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Attempted to detect Earth’s motion through the luminiferous aether by comparing light travel times in perpendicular arms of an interferometer. Classical aether theory predicted a fringe shift of order \((v/c)^2\).
Outcome: Null result (no detectable shift). Speed of light is isotropic and independent of the motion of source/observer. This eliminated the preferred aether frame and led to Special Relativity.
Arm length \(L\), Earth speed \(v\), light speed \(c\), \(\beta=v/c\).
Round-trip time parallel to velocity:
\[ t_\parallel = \frac{L}{c-v} + \frac{L}{c+v} = \frac{2L}{c}\frac{1}{1-\beta^2} \]Perpendicular (using light path Pythagoras):
\[ t_\perp = \frac{2L}{c\sqrt{1-\beta^2}} \]Difference to order \(\beta^2\):
\[ \Delta t \approx \frac{L}{c}\beta^2 \]Path difference \(\delta = c\Delta t \approx L\beta^2\). Fringe shift \(N=\delta/\lambda\).
For \(L=11\) m, \(v=30\) km/s, \(\lambda=550\) nm: \(N\sim 0.2\). Observed \(N\approx 0\) within error.
Calculate the expected classical fringe shift for \(L=11\) m, \(v=3\times10^4\) m/s, \(\lambda=5.5\times10^{-7}\) m.
These two postulates replace the Galilean transformation with the Lorentz transformation and imply the relativity of simultaneity, time dilation, and length contraction.
Assume linear transformation (homogeneous spacetime, constant relative velocity \(v\) along \(x\)):
\[ x' = a(x-vt),\qquad t' = bx + dt \](or equivalently \(x'=\gamma(x-vt)\), \(t'=\gamma(t-\frac{vx}{c^2})\) after determining constants).
1. Light signal along +x: \(x=ct\) must map to \(x'=ct'\). Substituting yields consistency conditions.
2. Inverse transformation: Interchange frames and \(v\to-v\). Group property and reciprocity require \(\gamma(v)=\gamma(-v)\).
3. Invariance of the interval: Require \(c^2t'^2-x'^2=c^2t^2-x^2\) (light cone preserved). This forces
\[ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \]Final Lorentz transformation (boost along \(x\)):
\[ \begin{align} x' &= \gamma(x-vt)\\ y' &= y,\quad z'=z\\ t' &= \gamma\Bigl(t-\frac{vx}{c^2}\Bigr) \end{align} \]where \(\displaystyle\gamma=\frac{1}{\sqrt{1-\beta^2}}\), \(\beta=v/c\).
Inverse: replace \(v\) by \(-v\) (or interchange primed/unprimed).
In rapidity \(\phi\) (\(\tanh\phi=\beta\)) the transformation is a hyperbolic rotation:
\[ ct' = ct\cosh\phi - x\sinh\phi,\qquad x' = -ct\sinh\phi + x\cosh\phi \]The LT mixes space and time. Key immediate consequences:
In frame \(S\), events occur at \((ct_1=0,x_1=0)\) and \((ct_2=4\,\text{m},x_2=3\,\text{m})\). Frame \(S'\) moves at \(v=0.6c\). Find coordinates in \(S'\).
Interactive Lorentz transformation of a spacetime grid. Blue = S, orange = S′.
Light-clock argument: A clock consisting of two mirrors separated by proper distance \(d\) (perpendicular to motion). Proper period \(\tau_0=2d/c\).
In lab frame the light travels a longer zigzag path. Time for one tick:
\[ \Bigl(\frac{c\Delta t}{2}\Bigr)^2 = d^2 + \Bigl(\frac{v\Delta t}{2}\Bigr)^2 \implies \Delta t = \frac{2d}{c\sqrt{1-\beta^2}} = \gamma\tau_0 \]From LT: Two events at the same location in \(S'\) (\(x'=0\)) have \(\Delta t'=\Delta\tau\). Then \(\Delta t=\gamma\Delta\tau\).
