How to solve problems in special relativity using four-dimensional vectors (basic transformations of spacetime and collisions)

This article uses the plural Euclidean space. In fact, many books teach using the Minkowski metric, but due to personal preference, I only provide one method of derivation. The physical ideas are interconnected, and I hope this can also help students who adopt a different approach.

Additionally, if you need it, this article also provides slidev (you can also export it in formats like PDF).

During my study of physics competitions, I found that there is a severe lack of articles and even books in Chinese regarding four-dimensional vectors in special relativity (STR). In light of this, I hope to combine what I have learned to write a related article, aiming to provide a learning reference for students interested in this topic.

Why Use Four-Dimensional Vectors#

When studying STR, I often found myself troubled by the "eigen" states of various scenarios. Many exercises and even the problems themselves contain certain errors, which has led to a chaotic understanding of the scenarios. But why does using this "classical" transformation feel so counterintuitive? At its core, it is probably because our understanding of time and space is built on classical Galilean spacetime, but this does not work in Einstein's theory of special relativity, as its spacetime is not independent. This is based on the two postulates of special relativity:

The principle of the constancy of the speed of light: The speed of light in a vacuum is the same constant in all inertial reference frames, regardless of the motion of the light source and the observer.
The principle of relativity: The form of physical laws is the same in all inertial reference frames; there is no "absolute rest" inertial frame.

What conclusions can we draw from this?

From the invariance of the speed of light, we can derive \ x^2+y^2+z^2-c^2t^2=0

This conclusion can be derived simply by imagining the equations of the wavefronts of two beams of light over different time intervals (this will not be elaborated here, as it is not an introductory article on STR).

At this point, since the existing spacetime system has led to the complexity of understanding, is there a way to resolve this? The answer is yes; Minkowski proposed such a space because special relativity is essentially a theory of the invariants of the Lorentz group. Therefore, by constructing such a spacetime, we can elegantly solve this problem, which is the Minkowski spacetime.

Basic Four-Dimensional Vectors in Minkowski Spacetime#

From the above invariance

x^2+y^2+z^2-c^2t^2=0

we can obtain the first set of four-dimensional vectors

(x,y,z,ict)

So how do we derive the basic Lorentz transformation from it? That is, the relationship between (x,y,z,ict) and (x',y',z',ict').

\text{Starting from the general transformation}

\begin{pmatrix} x' \\ y' \\ z' \\ ict' \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ % Here a_31, a_32 have been corrected; if the original intention was a_31, a_31, it can be reverted a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ ict \end{pmatrix}

\text{Next, we consider the motion equation of the coordinate origin $O$ in the $S'$ system}

x' = a_{14} t \ , \quad y' = a_{24} t \ , \ z'=a_{34} t \ , \ t'=a_{44} t

\text{Thus we have}

v_x'=\frac{dx'}{dt'}=-v \ , \ v_y'=\frac{dy'}{dt'}=0 \ , \ v_z'=\frac{dz'}{dt'}=0

\text{For the coordinate origin O', there should be a velocity relationship}

\begin{gathered} % These closely related formulas can be placed in a gathered environment a_{11}v_x+a_{12}v_y+a_{13}v_z+a_{14}=0 \\ a_{21}v_x+a_{22}v_y+a_{23}v_z+a_{24}=0 \\ a_{31}v_x+a_{32}v_y+a_{33}v_z+a_{34}=0 \end{gathered}

\text{And there is the velocity}

v_x=v \ , \ v_y=0 \ , \ v_z=0

\text{Let us set}

a_{44}=\gamma

\text{Then from the above equations, we can obtain}

\begin{gathered} % These closely related formulas can be placed in a gathered environment x'=\gamma x+a_{12}y+a_{13}z-\gamma vt \\ y'=a_{22}y+a_{23}z \\ z'=a_{32}y+a_{33}z \\ t'=a_{41}x+a_{12}y+a_{13}z+\gamma t \end{gathered}

