APPENDICES
APPENDIX ISIMPLE DERIVATION OF THE LORENTZ TRANSFORMATION(SUPPLEMENTARY TO SECTION XI)For the relative orientation of the co-ordinate systems indicated in Fig. 2, thex-axes of both systems permanently coincide. In the present case we can divide the problem into parts by considering first only events which are localised on thex-axis. Any such event is represented with respect to the co-ordinate systemKby the abscissaxand the timet, and with respect to the systemK′by the abscissax′and the timet′. We require to findx′andt′whenxandtare given.A light-signal, which is proceeding along the positive axis ofx, is transmitted according to the equationx=ctorx–ct= 0 . . . . . (1).Since the same light-signal has to be transmitted relative toK′with the velocityc, the propagation relative to the systemK′will be represented by the analogous formulax′–ct′= 0 . . . . . (2)Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation(x′–ct′) = λ(x–ct) . . . (3).is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (x–ct) involves the disappearance of (x′–ct′).If we apply quite similar considerations to light rays which are being transmitted along the negativex-axis, we obtain the condition(x′+ct′) = (x + ct) . . . (4).By adding (or subtracting) equations (3) and (4), and introducing for convenience the constantsaandbin place of the constants λ and μ whereimage038andimage039we obtain the equationsimage040We should thus have the solution of our problem, if the constantsaandbwere known. These result from the following discussion.For the origin ofK′we have permanentlyx′= 0, and hence according to the first of the equations (5)image041If we callvthe velocity with which the origin ofK′is moving relative toK, we then haveimage042The same valuevcan be obtained from equations (5), if we calculate the velocity of another point ofK′relative toK, or the velocity (directed towards the negativex-axis) of a point ofKwith respect toK′. In short, we can designatevas the relative velocity of the two systems.Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference toK′must be exactly the same as the length, as judged fromK′, of a unit measuring-rod which is at rest relative toK. In order to see how the points of thex′-axis appear as viewed fromK, we only require to take a “snapshot” ofK′fromK; this means that we have to insert a particular value oft(time ofK),e.g.t= 0. For this value oftwe then obtain from the first of the equations (5)x′=axTwo points of thex′-axis which are separated by the distance Δx′= 1 when measured in theK′system are thus separated in our instantaneous photograph by the distanceimage043But if the snapshot be taken fromK′(t′= 0), and if we eliminatetfrom the equations (5), taking into account the expression (6), we obtainimage044From this we conclude that two points on thex-axis separated by the distance 1 (relative toK) will be represented on our snapshot by the distanceimage045But from what has been said, the two snapshots must be identical; hence Δxin (7) must be equal to Δx′in (7a), so that we obtainimage046The equations (6) and (7b) determine the constantsaandb. By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in Section XI.image047Thus we have obtained the Lorentz transformation for events on thex-axis. It satisfies the conditionx′2–c2t′2=x2–c2t2. . . . . . (8a).The extension of this result, to include events which take place outside thex-axis, is obtained by retaining equations (8) and supplementing them by the relationsimage048In this way we satisfy the postulate of the constancy of the velocity of lightin vacuofor rays of light of arbitrary direction, both for the systemKand for the systemK′. This may be shown in the following manner.We suppose a light-signal sent out from the origin ofKat the timet= 0. It will be propagated according to the equationimage049or, if we square this equation, according to the equationx2+y2+z2–c2t2= 0 . . . . . (10).It is required by the law of propagation of light, in conjunction with the postulate of relativity, that the transmission of the signal in question should take place—as judged fromK′—in accordance with the corresponding formular′=ct′or,x′2+y′2+z′2–c2t′2= 0 . . . . . . (10a).In order that equation (10a) may be a consequence of equation (10), we must havex′2+y′2+z′2–c2t′2= σ (x2+y2+z2–c2t2) (11).Since equation (8a) must hold for points on thex-axis, we thus have σ = 1. It is easily seen that the Lorentz transformation really satisfies equation (11) for σ = 1; for (11) is a consequence of (8a) and (9), and hence also of (8) and (9). We have thus derived the Lorentz transformation.The Lorentz transformation represented by (8) and (9) still requires to be generalised. Obviously it is immaterial whether the axes ofK′be chosen so that they are spatially parallel to those ofK. It is also not essential that the velocity of translation ofK′with respect toKshould be in the direction of thex-axis. A simple consideration shows that we are able to construct the Lorentz transformation in this general sense from two kinds of transformations, viz. from Lorentz transformations in the special sense and from purely spatial transformations. which corresponds to the replacement of the rectangular co-ordinate system by a new system with its axes pointing in other directions.Mathematically, we can characterise the generalised Lorentz transformation thus:It expressesx′, y′, x′, t′, in terms of linear homogeneous functions ofx, y, x, t, of such a kind that the relationx′2+y′2+z′2–c2t′2=x2+y2+z2–c2t2(11a).is satisficd identically. That is to say: If we substitute their expressions inx, y, x, t, in place ofx′, y′, x′, t′, on the left-hand side, then the left-hand side of (11a) agrees with the right-hand side.
