Derivation of the Lorentz transformation without the use of Einstein's second postulate

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by  Andrei Galiautdinov

Derivation of the Lorentz transformation without the use of Einstein’s second
postulate
Andrei Galiautdinov
Department of Physics and Astronomy, University of Georgia, Athens, GA 30602, USA
(Dated: January 3, 2017)
Derivation of the Lorentz transformation without the use of Einstein’s Second Postulate is pro-
vided along the lines of Ignatowsky, Terletskii, and others. This is a write-up of the lecture first
deliveredinPHYS4202E&MclassduringtheSpringsemesterof2014attheUniversityofGeorgia.
The main motivation for pursuing this approach was to develop a better understanding of why the
faster-than-light neutrino controversy (OPERA experiment, 2011) was much ado about nothing.
7
1 Special relativity as a theory of space and time
Capsule
0
2

n All physical phenomena take place in space and time.
a The theory of space and time (in the absence of gravity)
J
is called the Special Theory of Relativity.
1
Floating Ball
We do not get bogged down with the philosophical
]
h problems related to the concepts of space and time. We
p simply acknowledge the fact that in physics the notions
- of space and time are regarded as basic and cannot be
s
s reduced to something more elementary or fundamental.
a We therefore stick to pragmatic operational definitions:
l Touch-sensitive
c
. Time is what clocks measure. Space is what measuring surface
s
c rods measure.
i
s FIG.1: (Coloronline.) Afloating-ballinertialdetector. After
y Inordertostudyandmakeconclusionsabouttheprop- [1].
h erties of space and time we need an observer. A natural
p choice is an observer who moves freely (the one who is
[
free from any external influences). An observer is not a
Definition of clock synchronization

