Momentum & Energy in Relativity

Velocity-Addition Formula in Relativity

From Lorentz transformation, we know that the coordinates in changing between inertial frames are given by $$\begin{cases} x' = \gamma(x - vt) \\[2ex] y' = y \\[2ex] z' = z\\[2ex] t' = \gamma(t - \frac{vx}{c^2})\\[2ex] \end{cases}$$ Given that an object is travelling at $$\begin{cases} u_x = \frac{dx}{dt} \\[2ex] u_y = \frac{dy}{dt} \\[2ex] u_z = \frac{dz}{dt} \\[2ex] \end{cases}$$ in an inertial frame, ${S}$. In another inertial frame, ${S'}$ that is moving at speed $v$ relative to the previous inertial frame, the speed of the object observed is $$\begin{align} u_x' &= \frac{dx'}{dt} \\ &= \frac{dx'}{dt}\cdot \frac{dt}{dt'} \\ &= \gamma \left( \frac{dx}{dt} - v \right)\left( \gamma\left(1 - \frac{v}{c^2}\frac{dx}{dt}\right) \right)^{-1} \\ &= \frac{u_x-v}{1 - \frac{vu_x}{c^2}} \end{align}$$ Similarly for $u_y'$ and $u_z'$, we have $$\begin{cases} u_x' = \frac{u_x-v}{1 - \frac{vu_x}{c^2}} \\[2ex] u_y' = \frac{u_y}{\gamma(1-\frac{vu_x}{c^2})} \\[2ex] u_z' = \frac{u_z}{\gamma(1-\frac{vu_x}{c^2})} \\[2ex] \end{cases}$$

Relativistic Momentum

Classical momentum is defined as mass times velocity $$\vec{p} = m\vec{v}$$ Observing a closed system without any external force in ${S}$, the sum of all momentum is invariant, i.e. $$m_1 u_{1_\text{ini}} + m_xu_{2_\text{ini}} = m_1 u_{1_\text{fin}} + m_xu_{2_\text{fin}}$$ In ${S'}$, according to the velocity addition rule, the initial momentum is $$m_1\left(\frac{u_{1\,\text{ini}}-v}{1-\frac{vu_{1\text{ini}}}{c^2}} \right) + m_2\left(\frac{u_{2\,\text{ini}}-v}{1-\frac{vu_{2\text{ini}}}{c^2}} \right)$$ The final momentum after internal collision is $$m_1\left(\frac{u_{1\,\text{fin}}-v}{1-\frac{vu_{1\text{fin}}}{c^2}} \right) + m_2\left(\frac{u_{2\,\text{fin}}-v}{1-\frac{vu_{2\text{fin}}}{c^2}} \right)$$ which is not necessarily conserved. This leads to the catastrophe that whether there is energy change depends on which inertial frame is chosen.
Relativistic momentum for massive object is defined as the derivative of position w.r.t proper time $$\begin{align} \vec{p} &\equiv m \left(\frac{d\vec{x}}{d\tau} + \frac{d\vec{y}}{d\tau} + \frac{d\vec{z}}{d\tau} \right) \\ &= m \left(\frac{d\vec{x}}{dt}\frac{dt}{d\tau} + \frac{d\vec{y}}{dt}\frac{dt}{d\tau} + \frac{d\vec{z}}{dt}\frac{dt}{d\tau} \right) \\ &= m \gamma \vec{u} \end{align}$$ As we can see the momentum tends to infinity as speed approaches $c$, a massive object cannot be accelerated to $c$. Relativistic force is defined as the derivative of relativistic momentum w.r.t. proper time $$\vec{F} = \frac{d\vec{p}}{d\tau}$$

Checkpoint

In frame $S$, ball B with speed $v$ along x-axis and $u_0$ along y-axis collides with another ball A which is initially travelling at speed $u_0$, as shown in the image below. After that ball B rebounces at speed $u_0$ along y-axis and reserving speed along x-axis, while ball A bounces backward with speed $u_0$ along z-axis. Ball A and ball B are idential, that is to say they have the same mass $m$ and size. Show that the relativistic momentum along y-axis is conserved in both $S$ frame and a frame $S'$ which is moving towards right with speed $v$ relative to $S$.

