Minkowski spacetime

Minkowski spacetime is a 4-dimensional pseudo-Riemannian manifold equipped with a flat Lorentzian metric. The coordinates of an event can be written as $x^{\mu }=(ct,x,y,z)=(ct,\mathbf{x})$. Using the metric signature $(+,-,-,-)$, lowering the index gives $x_{\mu }=(ct,-x,-y,-z)$.

The metric of the Minkowski space is (spacetime interval) defined as $s^2=c^2(\Delta t{)}^2-(\Delta x{)}^2-(\Delta y{)}^2-(\Delta z{)}^2$

For a positive constant $a^2$, the set of events with a fixed Minkowski norm is a two-sheeted timelike hyperboloid: (an equal-proper-time surface)

$$x^{\mu }{x}_{\mu }=c^2{t}^2-\Vert \mathbf{x}{\Vert }^2=a^2(a>0)$$ $$\frac{c^2{t}^2}{a^2}-\frac{\Vert \mathbf{x}{\Vert }^2}{a^2}=1$$

In the limiting case $a=0$, we get the light cone:

$$c^2{t}^2-\Vert \mathbf{x}{\Vert }^2=0⟹\Vert \mathbf{x}\Vert =c|t|$$

The hyperboloid approaches the light cone as $a$ becomes smaller.

Lorentz transformation

An orthochronous Lorentz transformation preserves the future and past sheets separately. A general Lorentz transformation need not do so, since time reversal exchanges the two sheets.

Proper Time

For two infinitesimally separated events on a timelike worldline, the proper time $d\tau$ is the time measured by a clock moving along that worldline. It is defined by

$$c^{2}d{\tau }^2=d{s}^2=c^{2}d{t}^2-\Vert d\mathbf{x}{\Vert }^2.$$

Since $\mathbf{v}=d\mathbf{x}/dt$, we have

$$d\tau =dt\sqrt{1-\frac{\Vert \mathbf{v}{\Vert }^2}{c^2}}=\frac{dt}{\gamma }.$$

Proper time is Lorentz invariant. It is defined along timelike worldlines; along a null worldline, $d\tau =0$.

Velocity

For a particle following a timelike worldline, we define the 4-velocity using its proper time $\tau$:

$$U^{\mu }=\frac{d{x}^{\mu }}{d\tau }=\gamma (c,v_x,v_y,v_z)=(\gamma c,\gamma \mathbf{v}).$$

With our metric signature, its invariant norm is $U^{\mu }{U}_{\mu }=c^2$. Solving for $\gamma$ gives

$$U^{\mu }{U}_{\mu }={\gamma }^2\left(c^2-\Vert \mathbf{v}{\Vert }^2\right)=c^2⟹\gamma =\frac{1}{\sqrt{1-\frac{\Vert \mathbf{v}{\Vert }^2}{c^2}}}.$$

Momentum

For a massive particle with invariant mass $m$, we define the 4-momentum as

$$p^{\mu }=m{U}^{\mu }=(\gamma mc,\gamma m\mathbf{v})=\left(\frac{E}{c},p_x,p_y,p_z\right).$$ $$p^{\mu }{p}_{\mu }=m^2{c}^2⟹\frac{E^2}{c^2}-\Vert \mathbf{p}{\Vert }^2=m^2{c}^2⟹E^2=\Vert \mathbf{p}{\Vert }^2{c}^2+m^2{c}^4$$

In the particle’s rest frame, $\mathbf{p}=0$. Choosing the positive-energy branch gives $E=m{c}^2$. For a massive particle with positive energy, in general we have $E=\gamma m{c}^2$:

$$E^2=\Vert \gamma m\mathbf{v}{\Vert }^2{c}^2+m^2{c}^4=m^2{c}^4\left({\gamma }^2\frac{\Vert \mathbf{v}{\Vert }^2}{c^2}+1\right)=m^2{c}^4{\gamma }^2⟹E=\gamma m{c}^2.$$

Massless Particles

A massless particle follows a null worldline, for which $d{s}^2=0$ and $d\tau =0$. Therefore its 4-velocity $U^{\mu }=d{x}^{\mu }/d\tau$ is not defined, and its 4-momentum cannot be constructed as $p^{\mu }=m{U}^{\mu }$.

Its 4-momentum is nevertheless well-defined and satisfies the mass-shell condition

$$p^{\mu }{p}_{\mu }=0.$$

Therefore,

$$\frac{E^2}{c^2}-\Vert \mathbf{p}{\Vert }^2=0⟹E=c\Vert \mathbf{p}\Vert ,$$

where the positive-energy branch has been chosen.