For the relative orientation of the co-ordinate systems indicated in Fig. 2, thex-axes of both systems permanently coincide. In the present case we can divide the problem into parts by considering first only events which are localised on thex-axis. Any such event is represented with respect to the co-ordinate systemKby the abscissaxand the timet, and with respect to the systemK′by the abscissax′and the timet′. We require to findx′andt′whenxandtare given.
A light-signal, which is proceeding along the positive axis ofx, is transmitted according to the equation
x=ct
or
x–ct= 0 . . . . . (1).
Since the same light-signal has to be transmitted relative toK′with the velocityc, the propagation relative to the systemK′will be represented by the analogous formula
x′–ct′= 0 . . . . . (2)
Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation
(x′–ct′) = λ(x–ct) . . . (3).
is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (x–ct) involves the disappearance of (x′–ct′).
If we apply quite similar considerations to light rays which are being transmitted along the negativex-axis, we obtain the condition
(x′+ct′) = (x + ct) . . . (4).
By adding (or subtracting) equations (3) and (4), and introducing for convenience the constantsaandbin place of the constants λ and μ where
image038
and
image039
we obtain the equations
image040
We should thus have the solution of our problem, if the constantsaandbwere known. These result from the following discussion.
For the origin ofK′we have permanentlyx′= 0, and hence according to the first of the equations (5)
image041
If we callvthe velocity with which the origin ofK′is moving relative toK, we then have
image042
The same valuevcan be obtained from equations (5), if we calculate the velocity of another point ofK′relative toK, or the velocity (directed towards the negativex-axis) of a point ofKwith respect toK′. In short, we can designatevas the relative velocity of the two systems.
Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference toK′must be exactly the same as the length, as judged fromK′, of a unit measuring-rod which is at rest relative toK. In order to see how the points of thex′-axis appear as viewed fromK, we only require to take a “snapshot” ofK′fromK; this means that we have to insert a particular value oft(time ofK),e.g.t= 0. For this value oftwe then obtain from the first of the equations (5)
x′=ax
Two points of thex′-axis which are separated by the distance Δx′= 1 when measured in theK′system are thus separated in our instantaneous photograph by the distance
image043
But if the snapshot be taken fromK′(t′= 0), and if we eliminatetfrom the equations (5), taking into account the expression (6), we obtain
image044
From this we conclude that two points on thex-axis separated by the distance 1 (relative toK) will be represented on our snapshot by the distance
image045
But from what has been said, the two snapshots must be identical; hence Δxin (7) must be equal to Δx′in (7a), so that we obtain
image046
The equations (6) and (7b) determine the constantsaandb. By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in Section XI.
image047
Thus we have obtained the Lorentz transformation for events on thex-axis. It satisfies the condition
x′2–c2t′2=x2–c2t2. . . . . . (8a).
The extension of this result, to include events which take place outside thex-axis, is obtained by retaining equations (8) and supplementing them by the relations
image048
In this way we satisfy the postulate of the constancy of the velocity of lightin vacuofor rays of light of arbitrary direction, both for the systemKand for the systemK′. This may be shown in the following manner.
We suppose a light-signal sent out from the origin ofKat the timet= 0. It will be propagated according to the equation
image049
or, if we square this equation, according to the equation
x2+y2+z2–c2t2= 0 . . . . . (10).
It is required by the law of propagation of light, in conjunction with the postulate of relativity, that the transmission of the signal in question should take place—as judged fromK′—in accordance with the corresponding formula
r′=ct′
or,
x′2+y′2+z′2–c2t′2= 0 . . . . . . (10a).
In order that equation (10a) may be a consequence of equation (10), we must have
x′2+y′2+z′2–c2t′2= σ (x2+y2+z2–c2t2) (11).
Since equation (8a) must hold for points on thex-axis, we thus have σ = 1. It is easily seen that the Lorentz transformation really satisfies equation (11) for σ = 1; for (11) is a consequence of (8a) and (9), and hence also of (8) and (9). We have thus derived the Lorentz transformation.
The Lorentz transformation represented by (8) and (9) still requires to be generalised. Obviously it is immaterial whether the axes ofK′be chosen so that they are spatially parallel to those ofK. It is also not essential that the velocity of translation ofK′with respect toKshould be in the direction of thex-axis. A simple consideration shows that we are able to construct the Lorentz transformation in this general sense from two kinds of transformations, viz. from Lorentz transformations in the special sense and from purely spatial transformations. which corresponds to the replacement of the rectangular co-ordinate system by a new system with its axes pointing in other directions.
Mathematically, we can characterise the generalised Lorentz transformation thus:
It expressesx′, y′, x′, t′, in terms of linear homogeneous functions ofx, y, x, t, of such a kind that the relation
x′2+y′2+z′2–c2t′2=x2+y2+z2–c2t2(11a).
is satisficd identically. That is to say: If we substitute their expressions inx, y, x, t, in place ofx′, y′, x′, t′, on the left-hand side, then the left-hand side of (11a) agrees with the right-hand side.