9.2 Relativistic kinematics

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Special Lorentz transformations

From time to time we will call upon specific types ofLorentz transformations. The following two examples present the most commonly used types.

Example 9.1

Time-preserving Lorentz transformations have $t^{\prime }=t$ , or equivalently $x^{\prime 4}=x^4$ . Such transformations have $L_4^4=1,L_i^4=0\mathrm{for}i=1,2,3$ , and substituting in Eq. (9.4) with $\rho =\sigma =4$ gives

$$\sum _{i^{\prime }=1}^3{\left(L_4^{i^{\prime }}\right)}^2-{\left(L_4^4\right)}^2=-1⟹\sum _{i^{\prime }=1}^3(L_4^{i^{\prime }}{)}^2=0.$$

This can only hold if $L_4^{i^{\prime }}=0$ for $i^{\prime }=1$ , 2, 3. Hence

$$\mathsf{L}=[L_{\nu }^{{\mu }^{\prime }}]=\left(\begin{array}{cccc}& & & 0\\ & [a_{ij}]& & 0\\ & & & 0\\ 0& 0& 0& 1\end{array}\right)(i,j,\cdots =1,2,3)\text{\hspace{1em}(9.8)}$$

where $\mathsf{A}=[a_{ij}]$ is an orthogonal $3\times 3$ matrix, ${\mathsf{A}}^T\mathsf{A}=\mathsf{I}$ , which follows on substituting $L$ in Eq. (9.5). If $\det$ $\mathsf{L}=\det \mathsf{A}=+1$ these transformations are spatial rotations, while if $\det$ $\mathsf{A}=$ 1 they are space reflections.

Example 9.2

Lorentz transformations that leave the $y$ and $z$ coordinates unchanged are of the form

$$\mathsf{L}=[L_{\nu }^{{\mu }^{\prime }}]=\left(\begin{array}{cccc}{L}_1^1& 0& 0& L_4^1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ L_1^4& 0& 0& L_4^4\end{array}\right).$$

Substituting in Eq. (9.4) gives

$$L_1^1{L}_1^1-L_1^4{L}_1^4=g_{11}=1,\text{\hspace{1em}(9.9)}$$

$$L_1^1{L}_4^1-L_1^4{L}_4^4=g_{14}=g_{41}=0,\text{\hspace{1em}(9.10)}$$

$$L_4^1{L}_4^1-L_4^4{L}_4^4=g_{44}=-1.\text{\hspace{1em}(9.11)}$$

From Eq. (9.11), we have $\begin{array}{r}(L_4^4{)}^2=1+{\sum }_{i=1}^3(L_4^i{)}^2\ge 1\end{array}$ , and assuming $L_4^4\ge 1$ it is possible to set $L_4^4=\cosh \alpha$ for some real number $\alpha .$ . Then $L_4^1=\pm \sqrt{{\cosh }^2\alpha -1}=$ $\sinh$ $\alpha$ on choosing $\alpha$ with the appropriate sign. Similarly, Eq. (9.9) implies that $L_1^1=\cosh \beta ,L_1^4=\sinh \beta$ and Eq. (9.10) gives

$$0=\sinh \alpha \cosh \beta -\cosh \alpha \sinh \beta =\sinh (\alpha -\beta )⟹\alpha =\beta .$$

Let $v$ be the unique real number defined by

$$\tanh \alpha =-\frac{v}{c},$$

then trigonometric identities give that $|v|<c$ and

$$\cosh \alpha =\gamma ,\sinh \alpha =-\gamma \frac{v}{c}$$

where

$$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}.\text{\hspace{1em}(9.12)}$$

The resulting Lorentz transformations have the form

$$\mathsf{L}=[L_v^{{\mu }^{\prime }}]=\left(\begin{array}{cccc}\gamma & 0& 0& -\gamma \frac{v}{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\gamma \frac{v}{c}& 0& 0& \gamma \end{array}\right),\text{\hspace{1em}(9.13)}$$

and are known as boosts with velocity $v$ in the $x$-direction. Written out explicitly in $x$, $y$, $z$, $t$ coordinates they read

$$x^{\prime }=\gamma (x-vt),y^{\prime }=y,z^{\prime }=z,t^{\prime }=\gamma \left(t-\frac{v}{c^2}x\right).\text{\hspace{1em}(9.14)}$$

The inverse transformation is obtained on replacing $v$ by $v$

$$x=\gamma (x^{\prime }+v{t}^{\prime }),y=y^{\prime },z=z^{\prime },t=\gamma \left(t^{\prime }+\frac{v}{c^2}{x}^{\prime }\right).\text{\hspace{1em}(9.15)}$$

The parameter $v$ plays the role of a relative velocity between the two frames since the spatial origin $(x^{\prime }=0,y^{\prime }=0,z^{\prime }=0)$ in the primed frame satisfies the equation $x=vt$ in the unprimed frame. As the relative velocity $v$ must always be less than $c$ we have the first indication that according to relativity theory, the velocity of light $c$ is a limiting velocity for material particles.

