The rest energy and total energy, respectively, of three particles, expressed in terms of a basic amount A are (1) A, 2A; (2) A, 3A; (3) 3A, 4A. Without written calculation, rank the particles according to their (a) mass, (b) kinetic energy, (c) Lorentz factor, and (d) speed, greatest first.

a)

An object at rest has some energy called rest mass energy and is expressed as

$E=m{c}^2$

Lets use 1,2, and 3 subscripts for three particles respectively. The rest mass energy for three particles are given below

$\begin{array}{l}{E}_1=A\\ E_2=A\\ E_3=3A\end{array}$width="66">E1=AE2=AE3=3A

As rest mass energy is only dependent on mass with $c^2$ term being constant. Therefore, the particle with greatest rest energy will have greatest mass, here it is the third particle. As 1^st and 2^nd particle has same rest energy will have same mass.

$m_3>m_1=m_2$

(b)

Kinetic energy can be expressed as total energy minus rest mass energy.

$k=T-E$

Here, in the problem the total energies and rest energies are given, therefore the kinetic energies of the three particles are

$\begin{array}{l}{K}_1=2A-A=A\\ K_2=3A-A=2A\\ K_4=4A-3A=A\end{array}$

Therefore,

$K_2>K_1=K_3$

The result of 2^nd postulate of the special theory of relativity is that the clocks run slower for a moving object when measured from a rest frame. The factor by which the clock is running differently is called the Lorentz factor.

The expression for Lorentz factor is

$Y=\frac{1}{\sqrt{1-{\beta }^2}}$

Here $\beta$is the speed parameter $v/c$.

The expression for total energy of an object is

$T=\gamma m{c}^2$

The total energies of the three particles is

$\begin{array}{l}{T}_1=2A\\ T_2=3A\\ T_3=4A\end{array}$

Rankings will be

$$\begin{gathered} T_3>T_2>T_1 \\ {\gamma }_3{m}_3{c}^2>{\gamma }_2{m}_2{c}^2>{\gamma }_1{m}_1{c}^2 \\ {\gamma }_3{m}_3>{\gamma }_2{m}_2>{\gamma }_1{m}_1 \end{gathered}$$

Let’s consider particle 1 and 2

${\gamma }_2{m}_2>{\gamma }_1{m}_1$

As $m_1=m_2$, therefore, ${\gamma }_2>{\gamma }_1$.

And for particle 3, the mass and total energy, both are greatest, therefore will be greatest.

${\gamma }_3>{\gamma }_2>{\gamma }_1$

(d)

And, as Lorentz factor is related to speed with relation

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\beta }^2}} \\ \beta =\sqrt{1-\frac{1}{{\gamma }^2}} \\ v=c\left(\sqrt{1-\frac{1}{{\gamma }^2}}\right) \end{gathered}$$

Therefore,

$V_3>V_2>V_1$

Hence, greatest $\gamma$will have greatest speed.