $y<0$ → quadrant IV (down-right); here the raw ${\tan }^{-1}$ is valid: $\theta ={\tan }^{-1}\left(\frac{-15}{20}\right)\approx -36.87^{\circ }\text{ from }+x.$ Why quadrant IV needs no fix: ${\tan }^{-1}$ natively returns quadrant-I and quadrant-IV angles, and the sign pair $(+,-)$ confirms quadrant IV.

Apply $u=u^{\prime }+V$ twice, working outward from the mouse: $u_{\text{trolley/ground}}=V_2+V_1=3+4=7\text{ m/s east},$ $u_{\text{mouse/ground}}=u^{\prime \prime }+u_{\text{trolley/ground}}=2+7=9\text{ m/s east}.$ Why it chains cleanly: all motions are collinear (east), so each Galilean shift is a simple add; the frame velocities stack.

With $\vec{V}=\vec{0}$: ${\vec{r}}^{\prime }=\vec{r}-\vec{0}\cdot t=\vec{r},{\vec{u}}^{\prime }=\vec{u},{\vec{a}}^{\prime }=\vec{a}.$ Every measurement agrees. Physically: two observers with no relative motion share the same inertial frame — the transformation is the identity. This is the sanity-check limiting case: no relative motion, nothing to translate.

Start from ${\vec{u}}^{\prime }=\vec{u}-\vec{V}(t)$ and differentiate. Now $\vec{V}$ is not constant, so its derivative survives: ${\vec{a}}^{\prime }=\vec{a}-\frac{d\vec{V}}{dt}=\vec{a}-\vec{A}.$ Acceleration is no longer invariant. Rearranging, $m{\vec{a}}^{\prime }=m\vec{a}-m\vec{A}={\vec{F}}_{\text{real}}-m\vec{A}$; the extra $-m\vec{A}$ acts like a force with no source — a pseudo-force. Why this matters: constant $\vec{V}$ was the magic ingredient (see Inertial and non-inertial frames). Drop it and Newton's second law gains fictitious terms.

Compute the correction denominator: $\frac{u^{\prime }V}{c^2}=\frac{300\times 300}{(3\times 10^8{)}^2}=\frac{9\times 10^4}{9\times 10^{16}}=10^{-12}.$ The relativistic answer is $u=\frac{600}{1+10^{-12}}\approx 600\times (1-10^{-12})$, differing from $600\text{ m/s}$ by about $6\times 10^{-10}\text{ m/s}$ — one part in $10^{12}$. Why Galileo is safe here: the correction scales as $u^{\prime }V/c^2$, absurdly tiny at everyday speeds. Only when speeds approach $c$ does the $t^{\prime }=t$ assumption fail and you must switch to the Lorentz transformation.

Why does the relative shift vanish for acceleration but not for velocity? ::: Because $\vec{V}$ is constant: velocity differs by $\vec{V}$, but $d\vec{V}/dt=0$, so acceleration differs by nothing. Each derivative removes one power of $t$ from the shift $\vec{V}t\to \vec{V}\to 0$.