9.3 Particle dynamics

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World-lines and proper time

Let $I=[a,b]$ be any closed interval of the real line $R$. A continuous map $\sigma :IM$ is called a parametrized curve in Minkowski space $(M,V)$ . In an inertial frame generated by a basis $e_{\mu }$ of $V$ such a curve may be written as four real functions $x^{\mu }\circ \sigma :I\mathbb{R}$ . We frequently write these functions as $x^{\mu }(\lambda )(a\le \lambda \le b)$ in place of $x^{\mu }(\sigma (\lambda ))$ ), and generally assume them to be differentiable.

If the parametrized curve $\sigma$ passes through the event $p$ having coordinates $p^{\mu }$ , so that $p^{\mu }=x^{\mu }({\lambda }_0)\equiv x^{\mu }(\sigma ({\lambda }_0))$ for some ${\lambda }_0\in I,$ , define the tangent 4-vector to the curve at $p$ to be the 4-vector $U$ given by

$$U=U^{\mu }{e}_{\mu }\in V\text{ where }{U}^{\mu }=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}\lambda }{|}_{\lambda ={\lambda }_0}.$$

This definition is independent of the choice of orthonormal basis $e_{\mu }$ on $V$ , for if $e_{{\mu }^{\prime }}^{\prime }$ is a second o.n. basis related by a Lorentz transformation $e_{\nu }=L_{\nu }^{{\mu }^{\prime }}{e}_{{\mu }^{\prime }}^{\prime }$ , then

$$U^{\prime }=U^{\prime {\mu }^{\prime }}{e}_{{\mu }^{\prime }}^{\prime }=\frac{\mathrm{d}{x}^{\prime {\mu }^{\prime }}(\lambda )}{\mathrm{d}\lambda }{|}_{\lambda ={\lambda }_0}{L}_{{\mu }^{\prime }}^{\mu }{e}_{\mu }=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}\lambda }{|}_{\lambda ={\lambda }_0}{e}_{\mu }=U.$$

The parametrized curve $\sigma$ is called timelike, spacelike or null at $p$ if its tangent 4- vector at $p$ is timelike, spacelike or null, respectively. The path of a material particle wil be assumed to be timelike at all events through which it passes, and is frequently referred to as the particle’s world-line (see Fig. 9.2). This assumption amounts to the requirement that the particle’s velocity is always less than $c$, for

$$0>g(U(\lambda ),U(\lambda ))=g_{\mu \nu }\frac{\mathrm{d}{x}^{\mu }(\lambda )}{\mathrm{d}\lambda }\frac{\mathrm{d}{x}^{\nu }(\lambda )}{\mathrm{d}\lambda }={\left(\frac{\mathrm{d}t}{\mathrm{d}\lambda }\right)}^2\left(\sum _{i=1}^3{\left(\frac{\mathrm{d}{x}^i(t)}{\mathrm{d}t}\right)}^2-c^2\right),$$

Figure 9.2 World-line of a material particle

on setting $t=x^4/c=t(\lambda )$ . Hence

$$v^2=\sum _{i=1}^3{\left(\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\right)}^2<c^2.$$

For two neighbouring events on the world-line, $x^{\mu }(\lambda )$ and $x^{\mu }(\lambda +\Delta \lambda )$ , set

$$\Delta {\tau }^2=-\frac{1}{c^2}\Delta s^2=-\frac{1}{c^2}{g}_{\mu \nu }\Delta x^{\mu }\Delta x^{\nu }>0,$$

where

$$\Delta x^{\mu }=x^{\mu }(\lambda +\Delta \lambda )-x^{\mu }(\lambda ).$$

In the limit $\Delta \lambda 0$

$$\Delta {\tau }^2\to -\frac{1}{c^2}{g}_{\mu \nu }\frac{\mathrm{d}{x}^{\mu }(\lambda )}{\mathrm{d}\lambda }\frac{\mathrm{d}{x}^{\nu }(\lambda )}{\mathrm{d}\lambda }(\Delta \lambda {)}^2=-\frac{1}{c^2}(v^2-c^2){\left(\frac{\mathrm{d}t}{\mathrm{d}\lambda }\right)}^2\Delta {\lambda }^2.$$

