# Time Dilation and Time Travel

In this note, we discuss one of the relativistic effects called Time Dilation namely a clock that is moving relative to an observer will be measured to tick slower than a clock that is at rest in the observer’s reference frame. This is pretty intriguing for those who are familiar with Newtonian notion of time as being a universal parameter for motions. Let us do a thought experiment. Let us consider a frame $K$ at rest and suppose that a light ray is emitted by the light source $Q$ and after reflection by the mirror $S$ is received at $E$. See Figure 1.

Figure 1. Time Dilation

The measured time interval in the frame $K$ is $\Delta t=t_2-t_1=\frac{2l}{c}$. Now consider a frame $K’$ moving at a constant speed $v$ to the right. An observer at rest in $K’$ sees the light ray emerging from $Q$, hitting the mirror (at rest in $K$) at $M$ and reaching the $x’$-axis again at $E$. The observer measures a longer time interval as the light has to travel a longer path to reach the receiver but the speed of light is remained the same according to Einstein’s postulate. How much longer? The time $\Delta t’$ measured by an observer at rest in the frame $K’$ can be easily calculated using the Pythagorean law applied to the isosceles triangle seen in Figure 1. We find
$$\left(\frac{c\Delta t’}{2}\right)^2=l^2+\left(\frac{v\Delta t’}{2}\right)$$
Solving this for $\Delta t’$ we find

\label{eq:timedilation}
\Delta t’=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}

Note that \eqref{eq:timedilation} amounts to the Lorentz transformation into the system $K’$
$$\Delta t’=t_2′-t_1′,$$
where
$$t_i’=\frac{t_i-\frac{v}{c^2}x_i}{\sqrt{1-\frac{v^2}{c^2}}},\ i=1,2$$
Since $x_1=x_2$, we obtain \eqref{eq:timedilation}. In case this whole frame thing is confusing, let us imagine that you are sitting in a train that is running at a constant speed. Since there is no acceleration, you do not feel that you are moving. So inside the train you are at rest (frame $K$). For an observer outside you are moving (frame $K’$) and the observer would measure the time ($\Delta t’$) on your clock ticking slower than what you would measure it ($\Delta t$). In physics $\Delta t$ is called proper time. Simply speaking proper time is the time measured by a clock that is moving along with inertial frame. Mathematically, proper time can be calculated from the arc length $ds^2$ of a worldline, the trajectory of a moving particle or an object in spacetime. Denote by $\tau$ the proper time. Then since the worldline is timelike (meaning leaning more toward time), $ds^2=-c^2d\tau^2$. So the proper time interval is given by

\begin{aligned}
\Delta\tau&=\frac{1}{c}\int\sqrt{-ds^2}\\
&=\frac{1}{c}\int\sqrt{c^2dt^2-dx^2-dy^2-dz^2}\\
&=\int\sqrt{1-\frac{v(t)^2}{c^2}}dt
\end{aligned}\label{eq:propertime}

If $v(t)$ is constant speed $v$, \eqref{eq:propertime} becomes \eqref{eq:timedilation}.
The time dilation effect in \eqref{eq:timedilation} hints us that a time travel to the future may be possible. Here is how. The exoplanet Proxima b is interesting because it is orbiting within the habitable zone of the red dwarf star Proxima Centauri which is a part of triple star system Alpha Centauri in the Constellation of Centaurus, and also because it is relatively close to our world. It is located about 4.2 light-years or 40 trillion km from Earth. In fact, it is the closest known exoplanet to the Solar System.

