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

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{equation}
\begin{aligned}
t’&=t\cosh\phi-x\sinh\phi\\
x’&=-t\sinh\phi+x\cosh\phi
\end{aligned}\label{eq:boost1}
\end{equation}
$t’$-axis ($x’=0$) is moving at a constant speed
\begin{equation}
\label{eq:velocity1}
v=\frac{x}{t}=\frac{\sinh\phi}{\cosh\phi}=\tanh\phi\end{equation}
while $x’$-axis ($t’=0$) is moving at a constant speed
\begin{equation}
\label{eq:velocity2}
v=\frac{x}{t}=\frac{\cosh\phi}{\sinh\phi}=\coth\phi
\end{equation}
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{equation}
\begin{aligned}
t’&=\frac{t-vx}{\sqrt{1-v^2}}\\
x’&=\frac{x-vt}{\sqrt{1-v^2}}
\end{aligned}\label{eq:boost2}
\end{equation}
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{equation}
\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}
\end{equation}
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{equation}
\begin{aligned}
t’&=t\\
x’&=x-vt
\end{aligned}\label{eq:galilean}
\end{equation}
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.

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
\begin{equation}
\label{eq:lo}
A^t\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}A=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
\end{equation}
is called an Lorentz orthogonal matrix. More generally a $4\times 4$ Lorentz orthogonal matrix $A$ satisfies
\begin{equation}
\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}
\end{equation}
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{equation}
\begin{aligned}
-a^2+c^2&=-1\\
-ab+cd&=0\\
-b^2+d^2&=1\\
ad-bc&=1
\end{aligned}
\label{eq:slo}
\end{equation}
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
\begin{equation}
\label{eq:boost}
\begin{pmatrix}
\cosh\phi & -\sinh\phi\\
-\sinh\phi & \cosh\phi
\end{pmatrix}
\end{equation}
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.