# Relativistic Dynamics. Mass, Impulse and Energy in Relativity

Relativity Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
18.6Relativistic Dynamics. Mass, Impulse and Energy in Relativity

In this Physics tutorial, you will learn:

• How to find the mass of particles in relativistic events?
• How to find the impulse of particles in relativistic events?
• What is the rest mass?
• How to find the particles energy in relativistic events?
• What is relativistic energy of motion?
• What happens to energy of particles when switching from one inertial system of reference to another?

## Introduction

So far, we have discussed a list of topics that belong to relativistic kinematics. They include the spacetime coordinates of physical events and the relationship between them, such as changes in length and time, changes in the observable frequency, etc.

Now, we will deal with concepts and quantities belonging to relativistic dynamics, i.e. situations involving mass, impulse and energy in relativistic events. In this tutorial, we will see what radical changes the Einstein theory brought in this regard.

## Mass in Relativistic Events

Let's start by analyzing the mass of objects - a parameter that characterizes the inertia of objects in Newtonian system. In the (classical) Newtonian system, mass is considered as absolute (unchangeable) in all inertial systems of reference.

Let's suppose we have a particle having a Newtonian mass m which moves at constant velocity v in a fixed inertial system S. We analyze this motion in another inertial system S'. From previous articles, it is known that when S' moves at high velocities comparable to that of light relative to S, the velocity of particle in S' is different from that measured in S. However, we have only dealt with changes in velocity - a kinematic parameter; nothing has been said about any change in mass so far. In other words, is it possible to accept the idea of an absolute (unchangeable) mass, independent from the system of reference we choose? Our experience with spacetime related quantities urge us to reject this claim.

From previous articles, it is known that the speed of light c is unreachable by material particles. A material particle can only get closer to the value of c but is can never equal or overcome it. In the actual stage of learning, we already consider the Newtonian system as outdated in relativistic events but we still accept the Newton's First Law of Motion as absolute. In the practical sense, this law is also valid in relativistic events, as we known that a free particle is at rest or in uniform motion in all inertial systems.

What about the Newton's Second Law of Motion? Doubting about the concept of absolute mass means putting in discussion the truthfulness of Newton's Second Law. This law in the form F = m ∙ a is valid in inertial systems but always in the framework of absolute time and for a fixed mass. We will try to prove this law is valid in relativistic events as well.

Let's suppose a constant force F is exerted on a particle in the X-direction of the inertial system S. At t = 0, the particle was at rest in the origin O of the system. Since there is a constant force is acting on the particle, there is a constant acceleration a as well, based on the Newton's Second Law of Motion. In the classical approach, the constant acceleration brings a constant increase in velocity by a · t, which means that theoretically, the velocity can take very large values depending on the duration of event. These values can overcome the limit velocity of light, which results in a paradox - now it is proven that nothing can overcome the value of light speed c.

Since this cannot occur, we believe something has happened to the inertia of particle. When v → c, the particle resists more and more to any change in velocity. This means its mass increases with the increase in its velocity. If this situation is handled mathematically, we will see that when v → c, then a → 0. Eventually, since F is constant, then m → ∞ because only when we multiply zero by infinity we have any chance to get a fix number (from theory of limits explained in mathematics).

Let's enforce this reasoning using another approach. We have explained that Newton's Second Law of Motion is also expressed in terms of impulse and time as

F = ∆p/∆t

where p = m ∙ v is the instantaneous momentum of particle (impulse is a change in momentum). When F is constant, the above rate is constant too. This means the particle gains a constant impulse regardless the velocity in reality increases less and less when approaching the value of c. This can occur only when mass increases more and more. In other words, in relativistic events mass does not remain constant but increases with the increase in velocity according the function m = m(v) such that m → ∞ when v → c.

## Impulse in Relativistic Events

Despite that the Newton's Second Law written as

F = ∆p/∆t

is OK, we cannot rely anymore on the Newtonian relation p = m ∙ v which assumes the mass as constant. For this reason, we must find a new form of the impulse expression. This is found by considering the process of elastic collision between two small spheres. From tutorial 6.6 "Collision and Impulse. Types of Collision", we know that in (perfectly) elastic collisions, the kinetic energy and impulse are both conserved, i.e. the impulse (and kinetic energy) of the system before and after the collision is equal. This concept is valid in the relativistic events as well. If we analyze the law of conservation of impulse, we find that a constant Newtonian mass does not necessarily result in a conservation of impulse.