Artist’s depiction of Proxima b

Let us say we are sending a manned spaceship to Proxima b. Also let us assume that the spaceship can travel at 90% of the speed of light. (It is actually impossible to achieve this due to a physical limitation. I will discuss this in my other note at a later time. In reality, the best we can achieve using nuclear propulsion is about 0.067% of the speed of light.) For people on Earth it would take $\Delta t’=\frac{4\times 10^{13}\mbox{km}}{2.7\times 10^5\mbox{km/sec}}=1.\overline{481}\times 10^8\mbox{sec}$ for the spaceship to get to Proxima b. Since $1\mbox{sec}=3.17\times 10^{-8}\mbox{years}$, it is 4.7 years. Since it would take the same time from Proxima b to Earth, the overall travel time for people on Earth is 9.4 years. In reality, we will have to take some factors into consideration: it takes time for the spaceship to accelerate to reach 90% of the speed of light, once the spaceship is near Proxima b it will have to slow down for stopping or U-turning, etc. But for the sake of simplicity we will disregard those factors. For the crew memebers it took only
\begin{align*}
\Delta t&=\sqrt{1-\frac{v^2}{c^2}}\Delta t’\\
&=\sqrt{1-(0.9)^2}\cdot 1.\overline{481}\times 10^8\mbox{sec}\\
&\approx 0.65\times 10^8\mbox{sec}\\
&\approx 2\mbox{yrs}
\end{align*}
to get to Proxima b. So when they come back home, it’s like they traveled more then 5 years forward in time. I know it is not what you probably think and yes I admit that this is a kind of boring time travel. Can one travel backward in time? This is one of the most intriguing questions. I will come back to this question at another time.

I will finish this note with an example as an application of \eqref{eq:timedilation}. This example was taken from [1].

Example. Muon Decay

The Earth is surrounded by an atmosphere of about 30 km thickness screening us off from cosmic radiation. If a proton from the consmic radiation hits the atmosphere, $\pi$-mesons are produced and several of them decay further into a muon and a neutrino. The muon has a mean lifetime of $\Delta t=2\times 10^{-6}\mbox{sec}$ in its rest system. Classically it would travel even with the speed of light (only massless particles can travel at the speed of light)
\begin{align*}
s&=c\Delta t\\
&=3\times 10^5\mbox{km/sec}\cdot 2\times 10^{-6}\mbox{sec}\\
&=0.6\mbox{km}
\end{align*}
or 600m. If this were true, muon particles would never reach the surface, but they are detected on the surface. In the relativistic approach,
$$s’=v\Delta t’=\frac{v\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$
Muons at rest have a mass of $m_0c^2=10^8$eV (I know it is actually energy but due to mass-energy equivalence physicists customarily call it mass.) The cosmic muons are created at an altitude of about 10km with a total energy of $E=5\times 10^9$eV. In order to apply this information we rewite $S’$ as
\begin{align*}
S’&=\frac{vm_0c^2}{m_0c^2\sqrt{1-\frac{v^2}{c^2}}}\Delta t\\
&=\frac{v}{m_0c^2}E\Delta t\\
&\leq\frac{c}{m_0c^2}E\Delta t\\
&=\frac{3\times 10^5\mbox{km/sec}}{10^8\mbox{eV}}\cdot 5\times 10^9\mbox{eV}\cdot 2\times 10^{-6}\mbox{sec}\\
&=30\mbox{km}
\end{align*}
Here we used $E=mc^2=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$. We will discuss this later in another post. The actual measurement gives a value of 38km.

References:

[1] Walter Greiner, Classical Mechanics, Point Particles and Relativity, Springer, 2004

[2] Paul A. Tipler and Ralph A. Llewellyn, Modern Physics, 5th Edition, W. H. Freeman and Company, 2008

# Lorentz Invariance of Relativistic Equations

Relativistic equations are the equations whose solutions describe certain relativistic motions. Such equations include wave equation, Klein-Gordon equation, Dirac equation etc. A relativistic equation must describe the same physical motion independent of frames i.e. whether an observer is in a frame at rest or in a frame moving at the constant speed $v$. For this reason, all those relativistic equations are required to be invariant under the Lorentz transformation. We show that the wave equation

\label{eq:waveeq}
-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=0

is Lorentz invariant. Here we consider only 1-dimensional wave equation for simplicity. Wave equation has two kinds of solutions. Given boundary conditions its solution describes a vibrating string in which case the boundary conditions are the ends of the string that are held fixed. This is called Fourier’s solution. The other type can be obtained by not imposing any boundary conditions. The resulting solution would describe a propagating wave in vacuum spacetime. Such a propagating wave includes electromagnetic waves. Light is also an electromagnetic wave. This is called a d’Alembert’s solution. The proof is easy. All that’s required is the chain rule.

