notes/docs/physics/spacetime/special-formalism.md

Special formalism of spacetime

The assumption had been that the three-dimensional geometry of the universe was distinct from time. However, with the uncovering of the dynamics of electromagnetism, light has been found to have a constant propagation speed c in vacuum. This finding, together with the principle of relativity, refutes the absoluteness and independence of space and time. We, thus, require to fuse space and time into four-dimensional spacetime.

In the special formalism of spacetime we only consider spacetime in its non-perturbed (flat) state.

Geometric postulates

Special postulate: Flat spacetime is described by a differential manifold (\mathrm{M},\bm{\eta}) with a metric

\bm{\eta} = c^2 \mathbf{d}t \otimes \mathbf{d} t - \sum_i \mathbf{d}x^i \otimes \mathbf{d}x^i

and \dim \mathrm{M} = 4 called Minkowski spacetime and contains points called events.

We may now state that the evolution of a body in Minkowski spacetime is a sequence of events constituting a smooth curve in \mathrm{M} that we call the world-line of that body.

Such a world-line \gamma: \mathscr{D}(\gamma) \to \mathrm{M}: \lambda \mapsto \gamma(\lambda) is thus a smooth curve on the manifold parameterized by an open interval \mathscr{D}(\gamma) \subset \mathbb{R}. If \gamma is a geodesic on \mathrm{M} then the attached reference system to \gamma may be denoted as an inertial reference system.

Princple of relativity: All physical axioms are of identical form in all inertial reference systems.

Generally there is a second statement considering the constancy of the speed of light c in all inertial reference systems. But this statement is already implicitly included in the principle of relativity.

Lorentz transformation

Theorem 1: Let X and Y be two inertial reference systems. The transformation \psi: X \to Y obeys the principle of relativity and the special postulate if \psi \in \mathrm{O}(1,3), i.e. the transformation is in the Lorentz group.

??? note "Proof:"

Let $X$ and $Y$ be two inertial reference systems and $\psi: X \to Y$ a transformation. The *principle of relativity* requires that

$$
    \forall \mathbf{x} \in X: \|\psi \mathbf{x}\| = \|\mathbf{x}\|.
$$

Using the *special postulate* we may write

$$
    \begin{cases}
        \|\psi \mathbf{x}\| &= \eta_{\mu\nu} \psi^\mu_\rho x^\rho \psi^\nu_\sigma x^\sigma,\\
        \|\mathbf{x}\| &= \eta_{\mu\nu} x^\mu x^\nu,
    \end{cases}
$$

and thus

$$
    \psi^\rho_\mu \eta_{\rho\sigma} \psi^\sigma_\nu = \eta_{\mu\nu},
$$

or as a matrix relation

$$
    \psi^T \eta \psi = \eta.
$$

Defines itself as the orthogonal 1,3 dimensional group $\mathrm{O}(1,3)$.

Note that the full Lorentz group is not $\mathrm{SO}(1,3)$ as 

$$
    -1 = \det(\eta) = \det(\psi^T \eta \psi) = - \big(\det(\psi)\big)^2 \implies \det(\psi) = \pm 1.
$$

The transformations in the (non-abelian) Lorentz group are called Lorentz transformations and they enable us to relate the inertial reference systems to each other. The Lorentz group consists of rotations, reflections and boosts. For translations we need to extend the Lorentz group to the Poincaré group.

By removing the reflections from the Lorentz group we obtain the proper Lorentz group \mathrm{SO}(1,3) with

\forall \psi \in \mathrm{SO}(1,3): \det(\psi) = 1.

Rotation

Let \psi_\theta denote a rotation in the x,y plane with respect to an angle \theta. The indices of \psi_\theta may then be given by

(\psi_\theta)_\nu^\mu = \begin{pmatrix} 1 & 0 & 0 & 0\ 0 & \cos \theta & \sin \theta & 0\ 0 & -\sin\theta & \cos\theta & 0\ 0 & 0 & 0 & 1\end{pmatrix}.

Reflection

Let \psi denote a time reversal, the indices of \psi may then be given by

\psi_\nu^\mu = \begin{pmatrix} -1 & 0 & 0 & 0\ 0 & 1 & 0 & 0\ 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\end{pmatrix}.

Boost

Let \psi_\xi denote a boost in the x direction of speed v = \tanh\xi. The indices of \psi_\xi may then be given by

(\psi_\xi)^\mu_\nu = \begin{pmatrix} \cosh\xi & -\sinh\xi & 0 & 0\ -\sinh\xi & \cosh\xi & 0 & 0\ 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\end{pmatrix}.