Since the length of objects in relativistic events experiences contraction, there must be an increase in mass at the same rate (recall that the shape of objects does not change - a sphere will still be a sphere, a cube will still be a cube after relativistic contraction and so on). The rate of mass increase is the equal to that of length decrease. Therefore, we obtain the following equation for the relativistic mass

m = m0/1 - v2/c2

where m is the relativistic mass of particle, m0 is the classical mass (rest mass), and v is the velocity of particle in the system S. This means we obtain for the relativistic impulse

p = m0 ∙ v/1 - v2/c2

This formula still contains the relationship between impulse, mass and velocity but now this relationship is not anymore linear. The above two formulae are closely related to each other and are known as the relativistic formulae of mass and impulse. It is worth to mention here that these formulae are valid for all objects whether microscopic or macroscopic. (In the practical sense, many microscopic particles move very often at very high speeds, comparable to the speed of light). The concept of Newtonian mass m0 is very easy to understand. If v = 0, then m0 = m. This means the Newtonian mass m0 (otherwise known as the rest mass m0) is nothing else but the relativistic mass of particle in an inertial system where the particle is at rest. For v → c the mass and impulse point towards infinity (m → ∞, p → ∞). Hence, the velocity v = c is as a barrier for material particles.

### Example 1

A proton has a rest mass m0 and a velocity of v = 0.5c. How much percent does the mass increase during this motion when compared to the rest mass?

### Solution 1

We must find the relativistic mass m given the rest mass m0 and velocity v.

Using the equation of relativistic mass

m' = m0/1 - v2/c2

we obtain for the mass m of particle after substituting the values:

m = m0/1 - (0.5c)2/c2
= m0/1 - 0.25
= m0/0.866

Thus, the new mass is

m/m0 × 100%
= m0/0.866/m0 × 100%
= 1/0.866 × 100%
= 1.155 × 100%
= 115.5%

of the original mass. This results in an increase of

115.5% - 100%
= 15.5%

in the mass of particle.

## Energy in Relativistic Events. Relativistic Energy of Motion

So far, we have found how the formula of relativistic mass and impulse - two quantities related to each other. The mass m0 in the system S connected to the Earth (rest mass) is a constant parameter in all inertial systems; it represents the rest mass of a particle or object. When we see in tables the value of proton or electron mass, we must know that value represents the rest mass of the given particle. When the particle is moving at constant velocity v relative to the system S, then we use the formula

m = m0/1 - v2/c2

(You must not confuse the velocity v of particle in the system S to the relative velocity V at which the system S' moves in respect to S, which here does not appear in the formula.)

On the other hand, we must see what happens to the kinetic energy of a particle in relativistic events. For this, we recall once again the equation of kinetic energy in classical approach:

KE = m0 ∙ v2/2

Moreover, any work done by an external force results in a change in the kinetic energy of particle (work - kinetic energy theorem), i.e.

Wext = Fext ∙ ∆x

It is known that a forward constant force causes a constant acceleration. Therefore, the average force is the arithmetic mean of minimum and maximum force acting on the system. If the minimum force is assumed as zero, then we take as average force half of the external force in the above formula. Thus, we write

Wext = 1/2 Fext ∙ ∆x
= 1/2 ∆p/∆t ∙ ∆x
= 1/2 m0 ∙ ∆v/∆t ∙ ∆x
= 1/2 m0 ∙ ∆v∆x/∆t
= 1/2 m0 ∙ ∆v ∙ ∆v
= m0 ∙ (∆v )2/2
= m0 ∙ v2/2 - m0 ∙ v20/2

This relation (as explained in tutorial 15.2) is known as the "work - kinetic energy theorem."

Now, in the light of relativistic approach, we already know that the Newton's Second Law of motion is still valid, but it is mostly expressed in the form

F = ∆p/∆t

instead of the traditional form F = m ∙ a. However, the impulse does not have the old form anymore. We would use the same approach to derive the relativistic kinetic energy from the relativistic impulse as we did earlier for the classical form; however, there is a much easier method for this. We raise both sides of the expression

p = m ∙ v = m0 ∙ v/1 - v2/c2

in power two, and in this way, we obtain

(p )2 = (m0 ∙ v/1 - v2/c2)2
= m20 ∙ v2/1 - v2/c2
= γ2 ∙ m20 ∙ v2
= γ2 ∙ β2 ∙ m20 ∙ c2

where

γ = 1/1 - v2/c2 and β = v/c

as discussed in tutorial 12.7.

Let's try to prove the equation

γ2 - β2 ∙ γ2 = 1

which we use later when finding the formula of energy in relativistic events. Thus, we have

γ2 - β2 ∙ γ2 = γ2 ∙ (1 - β2 )
= (1/1 - v2/c2)2 ∙ (1 - v2/c2 )
= 1/1 - v2/c2 ∙ (1 - v2/c2 )
= 1

This equation means that when we move from a system S in which the particle moves at velocity v to another inertial system S' in which the particle moves at velocity v', we must have

γ2 (v) - β2 (v) ∙ γ2 (v) = γ2 (v') - β2 (v') ∙ γ2 (v' ) = 1

This means the above difference does not depend on the inertial system of reference we choose; it is always 1.