In frame $S$, initially the velocity of ball A is $$\begin{align} u_{Ax0} &= 0 \\ u_{Ay0} &= u_0. \\ \end{align}$$ After collision, the velocity of ball A is $$\begin{align} u_{Ax1} &= 0 \\ u_{Ay1} &= -u_0. \end{align}$$ The change of relativistic momentum along y-axis of ball A after collision is $$\Delta p_{Ay} = \frac{-2mu_0}{\sqrt{1-\frac{u_0^2}{c^2}}}.$$ Initially the velocity of ball B is $$\begin{align} u_{Bx0} &= v \\ u_{By0} &= -u_0. \\ \end{align}$$ After the collision, the velocity of ball B is $$\begin{align} u_{Bx1} &= v \\ u_{By1} &= u_0. \end{align}$$ The change of relativistic momentum along y-axis of ball B after collision is $$\Delta p_{By} = \frac{2mu_0}{\sqrt{1-\frac{u_0^2}{c^2}}}.$$ Therefore, the total change of relativistic momentum along y-axis of the whole system is $$\Delta p_{Ay} + \Delta p_{By} = 0.$$ In frame $S'$, initially the velocity of ball A is $$\begin{align} u_{Ax0}' &= -v \\ u_{Ay0}' &= \frac{u_0}{\gamma(1-v(0)/c^2)} = u_0\sqrt{1-\frac{v^2}{c^2}}. \\ \end{align}$$ After the collision, the velocity of ball A is $$\begin{align} u_{Ax1}' &= -v \\ u_{Ay1}' &= -u_0\sqrt{1-\frac{v^2}{c^2}}. \\ \end{align}$$ The speed of ball A before and after the collision is $$u_A' = \sqrt{v^2 + u_0^2(1-\frac{v^2}{c^2})}$$ The change of relativistic momentum along y-axis of ball A after collision is $$\begin{align} \Delta p_{Ay}' &= \frac{-2mu_0\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{1-\frac{1}{c^2}(v^2 + u_0^2(1-\frac{v^2}{c^2}))}} \\ &= -\frac{2mu_0\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{\left(1-\frac{v^2}{c^2}\right)\left(1-\frac{u_0^2}{c^2}\right)}}\\ &= -\frac{2mu_0}{\sqrt{1-\frac{v^2}{c^2}}} \end{align}$$ Initially the velocity of ball B is $$\begin{align} u_{Bx0}' &= 0 \\ u_{By0}' &= -\frac{u_0}{\sqrt{1-\frac{v^2}{c^2}}}. \\ \end{align}$$ After the collision, the velocity of ball B is $$\begin{align} u_{Bx1}' &= 0 \\ u_{By1}' &= \frac{u_0}{\sqrt{1-\frac{v^2}{c^2}}}. \end{align}$$ The speed of ball B before and after the collision is $$u_B' = \frac{u_0}{\sqrt{1-\frac{v^2}{c^2}}}.$$ The change of relativistic momentum along y-axis of ball B after collision is $$\begin{align} \Delta p_{By}' &= \frac{2mu_0/\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{1-\frac{1}{c^2}\left(\frac{u_0^2}{1-v^2/c^2}\right)}}\\ &= \frac{2mu_0}{\sqrt{1-\frac{v^2}{c^2}}\sqrt{\frac{1-\frac{u_0^2}{c^2}}{1-\frac{v^2}{c^2}}}}\\ &= \frac{2mu_0}{\sqrt{1-\frac{u_0^2}{c^2}}} \end{align}$$ Therefore, the total change of relativistic momentum along y-axis of the whole system is $$\Delta p_{Ay}' + \Delta p_{By}' = 0.$$

Relativistic Kinetic Energy

Kinetic energy is defined as the work done to accelerate an object from zero velocity to velocity $\vec{u}$ $$\begin{align} \vec{F}\cdot d\vec{x} &= \frac{d\vec{p}}{d\tau}\cdot \vec{u} d\tau \\ &= \vec{u}\cdot d\vec{p} \end{align}$$ Substitute the relativistic momentum definition, $$\begin{align} \int \vec{u}\cdot d\vec{p} &= \int \vec{u}\cdot d(\gamma m\vec{v}) \\ &= \int d(\vec{u}\cdot(\gamma m \vec{u})) - \int \gamma m \vec{u}\cdot d\vec{u} \\ &= \gamma m \vec{u}\cdot\vec{u} - \frac{m}{2}\int \gamma du^2 \\ &= \gamma mu^2 - \frac{m}{2}(-c^2)\int\gamma d\gamma^{-2} \\ &= \gamma mu^2 + mc^2 \gamma ^{-1} + C \\ &= \gamma mc^2 + C \\ \end{align}$$ where $C$ is a constant. Substitute $u=0$, the work done to accelerate an object from zero velocity to zero velocity is zero $$\begin{align} 0 &= 0 + mc^2(1-0) + C \\ C &= -mc^2 \end{align}$$ which means even an object is at rest, it carries the rest energy $$\bbox[5px,border:2px solid #666] { E = mc^2 }$$ Thus, the kinetic energy is $$\gamma m c^2 -mc^2$$ where $${ E_{\text{tot}} = \gamma mc^2 }$$ is the total energy of the massive object.

Energy-momentum Relation

The total energy of an object is $$E = \gamma mc^2$$ where $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ in which $v$ is the speed of the object. Thus, square of the total energy is $$\begin{align} E^2 &= \frac{(mc^2)^2}{1-\frac{v^2}{c^2}} \\ E^2\left(1- \frac{v^2}{c^2} \right) &= (mc^2)^2 \\ E^2 &= E^2\frac{v^2}{c^2} + (mc^2)^2 \\ &= \gamma^2(mc^2)^2 \left(\frac{v^2}{c^2} \right) + (mc^2)^2 \\ &= ((\gamma mv)c)^2 + (mc^2)^2 \end{align}$$ As the momentum is $$p = \gamma mv$$ we have $$\bbox[5px,border:2px solid #666] { E^2 = (pc)^2 + (mc^2)^2 }$$