Poincaré transformation

Theorem 2: Let X and Y be two inertial reference systems. The transformation \psi: X \to Y obeys the principle of relativity and the special postulate if \psi \in \mathbb{R}^{1,3} \times \mathrm{O}(1,3), i.e. the transformation is in the Poincaré group.

??? note "Proof:"

Proof follows from Lorentz transformation, this is just a semidirect product with the spacetime translations group $\mathbb{R}^{1,3}$. 

The Poincaré group extends the Lorentz group with translations on \mathrm{M}, denoted with the semidirect product of the spacetime translations group \mathbb{R}^{1,3} and the Lorentz group \mathrm{O}(1,3). The transformations in the Poincaré group are called Poincaré transformations.

Translation

Let \psi denote a translation of \mathbf{a} \in X, then

\psi(\mathbf{x}) = \mathbf{x} + \mathbf{a},

for all \mathbf{x} \in X.

Causal structure

Consider a light cone in the origin of an inertial reference system X. Then the subsystem A \subset X inside the cone has the property that

\forall \mathbf{a} \in A: |\mathbf{a}|^2 > 0,

such that all events in A are timelike. The world-lines of massive bodies that intersect the origin are always in A.

The subsystem B \subset X on the surface of the cone has the property that

\forall \mathbf{b} \in B: |\mathbf{b}|^2 = 0,

such that all events in B are lightlike. The world-lines of massless bodies that intersect the origin are always in B.

Finally, the subsystem C \subset X outside the cone has the property that

\forall \mathbf{c} \in C: |\mathbf{c}|^2 < 0,

such that all event in C are spacelike.

A trivial remark is that A \cup B \cup C = X and A \cap B \cap C = \empty.

For timelike events we may define a proper time \tau. The proper time between two timelike events \mathscr{A} and \mathscr{B} measures the time as observed by an inertial observer on the geodesic connecting \mathscr{A} and \mathscr{B} such that

\tau_{\mathscr{B}\mathscr{A}} = |\mathscr{B} - \mathscr{A}|.

Mass, energy and momentum

In the consideration of the world-lines of massive bodies it is convenient to use the proper time \tau as the parameter. By choosing a suitable \lambda we may invert and express \lambda in terms of the proper time \lambda: \tau \mapsto \lambda(\tau), such that our world-line \gamma can be expressed in terms of \tau, i.e. \gamma: \tau \mapsto \gamma(\tau).

The vector tangent to this world-line is known as the four-velocity

\mathbf{u} = d_\tau \gamma \overset{\mathrm{def}}{=} \mathbf{d}_\tau,

with

|\mathbf{u}| = c.

This absolute normalization is a reflection of the fact that the four-velocity is a velocity in spacetime, through which one always travels at the same rate.

A related concept is the four-momentum which in terms of the action S may be defined as

\mathbf{p} = \mathbf{d}S,

with

|\mathbf{p}|^2 = \Bigg(\frac{E}{c}\Bigg)^2 - \sum_i p_i p_i.

The four-momentum thus extends the notion of momentum to spacetime by the consideration of an inherent energy E tied to the constant rate at which one travels through spacetime. A second definition thus follows (linear approximation) in terms of the four-velocity

\mathbf{p} = m \mathbf{u},

with m the (rest) mass of the body.

Theorem 3: The mass m, energy E and momentum p_i of a massive body is related by

E^2 = \big(mc^2\big)^2 + \sum_i p_i p_i.

??? note "Proof:"

We have from the definition of the four-momentum in terms of the mass and the four-velocity that

$$
    \begin{align*}
        \|\mathbf{p}\| &= \|m\mathbf{u}\|,\\
        &= mc,
    \end{align*}
$$

and from the definition of the four-momentum in terms of the action (energy and momentum) that

$$
    \|\mathbf{p}\|^2 = \Bigg(\frac{E}{c}\Bigg)^2 - \sum_i p_i p_i,
$$

thus

$$
    \Big(mc\Big)^2 = \Bigg(\frac{E}{c}\Bigg)^2 - \sum_i p_i p_i,
$$

imposes

$$
    E^2 = \big(mc^2\big)^2 + \sum_i p_i p_i.
$$

In the particular case that the momentum of a body is zero we obtain the (reasonably famous) mass energy equivalence

E = mc^2.