Exercise

Verify that performing two Lorentz transformations with velocities $v_1$ and $v_2$ in the $x-$ directions in succession is equivalent to a single Lorentz transformation with velocity

$$v=\frac{v_1+v_2}{1+v_1{v}_2/c^2}.$$

Relativity of time, length and velocity

Two events $p=(x_1,y_1,z_1,c{t}_1)$ and $q=(x_2,y_2,z_2,c{t}_2)$ are called simultaneous with respect to an inertial frame $K$ if $\Delta t=t_2-t_1=0$ . Consider a second frame $K^{\prime }$ related to $K$ by a boost, Eq. (9.14). These equations are linear and therefore apply to coordinate differences,

$$\Delta x^{\prime }=\gamma (\Delta x-v\Delta t),\Delta t^{\prime }=\gamma (\Delta t-\frac{v}{c^2}\Delta x).$$

Hence,

$$\Delta t=0⟹\Delta t^{\prime }=-\gamma \frac{v}{c^2}\Delta x\ne 0\text{ if }{x}_1\ne x_2,$$

demonstrating the effect known as relativity of simultaneity: simultaneity of spatially separated points is not an absolute concept.

Consider now a clock at rest in $K^{\prime }$ marking offsuccessive ‘ticks’ at events $(x^{\prime },y^{\prime },z^{\prime },c{t}_1^{\prime })$ and $(x^{\prime },y^{\prime },z^{\prime },c{t}_2^{\prime })$ . The time difference according to $K$ is given by Eq. (9.15),

$$\Delta t=\gamma \left(\Delta t^{\prime }+\frac{v}{c^2}\Delta x^{\prime }\right)=\gamma \Delta t^{\prime }\text{ if }\Delta x^{\prime }=0.$$

That is,

$$\Delta t=\frac{\Delta t^{\prime }}{\sqrt{1-\frac{v^2}{c^2}}}\ge \Delta t^{\prime },\text{\hspace{1em}(9.16)}$$

an effect known as time dilatation – a moving clock appears to slow down. Equivalently, a stationary clock in $K$ appears to run slow according to the moving observer $K^{\prime }$

Now consider a rod of length $\ell =\Delta x$ at rest in $K$. Again, using the inverse boost transformation Eq. (9.15) we have

$$\ell =\Delta x=\gamma (\Delta x^{\prime }+v\Delta t^{\prime })=\gamma \Delta x^{\prime }\mathrm{if}\Delta t^{\prime }=0.$$

The rod’s length with respect to $K^{\prime }$ is determined by considering simultaneous moments $t_1^{\prime }=t_2^{\prime }$ at the end points,

$${\ell }^{\prime }=\Delta x^{\prime }=\frac{\ell }{\gamma }=\sqrt{1-\frac{v^2}{c^2}}\ell \le \ell .\text{\hspace{1em}(9.17)}$$

The common interpretation of this result is that the length of a rod is contracted when viewed by a moving observer, an effect known as the Lorentz–Fitzgerald contraction. By reversing the roles of $K$ and $K^{\prime }$ it is similarly found that a moving rod is contracted in the direction of its motion. The key to this effect is that, by the relativity of simultaneity, pairs of events on the histories of the ends of the rod that are simultaneous with respect to $K^{\prime }$ differ from simultaneous pairs in the frame $K.$ . Since there is no contraction perpendicula to the motion, a moving volume $V$ will undergo a contraction

$$V^{\prime }=\sqrt{1-\frac{v^2}{c^2}}V.\text{\hspace{1em}(9.18)}$$

This is the most useful application of the Lorentz–Fitzgerald contraction.

Exercise

Give the reverse arguments to the above; that a clock at rest runs slow relative to a moving observer, and that a moving rod appears contracted.

