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 concepts are interconnected, and I hope this can be helpful for students using another 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 Chinese materials on the four-dimensional vectors of special relativity (STR) are extremely scarce. Therefore, 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 struggled with the "eigen" states of various scenarios. Many exercises and even the problems themselves contain certain errors, which led to a chaotic understanding of the scenarios. But why does using this "classical" transformation feel so counterintuitive? The root cause is that our understanding of time and space is based on the classical Galilean spacetime, which does not work in Einstein's theory of special relativity because its spacetime is not independent. This is based on the two postulates of special relativity:

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

This conclusion can be derived 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 framework leads to complexity in understanding, is there a way to resolve this? The answer is affirmative; Minkowski proposed such a space because special relativity is essentially the theory of 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 and a_32 have been corrected; if the original intention was a_31, a_31, it can be changed back a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ ict \end{pmatrix}

\text{Next, 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{Also, we have 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 coefficient comparison, we can see}

\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 in 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 \text{ 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
$k_p=(k_x \ ,\ k_y \ , \ k_z \ , \ i\frac{\omega}{c})$
It can be understood through
$kr-wt=0$
Four-Dimensional Current Density Vector
$j_p=(j_x \ ,\ j_y \ , \ j_z \ , \ ic\rho)$
It 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, leading to
$U_\mu = \frac{dx_\mu}{d\tau}=\gamma(u,ic)$
It is not difficult to see that
$p_\mu = mU_\mu \ , \ j_p=\rho U_p$
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. Let me introduce a few 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 invariant under Lorentz transformations.} \\[1em] \text{Let's derive it.} \\[1em] \text{Let the four-dimensional vector } X \text{ be represented as a column vector in the } S \text{ frame.} \\[1em] \text{In } S' \text{ it is represented as } X'. \\[1em] \text{And they are related by the Lorentz transformation matrix } L \text{ such that } 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 the 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 in the imaginary time coordinate system } L^T L = I\text{,} \\[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 magnitude of four-dimensional vectors).} \\[1em] \text{We do not have to worry that choosing different inertial reference frames will lead to numerical changes.} \\[1em] \text{For example, the magnitude 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*}

Inner Products and Conservation of Some Special 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
$E_{tot}^2 = E_0^2 + p^2c^2$

Thus, for photons, we have

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

The Clever Use of Four-Dimensional Vectors to Solve 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 apply:

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:
$P_3=\frac{hf_1}{c} \cdot (0 \ , \ 1 \ , \ 0 \ , \ i)$
Diagonally 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 derive:

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}

Thus, 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 apply:

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 equivalent to >> 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 involved particles 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 apply:

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
$m_1^2 + 2 \gamma_1 \gamma_2m_1m_2 \cdot \frac{(c^2-v_1v_2)}{c^2} + m_2^2 = m_3^2$
This allows us to find
$m_3$
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 apply:

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 derive

\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}

This allows us to solve for

From the conservation of energy and momentum, we can derive

\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 apply:

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 apply:

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 the clever use of four-dimensional vectors goes far beyond this. Due to space limitations, I will conclude this article here (perhaps I can start a series 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 high school physics competitions, my current experience is that they can be attempted for problems you are very confident about, but if you make a mistake, do not expect to receive partial credit for the process. 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).