# 15.25: Addition of Kinetic Energies

I want now to consider two particles moving at nonrelativistic speeds – by which I mean that the kinetic energy is given to a sufficient approximation by the expression \( \frac{1}{2}mu^{2}\) and so that parallel velocities add linearly.

Consider the particles in figure XV.37, in which the velocities are shown relative to laboratory space.

Referred to laboratory space, the kinetic energy is \( \frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\). However, the centre of mass is moving to the right with speed \( V=\frac{(m_{1}u_{1}+m_{2}u_{2})}{(m_{1}+m_{2})}\), and, referred to centre of mass space, the kinetic energy is \( \frac{1}{2}m_{1}(u_{1}-V)^{2}+\frac{1}{2}m(u_{2}+V)^{2}\). On the other hand, if we refer the situation to a frame in which \( m_{1}\) is at rest, the kinetic energy is \( \frac{1}{2}m_{2}(u_{1}+u_{2})^{2}\), and, if we refer the situation to a frame in which \( m_{2}\) is at rest, the kinetic energy is \( \frac{1}{2}m_{1}(u_{1}+u_{2})^{2}\).

All we are saying is that the kinetic energy depends on the frame to which speed are referred – and this is not something that crops up only for relativistic speeds.

Let us put some numbers in. Let us suppose, for example that

\( m_{1} = 3\)kg \( u_{1} = 4\)m s^{-1}

\( m_{2} = 2\)kg \( u_{3} = 4\)m s^{-1}

so that

\( V = 1.2\)m s^{-1}.

In that case, the kinetic energy

referred to laboratory space is 33 J,

referred to centre of mass space is 29.4 J,

referred to \( m_{1}\) is 49 J,

referred to \( m_{2}\) is 73.5 J.

In this case the kinetic energy is least when referred to centre of mass space, and is greatest when referred to the lesser mass.

*Exercise.* Is this always so, whatever the values of *m*_{1}, *m*_{2} , *u*_{1 }and *u*_{2}?

It may be worthwhile to look at the special case in which the two masses are equal (*m*) and the two speeds(whether in laboratory or centre of mass space) are equal (*u*).

In that case the kinetic energy in laboratory or centre of mass space is *mu*^{2}, while referred to either of the masses it is 2*mu*^{2}.

We shall now look at the same problem for particles travelling at relativistic speeds, and we shall see that the kinetic energy referred to a frame in which one of the particles is at rest is very much greater than (not merely twice) the energy referred to a centre of mass frame.

If two particles are moving towards each other with “speeds” given by g_{1} and g_{2} in centre of mass space, the g of one relative to the other is given by equation 15.16.14, and, since *K* = g - 1, it follows that if the two particles have kinetic energies *K*_{1} and *K*_{2} in centre of mass space (in units of the *m*_{0}c^{2} of each), then the kinetic energy of one relative to the other is

\[ K=K_{1} \oplus K_{2} = K_{1}+K_{2}+K_{1}K_{2}+\sqrt{K_{1}K_{2}(K_{1}+2)(K_{2}+2)}. \label{12.25.1}\]

If two identical particles, each of kinetic energy \( K_{1}\) times \( m_{0}c^{2}\), approach each other, the kinetic energy of one relative to the other is

\[ K=2K_{1}(K_{1}+2). \label{15.25.2}\]

For nonrelativistic speeds as \( K_{1}\rightarrow 0\), this tends to \( K=4K_{1}\), as expected.

Let us suppose that two protons are approaching each other at 99% of the speed of light in centre of mass space (\( K_{1}\) = 6.08881). Referred to a frame in which one proton is at rest, the kinetic energy of the other will be \( K\) = 98.5025, the relative speeds being 0.99995 times the speed of light. Thus \( K=16K_{1}\) rather than merely \( 4K_{1}\) as in the nonrelativistic calculation. For more energetic particles, the ratio \( \frac{K}{K_{1}}\) is even more. These calculations are greatly facilitated if you wrote, as suggested in Section 15.3, a program that instantly connects all the relativity factors given there.

Exercise \(\PageIndex{1}\)

Two protons approach each other, each having a kinetic energy of 500 GeV in laboratory or centre of mass space. (Since the two rest masses are equal, these TWO spaces are identical.) What is the kinetic energy of one proton in a frame in which the other is at rest?

(Answer: I make it 535 TeV.)

The factor \( K\) (the kinetic energy in units of \( m_{0}c^{2}\)) is the last of several factors used in this chapter to describe the speed at which a particle is moving, and I take the opportunity here of summarising the formulas that have been derived in the chapter for combining these several measures of speed. These are

\[ \beta_{1}\oplus\beta_{2}=\frac{\beta_{1}+\beta_{2}}{1+\beta_{1}\beta_{2}}. \label{15.16.7}\tag{15.16.7}\]

\[ \gamma_{1}\oplus\gamma_{2}=\gamma_{1}\gamma_{2}+\sqrt{(\gamma_{1}^{2}-1)(\gamma_{2}^{2}-1)}. \label{15.16.14}\tag{15.16.14}\]

\[ k_{1}\oplus k_{2}=k_{1}k_{2} \label{15.18.11}\tag{15.18.11}\]

\[ z_{1}\oplus z_{2}=z_{1}z_{2}+z_{1}+z_{2}. \label{15.18.12}\tag{15.18.12}\]

\( K=K_{1} \oplus K_{2} = K_{1}+K_{2}+K_{1}K_{2}+\sqrt{K_{1}K_{2}(K_{1}+2)(K_{2}+2)}\).

\[ \frac{\phi_{1}}{\phi_{2}}=\phi_{1}+\phi_{2} \label{15.16.11}\tag{15.16.11}\]

If the two speeds to be combined are equal, these become

\[ \beta_{1}\oplus\beta_{1}=\frac{2\beta_{1}}{1+\beta_{1}^{2}}. \label{12.25.3}\]

\[ \gamma_{1}\oplus\gamma_{1}=2\gamma_{1}^{2}-1 \label{12.25.4}\]

\[ \frac{k_{1}}{k_{1}}=k_{1}^{2} \label{12.25.5}\]

\[ z_{1}\oplus z_{1}=z_{1}(z_{1}+2) \label{12.25.6}\]

\[ K_{1}\oplus K_{1}=2K_{1}(K_{1}+2). \label{12.25.7}\]

\[ \phi_{1}\oplus\phi_{1}=2\phi. \label{12.25.8}\]

These formulas are useful, but for numerical examples, if you already have a program for interconverting between all of these factors, the easiest and quickest way of combinng two “speeds” is to convert them to \( \phi\). We have seen examples of how this works in Sections 15.16 and 15.18. We can do the same thing with our example from the present section when combining two kinetic energies. Thus we were combining two kinetic energies in laboratory space, each of magnitude \( K_{1}\) = 6.08881 (\( \phi_{1}\) = 2.64665). From this, \( \phi\) = 5.29330, which corresponds to \( K\) = 98.5025.