Now we want to figure out how the values of some physical quantities measured in a reference frame change when they are measured in another reference frame. The issue of time intervals has already been addressed in the previous section.
First, we want to find the transformation rule for the Cartesian coordinates, namely the relativistic generalization of the Galilean transformation in (1.35). Let{xμ}be the Cartesian coordinates of the first inertial reference frame and let {xμ}be the Cartesian coordinates of the second inertial reference frame moving with constant velocity v=(v,0,0)with respect to the former. If the two Cartesian coordinate systems coincide at the timet=t=0, we have
xμ→xμ=Λxμ
νxν, (2.40)
as follows by the simple integration of Eq. (2.15) and where Λxμ
ν is given by Eq. (2.28). The extension to the case in which the two reference frames have a dif- ferent relative velocity is straightforward.
Let us now consider the case of length measurements. We have again two inertial reference systems, the first has Cartesian coordinates (ct,x,y,z)and the second,
moving with constant velocityv=(v,0,0)with respect to the former, has Cartesian coordinates(ct,x,y,z). We have a rod at rest in the reference frame(ct,x,y,z) and parallel to thex-axis. If the two ends of the rod have coordinatesx1 andx2, the proper lengthof the rod, i.e. the length in the rest frame of the rod, isl0=x2 −x1. From the coordinate transform in Eq. (2.40), we find that the two ends of the rod have coordinatesx1 andx2
x1 = −γvt+γx1, x2= −γvt+γx2. (2.41) Since the observer with Cartesian coordinates(ct,x,y,z)must measure the length of the rod by measuring the two ends at the same timet, we have
l0=x2−x1 =γ (x2−x1)=γl, (2.42) wherelis the length measured in the system(ct,x,y,z). We thus see that the proper length is larger than the length measured in another reference frame
l=l0
γ . (2.43)
This phenomenon is calledLorentz contraction. If we consider a volume and the fact that y=yandz=z, we find the relation between the proper volumeV0 and the volume in another reference frame
V = V0
γ . (2.44)
Let us now consider the trajectory of a particle. The Cartesian coordinates of the particle in the first inertial reference frame are (ct,x,y,z). In the second inertial reference frame, the Cartesian coordinates of the particle are (ct,x,y,z). The infinitesimal displacements of the particle in the two coordinate systems are related by the following equations
dt=γdt−γvd x c2 , d x= −γvdt+γd x, d y=d y,
d z=d z, (2.45)
The velocity of the particle in reference frames (ct,x,y,z)and(ct,x,y,z)is, respectively,
w=(d x/dt,d y/dt,d z/dt) , w=(d x/dt,d y/dt,d z/dt) . (2.46) We thus have
2.5 Transformation Rules 41
˙ x= d x
dt = −γvdt+γd x
γdt−γvd x/c2 = −v+ ˙x 1−vx/c˙ 2,
˙ y= d y
dt = d y
γdt−γvd x/c2 = y˙ γ
1−vx/c˙ 2,
˙ z= d z
dt = d z
γdt−γvd x/c2 = z˙ γ
1−vx/c˙ 2. (2.47) Let us consider a particle moving at the speed of light in the inertial reference frame (ct,x,y,z). Without loss of generality, we can rotate the reference frame to have the motion of the particle along thex axis. We thus havex˙ =candy˙= ˙z=0. In the inertial reference frame(ct,x,y,z)moving with constant velocityv=(v,0,0) with respect to the former, we have
˙
x= −v+c
1−v/c =c, y˙=0, ˙z=0. (2.48) If x˙ = −candy˙= ˙z=0, then x˙= −c. The particle moves at the speed of light in both reference frames, as is expected becauseds2is an invariant. If| ˙x|<c, then
| ˙x|<cas no particle can exceed the speed of light just because we are changing reference frame.
We now have a particle moving in the x y plane in the reference frame with coordinates(ct,x,y,z). Its velocity isx˙=wcosΘandy˙=wsinΘ, so the absolute velocity is w andΘ is the angle between the x axis and the particle velocity. In the reference frame with coordinates(ct,x,y,z)moving with constant velocity v=(v,0,0)with respect to the former, the particle has the velocityx˙=wcosΘ and y˙=wsinΘ, where w andΘ are, respectively, the absolute velocity and the angle as measured in the reference frame with coordinates(ct,x,y,z). The relation betweenΘandΘcan be obtained by calculating the ratio betweeny˙and
˙
xfrom Eq. (2.47)
˙ y
˙
x =tanΘ= y˙
γ (x˙−v)= wsinΘ
γ (wcosΘ−v). (2.49) In the case where the particle moves at the speed of light (w=c), we have
tanΘ= sinΘ
γ (cosΘ−v/c). (2.50)
The fact thatΘ =Θis the phenomenon known asaberration of light.
Fig. 2.2 The distance between the observer and the galaxy isD. At timet=0, some material is ejected with velocityvfrom the center of the galaxy, and thus the distance between the ejected material and the center of the galaxy attisR=vt. The radiation emitted by the ejected material at tand detected by the observer at timettravels a distanceL.ϕis the angle between the trajectory of the ejected material and the line of sight of the galaxy with respect to the distant observer
2.5.1 Superluminal Motion
Superluminal motion refers to the phenomenon of observation of some material ejected by certain galactic nuclei that apparently moves faster than the speed of light.
Actually there is no violation of the Einstein Principle of Relativity. As illustrated in Fig.2.2, we have an observer at the distanceDfrom a galaxy. Relativistic material is ejected at timet=0 from the galactic nucleus with velocityvandϕis the angle between the trajectory of the ejected material and the line of sight of the observer. At the timet, the ejected material is at a distanceR=vtfrom the galactic nucleus, and emits some radiation. The latter travels a distanceLand reaches the distant observer at timet.
Considering that RD, we can calculate the distance L between the position of the ejected material at timetand the observer
L =
(D−vtcosϕ)2+v2t2sin2ϕ=
D2−2Dvtcosϕ+v2t2
=D−vtcosϕ+O v2t2
D2 . (2.51)
Since the radiation emitted by the ejected material moves at the speed of lightc, we also have
L=c t−t
. (2.52)
Combining Eqs. (2.51) and (2.52), we can write
2.5 Transformation Rules 43
t= D
c +t(1−βcosϕ) , (2.53)
where β =v/c. The apparent velocity of the ejected material along the yaxis as measured by the distant observer is
vy= d y dt =d y
dt dt
dt = vsinϕ
1−βcosϕ. (2.54)
The angleϕmaxfor which we have the highest apparent velocityvyis given by
∂vy
∂ϕ =0,
vcosϕmax(1−βcosϕmax)−vβsin2ϕmax=0, vcosϕmax−vβ =0,
cosϕmax=β . (2.55)
sinϕmaxis
sinϕmax=
1−β2 = 1
γ . (2.56)
If we plug cosϕmaxand sinϕmaxinto Eq. (2.54), we find
vy=vγ , (2.57)
and we can see thatvycan easily exceedcfor relativistic matter even if there is no violation of the Einstein Principle of Relativity.