15.1 Why Newtonian momentum must be modified

The Lorentz transformation (Chapter 14) changes the rules for adding velocities, and with them the bookkeeping of momentum. In Newtonian mechanics the momentum \(m\mathbf v\) of an isolated system is conserved in every inertial frame, because the Galilean transformation adds the same constant to every velocity. Under the Lorentz transformation velocities no longer add linearly, and it is easily shown that \(\sum m\mathbf v\) cannot be conserved in all inertial frames at once. Since the principle of relativity requires that a law of conservation, if it holds in one inertial frame, hold in all, either momentum conservation must be abandoned or the definition of momentum must be changed. We keep the conservation law and look for the correct momentum. In this book \(m\) always denotes the rest mass of a particle, the mass measured in a frame in which it is at rest; it is a property of the particle and is the same for all observers.

(a) A glancing collision seen from two frames

The classic argument, due to Lewis and Tolman, considers a glancing collision between two identical balls \(A\) and \(B\), each of rest mass \(m\). Frame \(S'\) moves relative to frame \(S\) with velocity \(v\hat{\mathbf i}\). An observer in \(S\) throws ball \(A\) along the \(y\)-axis with velocity \(-v_A\hat{\mathbf j}\), and an observer in \(S'\) throws ball \(B\) in exactly the same way along the \(y'\)-axis with velocity \(v'_B\hat{\mathbf j}\) (Figure 15.1). The balls start a transverse distance \(Y\) apart, and by symmetry each takes the same proper time \(T_0\) to cover that distance in its own frame, so that

\[v_A=v'_B=\frac{Y}{T_0}.\]

The balls meet in a glancing collision, and by the symmetry between the two frames each ball simply reverses its \(y\)-velocity in its own frame. Seen from \(S\), ball \(A\) moves almost vertically down and back up, while ball \(B\) moves along the arc shown in Figure 15.2.

Transverse distances are not changed by a boost along \(x\), but the time that \(B\) takes to cover the distance \(Y\) is dilated in \(S\) to \(T=\gamma T_0\), where \(\gamma=1/\sqrt{1-v^2/c^2}\). The velocity of \(B\) in \(S\) is therefore

\[\mathbf v_{B}=v\hat{\mathbf i}+v_B\hat{\mathbf j},\qquad v_B=\frac{Y}{T}=\frac{Y}{T_0}\sqrt{1-\frac{v^2}{c^2}}=v'_B\sqrt{1-\frac{v^2}{c^2}},\]

which is also what the velocity transformation \(u_y=u'_y/[\gamma(1+vu'_x/c^2)]\) gives with \(u'_x=0\). Let \(m_A\) and \(m_B\) be the coefficients by which the velocities must be multiplied to give conserved momenta in \(S\). Each ball reverses its \(y\)-velocity in the collision, so the \(y\)-momentum changes from \(-m_Av_A+m_Bv_B\) to \(m_Av_A-m_Bv_B\). Conservation requires \(m_Av_A=m_Bv_B\), that is,

\[m_A\frac{Y}{T_0}=m_B\frac{Y}{T_0}\sqrt{1-\frac{v^2}{c^2}}\quad\Longrightarrow\quad m_B=\frac{m_A}{\sqrt{1-v^2/c^2}}.\]

If the two coefficients were equal, \(y\)-momentum would not be conserved. Now let the transverse speed \(v_A\) tend to zero. Ball \(A\) is then at rest in \(S\), so \(m_A\) becomes the rest mass \(m\), while ball \(B\) moves with speed \(v\). The coefficient that multiplies the velocity of a particle moving with speed \(v\) is therefore

\[\boxed{m_{\text{rel}}(v)=\frac{m}{\sqrt{1-v^2/c^2}}=\gamma m}.\]

(b) Relativistic mass: a historical concept

In the early literature the quantity \(\gamma m\) was called the relativistic mass, and the statement “the mass of a body moving with speed \(v\) relative to an observer is larger than its mass at rest” was taken literally. The rest mass was written \(m_0\) and the relativistic mass \(m\). Figure 15.3 shows how \(\gamma m\) grows without limit as \(v\to c\). Many older examination questions use this language, and a symbol \(m\) in a question must be read with care: in such questions it may mean \(\gamma m\).

The modern view is different, and it is the one adopted in this book. The factor \(\gamma\) belongs to the kinematics, not to the particle: the rest mass \(m\) is an invariant, and the momentum is \(\gamma m\mathbf v\). We shall see in Section 15.4 that \(\gamma mc^2\) is the total energy \(E\), so the relativistic mass is nothing but \(E/c^2\), and a separate name for it is unnecessary. Moreover \(\gamma m\) is not the inertia of the particle in any simple sense, because the response of a moving particle to a force depends on the direction of the force (Section 15.2(b)). Whenever we use the relativistic mass below, we write it explicitly as \(\gamma m\).