\text{Substituting the above into}

x'^2+y'^2+z'^2-c^2t'^2=0

\text{Adding the anisotropy of spatial position gives}

x'^2+y'^2+z'^2-c^2t'^2=x^2+y^2+z^2-c^2t^2

\text{From the coefficient comparison, we can see that}

\begin{gathered} % These closely related formulas can be placed in a gathered environment \gamma = \pm \frac{1}{\sqrt{1-\beta^2}} \ , \ \beta=\frac{v}{c} \\ a_{41}=\frac{\gamma v}{c^2} \ , \ a_{12}=a_{13}=a_{42}=a_{43}=0 \\ a_{22}^2+a_{23}^2=a_{32}^2+a_{33}^2=1 \ , \ a_{22}a_{23}+a_{32}a_{33}=0 \end{gathered}

\text{Also, since the transformation of the y-z space is an identity transformation, we should have}

a_{22}=a_{33}=1 \ , \ a_{23}=a_{32}=0

\text{From the transformation with v=0 being an identity transformation, we can conclude that } \gamma \text{ is positive.}

\text{In summary, we can obtain}

\begin{pmatrix} x' \\ y' \\ z' \\ ict' \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & i \gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -i \gamma \beta & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ ict \end{pmatrix}

This is the most basic four-dimensional vector of spacetime coordinates. What other basic four-dimensional vectors are there?

Four-dimensional wave vector
$k_p=(k_x \ ,\ k_y \ , \ k_z \ , \ i\frac{\omega}{c})$
This can be understood through
$kr-wt=0$
Four-dimensional current density vector
$j_p=(j_x \ ,\ j_y \ , \ j_z \ , \ ic\rho)$
This can be understood through
$\nabla \cdot \mathbf{j} + \frac{\partial \rho}{\partial t}=0$
Four-dimensional momentum vector
$p_\mu = (p_x \ ,\ p_y \ , \ p_z \ , \ i\frac{W}{c})$

In fact, we can use these quantities to derive many four-dimensional vectors.
For example, the four-dimensional velocity is obtained by differentiating the spacetime coordinates, which is not difficult to derive:
$U_\mu = \frac{dx_\mu}{d\tau}=\gamma(u,ic)$
It is not difficult to see that
$p_\mu = mU_\mu \ , \ j_p=\rho U_p$
As for dynamics, differentiating P can yield the four-dimensional force vector K, which will not be elaborated here (by the way, it can also be derived using four-dimensional acceleration; interested readers can derive it themselves).

Properties of Four-Dimensional Vectors#

After discussing so much, we still haven't introduced the properties of four-dimensional vectors well, but this is key to solving practical problems later. Below, I will introduce several properties of four-dimensional vectors.

Lorentz Covariance#

This is an important property of four-dimensional vectors. Let's derive it.

\begin{gather*} \text{This is an important property of four-dimensional vectors.} \\[1em] \text{In special relativity, the form of physical laws is the same in all inertial reference frames.} \\[1em] \text{This means that physical quantities should maintain some invariance under Lorentz transformations.} \\[1em] \text{The introduction of four-dimensional vectors is precisely to mathematically realize this invariance.} \\[1em] \text{For a four-dimensional vector } X=(x, y, z, ict) \text{ and another four-dimensional vector } Y=(x', y', z', ict'). \\[1em] \text{Their inner product is defined as:} \\[1em] X \circ Y = x x' + y y' + z z' + (ict)(ict') = x x' + y y' + z z' - c^2 tt' \\[1em] \text{This inner product is a scalar.} \\[1em] \text{It has the same value in all inertial reference frames.} \\[1em] \text{Thus, it remains unchanged under Lorentz transformations.} \\[1em] \text{Let's derive it.} \\[1em] \text{Let the four-dimensional vector } X \text{ be represented as a column vector } X \text{ in the } S \text{ system.} \\[1em] \text{In } S' \text{ it is represented as } X'. \\[1em] \text{And they are related by the Lorentz transformation matrix } L: X' = L X\text{.} \\[1em] \text{Similarly, } Y' = LY\text{.} \\[1em] \text{(It is worth noting that in the coordinate system using the imaginary time component } ict \text{, the Lorentz transformation matrix } L \text{ is an orthogonal matrix.)} \\[1em] \text{That is, it satisfies } L^T L = I\text{, where } I \text{ is the identity matrix.} \\[1em] \text{This property ensures that the numerical value of the inner product remains unchanged after transformation.)} \\[1em] \begin{gathered} \text{Then, the inner product of the transformed four-dimensional vectors } X' \text{ and } Y' \text{ is:} \\[8pt] X' \circ Y' = (X')^T Y' = (L X)^T (L Y) = X^T L^T L Y \end{gathered} \\[1em] \begin{gathered} \text{Since the Lorentz transformation matrix } L \text{ satisfies the orthogonal condition } L^T L = I\text{ in the imaginary time coordinate system,} \\[8pt] \text{the above expression becomes:} \\ X^T L^T L Y = X^T I Y = X^T Y = X \circ Y \end{gathered} \\[1em] \text{This proves Lorentz covariance:} \\[1em] \text{After a Lorentz transformation, the numerical value of the inner product of four-dimensional vectors remains unchanged.} \\[1em] \text{This property allows us to simplify the treatment of physical quantities by constructing Lorentz invariants (such as the length of four-dimensional vectors).} \\[1em] \text{Without worrying that choosing different inertial reference frames will lead to numerical changes.} \\[1em] \text{For example, the length of the four-dimensional vector } X \circ X = x^2+y^2+z^2-c^2t^2 \text{ is invariant in any reference frame.} \end{gather*}