When we multiply both sides of the above equation (in S) by m2 ∙ c4 (which is a constant) to obtain

m20 ∙ c4 ∙ (γ2 - β2 ∙ γ2 ) = m20 ∙ c4
m20 ∙ c4 ∙ γ2 - m20 ∙ c4 ∙ β2 ∙ γ2 = m20 ∙ c4
m20 ∙ c4 ∙ γ2 - m20 ∙ c4v2/c21/1 - v2/c2 = m20 ∙ c4
m20 ∙ c4 ∙ γ2 - m20 ∙ v2 ∙ c2/1 - v2/c2 = m20 ∙ c4
m20 ∙ c4 ∙ γ2 - p2 ∙ c2 = m20 ∙ c4

The m2 ∙ c4 term is independent from the reference system we choose. Therefore, it is easy to see that m2 ∙ c4 ∙ γ2 - p2 ∙ c2 is also independent from the system of reference (i.e. it is constant for the same event).

We already know what the term p2 ∙ c2 does represent. Now, let's see what does the other term represent. We have

m0 ∙ c2 ∙ γ = m0 ∙ c2/1 - v2/c2

Earlier we found that the square of this term minus p2 ∙ c2 is independent from the system of reference. Obviously, all the above equations are true, even for v << c. Therefore, using the mathematical approximation valid for a << b

(1 - a/b) - 1/2 ≈ 1 + 1/2a/b

we write the last equation for v << c

(1 - v2/c2 )-1/2 ≈ 1 + 1/2v2/c2

Hence, we obtain

m0 ∙ c2/1 - v2/c2 ≈ m0 ∙ c2 ∙ (1 + 1/2v2/c2 )
= m0 ∙ c2 + m0 ∙ v2/2

We see that for v << c there is a m0 ∙ c2 term, which is the same for all inertial systems, and another term (m0 · v2 / 2) which is equal to the classical kinetic energy of particle. Since the addition operation has the closure property (addends and sum must belong to the same type of quantity), then the m0 ∙ c2 term must be a form of energy as well. Obviously, the sum of these two terms must represent the kinetic energy in relativistic events. We denote this special kinetic energy by ε to avoid confusion with the classical kinetic energy we have discussed in Section 5. Thus, we have

ε = m0 ∙ c2/1 - v2/c2

or

ε = m ∙ c2

where m is the relativistic mass of particle.

The above two formulae are the famous equations of energy of motion for a particle in the system S, in which the particle has the velocity v. If the particle is at rest in the system S, it does not have zero kinetic energy as expected in classical physics; rather, it has an energy of ε = m0 ∙ c2 known as the rest energy of particle. We can write for the system S:

ε2 - p2 ∙ c2 = m20 ∙ c4

When the particle is observed in the system S', we obtain

(ε')2 - (p')2 ∙ c2 = ε2 - p2 ∙ c2

From the above equation, we conclude that energy and impulse are relativistic quantities; their values depend on the system of reference used to study the motion. However, the difference of squares of energy and the corresponding p · c terms is constant. This helps us find the energy and impulse formulae when switching from one inertial system to another, as we will discuss in the next section.

### Example 2

A neutron has a rest mass of 1.675 × 10-27 kg. It is set in motion at 75 000 km/s through a long and straight tunnel. What is the energy of this neutron?

### Solution 2

Clues:

m0 = 1.675 × 10-27 kg
v = 75 000 km/s = 1/4 c
ε = ?

First, we find the relativistic mass of neutron. We have

m = m0/1 - v2/c2
= m0/1 - 1/4 c)2/c2
= m0/1 - 1/16
= m0/15/16
= 4m0/15
= 4m0/3.873
= 1.033 m0

Thus, the value of relativistic impulse of proton is

p = m ∙ v
= 1.033 m0 ∙ 0.25 c
= 0.25825 m0 ∙ c

Now, using the equation

ε2 - p2 ∙ c2 = m20 ∙ c4

we obtain for the energy square of energy ε2 of neutron after substitutions

ε2 = m20 ∙ c4 + p2 ∙ c2
= m20 ∙ c4 + (0.25825 m0 ∙ c)2 ∙ c2
= m20 ∙ c4 + 0.0667 m20 ∙ c4
= 1.0667m20 ∙ c4
= 1.0667 ∙ (1.675 × 10-27 kg)2 ∙ (3 × 108 m/s)4
= 2.3523 × 10-20 J2

Therefore, the energy of neutron is

ε = √2.3523 × 10-20 J2
= 1.534 × 10-10 J

## Whats next?

Enjoy the "Relativistic Dynamics. Mass, Impulse and Energy in Relativity" physics tutorial? People who liked the "Relativistic Dynamics. Mass, Impulse and Energy in Relativity" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Relativity Revision Notes: Relativistic Dynamics. Mass, Impulse and Energy in Relativity. Print the notes so you can revise the key points covered in the physics tutorial for Relativistic Dynamics. Mass, Impulse and Energy in Relativity
3. Relativity Practice Questions: Relativistic Dynamics. Mass, Impulse and Energy in Relativity. Test and improve your knowledge of Relativistic Dynamics. Mass, Impulse and Energy in Relativity with example questins and answers
4. Check your calculations for Relativity questions with our excellent Relativity calculators which contain full equations and calculations clearly displayed line by line. See the Relativity Calculators by iCalculator™ below.
5. Continuing learning relativity - read our next physics tutorial: Relativity. Galilean Transformations. Einstein's Postulates and Newton's Laws