Relativistic Momentum and Relativistic Mass
Introduction
In special relativity, based on the classical definition for momentum and mass, the conservation of momentum and the conservation of mass would not be valid. This means that the classical definition of momentum and mass might need to be adjusted. The concept of relativistic mass was introduced, as opposed to the classical invariant rest mass. Rest mass is a constant quantity that is the same in all inertial reference frames and can be measured when it remains rest by the observer in the reference frame. Relativistic mass, however, is dependent on the velocity relative to the observer in the reference frame.
In this blog post, I would like to talk about conservation of mass, conservation of momentum, how people created the concept of relativistic mass which is dependent on the velocity, how the relativistic mass will change with respect to velocity, relativistic momentum, and why people sometimes deliberately tries to avoid using the concept of relativistic mass.
Postulates of Special Relativity
I will reiterate the two postulates of special relativity again since they are important for our derivations and discussions.
First Postulate (Principle of Relativity)
The laws of physics take the same form in all inertial frames of reference.
Second Postulate (Invariance of Light Speed)
As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity $c$ that is independent of the state of motion of the emitting body. It is also equivalent to say, the speed of light in free space has the same value $c$ in all inertial frames of reference.
Reference Frames
I will also reiterate the two reference frames used in our derivation and discussion again.
In the spacetime, we have two reference frames, a reference frame $S$ and another reference frame $S’$ moving at a velocity $v$ with respect to it. So the two reference frames in this scenario are inertial reference frame. The coordinate axes in each reference frame are parallel, i.e., the $x$ and $x’$ axes are parallel, the $y$ and $y’$ axes are parallel, and the $z$ and $z’$ axes are parallel, and remain mutually perpendicular. We assume the relative motion is along the coincident $xx’$ axes. At $t = t’ = 0$, the origins of both coordinate systems are the same, $(x,y,z) = (x’,y’,z’) = (0, 0, 0)$.
An event in the time space could be observed and recorded by the observers on the two reference frames using spacetime coordinates $(t,x,y,z)$ in the reference frame $S$ and $(t’,x’,y’,z’)$ in the reference frame $S’$, respectively.
Relativistic Velocity Transformation
The motion is a sequence of continuous events in spacetime. It could go along any directions. In the reference frame $S$, we use $\mathbf{u} = \{u_x, u_y, u_z\}$ to describe the velocity of motion. In the reference frame $S’$, similarly, we use $\mathbf{u^{\prime}} = \{u_x^{\prime}, u_y^{\prime}, u_z^{\prime}\}$ to describe the velocity of motion. Here, $u_x$ and $u_x^{\prime}$, $u_y$ and $u_y^{\prime}$, $u_z$ and $u_z^{\prime}$, represent the portions of velocities along the $x$, $y$, $z$ axes, respectively. If the quantity is positive, it means that direction of the portion of the velocity is toward the positive axis; if the quantity is negative, it means that direction of the portion of the velocity is toward the negative axis.
We have derived the relativistic velocity transformation in my previous blog post “Relativistic Velocity Transformation”. The conclusions would be directly used in our derivations for relativistic mass.
The transformation for the relativistic velocity observed from the reference frame $S$ to the reference frame $S’$ is
$$
\begin{align}
u_x^{\prime} &= \frac{u_x - v}{1 - \frac{v}{c^2}u_x} \\
u_y^{\prime} &= \frac{u_y}{\gamma (1 - \frac{v}{c^2}u_x)} \\
u_z^{\prime} &= \frac{u_z}{\gamma (1 - \frac{v}{c^2}u_x)} \\
\end{align}
$$
The transformation for the relativistic velocity observed from the reference frame $S’$ to the reference frame $S$ is
$$
\begin{align}
u_x &= \frac{u_x^{\prime} + v}{1 + \frac{v}{c^2}u_x^{\prime}} \\
u_y &= \frac{u_y^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\
u_z &= \frac{u_z^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\
\end{align}
$$
Thought Experiment
Think of the following thought experiment. There is a resting particle in the reference frame $S’$ and it splits to two smaller particles. After the split, one particle $A$ is traveling along the $x’$ axis to the negative side with a velocity denoted as $u_A^{\prime}$, the other particle $B$ is traveling along the $x’$ axis to the positive side with a velocity denoted as $u_B^{\prime}$. There are no velocities along the $yy’$ and $zz’$ axes for the particles $A$ and $B$. In the reference frame $S’$, the observer measured that $u_A^{\prime} = -v$ and $u_B^{\prime} = -v$.