Some Special Inner Products and Conservations of Four-Dimensional Vectors#

Inner product of four-dimensional velocity
$U \circ U =U \circ U =-c^2$
Inner product of four-dimensional momentum

P \circ U = m(U \circ U)=-mc^2=-E_0 \ , \ P \circ P = -m^2c^2

Four-dimensional momentum is conserved.

Have you noticed that due to Lorentz covariance, we can obtain

P \circ P = -(\frac{E_{tot}}{c})^2 + p^2 \ , \ P' \circ P' = -(\frac{E_0}{c})^2

This is the famous relativistic energy-momentum relation
$E_{tot}^2 = E_0^2 + p^2c^2$

In this case, for photons, we have

E_{tot}=-p \cdot c \ , \ P \circ P = 0

The Wonderful Use of Four-Dimensional Vectors in Solving Various Spacetime and Collision Problems#

Doppler Effect in Relativistic Situations#

From the four-dimensional wave vector, we can obtain

k'_x = \gamma\left(k_x - \frac{v}{c^2}\omega\right) \ , \ k'_y = k_y \ , \ k'_z = k_z \ , \ \omega' = \gamma(\omega - v k_x)

Thus we have
$\omega' = \gamma\omega(1 - \frac{v}{c}cos\theta)$
This is the Doppler effect in relativistic situations.

Particle Collision Annihilation Producing Photons#

pair_annihilation_hd

For each particle, the following four-dimensional momentum will be present:

Incident particle:
$P_1=\gamma m \cdot (v \ , \ 0 \ , \ 0 , \ ic)$
Stationary particle:
$P_2=M \cdot (0 \ , \ 0 \ , \ 0 , \ ic)$
Perpendicular emitted photon beam:
$P_3=\frac{hf_1}{c} \cdot (0 \ , \ 1 \ , \ 0 \ , \ i)$
Obliquely emitted photon beam:
$P_4=\frac{hf_2}{c} \cdot (cos(\theta) \ , \ -sin(\theta) \ , \ 0 \ , \ i)$
From the conservation of four-dimensional momentum, we have
$P_1 + P_2 = P_3 + P_4$
We can obtain:

E_1 + E_2 = E_3 + E_4\\ E_1 \cdot \frac{v}{c} = E_4 \cdot cos(\theta) \\ E_3 = -E_4 \cdot sin(\theta) \end{gathered}

So we can have

\begin{gathered} E_4^2 - E_3^2=E_1^2 \cdot \frac{v^2}{c^2} \\ (E_1 + E_2) \cdot (E_4 - E_3)=E_4^2 - E_3^2 \\ E_4 - E_3 = \frac{E_1^2 \cdot \frac{v^2}{c^2}}{E_1 + E_2} \end{gathered}

We also have
$\frac{v^2}{c^2} = 1 - \frac{1}{\gamma^2} = 1 - \frac{E_2^2}{E_1^2} = \frac{E_1^2 - E_2^2}{E_1^2}$
Thus we obtain
$2 \cdot E_4 = E_1 + E_2 + \frac{E_1^2}{E_1+E_2} \cdot \frac{E_1^2-E_2^2}{E_1^2} = E_1 + E_2 + E_1 - E_2 = 2 \cdot E_1$
So we have

