Exercises — Lorentz transformation — derivation

What: plug into $\gamma =1/\sqrt{1-v^2/c^2}$. Why: $\gamma$ is the single number that controls how much space and time stretch; every relativity problem starts here. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=\mathrm{1.25.}$ Answer: $\gamma =1.25$.

The extra piece is the $-\frac{vx}{c^2}$ term. What it does: it makes the time-shift depend on where the event happens ($x$). Two events at different places but the same lab time $t$ get different primed times — this is the seed of Relativity of simultaneity. The overall $\gamma$ is the second departure (it multiplies everything).

$\frac{v^2}{c^2}=\frac{(1.5{)}^2}{(3\times 10^8{)}^2}=\frac{2.25}{9\times 10^{16}}=2.5\times 10^{-17}.$ This is unimaginably tiny, so $\gamma =1/\sqrt{1-2.5\times 10^{-17}}\approx 1$. Meaning: the Lorentz transform collapses back to Galileo, $x^{\prime }\approx x-vt$, $t^{\prime }\approx t$. The new theory contains the old one — as any correct generalization must.

What: substitute into both master equations. $x^{\prime }=\gamma (x-vt)=1.25(2-0.6\cdot 5)=1.25(2-3)=1.25\times (-1)=-1.25\text{ls}.$ $t^{\prime }=\gamma \left(t-\frac{vx}{c^2}\right)=1.25(5-0.6\cdot 2)=1.25(5-1.2)=1.25\times 3.8=4.75\text{s}.$ Why the numbers: $vx/c^2=0.6\times 2=1.2$ s because $c=1$. Answer: $(x^{\prime },t^{\prime })=(-1.25\text{ls},4.75\text{s})$. The event is behind the moving origin ($x^{\prime }<0$) and happens later in $S^{\prime }$.

What: the clock is at the moving origin, so use the inverse time equation with $x^{\prime }=0$: $t=\gamma \left(t^{\prime }+\frac{v{x}^{\prime }}{c^2}\right)\stackrel{x^{\prime }=0}{\to }\Delta t=\gamma \Delta t^{\prime }=1.25\times 1=1.25\text{s}.$ Why $x^{\prime }=0$ matters: Time dilation compares the same clock's ticks; that clock is stationary in $S^{\prime }$, so its position never changes there. Answer: $\Delta t=1.25$ s. The lab sees the moving clock tick slowly (more lab time per tick).

What: measuring length in the lab means noting both ends simultaneously in $S$, so $\Delta t=0$. Use $x^{\prime }=\gamma (x-vt)$ as a difference: $\Delta x^{\prime }=\gamma (\Delta x-v\Delta t)\stackrel{\Delta t=0}{\to }{L}_0=\gamma L\Rightarrow L=\frac{L_0}{\gamma }=\frac{4}{1.25}=3.2\text{ls}.$ Why $\Delta t=0$: if you noted the ends at different times the moving rod would have shifted, corrupting the measurement. Answer: $L=3.2$ ls (contracted). See Length contraction.

What: transform each event's time. $t_1^{\prime }=\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)=1.25(0-0)=0.$ $t_2^{\prime }=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)=1.25(0-0.6\cdot 3)=1.25\times (-1.8)=-2.25\text{s}.$ Reading it: $t_2^{\prime }=-2.25$ s is earlier than $t_1^{\prime }=0$, so in $S^{\prime }$ the far explosion (event 2) happens first, by $2.25$ s. Why: the $-vx/c^2$ term punishes larger $x$ with a bigger negative time-shift — simultaneity is not absolute. See Relativity of simultaneity. Look at the tilted line of simultaneity in the figure below.

