Intervals

• $(ct,x,y,z)$: location in spacetime (event)

$$\Delta s^2=-(c\Delta t{)}^2+(\Delta x{)}^2+(\Delta y{)}^2+(\Delta z{)}^2$$

• if only $x$, can visualize events $A,B,C,D,E$ as

• $C-D$: $\Delta s^2=0⟹$ null

• $A-B$: $\Delta s^2>0⟹$ time-like

• $C-D$: $\Delta s^2<0⟹$ space-like

• insert cone diagram

4-vector §

• can represent event as

$$\tilde{x}=(ct,x,y,z)=(x_0,x_1,x_2,x_3)$$

• or

$$x_{\mu },u=\{0,1,2,3\}$$

• or

$$A_{\mu }=(A_0,A_1,A_2,A_3)$$

• the magnitude is invariant

$$|\tilde{A}{|}^2=-A_0^2+A_1^2+A_2^2+A_3^2\to \text{invariant}$$

• and so is the dot product

$$\tilde{A}\cdot \tilde{B}.$$

Scalars in Special Relativity §

• mass
• proper time $\tau$
• velocity
  • $\frac{d}{dt}\tilde{x}$ is frame dependent because of $dt$
  • so instead

$$\frac{\Delta \tilde{x}}{\Delta \tau }\to \underset{\Delta \tau \to 0}{\lim }\frac{\Delta \tilde{x}}{\Delta \tau }\to \frac{d\tilde{x}}{d\tau }=\tilde{v}$$

• so

$$\tilde{v}=\left(c\frac{dt}{d\tau },\frac{dx}{d\tau },\dots \right)$$

• since $\frac{dt}{d\tau }=\gamma$,

$$\tilde{v}=(c\gamma ,v_x\gamma ,\dots )$$

• what is $|\tilde{v}|$?
• value is invariant, so we can choose the moving object’s frame
• $⟹$ $v_x=v_y=v_z=0⟹\gamma =1$
• $\tilde{v}=(c,0,0,0)\to |\tilde{v}{|}^2=-c^2$
• $⟹$ always moving through spacetime at $c$; if at rest, moving through spacetime at $c$, if moving spatially, move through time slower and space faster.

4-momentum §

$$\tilde{p}=m\tilde{v}$$

• $m$ is rest mass of object

$$\begin{array}{rl}\vec{p}& =(p_1,p_2,p_3)\\ & =m\tilde{v}\\ & =m\vec{v}\gamma \end{array}$$

• $p_0=mc\gamma$
• $E_0=p_{0}c=m{c}^2\gamma$
• $KE=E_0-m{c}^2=m{c}^2(\gamma -1)$
  • limit $v\ll c$, $\beta \ll 1$
  • so
• $KE\approx m{c}^2(\frac{1}{\sqrt{1-{\beta }^2}}-1)=\frac{1}{2}m{c}^2{\beta }^2$
• if $v\to c$, $|\vec{p}|,E_0\to \infty$
• we know $\tilde{p}=(mc\gamma ,m{v}_x\gamma ,\dots )$
• so

$$\begin{array}{rl}|\tilde{p}{|}^2& =-(mc\gamma {)}^2+(m\vec{v}\gamma {)}^2\\ & =-(mc{)}^2{\gamma }^2+m^2|\vec{v}{|}^2{\gamma }^2\\ & =-m^2{g}^2(c^2+1)\\ & =\frac{|\tilde{p}{|}^2}{c^2}=-m^2\\ ⟹E^2& =p^2+m^2\end{array}$$