Special Relativity¶

Special relativity is the extension of classical mechanics to extremely high energy situations, and is based around just two axioms.

[Event] An event is something which happens at a specific place at a particular instant in time.

[Inertial Reference Frame] An inertial reference frame is a means of assigning a position to an event. They have no acceleration, such that an inertial reference fram is a reference frame with respect to which Newton’s Second Law holds.

The axioms of Special Relativity¶

All inertial reference frames are equivalent for the performance of all physical experiments.
The speed of light has the same, constant value in all reference frames.

The first axiom is more commonly known as the principle of relativity.

Minkowski Diagrams¶

Any event can be described by four coordinates, \((t;x,y,z)\), but, by choosing an appropriate reference frame we can describe this with just two coordinates, \((t;x)\); a diagram of this situation is shown in figure [fig:minksimple].

[scale=1.5] (.5, .5) circle (0.05); (-1,-1) – (1.7,-1) node [below] space, \(x\); (-1, -1) – (-1, 1.2) node [right] time, \(t\);

Invariant Interval¶

In Euclidean space there is a concept of distance which is invariant; \(x^2 + y^2 + z^2\) is the same under all transformations. In spacetime there is a generalisation of this concept in the invariant interval:

\[ \begin{align}\begin{aligned} \label{eq:invariantinter} s^2 = \Delta t^2 - \Delta x^2\\The value of :math:`s^2` provides a simplified understanding of\end{aligned}\end{align} \]

causality.

\(s^2 > 0\)—time-like separation.
\(s^2 = 0\)—null separation (light-like).
\(s^2 < 0\)—space-like separation.

[scale=1.7] (-1,-1) – (1,1) node [above] \(s^2 = 0\); (-1,-1) – (1,1) – (-1,1) –cycle; (-.5,0) node \(s^2 > 0\); (-1,-1) – (1,-1) – (1,1) –cycle; (.5,0) node \(s^2 < 0\); (-1,-1) – (1.7,-1) node [below] space, \(x\); (-1, -1) – (-1, 1.2) node [right] time, \(t\);

Lorentz Transformations¶

Consider two frames moving relative to one another, \(S\) and \(S^\prime\). The frames are in standard configuration, i.e. an event in \(S\) at \((0,0)\) also occurs at \((0,0)\) in \(S^\prime\). To convert from an event occuring in \(S\) to one in \(S^\prime\) we use a Lorentz transform. These have the form

\[\begin{split}\begin{aligned} x^\prime &= x \cosh \phi - t \sinh \phi \\ t^\prime &= -x \sinh \phi + t \cosh \phi\end{aligned}\end{split}\]

and with

\[ \begin{align}\begin{aligned}\phi(v) = \tanh[-1](v)\\These can also be written\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} t^\prime &= \gamma (t-vx) \\ x^\prime &= \gamma (x-vt) \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} t &= \gamma (t^\prime + v x^\prime) \\ x &= \gamma (x^\prime + v t^\prime) \end{aligned}\end{split}\]

Adding velocities¶

Since \(e^{\pm \phi} = \cosh(\phi) \pm \sinh(\phi)\),

\[\begin{split}\begin{aligned} t^\prime - x^\prime &= e^{\phi} (t-x) \\ t^\prime + x^\prime &= e^{-\phi} (t+x) \end{aligned}\end{split}\]

Proper Time¶

In the frame \(S\) there are events at \((0,0)\) and \((t,x)\). A clock is moving at a constant speed \(v\), so that it is present at both events. Pick \(S^{\prime \prime}\), the clock’s rest frame; :math:`t^{prime

prime}` is thus the reading on the clock. This is denoted

\(\tau\), the proper time. In any other frame in standard configuration we can then write

\[ \begin{align}\begin{aligned} \label{eq:1} \tau = \gamma(t-vx)\\The proper time will be agreed upon by all observers, and as such is a\end{aligned}\end{align} \]

Lorentz scalar. Since the velocity is \(v = \frac{x}{t}\),

\[t^2 - x^2 = \tau^2 = t^{\prime 2} - x^{\prime 2}\]

Four-vectors¶

In \(4\)-dimensional spacetime the prototype displacement vector has the form

\[ \begin{align}\begin{aligned}\qty( \Delta t , \Delta x, \Delta y, \Delta z )\\and the transformation from one :math:`4`-vector to another is\end{aligned}\end{align} \]

\[\begin{split}\label{eq:2} \begin{bmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix} = \begin{bmatrix} \gamma & + \gamma v & 0 & 0 \\ + \gamma v & \gamma &0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \Delta t^{\prime} \\ \Delta x^{\prime} \\ \Delta y^{\prime} \\ \Delta z^{\prime} \end{bmatrix}\end{split}\]

\[\begin{split}\label{eq:2} \begin{bmatrix} \Delta t^{\prime} \\ \Delta x^{\prime} \\ \Delta y^{\prime} \\ \Delta z^{\prime} \end{bmatrix} = \begin{bmatrix} \gamma & - \gamma v & 0 & 0 \\ - \gamma v & \gamma &0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}\end{split}\]

It is more normal to denote each component of the 4-vector as \(x^n\), \(n \in \{0,1,2,3 \}\), or collectively, \(x^{\mu}\).

Four-vector inner products¶

The inner product of two four-vectors is somewhat less straightforward that a three-vector scalar product.

[Four-vector inner product]

\[ \begin{align}\begin{aligned} \label{eq:3} \braket{A}{B} = \sum_{\mu, \nu} \eta_{\mu \nu} A^{\mu} B^{\nu}\\for :math:`\eta_{\mu \nu} = \diag(1, -1, -1, -1)` which is the metric\end{aligned}\end{align} \]

tensor for a non-accelerating reference frame.