First let us recall the Lorenz transformation
$$t’=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}},\ x’=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$
Using the chain rule we find
\begin{align*}
\frac{\partial}{\partial x}&=\frac{\partial}{\partial x’}\frac{\partial x’}{\partial x}+\frac{\partial}{\partial t’}\frac{\partial t’}{\partial x}\\
&=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial x’}-\frac{v}{c^2\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial t’}\\
\frac{\partial}{\partial t}&=\frac{\partial}{\partial x’}\frac{\partial x’}{\partial t}+\frac{\partial}{\partial t’}\frac{\partial t’}{\partial t}\\
&=-\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial x’}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial t’}
\end{align*}
Applying the chain rule again,
\begin{align*}
\frac{\partial^2}{\partial x^2}&=\frac{1}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {x’}^2}+\frac{v^2}{c^4\left(1-\frac{v^2}{c^2}\right)}\frac{\partial^2}{\partial {t’}^2}-2\frac{v}{c^2\left(1-\frac{v^2}{c^2}\right)}\frac{\partial^2}{\partial t’\partial x’}\\
\frac{\partial^2}{\partial t^2}&=\frac{v^2}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {x’}^2}+\frac{1}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {t’}^2}-2\frac{v}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial t’\partial x’}
\end{align*}
It follows that
$$-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=-\frac{1}{c^2}\frac{\partial^2\psi}{\partial {t’}^2}+\frac{\partial^2\psi}{\partial {x’}^2}$$
Therefore, the wave equation is Lorentz invariant.

Would the wave equation be invariant under the Galilean transformation? The answer is no. Recall the Galilean transformation
$$t’=t,\ x’=x-vt$$
We find that under the Galiean transformation
$$-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=-\frac{1}{c^2}\frac{\partial^2\psi}{\partial {t’}^2}+\left(1-\frac{v^2}{c^2}\right)\frac{\partial^2\psi}{\partial {x’}^2}+\frac{2v}{c^2}\frac{\partial^2\psi}{\partial t’\partial x’}$$
Hence obvisouly the wave equation is not invariant under the Galiean transformation. This implies that there is no light in Euclidean space.

Food for Thought. You can also show that the heat equation (1-dimensional)
$$-\frac{\partial u}{\partial t}+\alpha\frac{\partial^2 u}{\partial x^2}=0$$
is not Lorentz invariant. Is there any relativistic version of the heat equation? There are models of relativistic heat conduction but in my opinion they are more like mathematically augmented equations rather than they are derived in a physically meaningful way. So my question is can we derive a physically meaningful equation of relativistic heat conduction? One may wonder if there is actually any physical phenomenon that exhibits a relativistic heat conduction. As far as I know there isn’t any observed one yet. I speculate though that one may observe a relativistic heat conduction from an extreme physical phenomenon such as a quasar jet.

Update: Of course the Lorentz invariance can be also shown using the Lorentz transformation \begin{align*}t’&=\cosh\phi t-\sinh\phi x\\x’&=-\sinh\phi t+\cosh\phi x\end{align*}

Update: For 3-dimensional case the wave equation is given by $$\Box\psi=0$$
where $\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ is the d’Alembert’s operator. This case is actually simpler to show its Lorentz invariance. Note $\Box=\nabla\cdot \nabla$ where $\nabla=\frac{1}{c}\frac{\partial}{\partial t}\hat e_0+\frac{\partial}{\partial x}\hat e_x+\frac{\partial}{\partial y}\hat e_2+\frac{\partial}{\partial z}\hat e_3$. Since $\nabla$ is a 4-vector (rigorously it is not a vector but an operator but can be treated as a vector), its squared norm $\Box$ has to be Lorentz invariant.

# Lorentz Transformation 3

Let $\begin{pmatrix} t\\ x \end{pmatrix}$ be a vector in $tx$-plane and $\begin{pmatrix} t’\\ x’ \end{pmatrix}$ denote its rotation by a hyperbolic angle $\phi$. Then as we have seen here we have:

\begin{aligned}
t’&=t\cosh\phi-x\sinh\phi\\
x’&=-t\sinh\phi+x\cosh\phi
\end{aligned}\label{eq:boost1}

$t’$-axis ($x’=0$) is moving at a constant speed

\label{eq:velocity1}
v=\frac{x}{t}=\frac{\sinh\phi}{\cosh\phi}=\tanh\phi
while $x’$-axis ($t’=0$) is moving at a constant speed

\label{eq:velocity2}
v=\frac{x}{t}=\frac{\cosh\phi}{\sinh\phi}=\coth\phi

This means that the time and spacial axes scissor together and hence the speed of light remains the same regardless of the coordinate transformation as illustrated in Figure 1. The picture in Figure 1 is called the spacetime diagram.