In $S$: $\Delta t=0$, $\Delta x=3$ ls. With $c=1$: $s^2=(1{)}^2(0{)}^2-(3{)}^2=-9{\text{ls}}^2.$ In $S^{\prime }$: from L3.1, $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }=-2.25$ s. And $\Delta x^{\prime }=\gamma (\Delta x-v\Delta t)=1.25(3-0)=3.75$ ls. $s^{\prime 2}=(1{)}^2(-2.25{)}^2-(3.75{)}^2=5.0625-14.0625=-9{\text{ls}}^2.$ Answer: both give $s^2=-9{\text{ls}}^2$. Why it's negative: a negative $s^2$ means the events are space-like separated — no signal can connect them, which is exactly why their time-order can flip between frames. See Spacetime interval.

What: use the relativistic addition rule (the inverse form, adding $v$): $u=\frac{u^{\prime }+v}{1+u^{\prime }v/c^2}=\frac{0.5c+0.6c}{1+(0.5)(0.6)}=\frac{1.1c}{1.3}\approx 0.846c.$ Why not $1.1c$: naive addition would exceed $c$! The denominator $1+u^{\prime }v/c^2=1.3$ tames it. Answer: $u\approx 0.846c$, still below $c$.

What: the muon's lifetime is a proper time (measured on its own clock, $\Delta x^{\prime }=0$). The lab sees it dilated: $\gamma =\frac{1}{\sqrt{1-0.98^2}}=\frac{1}{\sqrt{1-0.9604}}=\frac{1}{\sqrt{0.0396}}\approx \mathrm{5.025.}$ $\Delta t_{\text{lab}}=\gamma \tau =5.025\times 2.2\mu \text{s}\approx 11.06\mu \text{s}.$ Lab distance: $d=v\Delta t_{\text{lab}}=0.98\times (3\times 10^8)\times 11.06\times 10^{-6}\text{m}\approx 3251\text{m}\approx 3.25\text{km}.$ Naive distance (ignoring dilation): $v\tau =0.98\times 3\times 10^8\times 2.2\times 10^{-6}\approx 647$ m. Why the difference: without time dilation the muon "shouldn't" reach the ground; the factor $\gamma \approx 5$ lets it travel $\sim 5\times$ farther. This is a real observed effect. Answer: $\approx 3.25$ km (vs $\approx 0.65$ km naive).

Step 1 (${\gamma }_1=1.25$): $x^{\prime }=1.25(4-0.6\cdot 5)=1.25(4-3)=1.25\text{ls},t^{\prime }=1.25(5-0.6\cdot 4)=1.25(5-2.4)=3.25\text{s}.$ Step 2 (same $\gamma =1.25$, applied to primed coords): $x^{\prime \prime }=1.25(1.25-0.6\cdot 3.25)=1.25(1.25-1.95)=1.25(-0.7)=-0.875\text{ls},$ $t^{\prime \prime }=1.25(3.25-0.6\cdot 1.25)=1.25(3.25-0.75)=1.25(2.5)=3.125\text{s}.$ (b) Combined velocity: $v_{\text{tot}}=\frac{0.6c+0.6c}{1+0.36}=\frac{1.2c}{1.36}\approx 0.88235c,{\gamma }_{\text{tot}}=\frac{1}{\sqrt{1-0.88235^2}}\approx \mathrm{2.125.}$ Single boost from lab directly: $x^{\prime \prime }=2.125(4-0.88235\cdot 5)=2.125(4-4.41176)=2.125(-0.41176)\approx -0.875\text{ls},$ $t^{\prime \prime }=2.125(5-0.88235\cdot 4)=2.125(5-3.52941)=2.125(1.47059)\approx 3.125\text{s}.$ Answer: both routes give $(x^{\prime \prime },t^{\prime \prime })\approx (-0.875\text{ls},3.125\text{s})$. Why: Lorentz boosts along one axis form a group — two in a row equal one bigger boost, whose speed is set by velocity addition (not $1.2c$!).