Before the split, the observed mass of the original particle is denoted as $M$ and $M’$ in the reference frame $S$ and $S’$, respectively; the observed mass of the particle $A$ is denoted as $m_A$ and $m_A^{\prime}$ in the reference frame $S$ and $S’$, respectively; the observed mass of the particle $B$ is denoted as $m_B$ and $m_B^{\prime}$ in the reference frame $S$ and $S’$, respectively. Note that we ignore how to technically measure the motion matters for now, and the mass of the matter means the mass of the matter at the state when the mass of the matter is being measured.
Conservation of Mass
The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system’s mass cannot change, so the quantity can neither be added nor be removed. Therefore, the quantity of mass is conserved over time.
In our thought experiment, according to the first postulate of special relativity, the principle of relativity, conservation of mass should hold in both reference frame $S$ and $S’$.
In the reference frame $S$, we have mass conservation.
$$
M = m_A + m_B
$$
In the reference frame $S’$, we have mass conservation.
$$
M^{\prime} = m_A^{\prime} + m_B^{\prime}
$$
Conservation of Momentum
In Newtonian mechanics, momentum is the product of the mass and velocity of an object.
$$
\begin{align}
\mathbf{p} &= m \mathbf{v} \\
\end{align}
$$
where $m$ is the mass of matter and is a constant in all inertial reference frames.
In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum is constant. This fact, known as the law of conservation of momentum, is implied by Newton’s laws of motion. Conservation of momentum has been verified in classical physics experiments when the velocities of the matters are relatively low to the observer.
In our thought experiment, according to the first postulate of special relativity, the principle of relativity, conservation of momentum should hold in both reference frame $S$ and $S’$.
In the reference frame $S$, we have momentum conservation.
$$
Mv = m_A u_A + m_B u_B
$$
In the reference frame $S’$, we have momentum conservation.
$$
0 = m_A^{\prime} u_A^{\prime} + m_B^{\prime} u_B^{\prime}
$$
Relativistic Velocity Transformation
In the reference frame $S$, the observer could also measure the velocities of the particles $A$ and $B$. Using the relativistic velocity transformation we derived, we have
$$
\begin{align}
u_A &= \frac{u_A^{\prime} + v}{1 + \frac{v}{c^2} u_A^{\prime}} = \frac{-v + v}{1 - \frac{v^2}{c^2}} = 0\\
u_B &= \frac{u_B^{\prime} + v}{1 + \frac{v}{c^2} u_B^{\prime}} = \frac{v + v}{1 + \frac{v^2}{c^2}} = \frac{2v}{1 + \frac{v^2}{c^2}}\\
\end{align}
$$
Mass is Not Constant With Respect to Velocity
In the reference frame $S’$, because $u_A^{\prime} = -v$ and $u_B^{\prime} = v$, and conservation of momentum $0 = m_A^{\prime} u_A^{\prime} + m_B^{\prime} u_B^{\prime}$. We could get $m_A = m_B$. We further denote $m_A = m_B = m$.
Let’s assume mass is a constant with respect to velocity first, i.e., mass is just the rest mass regardless of the reference frame. We should have $m_A = m_B = m_A^{\prime} = m_B^{\prime} = m$.
In the reference frame $S$, we also expect that conservation of mass and conservation of momentum are also valid.
Based on the conservation of mass,
$$
M = m_A + m_B = 2m
$$
The momentum before split
$$
Mv = 2mv
$$
The momentum after the split
$$
m_A u_A + m_B u_B = m \times 0 + m \frac{2v}{1 + \frac{v^2}{c^2}} = \frac{2mv}{1 + \frac{v^2}{c^2}}
$$
We found the conservation of momentum is no longer valid in the reference frame $S$!
$$
Mv \neq m_A u_A + m_B u_B
$$
This means that our assumption mass is a constant with respect to velocity is not correct.
Relativistic Mass
To make sure conservation of mass and conservation of momentum are valid in the reference frame $S$, we would have to solve the following equations.
$$
\begin{gather}
M = m_A + m_B\\
Mv = m_A u_A + m_B u_B = m_B \frac{2v}{1 + \frac{v^2}{c^2}}\\
\end{gather}
$$
We want to describe $m_B$ using $m_A$ and $u_B$ based on the above equations.
It is actually a little bit tricky to derive the desired form of expression, although it looks like an extremely simple math problem. I would show my derivation here.