Velocity and Acceleration Vectors¶

The displacement 4-vector, \(\Delta x^{\mu}\), transforms properly, as does the infinitessimal displacement, \(\dd{x^{\mu}}\). The proper time, \(\tau\), is the Lorentz scalar, so we can divide each component of \(\dd{x^{\mu}}\) by \(\dd{\tau}\) and get a new vector, the velocity,

\[ \begin{align}\begin{aligned}U = \qty( \dv{x^0}{\tau}, \dv{x^1}{\tau}, \dv{x^2}{\tau}, \dv{x^3}{\tau} )\\and likewise, the acceleration,\end{aligned}\end{align} \]

\[A = \qty( \dv[2]{x^0}{\tau}, \dv[2]{x^1}{\tau}, \dv[2]{x^2}{\tau}, \dv[2]{x^3}{\tau} )\]

Now, since \(t = \gamma \tau\),

\[U^{\mu} = \qty( \gamma c , \gamma \vec{v} )\]

Momentum¶

Given the four-velocity we can find a four-momentum as well.

\[ \begin{align}\begin{aligned}p^{\mu} = (\gamma m c, \gamma m \vec{v} ) = \qty(\frac{E}{c}, \vec{p} )\\The norm of any four-vector will be Lorentz invariant, so\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}p^2 = \frac{E^2}{c^2} - \vec{p}^2\\In the particle’s rest frame,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}p^{\mu} = (mc, \vec{0}) \therefore p^2 = m^2 c^2\\and since it is Lorentz invariant, it must be the case in all frames.\end{aligned}\end{align} \]

Thus

\[\frac{E^2}{c^2} - \vec{p}^2 = m^2 c^2 \quad \therefore \quad E^2 - \vec{p}^2 c^2 = m^2 c^4\]

Invariant Mass¶

In scattering interactions the invariant mass is a useful quantity. Given a pair of particles with momenta \(p_1\) and \(p_2\), the squared invariant mass, when working in natural units, is

\[\begin{split}\begin{aligned} m_{12}^2 & = (p_1 + p_2)^2 \nonumber \\ & = (E_1+E_2, \vec{p}_1 + \vec{p}_2)^2 \nonumber \\ & = (E_1+E_2)^2 - (\vec{p}_1 + \vec{p}_2) \cdot (\vec{p}_1 + \vec{p}_2) \nonumber \\ & = E_1^2 - \vec{p}_1^2 + E_2^2 - \vec{p}_2^2 + 2E_1 E_2 - 2 \vec{p}_1 \cdot \vec{p}_2 \nonumber \\ & = m_1^2 + m_2^2 + 2(E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2) \end{aligned}\end{split}\]

[Two particles, one at rest] \(\vec{p}_2 = \vec{0}\), and \(E_2 = m_2\). Thus

\[m_{12}^2 = m_1^2 + m_2^2 + 2 m_2 E_1\]

[The Centre of Momentum Frame] The total 3-mometum in this frame is \(\vec{0}\), so,

\[ \begin{align}\begin{aligned}\vec{p}^{\text{com}}_1 = - \vec{p}^{\text{com}}_2\\then\end{aligned}\end{align} \]

\[m_{12}^2 = \qty( E_1^{\text{com}} + E_2^{\text{com}} )^2\]

Thus, the invariant mass is the total energy in the centre of mass frame.

Particle Collisions¶

The vast majority of information about fundamental particles comes from collision experiments. There are two types of collider used in physics,

Fixed Target: The target is at rest, while high-energy particles are accelerated and fired into it. Examples include the Bevatron, and the SLAC
Colliding Beam: Two accelerated beams are focussed and fired into one another, so that the particles collide, usually in the centre of mass frame. The LHC is an example.

In all particle interactions the 4-momentum is conserved, so

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} p^{\mu}_1 + p^{\mu}_2 &= p^{\mu}_3 + p^{\mu}_4 \\ (p_1 + p_2)^2 &= (p_3 + p_4)^2\end{aligned}\end{split}\\During an interaction the original particles are annihilated, and then\end{aligned}\end{align} \]

new particles are created from the resulting energy. As a result, for a final state to be achieved there must be enough energy available to create the appropriate quantity of mass. Defining the Lorentz invariant,

\[ \begin{align}\begin{aligned}s = (\vec{p}_1 + \vec{p}_2)^2\\For a desired pair of particles to be produced with masses :math:`m_3`\end{aligned}\end{align} \]

and \(m_4\) we then require

\[\sqrt{s} \ge m_3 + m_4\]

[Creating two particles] We wish to create two particles, each with mass \(2 \giga \electronvolt\) by colliding two protons (each \(1 \giga \electronvolt\)).

The minimum energy in the case of a fixed-target experiment must come entirely from the proton in the beam, so

\[E_1 = \frac{m_{12}^2 - m_1^2 - m_2^2}{2 m_2} = \frac{(m_3+m_4)^2 - m_1^2 - m_2^2}{2 m_2}\]

thus, \(E_1 = 7 \giga \electronvolt\). In the case of colliding beams both particles collide in the centre-of-mass frame, and so

\[E = \frac{m_1+m_2}{2} \ge \frac{m_3+m_4}{2} = 2 \giga \electronvolt\]

Clearly a much higher beam energy is required in the case of a fixed target to get the same results, but the engineering challenges involved with colliding two beams have, in the past, made fixed-target experiments more feasible.