What: velocity is a ratio of differentials, so take differentials of both transforms. $d{x}^{\prime }=\gamma (dx-vdt),d{t}^{\prime }=\gamma \left(dt-\frac{vdx}{c^2}\right).$ Why differentials: $u^{\prime }=d{x}^{\prime }/d{t}^{\prime }$ by definition, and the $\gamma$'s will cancel in the ratio — that's the elegance. $u^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{\gamma (dx-vdt)}{\gamma \left(dt-\frac{vdx}{c^2}\right)}=\frac{dx-vdt}{dt-\frac{vdx}{c^2}}.$ Now divide top and bottom by $dt$ (so $dx/dt=u$ appears): $u^{\prime }=\frac{\frac{dx}{dt}-v}{1-\frac{v}{c^2}\frac{dx}{dt}}=\boxed{\frac{u-v}{1-uv/c^2}}.$ Check: set $u=c$: $u^{\prime }=\frac{c-v}{1-v/c}=\frac{c-v}{(c-v)/c}=c$. Light stays at $c$. $✓$

What: substitute the transforms and expand. $c^2{t}^{\prime 2}=c^2{\gamma }^2{\left(t-\frac{vx}{c^2}\right)}^2={\gamma }^2\left(c^2{t}^2-2vxt+\frac{v^2{x}^2}{c^2}\right).$ $x^{\prime 2}={\gamma }^2(x-vt{)}^2={\gamma }^2\left(x^2-2vxt+v^2{t}^2\right).$ Subtract — the cross terms $-2vxt$ cancel (that's why the interval survives): $c^2{t}^{\prime 2}-x^{\prime 2}={\gamma }^2\left[c^2{t}^2+\frac{v^2{x}^2}{c^2}-x^2-v^2{t}^2\right].$ Group by $t^2$ and $x^2$: $={\gamma }^2\left[c^2{t}^2\left(1-\frac{v^2}{c^2}\right)-x^2\left(1-\frac{v^2}{c^2}\right)\right]={\gamma }^2\left(1-\frac{v^2}{c^2}\right)(c^2{t}^2-x^2).$ Final blow: ${\gamma }^2(1-v^2/c^2)=1$ by definition of $\gamma$, leaving $c^2{t}^{\prime 2}-x^{\prime 2}=c^2{t}^2-x^2.✓$ This invariance is the geometric heart of the Minkowski diagram and Spacetime interval.

(a) $\gamma =1/\sqrt{1-0.64}=1/\sqrt{0.36}=1/0.6=1.667$. $L=\frac{L_0}{\gamma }=\frac{100}{1.667}=60\text{m}.$ (b) In the train frame the two doors open at the same time $\Delta t^{\prime }=0$, separated by $\Delta x^{\prime }=L_0=100$ m. Transform to the platform with the inverse time equation: $\Delta t=\gamma \left(\Delta t^{\prime }+\frac{v\Delta x^{\prime }}{c^2}\right)=\gamma \frac{v\Delta x^{\prime }}{c^2}.$ Plug numbers ($v=0.8c$, $c=3\times 10^8$): $\Delta t=1.667\times \frac{0.8\times 3\times 10^8\times 100}{(3\times 10^8{)}^2}=1.667\times \frac{0.8\times 100}{3\times 10^8}=1.667\times 2.667\times 10^{-7}\text{s}.$ $\Delta t\approx 4.44\times 10^{-7}\text{s}=0.444\mu \text{s}.$ Answer: platform measures length $60$ m and sees the doors open $\approx 0.444\mu \text{s}$ apart — the front and back events, simultaneous on the train, are not simultaneous on the platform. This is the engine behind every "pole-in-barn" paradox.

$\gamma$ for $v=0.6c$? ::: $1.25$ Term that breaks Galilean simultaneity? ::: $-vx/c^2$ Condition for time dilation? ::: same clock, $\Delta x^{\prime }=0\Rightarrow \Delta t=\gamma \Delta t^{\prime }$ Condition for length contraction? ::: simultaneous in lab, $\Delta t=0\Rightarrow L=L_0/\gamma$ Velocity of $u^{\prime }=0.5c$ inside $v=0.6c$ frame (lab)? ::: $\approx 0.846c$ Interval $s^2=c^2\Delta t^2-\Delta x^2$ is... ::: invariant across all inertial frames