Muons have proper lifetime \(\tau_0=2.2\,\mu\)s. They travel at \(v=0.995c\). (a) Lifetime in Earth frame? (b) Mean distance travelled?
A spaceship clock reads 5 s for an interval. The ship moves at \(0.6c\) relative to Earth. What time interval does an Earth observer measure for the same events (which occur at the same place on the ship)?
Proper length \(L_0\) is measured in the rest frame of the object (both ends at the same time \(t'\)).
In the lab frame the length \(L\) is the simultaneous (\(t=\)const) difference of the end coordinates. From the inverse LT:
\[ L_0 = x_2'-x_1' = \gamma(x_2-x_1)=\gamma L \implies L=\frac{L_0}{\gamma}=L_0\sqrt{1-\beta^2} \]Only the parallel dimension contracts; transverse lengths are unchanged.
A spaceship has proper length 100 m. It flies past Earth at \(0.8c\). What length does an Earth observer measure?
An arrow of rest length 1 m is launched at \(0.8c\). What length does a stationary observer measure?
From the LT: \(\Delta t'=\gamma(\Delta t - v\Delta x/c^2)\). Events simultaneous in \(S\) (\(\Delta t=0\)) have \(\Delta t'=-\gamma v\Delta x/c^2\neq0\) if \(\Delta x\neq0\).
Temporal order of spacelike-separated events can reverse between frames. Order of timelike-separated events is absolute (causality preserved).
Two lightning strikes are simultaneous (\(\Delta t=0\)) at \(x_A=0\) and \(x_B=1000\) m in \(S\). An observer in \(S'\) moves at \(0.8c\). Find \(\Delta t'\) and which strike occurs first in \(S'\).
Let an object have velocity \(u'=dx'/dt'\) in \(S'\). \(S'\) moves at \(v\) relative to \(S\).
\[ dx'=\gamma(dx-v\,dt),\qquad dt'=\gamma\Bigl(dt-\frac{v\,dx}{c^2}\Bigr) \] \[ u'=\frac{dx'}{dt'}=\frac{dx-v\,dt}{dt-v\,dx/c^2}=\frac{u-v}{1-uv/c^2} \]Solving for \(u\):
\[ u=\frac{u'+v}{1+u'v/c^2} \]General (3-D) components (boost along \(x\)):
\[ \begin{align} u_x&=\frac{u_x'+v}{1+u_x'v/c^2}\\ u_y&=\frac{u_y'}{\gamma(1+u_x'v/c^2)}\\ u_z&=\frac{u_z'}{\gamma(1+u_x'v/c^2)} \end{align} \]Special cases:
A spaceship moves at \(0.9c\) relative to Earth. It launches a probe forward at \(0.9c\) relative to itself. Speed of the probe relative to Earth?
Two spaceships approach each other, each with speed \(0.8c\) relative to Earth (opposite directions). What is their relative speed?
In \(S'\) a particle has velocity components \(u_x'=0.6c\), \(u_y'=0.6c\). \(S'\) moves at \(v=0.8c\) relative to \(S\). Find the velocity components in \(S\).
Show that if \(u'=c\) then \(u=c\) for any \(v<c\).
Classical (dashed) vs relativistic velocity addition. Relativistic result always stays below \(c\).
Invariant interval:
\[ \Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 \]Causality: A cause can influence an effect only if the separation is timelike or lightlike. This is preserved by Lorentz transformations.
Classify: (a) \(\Delta x=3\) m, \(c\Delta t=5\) m; (b) \(\Delta x=5\) m, \(c\Delta t=3\) m; (c) \(\Delta x=4\) m, \(c\Delta t=4\) m.
Four-momentum \(P^\mu=(E/c,\mathbf{p})\) with rest mass \(m_0\):
\[ E=\gamma m_0 c^2,\qquad \mathbf{p}=\gamma m_0\mathbf{v},\qquad E^2=p^2c^2+m_0^2c^4 \]Rest energy \(E_0=m_0c^2\). Kinetic energy \(K=E-m_0c^2=(\gamma-1)m_0c^2\).