Hence

$$\Delta \tau \to \sqrt{1-\frac{v^2}{c^2}}\Delta t=\frac{1}{\gamma }\Delta t.\text{\hspace{1em}(9.22)}$$

Since the velocity of the particle is everywhere less than $c,$ the relativistic law of transformation of velocities Eq. (9.19) can be used to find a combination of rotation and boost, Eq. (9.8) and Eq. (9.14), which transforms the particle’s velocity to zero at any given point $p=\sigma ({\lambda }_0)$ on the particle’s path. Any such inertial frame in which the particle is momentarily at res is known as an instantaneous rest frame or i.r.f. at $p.$ . The i.r.f. will of course vary from point to point on a world-line, unless the velocity is constant along it. Since $\mathbf{v}=0$ in an i.r.f. we have from Eq. (9.22) that $\Delta \tau /\Delta t\to 1\mathrm{as}\Delta t\to 0$ . Thus $\Delta \tau$ measures the time interva registered on an inertial clock instantaneously comoving with the particle. It is generally interpreted as the time measured on a clock carried by the particle from $x^{\mu }$ to $x^{\mu }+\Delta x^{\mu }$

The factor $1/\gamma$ in Eq. (9.22) represents the time dilatation effect of Eq. (9.16) on such a clock due to its motion relative to the external inertial frame. The total time measured on a clock carried by the particle from event $p$ to event $q$ is given by

$${\tau }_{pq}={\int }_p^q\mathrm{d}\tau ={\int }_{t_p}^{t_q}\frac{\mathrm{d}t}{\gamma },$$

and is called the proper time from $p$ to $q$ . If we fix the event $p$ and let $q$ vary along the curve then proper time can be used as a parameter along the curve,

$$\tau ={\int }_{t_p}^t\frac{\mathrm{d}t}{\gamma }=\tau (t).\text{\hspace{1em}(9.23)}$$

The tangent 4-vector $V=V^{\mu }{e}_{\mu }$ calculated with respect to this special parameter is called the 4-velocity of the particle,

$$V^{\mu }=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}\tau }=\gamma \frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}=\gamma (\mathbf{v},c).\text{\hspace{1em}(9.24)}$$

Unlike coordinate time $t$, proper time $\tau$ is a true scalar parameter independent of inertial frame; hence the components of 4-velocity $V^{\mu }$ transform as a contravariant 4-vector

$$V^{\prime {\mu }^{\prime }}=L_{\nu }^{{\mu }^{\prime }}{V}^{\nu }.$$

From Eq. (9.24) the magnitude of the 4-velocity always has constant magnitude $-c^2$

$$g(V,V)=V^{\mu }{V}_{\mu }=(v^2-c^2){\gamma }^2=-c^2.\text{\hspace{1em}(9.25)}$$

The 4-acceleration of a particle is defined to be the contravariant 4-vecto $A=A^{\mu }{e}_{\mu }$ with components

$$A^{\mu }=\frac{\mathrm{d}{V}^{\mu }}{\mathrm{d}\tau }=\frac{{\mathrm{d}}^2{x}^{\mu }(\tau )}{\mathrm{d}{\tau }^2}.\text{\hspace{1em}(9.26)}$$

Expressing these components in terms of the coordinate time parameter $t$ gives

$$A^{\mu }=\gamma \left(\frac{\mathrm{d}\gamma }{\mathrm{d}t}\mathbf{v}+\gamma \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t},c\frac{\mathrm{d}\gamma }{\mathrm{d}t}\right).\text{\hspace{1em}(9.27)}$$

The 4-vectors $A$ and $V$ are orthogonal to each other since

$$\frac{\mathrm{d}}{\mathrm{d}\tau }(g(V,V))=\frac{\mathrm{d}}{\mathrm{d}\tau }(-c^2)=0,$$

and expanding the left-hand side gives $g(A,V)+g(V,A)=2g(A,V)$ , so that

$$g(A,V)=A^{\mu }{V}_{\mu }=0.\text{\hspace{1em}(9.28)}$$

Exercise

Show that in an i.r.f. the components of 4-velocity and 4-acceleration are given by

$$V^{\mu }=(0,c),A^{\mu }=(\mathbf{a},0)\text{ where }\mathbf{a}=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t},$$

and verify that the 4-vectors $A$ and $V$ are orthogonal to each other.