Let a particle have velocity $\mathbf{u}=(u_x,u_y,u_z)$ with respect to $K$ , and ${\mathbf{u}}^{\prime }=(u_x^{\prime },u_y^{\prime },u_z^{\prime })$ with respect to $K^{\prime }$ . Setting

$$u_x=\frac{\mathrm{d}x}{\mathrm{d}t},u_x^{\prime }=\frac{\mathrm{d}{x}^{\prime }}{\mathrm{d}{t}^{\prime }},u_y=\frac{\mathrm{d}y}{\mathrm{d}t},\text{ etc. }$$

and using the Lorentz transformations Eq. (9.14), we have

$$u_x^{\prime }=\frac{u_x-v}{1-u_{x}v/c^2},u_y^{\prime }=\frac{u_y}{\gamma (1-u_{x}v/c^2)},u_z^{\prime }=\frac{u_z}{\gamma (1-u_{x}v/c^2)}.\text{\hspace{1em}(9.19)}$$

Comparing with the Newtonian discussion at the beginning of this chapter it is natural to call this the relativistic law of transformation of velocities. Similarly on using the

inverse Lorentz transformations Eq. (9.15), we arrive at the relativistic law of addition of velocities:

$$u_x=\frac{u_x^{\prime }+v}{1+u_x^{\prime }v/c^2},u_y=\frac{u_y^{\prime }}{\gamma (1+u_x^{\prime }v/c^2)},u_z=\frac{u_z^{\prime }}{\gamma (1+u_x^{\prime }v/c^2)}.\text{\hspace{1em}(9.20)}$$

The same result can be obtained from Eq. (9.19) by replacing $v$ by $v$ and interchanging primed and unprimed velocities.

For a particle moving in the $x-y$ plane set $u_x=u$ $\cos$ $\theta$, $u_y=u$ $\sin$ $\theta ,u_z=0$ and $u_x^{\prime }=u^{\prime }$ $\cos$ ${\theta }^{\prime },u_{\nu }^{\prime }=u^{\prime }$ $\sin$ $\theta$ , $u_z^{\prime }=0$ . If $u^{\prime }=c$ it follows from Eq. (9.20) that $u=c,$ , and the velocity of light is independent of the motion of the observer as required by Einstein’s principle of relativity. The second equation of Eq. (9.20) gives a relation between the $\theta$ and ${\theta }^{\prime }$ , the angles the light beam subtends with the $x$- and $x^{\prime }.$ -directions respectively:

$$\sin \theta =\frac{\sin {\theta }^{\prime }}{1+(v/c)\cos {\theta }^{\prime }}\sqrt{1-\frac{v^2}{c^2}}.\text{\hspace{1em}(9.21)}$$

This formula is known as the relativistic aberration of light. If $\frac{v}{c}\ll 1$ then

$$\delta \theta =\theta -{\theta }^{\prime }\approx -\frac{v}{c}\sin {\theta }^{\prime },$$

a Newtonian formula for aberration of light, which follows simply from the triangle addition law of velocities and was used by the astronomer Bradley nearly 300 years ago to estimate the velocity of light.

Problems

Problem 9.5

From the law oftransformation ofvelocities, Eq. (9.19), show that the velocity ofligh in an arbitrary direction is invariant under boosts.

Problem 9.6

If two intersecting light beams appear to be making a non-zero angle $\varphi$ in one frame $K$, show that there always exists a frame $K^{\prime }$ whose motion relative to $K$ is in the plane of the beams such that the beams appear to be directed in opposite directions.

Problem 9.7

A source of light emits photons uniformly in all directions in its own rest frame.

(a) If the source moves with velocity $v$ with respect to an inertial frame $K,$ , show the ‘headligh effect’: half the photons seem to be emitted in a forward cone whose semi-angle is given by $\cos$ $\theta =v/c$

(b) In films of the Star Wars genre, star fields are usually seen to be swept backwards around a rocket as it accelerates towards the speed of light. What would such a rocketeer really see as his velocity $vc?$

Problem 9.8

If two separate events occur at the same time in some inertial frame $S,$ prove tha there is no limit on the time separations assigned to these events in other frames, but that their space separation varies from infinity to a minimum that is measured in $S$. With what speed must an observe travel in order that two simultaneous events at opposite ends of $a$ 10-metre room appear to differ in time by 100 years?

Problem 9.9

A supernova is seen to explode on Andromeda galaxy, while it is on the western horizon. Observers $A$ and $B$ are walking past each other, $A$ at 5 $\mathrm{km}$/$h$ towards the east, $B$ at 5 $\mathrm{km}$/$h$ towards the west. Given that Andromeda is about a million light years away, calculate the difference in time attributed to the supernova event by $A$ and $B$. Who says it happened earlier?

Problem 9.10

Twin $A$ on the Earth and twin $B$ who is in a rocketship moving away from him at a speed of $\frac{1}{2}c$ separate from each other at midday on their common birthday. They decide to each blow out candles exactly four years from $B$’s departure.

(a) What moment in $B$’s time corresponds to the event $P$ that consists of $A$ blowing his candle out? And what moment in $A$’s time corresponds to the event $\varrho$ that consists of $B$ blowing her candle out?

(b) According to $A$ which happened earlier, $P$ or $Q$? And according to $B$?

(c) How long will $A$ have to wait before he sees his twin blowing her candle out?