Figure 1. Spacetime Diagram

From Euclidean perspective, $t’$ and $x’$ do not appear to be orthogonal. However, from Lorentzian perspective they are orthogonal. To see that let $e_0=\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $e_1=\begin{pmatrix} 0\\ 1 \end{pmatrix}$. Also let $e_0’$ and $e_1’$ be the rotations of $e_0$ and $e_1$ by the hyperbolic angle $\phi$, respectively. Then by \eqref{eq:boost1} we have
$$e_0’=\begin{pmatrix} \cosh\phi\\ -\sinh\phi \end{pmatrix},\ e_1’=\begin{pmatrix} -\sinh\phi\\ \cosh\phi \end{pmatrix}$$
Now \begin{align*}\langle {e_0}’,e_1’\rangle&=(\cosh\phi\ -\sinh\phi)\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
-\sinh\phi\\
\cosh\phi
\end{pmatrix}\\&=\cosh\phi\sinh\phi-\sinh\phi\cosh\phi=0\end{align*} So $e_0’$ and $e_1’$ are orthogonal.

\eqref{eq:boost1} with \eqref{eq:velocity1} can be written as
\begin{align*}
t’&=\cosh\phi(t-vx)\\
x’&=\cosh\phi(x-vt)
\end{align*}
Using the well-known identy $\cosh^2\phi-\sinh^2\phi=1$, we find $\cosh\phi=\frac{1}{\sqrt{1-v^2}}$. Therefore, \eqref{eq:boost1} can be written in terms of $t,x,t’,x’,v$ as

\begin{aligned}
t’&=\frac{t-vx}{\sqrt{1-v^2}}\\
x’&=\frac{x-vt}{\sqrt{1-v^2}}
\end{aligned}\label{eq:boost2}

Remember that we assumed the speed of light to be $c=1$ for simplicity. For the real $c$ value \eqref{eq:boost2} turns into

\begin{aligned}
t’&=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}\\
x’&=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}
\end{aligned}\label{eq:boost3}

In physics textbooks, \eqref{eq:boost3} is introduced as the Lorentz transformation. If $v\ll c$ meaning $v$ is significantly slower than the speed of light (such a motion is called a non-relativistic motion), \eqref{eq:boost3} is effectively the Galilean transformation

\begin{aligned}
t’&=t\\
x’&=x-vt
\end{aligned}\label{eq:galilean}

for Newtonian mechanics in Euclidean space. $t’=t$ means that time is independent of the relative motion of different observers and we already know this is the case of Newtonian mechanics. The Galilean transformation \eqref{eq:galilean} can be also obtained by taking the limit $c\to \infty$. This indicates that in Newtonian mechanics the speed of light is presumed to be infinity.

# What are Lorentz Transformations? 2

This time we consider spacetime $\mathbb{R}^{3+1}$ with the Minkowski metric $ds^2=-(dt)^2+(dx)^2+(dy)^2+(dz)^2$. (Here we set the speed of light in vacuum $c=1$). For each time slice i.e $t$=constant we get Euclidean 3-space and from what we know about Euclidean space there are rotations in the coordinate planes, the $xy$-plane, $yz$-plane, and $xz$-plane. What we have no clue about is rotations (from Euclidean perspective) in the $tx$-plane, $ty$-plane, and $tz$-plane. For that we consider $\mathbb{R}^{1+1}$ with metric $ds^2=-(dt)^2+(dx)^2$. Let $v,w$ be two tangent vectors to $\mathbb{R}^{1+1}$. Then $v,w$ are written as
\begin{align*}
v&=v_1\frac{\partial}{\partial t}+v_2\frac{\partial}{\partial x}\\
w&=w_1\frac{\partial}{\partial t}+w_2\frac{\partial}{\partial x}
\end{align*}
$ds^2$ acting on them results the inner product:
$$\langle v,w\rangle=ds^2(v,w)=-v_1w_1+v_2w_2=v^t\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}w$$
The last expression is obtained by considering $v$ and $w$ as column vectors (the reason we can do this is because every tangent plane to $\mathbb{R}^{1+1}$ is isomorphic to the vector space $\mathbb{R}^{1+1}$). Let $A$ be an isometry of $\mathbb{R}^{1+1}$. Let $v’=Av$ and $w’=Aw$. Then
\begin{align*}
\langle v’,w’\rangle&={v’}^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}w’\\
&=(Av)^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}(Aw)\\
&=v^tA^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}Aw
\end{align*}
In order for $A$ to be an isometry i.e. $\langle v’,w’\rangle=\langle v,w\rangle$ we require that $A^t\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}A=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}$. A $2\times 2$ matrix $A$ satisfying

