PHYS20040:
From Classical to Modern Physics
Relativity – Problem Set 1
Question 1. Assume a non-relativistic inertial frame \(\bar S\) that moves with respect to frame \(S\) at constant speed \(v\). Derive the Galilean transformations for the coordinates of point K, moving in frame \(S\) with velocity \(u\).
Solution
The Galilean transformations are:
\[ \begin{aligned} \bar x &= x - vt\\ \bar y &= y\\ \bar z &= z\\ \bar t &= t\\ \end{aligned} \]
The velocity of point K in the moving frame will be:
\[ \begin{aligned} \bar u &= \frac{d\bar x}{d\bar t} \\ &= \frac{d(x-vt)}{dt} \\ &= \frac{dx}{dt} -v \\ &= u -v \end{aligned} \]
Question 2. Fill in the table:
| Velocity | m/s | \(\beta\) | \(\gamma\) | |
|---|---|---|---|---|
| Usaine Bolt | 9.58s / 100m | 10.44 | 3.48E-08 | 1 |
| Bicycle | 30 km/h | 8.33 | 2.78E-08 | 1 |
| Car | 70 mph | 31.11 | 1.04E-07 | 1.00000000000001 |
| Bullet train | 320 km/h | 88.89 | 2.97E-07 | 1.00000000000004 |
| Plane | 800 km/h | 222.22 | 7.41E-07 | 1.00000000000027 |
| Sound | 1,235 km/h | 343.06 | 1.14E-06 | 1.00000000000065 |
| Concorde | 2,179 km/h | 605.28 | 2.02E-06 | 1.00000000000204 |
| Falcon-9 | 9.31 km/s | 9,310.00 | 3.11E-05 | 1.0000000004822 |
| LHC protons | 99.9999991% c | 299,792,455.30 | 9.99999991E-01 | 7,453.56012234555 |
Question 3. Using Python, make a graph of the Lorentz factor, \[\gamma = \frac{1}{\sqrt{1-v^2/c^2}},\] as a function of speed, up to the speed of light. Plot as points all moving objects from question 2.
Solution
Question 4. A ruler at rest has length X=30cm. What will its length be as measured from an inertial frame moving at speed \(5\%\cdot c\;\)?
Solution
The ruler is moving at \(5\%\cdot c\), so from Lorentz contraction:
\(\bar X = X\cdot\sqrt{1-v^2/c^2} = (30cm)\cdot(\sqrt{1-(0.05\cdot c)^2/c^2}) \Rightarrow \bar X = 29.96cm\)
Question 5. Synchronized clocks are stationed at regular intervals of a million km apart along a straight line. When the clock next to you reads 12 noon, what time do you see on the 90th clock down the line?
Solution
The clocks are synchronised, and there is no relative motion between us and the clocks. However, since the speed of light is not infinite, it takes time for the photon that bounced off the 90th clock to reach us.
The photon needs to travel distance \(l = 90\cdot10^6\;km\). This takes time \(t = {l}/{c}\).
So the photon we see from the 90th clock left
\[t = \frac{l}{c} = \frac{90\cdot10^6km}{300.000\;km/s} = \frac{900}{3}s = 300s\;\;ago.\]
The clock will read 11:55.
The Sun is 150 million km away, the light we see is 8-9 minutes old.
Question 6. At which velocity will the tick of a moving clock slows down by \(50\%\)?
We measure \((\Delta t)\) the time interval of a moving clock \((\Delta \bar t)\) to last 50% longer, i.e. \(\Delta t = 1.50 \cdot \Delta \bar t\).
From time dilation,
\[ \begin{aligned} \Delta t &= \frac{\Delta \bar t }{\sqrt{1-\frac{v^2}{c^2}}} \Rightarrow \\ 1.5 \cdot \Delta \bar t &= \frac{\Delta \bar t }{\sqrt{1-\frac{v^2}{c^2}}} \Rightarrow \\ \sqrt{1-\frac{v^2}{c^2}} &= \frac{1}{1.5} \Rightarrow \\ v &= 0.74\cdot c \end{aligned} \]