Relativistic particle dynamics

We assume each particle has a constant scalar $m$ attached to it, called its rest mass. This may be thought ofas the Newtonian mass in an instantaneous rest frame of the particle, satisfying Newton’s second law $\mathbf{F}=m\mathbf{a}$ for any imposed force $F$ in that frame. The 4-momentum of the particle is defined to be the 4-vector having components $P^{\mu }=m{V}^{\mu }$ where $V=V^{\mu }{e}_{\mu }$ is the 4-velocity of the particle,

$$P^{\mu }=\left(\mathbf{p},\frac{E}{c}\right)\text{\hspace{1em}(9.29)}$$

where

$$\mathbf{p}=m\gamma \mathbf{v}=\frac{m\mathbf{v}}{\sqrt{1-v^2/c^2}}=\text{ momentum },\text{\hspace{1em}(9.30)}$$

$$E=m\gamma c^2=\frac{m{c}^2}{\sqrt{1-v^2/c^2}}=\mathrm{energy}.\text{\hspace{1em}(9.31)}$$

For $v\ll c$ the momentum reduces to the Newtonian formula $\mathbf{p}=m\mathbf{v}$ and the energy can be written as $\begin{array}{r}E\approx m{c}^2+\frac{1}{2}m{v}^2+...\end{array}$ . The energy contribution $E=m{c}^2$ , which arises even when the particle is at rest, is called the particle’s rest-energy.

Exercise

Show the following identities:

$$g(P,P)=P^{\mu }{P}_{\mu }=-m^2{c}^2,E=\sqrt{p^2{c}^2+m^2{c}^2},\mathbf{p}=\frac{E\mathbf{v}}{c^2}.\text{\hspace{1em}(9.32)}$$

The relations Eq. (9.32) make sense even in the limit $vc$ provided the particle has zero rest mass, $m=0$ . Such particles will be termed photons, and satisfy the relations

$$E=pc,\mathbf{p}=\frac{E}{c}\mathbf{n}\text{ where }\mathbf{n}\cdot \mathbf{n}=1.\text{\hspace{1em}(9.33)}$$

Here $n$ is called the direction of propagation of the photon. The 4-momentum ofa photon has the form

$$P^{\mu }=\left(\mathbf{p},\frac{E}{c}\right)=\frac{E}{c}(\mathbf{n},1),$$

and is clearly a null vector, $P^{\mu }{P}_{\mu }=0$

In analogy with Newton’s law $\mathbf{F}=m\mathbf{a}.$ , it is sometimes useful to define a 4-force $F=F^{\mu }{e}_{\mu }$ having components

$$F^{\mu }=\frac{\mathrm{d}{P}^{\mu }}{\mathrm{d}\tau }=m{A}^{\mu }.\text{\hspace{1em}(9.34)}$$

By Eq. (9.28) the 4-force is always orthogonal to the 4-velocity. Defining 3-force $f$ in the usual way by

$$\mathbf{f}=\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}$$

and using $\frac{\mathrm{d}}{\mathrm{d}\tau }=\gamma \frac{\mathrm{d}}{\mathrm{d}t}$ we obtain

$$F^{\mu }=\gamma \left(\mathbf{f},\frac{1}{c}\frac{\mathrm{d}E}{\mathrm{d}t}\right).\text{\hspace{1em}(9.35)}$$

Problems

Problem 9.11

Using the fact that the 4-velocity $V^{\mu }=\gamma (u)(u_x,u_y,u_z,c)$ transforms as a 4-vector, show from the transformation equation for $V^{\prime 4}$ that the transformation of $u$ under boosts is

$$\frac{\gamma (u^{\prime })}{\gamma (u)}=\gamma (v)\left(1-\frac{v{u}_x}{c^2}\right).$$

From the remaining transformation equations for $V^{\prime }{}^{i^{\prime }}$ derive the law of transformation of velocities Eq. (9.19).