Force

Extending the consideration of the world-lines of massive bodies in spacetime we may define the four-acceleration in terms of the four-velocity as

\mathbf{a} = d_\tau \mathbf{u},

such that

0 = d_\tau \langle \mathbf{u}, \mathbf{u}\rangle = 2 \langle \mathbf{u}, \mathbf{a}\rangle \implies \langle \mathbf{u}, \mathbf{a}\rangle = 0.

The four-acceleration can only change the direction of the four-velocity, as the rate at which one travels through spacetime must always stay constant (\|\mathbf{u}\| = c).

A related concept is the four-force which in terms of the four-momentum may be defined as

\mathbf{f} = d_\tau \mathbf{p} \overset{\mathrm{def}}{=} m \mathbf{a}.

In the case that a constant force F in the x direction acts on a body we have that the four-force on that body adheres to

|\mathbf{f}|^2 = -F^2,

obtained by considering the instantaneous reference system of the body, and thus its four-acceleration should adhere to

|\mathbf{a}|^2 = -(F/m)^2.

Also called the proper acceleration of the body.

Using the properties of the four-velocity and four-acceleration we may obtain the evolution of the four-acceleration, four-velocity and the position of the accelerated body

\begin{align*} \mathbf{a} &= \xi c\sinh(\xi\tau) \bm{\partial}_t + \xi c \sinh(\xi\tau) \bm{\partial}_x,\ \mathbf{u} &= c \cosh(\xi\tau) \bm{\partial}_t + c \sinh(\xi\tau) \bm{\partial}_x,\ \mathbf{x} &= \frac{c}{\xi} \sinh(\xi\tau) \bm{\partial}_t + \frac{c}{\xi} \cosh(\xi\tau) \bm{\partial}_x, \end{align*}

with respect to an inertial reference system intersecting at \tau = 0 and \xi = \frac{F}{mc}.

We thus obtain hyperbolic motion as one accelerates with a constant proper acceleration, an interesting property of this motion is that it defines a horizon (Rindler horizon), i.e. no timelike or light like geodesics between events behind this horizon and the accelerating body exist.

??? note "Proof:"

We consider

$$
    \begin{align*}
        c^2 &= \|\mathbf{u}\|^2 &= (u^t)^2 - (u^x)^2,\\
        0 &= \langle \mathbf{u}, \mathbf{a}\rangle &= u^t a^t - u^x a^x,\\
        -(F/m)^2 &= \|\mathbf{a}\|^2 &= (a^t)^2 - (a^x)^2,
    \end{align*}
$$

then

$$
    \begin{align*}
        u^t a^t &= u^x a^x,\\
        (u^t a^t)^2 &= \big((u^t)^2 - c^2\big)\big((a^t)^2 + (F/m)^2\big),\\
        &= (u^t a^t)^2 - c^2 (a^t)^2 + (u^t)^2 (F/m)^2 - c^2 (F/m)^2,
    \end{align*}
$$

implies

$$
    \begin{align*}
        (a^t)^2 &= \xi^2 \big((u^t)^2 - c^2\big),\\
        &= \xi^2 (u^x)^2,
    \end{align*}
$$

with $\xi = \frac{F}{mc}$.

Obtains

$$
    \begin{align*}
        a^t &= \pm \xi u^x,\\
        a^x &= \pm \xi u^t,
    \end{align*}
$$

and since we are moving in the positive $x$ direction we only take $(+)$, now using the fact that $a^\mu = d_\tau u^\mu$ we have

$$
    d^2_\tau u^x = \xi^2 u^x.
$$

With as solution

$$
    u^x(\tau) = \chi_1 e^{\xi\tau} + \chi_2 e^{-\xi\tau},
$$

and using $u^t = \frac{1}{\xi} d_\tau u^x$ obtains

$$
    u^t(\tau) = \chi_1 e^{\xi\tau} - \chi_2 e^{-\xi\tau}.
$$

Applying the boundary conditions obtains

$$
    \begin{align*}
        u^x(0) = 0 &\implies \chi_1 = - \chi_2,\\
        \|\mathbf{u}\|^2 = c^2 &\implies \chi_1 = \frac{c}{2},
    \end{align*}
$$

Such that

$$
    \begin{cases}
        u^t(\tau) &= \frac{c}{2} \Big(e^{\xi\tau} + e^{-\xi\tau}\Big) &= c \cosh\xi\tau,\\
        u^x(\tau) &= \frac{c}{2} \Big(e^{\xi\tau} - e^{-\xi\tau}\Big) &= c \sinh\xi\tau,
    \end{cases}
$$

and the rest follows.