\label{eq:lo}
A^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}

is called an Lorentz orthogonal matrix. More generally a $4\times 4$ Lorentz orthogonal matrix $A$ satisfies

\label{eq:lo2}
A^t\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}

Let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ be a Lorentz orthogonal matrix. We also assume that $\det A=1$ i.e. $A$ is a special Lorentz orthogonal group. Then we get the following set of equations

\begin{aligned}
-a^2+c^2&=-1\\
-ab+cd&=0\\
-b^2+d^2&=1\\
\end{aligned}
\label{eq:slo}

Solving the equations in \eqref{eq:slo} simultaneously we obtain $A=\begin{pmatrix} a & b\\ b & a \end{pmatrix}$ with $a^2-b^2=1$. Hence one may choose $a=\cosh\phi$ and $b=-\sinh\phi$. Now we find a rotation matrix in the $tx$-plane

\label{eq:boost}
\begin{pmatrix}
\cosh\phi & -\sinh\phi\\
-\sinh\phi & \cosh\phi
\end{pmatrix}

A friend of mine, a retired physicist named Larry has to confirm everything by actually calculating. For being a lazy mathematician I try to avoid messy calculations as much as possible, instead try to confirm things indirectly (and more elegantly). In honor of my dear friend let us play Larry. Let $\begin{pmatrix} t\\ x \end{pmatrix}$ be a vector in the $tx$-plane and $\begin{pmatrix} t’\\ x’ \end{pmatrix}$ denote its rotation by an angle $\phi$. (This $\phi$ is not really an angle and it could take any real value. It is called a hyperbolic angle.)
$$\begin{pmatrix} t’\\ x’ \end{pmatrix}=\begin{pmatrix} \cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi \end{pmatrix}\begin{pmatrix} t\\ x \end{pmatrix}$$
i.e.
\begin{align*}
t’&=\cosh\phi t-\sinh\phi x\\
x’&=-\sinh\phi t+\cosh\phi x
\end{align*}
and
\begin{align*}
dt’&=\cosh\phi dt-\sinh\phi dx\\
dx’&=-\sinh\phi dt+\cosh\phi dx
\end{align*}
Using this we can confirm that
$$(ds’)^2=-(dt’)^2+(dx’)^2=-(dt)^2+(dx)^2=ds^2$$
i.e. \eqref{eq:boost} is in fact an isometry which is called a Lorentz boost. In spacetime $\mathbb{R}^{3+1}$ there are three boosts, one of which is
$$\begin{pmatrix} \cosh\phi & -\sinh\phi & 0 & 0\\ -\sinh\phi & \cosh \phi& 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$
This is a rotation (boost) in the $tx$-plane. In addition, there are regular rotations one of which is
$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \cos\theta & -\sin\theta & 0\\ 0 & \sin\theta & \cos\theta & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$
This is a rotation by the angle $\theta$ in the $xy$-plane. The three boosts and three rotations form generators of the Lorentz group $\mathrm{O}(3,1)$, the set of all $4\times 4$ Lorentz orthogonal matrices, which is a Lie group under matrix multiplication. As a topological space $\mathrm{O}(3,1)$ has four connected components depending on whether Lorentz transformations are time-direction preserving or orientation preserving. The component that contains the identity transformation consists of Lorentz transformations that preserve both the time-direction and orientation. It is denoted by $\mathrm{SO}^+(3,1)$ and called the restricted Lorentz group. There are 4 translations along the coordinate axes in $\mathbb{R}^{3+1}$. The three boosts, three rotations and four translations form generators of a Lie group called the Poincaré group. Like Euclidean motion group elements of the Poincaré group are affine transformations. Such an affine transformation would be given by $v\longmapsto Av+b$ where $A$ is a Lorentz transformation and $b$ is a fixed four-vector. The Lorentz group is not a (Lie) subgroup of the Poincaré group as the elements of the Poincaré group are not matrices. Note however that the Poincaré group acts on $\mathbb{R}^{3+1}$ in an obvious way and the Lorentz group fixes the origin under the group action, hence the Lorentz group is the stabilizer subgroup (also called the isotropy group) of the Poincaré group with respect to the origin.