Problem 9.12

Let $K^{\prime }$ be a frame with velocity $v$ relative to $K$ in the $x$-direction.

(a) Show that for a particle having velocity $u^{\prime },$ , acceleration $a^{\prime }$ in the $x^{\prime }.$ -direction relative to $K^{\prime },$ , its acceleration in $K$ is

$$a=\frac{a^{\prime }}{[\gamma (1+v{u}^{\prime }/c^2){]}^3}.$$

(b) A rocketeer leaves Earth at $t=0$ with constant acceleration $g$ at every moment relative to hi instantaneous rest frame. Show that his motion relative to the Earth is given by

$$x=\frac{c^2}{g}\left(\sqrt{1+\frac{g^2}{c^2}{t}^2}-1\right).$$

(c) In terms of his own proper time $\tau$ show that

$$x=\frac{c^2}{g}\left(\cosh \frac{g}{c}\tau -1\right).$$

(d) If he proceeds for 10 years of his life, decelerates with $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ for another 10 years to come to rest, and returns in the same way, taking 40 years in all, how much will people on Earth have aged on his return? How far, in light years, will he have gone from Earth?

Problem 9.13

A particle is in hyperbolic motion along a world-line whose equation is given by

$$x^2-c^2{t}^2=a^2,y=z=0.$$

Show that

$$\gamma =\frac{\sqrt{a^2+c^2{t}^2}}{a}$$

and that the proper time starting from $t=0$ along the path is given by

$$\tau =\frac{a}{c}{\cosh }^{-1}\frac{ct}{a}.$$

Evaluate the particle’s 4-velocity $V^{\mu }$ and 4-acceleration $A^{\mu }$ . Show that $A^{\mu }$ has constant magnitude.

Problem 9.14

For a system of particles it is generally assumed that the conservation of total 4- momentum holds in any localized interaction,

$$\sum _a{P}_{(a)}^{\mu }=\sum _b{Q}_{(b)}^{\mu }.$$

Use Problem 9.4 to show that the law ofconservation of4-momentum holds for a given system provided the law of energy conservation holds in all inertial frames. Also show that the law of conservation of momentum in all frames is sufficient to guarantee conservation of 4-momentum.

Problem 9.15

A particle has momentum $p$, energy $E$ in a frame $K$.

(a) If $K^{\prime }$ is an inertial frame having velocity $v$ relative to $K,$ , use the transformation law of the momentum 4-vecto $P^{\mu }=\left(\mathbf{p},\frac{\bar{E}}{c}\right)$ to show that

$$E^{\prime }=\gamma (E-\mathbf{v}\cdot \mathbf{p}),{\mathbf{p}}_{\perp }^{\prime }={\mathbf{p}}_{\perp }\text{ and }{\mathbf{p}}_{\parallel }^{\prime }=\gamma \left({\mathbf{p}}_{\parallel }-\frac{E}{c^2}\mathbf{v}\right),$$

where ${\mathbf{p}}_{\perp }$ and ${\mathbf{p}}_{\parallel }$ are the components of $p$ respectively perpendicular and parallel to $v$.

(b) If the particle is a photon, use these transformations to derive the aberration formula

$$\cos {\theta }^{\prime }=\frac{\cos \theta -v/c}{1-\cos \theta (v/c)}$$

where $\theta$ is the angle between $p$ and $v$.

Problem 9.16

Use $F^{\mu }{V}_{\mu }=0$ to show that

$$\mathbf{f}\cdot \mathbf{v}=\frac{\mathrm{d}E}{\mathrm{d}t}.$$

Also show this directly from the definitions Eq. (9.30) and Eq. (9.31) of $p$, $E$.