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Welcome to our Physics lesson on Energy in Relativistic Events. Relativistic Energy of Motion, this is the third lesson of our suite of physics lessons covering the topic of Relativistic Dynamics. Mass, Impulse and Energy in Relativity, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
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
(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:
Moreover, any work done by an external force results in a change in the kinetic energy of particle (work - kinetic energy theorem), i.e.
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
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
instead of the traditional form F⃗ = m ∙ a⃗. 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
in power two, and in this way, we obtain
where
as discussed in tutorial 12.7.
Let's try to prove the equation
which we use later when finding the formula of energy in relativistic events. Thus, we have
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
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
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
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
we write the last equation for v << c
Hence, we obtain
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
or
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:
When the particle is observed in the system S', we obtain
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.
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
Thus, the value of relativistic impulse of proton is
Now, using the equation
we obtain for the energy square of energy ε2 of neutron after substitutions
Therefore, the energy of neutron is
You have reached the end of Physics lesson 18.6.3 Energy in Relativistic Events. Relativistic Energy of Motion. There are 3 lessons in this physics tutorial covering Relativistic Dynamics. Mass, Impulse and Energy in Relativity, you can access all the lessons from this tutorial below.
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