E_1 = E_4 \\ E_2 = E_3 \end{gathered}

$cos(\theta) = \frac{v}{c} = \sqrt{1 - \frac{E_2^2}{E_1^2}}$

Light Illuminating Particles to Produce New Particles#

pair_production_hd

For each particle, the following four-dimensional momentum will be present:

Incident light:
$P_1=\frac{hf}{c} \cdot (1 \ , \ 0 \ , \ 0 \ , \ i)$
Stationary particle:
$P_2=M \cdot (0 \ , \ 0 \ , \ 0 \ , \ ic)$
New particle:
$P_3$
From the conservation of four-dimensional momentum, we can obtain
$P_1 + P_2 = P_3$
Squaring gives
$P_1 \circ P_1+2 P_1 \circ P_2+P_2 \circ P_2=P_3 \circ P_3$
If the original particles are significantly heavier than the new particles, we have:
$0 + 2 \cdot h \cdot f \cdot M + (M \cdot c)^2 \approx (M + 2 \cdot m)^2 \cdot c^2$
Simplifying gives
$h \cdot f \approx 2 \cdot m \cdot c^2 + 2 \cdot m^2 \cdot c^2 / M = 2 \cdot m \cdot c^2 \cdot \left(1 + \frac{m_0}{M}\right)$

Thus, it can be concluded that if the particles involved are electrons, the energy of the incident photons must be at least twice the rest energy of the produced particles.

Complete Inelastic Collision of Particles Generating a New Particle#

relativistic_collision_hd

For each particle, the following four-dimensional vectors will be present:

Incident particle:
$P_1$
Target particle:
$P_2$
Generated particle:
$P_3$
From the conservation of four-dimensional momentum, we have:
$P_1 + P_2 = P_3$
Squaring gives
$P_1 \circ P_1 +P_2 \circ P_2 + 2P_1 \circ P_2 = P_3 \circ P_3$
This leads to
$m_1^2 + 2 \gamma_1 \gamma_2m_1m_2 \cdot \frac{(c^2-v_1v_2)}{c^2} + m_2^2 = m_3^2$
From this, we can obtain
$m_3$
And from the conservation of four-dimensional momentum, we can obtain:
$v_3 = \frac{\gamma_1 \cdot m_1 \cdot v_1 + \gamma_2 \cdot m_2 \cdot v_2}{\gamma_1 \cdot m_1 + \gamma_2 \cdot m_2}$

Complete Elastic Collision of Particles#

elastic_collision_theta_alpha_hd

For each particle, the following four-dimensional vectors will be present:

Incident particle:
$P_1=\gamma_1 m \cdot (v \ , \ 0 \ , \ 0 , \ ic)$
Stationary particle:
$P_2=M \cdot (0 \ , \ 0 \ , \ 0 , \ ic)$
Generated particle 1:
$P_3 = \gamma_3 M \cdot (wcos(\theta) \ , \ wsin(\theta) \ , \ 0 \ , \ ic)$
Generated particle 2:
$P_4 = \gamma_4 m \cdot (rcos(\alpha) \ , \ rsin(\alpha) \ , \ 0 \ , \ ic)$
From the conservation of four-dimensional momentum, we have
$P_1 + P_2 = P_3 + P_4$
Squaring gives
$P_1 \circ P_1 +P_2 \circ P_2 + 2P_1 \circ P_2 = P_3 \circ P_3 +P_4 \circ P_4 + 2P_3 \circ P_4$
Also, since
$P_1 \circ P_3 + P_2 \circ P_3 = P_3 \circ P_3 + P_4 \circ P_3 = P_3 \circ P_3 + P_1 \circ P_2$
We can obtain

\begin{gathered} P_1 \circ P_3 = \gamma_1 \cdot \gamma_3 \cdot m \cdot M \cdot (c^2 - v \cdot w \cdot \cos(\alpha))\\ P_2 \circ P_3 = \gamma_3 \cdot M^2 \cdot c^2, P_3 \circ P_3 = M^2 \cdot c^2 \\ P_1 \circ P_2 = \gamma_1 \cdot m \cdot M \cdot c^2 \end{gathered}

From this, we can solve for

From energy conservation and momentum conservation, we can solve for

\begin{gathered} \gamma_4 = \gamma_1 + \frac{M}{m} - \frac{M}{m} \cdot \gamma_3 \\ r_x = \left(\gamma_1 \cdot v - \gamma_w \cdot \frac{M}{m} \cdot w_x\right)/\gamma_r \\ w_x = w \cdot cos(\alpha) \\ \beta = \sin^{-1}(r_x/r) \end{gathered}