Based on the conservation of momentum,
$$
\begin{gather}
(m_A + m_B) v = m_B \frac{2v}{1 + \frac{v^2}{c^2}} \\
\frac{v^2}{c^2}(m_A + m_B) = m_B - m_A \\
v = c \sqrt{\frac{m_B - m_A}{m_B + m_A}} \\
\end{gather}
$$
Based on the relativistic velocity transformation,
$$
\begin{align}
u_B &= \frac{2v}{1 + \frac{v^2}{c^2}} \\
&= \frac{2c \sqrt{\frac{m_B - m_A}{m_B + m_A}}}{1+\frac{m_B - m_A}{m_B + m_A}} \\
&= \frac{2c \sqrt{\frac{m_B - m_A}{m_B + m_A}}}{\frac{2m_B}{m_B + m_A}} \\
&= \frac{c}{m_B} \sqrt{(m_B - m_A)(m_B + m_A)} \\
\end{align}
$$
Then we could eliminate $v$.
$$
\begin{align}
vu_B &= c \sqrt{\frac{m_B - m_A}{m_B + m_A}} \frac{c}{m_B} \sqrt{(m_B - m_A)(m_B + m_A)} \\
&= \frac{c^2}{m_B} (m_B - m_A) \\
\end{align}
$$
$$
v = \frac{c^2}{m_B u_B} (m_B - m_A)
$$
Go back to the conservation of momentum,
$$
\begin{gather}
(m_A + m_B) v = m_B u_B \\
(m_A + m_B) \frac{c^2}{m_B u_B} (m_B - m_A) = m_B u_B \\
(m_B^2 - m_A^2) c^2 = m_B^2 u_B^2 \\
(c^2 - u^2) m_B^2 = c^2 m_A^2 \\
\frac{m_B^2}{m_A^2} = \frac{c^2}{c^2 - u_B^2} = \frac{1}{\frac{c^2 - u_B^2}{c^2}} = \frac{1}{1-\frac{u_B^2}{c^2}}\\
\end{gather}
$$
Because $\frac{m_B}{m_A} > 0$, we have the unique solution,
$$
m_B = \frac{m_A}{\sqrt{1 - \frac{u_B^2}{c^2}}}
$$
Because the particle $A$ is resting in the reference frame $S$, therefore $m_A$ is just the rest mass of particle $A$. Because when B is resting in the reference frame, i.e., $u_B = 0$, we have $m_B = m_A$. This means $m_A$ is also the rest mass for particle $B$. We denote the rest mass $m_A$ using $m_0$, use $m_r$ to denote $m_B$, and use $v$ to denote $u_B$ (this $v$ is different from the $v$ in our derivations). We got
$$
m_r = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
Where $m_0$ is the rest mass of matter, $v$ is the velocity of the matter we observed from a reference frame, and $m_r$ is the relativistic mass which is dependent on the velocity of the matter relative to the observer.
There is another version of the derivation for the relativistic mass from Feynman’s Lectures. The readers could read this additional material as well.
Relativistic Momentum
Looking back to the conservation of momentum, we used relativistic mass. Therefore, the correct momentum definition should be
$$
\begin{align}
\mathbf{p} &= m_r \mathbf{v} \\
&= \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}
\end{align}
$$
Where $m_0$ is the rest mass of matter, $v$ is the velocity of the matter we observed from a reference frame.
Fun Facts
Looking back at the mass conservation in the reference frame $S’$, $m_A = m_B = m$. Because particles $A$ and $B$ have the same magnitude of velocity, they should have the same rest mass as well. We denote this rest mass as $m_0$.
Based on mass conservation, we have
$$
M = m_A + m_B = \frac{2m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
Note that because $M$ is the rest mass so $M = M_0$, and $m_0$ is also rest mass for particles $A$ and $B$, we found
$$
M_0 > 2m_0
$$
This means that after a resting particle split, and somehow you catch the smaller split particles, measure the their resting masses, you would find the sum of the rest masses of the two smaller particles is not equal to the rest mass of the original particle!
Controversy
The concept of relativistic mass is not favored in modern physics. There are a lot of sophisticated arguments from physicists such as Adler and Okun. They argued that the mass should only have one definition which is the Newtonian rest mass, invariant to the reference frame. The physics lecturers should emphasize to students using relativistic mass is just something convenient during our correction to the description of momentum.
To a physics noob like me, I personally feel the concept of relativistic mass is awkward as well, simply because there is no way to directly measure the relativistic mass in motion.
In my later blog posts, we would see the famous mass-energy equivalence
$$
E = mc^2
$$
where $m$ is actually the relativistic mass instead of the rest mass that most of the people in the world are thinking of.
If the concept of the relativistic mass has been gone, the famous mass-energy equivalence would become
$$
E = \gamma mc^2
$$
where $m$ is the rest mass. This equation looks less confusing.
I finally think that the concept of relativistic mass was introduced to make the expression of some of the physical laws under the theory of relativity look more close to those in the classical Newtonian physics. But this might not just be necessary.
References
Relativistic Momentum and Relativistic Mass
https://leimao.github.io/blog/Relativistic-Momentum-Relativistic-Mass/