Massless particles (\(m_0=0\)): \(E=pc\), always travel at \(c\).
Historical “relativistic mass” \(m=\gamma m_0\) is now rarely used; prefer rest mass + energy-momentum.
Electron rest energy \(511\) keV, speed \(0.99c\). Find total energy, kinetic energy and momentum.
A four-vector \(A^\mu\) transforms exactly as the coordinates \(x^\mu=(ct,x,y,z)\). The Minkowski inner product \(A\cdot B=A^0B^0-\mathbf{A}\cdot\mathbf{B}\) is Lorentz invariant.
Conservation of four-momentum: In any collision or decay the total four-momentum is conserved. This unifies energy and momentum conservation and is frame-independent.
Show \(U\cdot U=c^2\).
A particle of rest mass \(M\) at rest decays into two identical particles of rest mass \(m\). Find the energy of each daughter.
Spacetime diagram with vertical \(ct\) and horizontal \(x\). Light worldlines are at \(45^\circ\). Worldlines of massive particles have \(|{\rm slope}|>1\). Simultaneous events in a given frame lie on horizontal (or boosted) lines of constant \(t\).
A Lorentz boost appears as a hyperbolic rotation of the axes. Hyperbolae \(c^2t^2-x^2={\rm const}\) are loci of constant proper time / interval.
Source emits frequency \(f_0\) in its rest frame. Relative velocity \(\beta\) (positive = receding).
Two effects combine: (1) classical Doppler from changing path length, (2) time dilation of the source clock. Result:
\[ f = f_0\sqrt{\frac{1-\beta}{1+\beta}}\quad\text{(receding)},\qquad f = f_0\sqrt{\frac{1+\beta}{1-\beta}}\quad\text{(approaching)} \]Transverse Doppler (source moving perpendicular to line of sight at closest approach): purely time-dilation redshift
\[ f = f_0/\gamma \]Aberration formula:
\[ \cos\theta=\frac{\cos\theta'+\beta}{1+\beta\cos\theta'} \](or \(\tan(\theta/2)=e^{-\phi}\tan(\theta'/2)\) with rapidity \(\phi\)).
A star emits light of proper frequency \(f_0\) and recedes at \(0.6c\). What frequency is observed?
Analysed most cleanly in the parent rest frame using four-momentum conservation. Lab lifetime is dilated: \(\tau_{\rm lab}=\gamma\tau_0\).
Charged pion \(\tau_0=26\) ns, \(m_\pi c^2=140\) MeV, moves with \(\gamma=10\). (a) Lab lifetime? (b) Mean distance? (c) Comment on muon energy.
For a boost along \(x\):
\[ \begin{align} E'_x&=E_x,& E'_y&=\gamma(E_y-vB_z),& E'_z&=\gamma(E_z+vB_y)\\ B'_x&=B_x,& B'_y&=\gamma\Bigl(B_y+\frac{v}{c^2}E_z\Bigr),& B'_z&=\gamma\Bigl(B_z-\frac{v}{c^2}E_y\Bigr) \end{align} \]Lorentz invariants: \(\mathbf{E}\cdot\mathbf{B}\) and \(E^2-c^2B^2\).
Because the field-strength tensor \(F^{\mu\nu}\) and the four-current \(J^\mu\) transform as tensors, Maxwell’s equations keep the same form in every inertial frame — historically a major motivation for Special Relativity.
In \(S\) there is a pure electric field \(\mathbf{E}=(0,E,0)\), \(\mathbf{B}=0\). What fields does an observer moving at velocity \(v\) along \(x\) measure?
One twin travels at high speed and returns; the travelling twin ages less. No paradox: the travelling twin changes inertial frames (accelerates at turnaround), breaking the symmetry. Only the stay-at-home twin remains in a single inertial frame throughout.
Twin A stays on Earth. Twin B travels at \(0.8c\) to a star 8 ly away (Earth frame) and returns at the same speed. How much does each twin age?