Compton Scattering#

compton_scattering_hd

For each particle, the following four-dimensional vectors will be present:

Incident light:
$P_1=\frac{hf}{c} \cdot (1 \ , \ 0 \ , \ 0 \ , \ i)$
Stationary particle:
$P_2=m \cdot (0 \ , \ 0 \ , \ 0 \ , \ ic)$
New particle 1:
$P_3 = \frac{hf'}{c} \cdot (cos(\theta) \ , \ sin(\theta) \ , \ 0 \ , \ i)$
New particle 2:
$P_4 = \gamma m \cdot ( \vec{v}\ , \ ic)$
From the conservation of four-dimensional momentum, we have
$P_1 + P_2 = P_3 + P_4$
Squaring gives
$P_1 \circ P_1 +P_2 \circ P_2 + 2P_1 \circ P_2 = P_3 \circ P_3 +P_4 \circ P_4 + 2P_3 \circ P_4$
This leads to
$P_1 \circ P_2 = P_3 \circ P_4$
Also, since
$P_1 \circ P_3 + P_2 \circ P_3 = P_3 \circ P_3 + P_4 \circ P_3 = P_1 \circ P_2$
This gives us
$\frac{h}{\lambda} \cdot \frac{h}{\lambda'}(1-\cos \theta)+m_0 \cdot c \cdot \frac{h}{\lambda'}=m_0 \cdot c \cdot \frac{h}{\lambda}$
We obtain
$\lambda'-\lambda = \frac{h}{m_0\cdot c} \cdot (1-\cos \theta)$
This is the classic conclusion of Compton scattering!

Inverse Compton Scattering#

inverse_compton_scattering_hd

For each particle, the following four-dimensional vectors will be present:

Incident light:
$P_1=\frac{hf}{c} \cdot (1 \ , \ 0 \ , \ 0 \ , \ i)$
High-speed particle:
$P_2 = \gamma_2 m \cdot (v \ , \ 0 \ , \ 0 \ , \ ic)$
New particle 1:
$P_3 = \frac{hf'}{c} \cdot (1 \ , \ 0 \ , \ 0 \ , \ i)$
New particle 2:
$P_4 = \gamma_4 m \cdot (v' \ , \ 0 \ , \ 0 \ , \ ic)$
From the conservation of four-dimensional momentum, we have
$P_1 + P_2 = P_3 + P_4$
Squaring gives
$P_1 \circ P_1 +P_2 \circ P_2 + 2P_1 \circ P_2 = P_3 \circ P_3 +P_4 \circ P_4 + 2P_3 \circ P_4$
This leads to
$P_1 \circ P_2 = P_3 \circ P_4$
Also, since
$P_1 \circ P_3 + P_2 \circ P_3 = P_3 \circ P_3 + P_4 \circ P_3 = P_1 \circ P_2$
This gives us
$\frac{2}{\gamma_2 m c^2} \cdot h \cdot f \cdot h \cdot f' + h \cdot f' \cdot \left(1-\frac{v}{c}\right) = h \cdot f \cdot \left(1+\frac{v}{c}\right) \approx 2 \cdot h \cdot f$
If v approaches c, we have
$h \cdot f' \cdot \left(\frac{1}{\gamma_2 m_0 c^2} \cdot h \cdot f + \frac{1}{2} \cdot \left(1-\frac{v}{c}\right)\right) = h \cdot f$
This gives us the relationship between f and f‘!

Thoughts on Four-Dimensional Vectors#

I hope this introductory article can provide you with some insights, but in fact, the wonderful use of four-dimensional vectors goes far beyond this. Due to space constraints, this article will conclude here (perhaps I can start a series to document it later: P). We see that four-dimensional vectors provide a unified and elegant mathematical framework for understanding relativistic phenomena. I hope this can inspire you and provoke thought; sometimes abstract mathematical tools can greatly advance our understanding of physics.

PS: Regarding whether four-dimensional vectors can be used in middle school physics competitions, my current experience is that they can be attempted for extremely confident problems, but once an error occurs, do not expect process points; after all, physics competitions are also exams, and the most basic and classical methods are often the most favored (but using four-dimensional vectors for verification is still very good: D).