I will finish this lecture with an interesting comment from geometry point of view. The spacetime $\mathbb{R}^{3+1}$ can be identified with the linear space $\mathscr{H}$ of all $2\times 2$ Hermitian matrices via the correspondence
\begin{align*}
v=(t,x,y,z)\longleftrightarrow\underline{v}&=\begin{pmatrix}
t+z & x+iy\\
x-iy & t-z
\end{pmatrix}\\
&=t\sigma_0+x\sigma_1+y\sigma_2+z\sigma_3
\end{align*}
where
$$\sigma_0=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix},\ \sigma_1=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix},\ \sigma_2=\begin{pmatrix} 0 & i\\ -i & 0 \end{pmatrix},\ \sigma_3=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$
are the Pauli spin matrices. The Lie group $\mathrm{SL}(2,\mathbb{C})$ acts on $\mathbb{R}^{3+1}$ isometrically via the group action:
$$\mu: \mathrm{SL}(2,\mathbb{C})\times\mathbb{R}^{3+1}\longrightarrow\mathbb{R}^{3+1};\ \mu(g,v)=gvg^\dagger$$
where $g^\dagger={\bar g}^t$. The action induces a double covering $\mathrm{SL}(2,\mathbb{C})\stackrel{2:1}{\stackrel{\longrightarrow}{\rho}}\mathrm{SO}^+(3,1)$. Since $\ker\rho=\{\pm I\}$, $\mathrm{PSL}(2,\mathbb{C})=\mathrm{SL}(2,\mathbb{C})/\{\pm I\}=\mathrm{SO}^+(3,1)$. $\mathrm{SL}(2,\mathbb{C})$ is simply-connected, so it is the universal covering of $\mathrm{SO}^+(3,1)$.

I will discuss some physical implications of Lorentz transformations next time.

# What are Lorentz Transformations? 1

When you study theory of relativity, one of the notions you will first stumble onto is about Lorentz transformations. It is actually a pretty big deal. Without knowing Lorentz transformations, you can’t study theory of relativity. So what are Lorentz transformations? A short answer is they are isometries of spacetime. If you haven’t heard of the word isometry, it is a linear isomorphism from an inner product space to another which also preserves inner product. Let $V$, $V’$ be inner product space. A linear map $\varphi: V\longrightarrow W$ is said to preserve inner product if $\langle v,w\rangle=\langle\varphi(v),\varphi(w)\rangle$ for all $v,w\in V$. Since the length between two vectors is measured by inner product, clearly inner product preserving map also preserves length, so we have the name isometry. An isometry is clearly conformal (angle-preserving). Before we discuss isometries of spacetime, which is usually denoted by $\mathbb{R}^{3+1}$, let us visit Euclidean space that we are most familiar with and consider isometries there. For a vector in Euclidean space there are three types of basic transformations: dilations, translations, and rotations. Dilations are linear but not isometries. Translations preserve the length of a vector but not linear (they are called affine transformations). Rotations are indeed isometries. Let us check that rotations are isometries by calculation. It suffices to consider 2-dimensional Euclidean space $\mathbb{R}^2$ with metric $ds^2=dx^2+dy^2$. This is not only because of simplicity but also because what we discuss for 2-dimensional case can be easily extended for higher dimensional cases such as 3-dimensional or 4-dimensional Euclidean space. The reason is simple. No matter in what dimensional space you are in a rotation takes place only in two dimensional space (plane). A rotation of a vector $\begin{pmatrix} x\\ y \end{pmatrix}$ by an angle $\theta$ is given by
$$\begin{pmatrix} x’\\ y’ \end{pmatrix}=\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$
that is,
\begin{align*}
x’&=\cos\theta x-\sin\theta y\\
y’&=\sin\theta x+\cos\theta y
\end{align*}
and
\begin{align*}
dx’&=\frac{\partial x’}{\partial x}dx+\frac{\partial x’}{\partial y}dy\\
&=\cos\theta dx-\sin\theta dy\\
dy’&=\frac{\partial y’}{\partial x}dx+\frac{\partial y’}{\partial y}dy\\
&=\sin\theta dx+\cos\theta dy
\end{align*}
With these you can easily check that
$$(ds’)^2=(dx’)^2+(dy’)^2=dx^2+dy^2=ds^2$$
i.e. the metric is preserved so rotations are isometries. Here I kind of cheated because I already know so well (as you would do also) about Euclidean space! Assume that the only thing we know about Euclidean space is its metric. Besides that we know nothing about its geometry whatsoever. How do we figure out things like rotations? This is important because once we can figure it out, we can apply the same method to figure out isometries of other spaces that we don’t know about. Let $v$ and $w$ be two tangent vectors to $\mathbb{R}^2$ (here $\mathbb{R}^2$ is regarded as a 2-dimensional differentiable manifold). Then they can be written as
\begin{align*}
v&=v_1\frac{\partial}{\partial x}+v_2\frac{\partial}{\partial y}\\
w&=w_1\frac{\partial}{\partial x}+w_2\frac{\partial}{\partial y}
\end{align*} We calculate
$$ds^2(v,w)=v_1w_1+v_2w_2=v^tw=v^t\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}w,$$
where the tangent vectors are identified with column vectors $v=\begin{pmatrix} v_1\\ v_2 \end{pmatrix}$ and $w=\begin{pmatrix} w_1\\ w_2 \end{pmatrix}$. (Recall that any
tangent plane $T_p\mathbb{R}^2$ is isomorphic to the vector space $\mathbb{R}^2$.) So, the Euclidean metric induces the usual dot product $\langle\ ,\ \rangle$. Let $A$ be a linear transformation from $\mathbb{R}^2$ to itself. Let $v’=Av$ and $w’=Aw$. Then
\begin{align*}
\langle v’,w’\rangle&={v’}^tw’\\
&=(Av)^t(Aw)\\
&=v^t(A^tA)w.
\end{align*}
Suppose that $A$ is also an isometry. In order for $A$ to be an isometry i.e. $\langle v’,w’\rangle=\langle v,w\rangle$ we require that $A^tA=I$. As you may have learned from linear algebra such a matrix is called an orthogonal matrix. Let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$. In addition to assuming that $A$ is orthogonal let us also assume that $\det A=1$. Such an orthogonal matrix is called a special orthogonal matrix. Then we obtain the equations
\begin{aligned}
a^2+b^2&=1\\
ab+cd&=0\\
$A=\begin{pmatrix} a & -c\\ c & a \end{pmatrix}$ with $a^2+c^2=1$. Hence we may choose $a$ and $c$ as $a=\cos\theta$ and $c=\sin\theta$. That is we obtained a rotational matrix only from the assumption that $A$ is an isometry without employing any familiar geometric intuition on Euclidean space. The set of all isometries of $\mathbb{R}^n$ is denoted by $\mathrm{O}(n)$. $\mathrm{O}(n)$ is the set of all $n\times n$ orthogonal matrices and it is a group with matrix multiplication. It is in fact more than an algebraic group. It is also a Lie group. Simply speaking a Lie group is a group which is also a differentiable manifold. Furthermore, the binary operation is a smooth map. The set of all isometries of $\mathbb{R}^n$ along with translations form a group with composition called the Euclidean motion group. The name obviously comes from Euclidean motions that comprise successive applications of rotations and translations. A Euclidean motion of a vector $v$ is given by $Av+b$ where $A$ is an orthogonal matrix and $b$ is a fixed vector i.e. it is an affine transformation.
Now we are ready to discuss Lorentz transformations i.e. isometries of spacetime $\mathbb{R}^{3+1}$ and we will continue this next time.