1 single person sitting at the origin of a rectangular coor-
v dinategrid. Rather, itisabunchoffriends(callitTeam
It is pretty clear how to measure distances: the team
0 K) equipped with identical clocks distributed through-
7 out the grid who record the events happening at their simply uses its rectangular grid of rods.
2 It is also clear how to measure time intervals at a par-
respective locations.
0 ticular location: the team member situated at that loca-
0
How do we know that this bunch of friends is free tion simply looks at his respective clock. What’s not so
.
1 from any external influences? We look around and make clear, however, is how the team measures time intervals
0 sure that nothing is pulling or pushing on any member between events that are spacially separated.
7
of the bunch; no strings, no springs, no ropes are at- A confusion about measuring this kind of time inter-
1
tachedtothem. Anevenbetterwayistouseacollection vals was going on for two hundred years or so, until one
:
v of “floating-ball detectors” (Fig. 1) distributed through- day Einstein said: “We need the notion of synchronized
Xi out the grid [1]. When detector balls are released, they clocks! Clock synchronization must be operationally de-
should remain at rest inside their respective capsules. If fined.”
r
a any ball touches the touch-sensitive surface of the cap- Theideathatclocksynchronizationand,consequently,
sule, the frame is not inertial. the notion of simultaneity of spacially separated events,
has to be defined (and not assumed apriori) is the single
In the reference frame associated with a freely moving most important idea of Einstein’s, the heart of special
observer(ourrectangularcoordinategrid),Galileo’sLaw relativity. Einstein proposed to use light pulses. The
of Inertia is satisfied: a point mass, itself free from any procedure then went like this:
external influences, moves with constant velocity. To be In frame K, consider two identical clocks equipped
able to say what “constant velocity” really means, and with light detectors, sitting some distance apart, at A
thus to verify the law of inertia, we need to be able to and B. Consider another clock equipped with a light
measuredistancesandtimeintervalsbetweeneventshap- emitter at location C which is half way between A and
pening at different grid locations. B(wecanverifythatC isindeedhalfwaybetweenAand
2
B withthehelpofthegridofrodsthathadalreadybeen Einsteinconstructedhistheoryofrelativityontheba-
put in place when we constructed our frame K). Then, sis of (1) The Principle of Relativity (laws of nature are
at some instant, emit two pulses from C in opposite di- the same in all inertial reference frames), and (2) The
rections, and let those pulses arrive at A and B. If the Postulate of the Constancy of the Speed of Light (the
clocksatAandB showsametimewhenthepulsesarrive speed of light measured by any inertial observer is in-
then the clocks there are synchronized, by definition. dependent of the state of motion of the emitting body).
Thelightpulsesusedinthesynchronizationprocedure [NOTE: This is not the same as saying that the speed
canbereplacedwithtwoidenticalballsinitiallysittingat of light emitted and measured in K is the same as the
C and connected by a compressed spring. The spring is speed of light emitted and measured in K(cid:48). This latter
released (say, the thread holding the spring is cut in the type of constancy of the speed of light is already implied
middle),theballsflyoffinoppositedirectionstowardsA by the principle of relativity.]
and B, respectively. Here we want to stick to mechanics and push the
How do we know that the balls are identical? Because derivation of the coordinate transformation as far as
Team K made them in accordance with a specific manu- possible without the use of the highly counterintuitive
facturing procedure. Einstein’s Second Postulate. The method that achieves
How do we know that all clocks at K are identical? this will be presented below and was originally due to
Because Team K made all of them in accordance with a Vladimir Ignatowsky [2].
specific manufacturing procedure. [DISCLAIMER: I never read Ignatowsky’s original
How do we know that a tic-toc of any clock sitting papers,buttheideaiswell-knownwithinthecommunity,
in frame K corresponds to 1 second? Because Team K often mentioned and discussed. Anyone with time to
called a tic-toc of a clock made in accordance with the burn can reproduce the steps without much difficulty.
manufacturing procedure “a second”. The derivation below consists of 14 steps. If you can
Similarly, clocks in K(cid:48) are regarded as identical and reduce that number, let me know.]
tick-tocking at 1 second intervals because in that frame
all of the clocks were made in accordance with the same
manufacturing procedure. Step 1: Galileo’s Law of Inertia for freely moving
Now, how do we know that the manufacturing proce- particles
dures in K and K(cid:48) are the same? (Say, how do we know
that a Swiss shop in frame K makes watches the same ...implies the linearity of the coordinate transforma-
way as its counterpart in frame K(cid:48)?) Hmm.... That’s tion between K and K(cid:48) (see Fig. 2),
an interesting question to ponder about.
x(cid:48) = α (v)x+α (v)y+α (v)z+α (v)t, (1)
11 12 13 14
When studying spacetime from the point of view of y(cid:48) = α (v)x+α (v)y+α (v)z+α (v)t, (2)
21 22 23 24
inertial frames of reference discussed above, people dis-
z(cid:48) = α (v)x+α (v)y+α (v)z+α (v)t, (3)
covered the following. 31 32 33 34
t(cid:48) = α (v)x+α (v)y+α (v)z+α (v)t. (4)
41 42 43 44
Properties of space and time:
Here we assumed that the origins of the two coordinate
1. At least one inertial reference frame exists. (Geo- systems coincide, that is event (0,0,0,0) in K has coor-
centric is OK for crude experiments; geliocentric is dinates (0,0,0,0) in K(cid:48).
better; the frame in which microwave background
radiation is uniform is probably closest to ideal).
K K’
y y’
2. Space is uniform (translations; 3 parameters). P (x, y, z, t)
3. Space is isotropic (rotations; 3 parameters). (x’, y’, z’, t’)

4. Time is uniform (translation; 1 parameter).
5. Space is continuous (down to ∼10−18 [m]). O O’ v x’
x
6. Time is continuous (down to ∼10−26 [s]).

7. Space is Euclidean (apart from local distortions,
whichweignore; cosmologicalobservationsputthe z z’
limitat∼1026 [m], thesizeofthevisibleUniverse;
FIG. 2: (Color online.) Two inertial reference frames (or-
this property is what makes rectangular grids of
thogonal grids of rods equipped with synchronized clocks) in
rods possible).
relative motion along the x-axis.
8. Relativity Principle (boosts; 3 parameters).
3
Step 2: Isotropy and homogeneity of space and Step 5: Inversion x˜=−x, y˜=−y, and x˜(cid:48) =−x(cid:48),
homogeneity of time y˜(cid:48) =−y(cid:48)
...imply that (i) x(cid:48) is independent of y and z, (ii) y(cid:48)
is independent of z, x, and t; (iii) z(cid:48) is independent of x, ~ ~
y, and t; (iv) t(cid:48) is independent of y and z, so K K’
~ ~
P (x, y, z, t)
x(cid:48) = α (v)x+α (v)t, (5)
11 14
y(cid:48) = α (v)y, (6) ( x~’ , y~’ , z’, t’)
22
z(cid:48) = α33(v)z, (7)
t(cid:48) = α (v)x+α (v)t. (8)
41 44 ~ ~
x’ O O’ v = − v
NOTE: The fact that y(cid:48) and z(cid:48) are independent of x
and t follows from the requirement that the x(cid:48)-axis (the x~
line y(cid:48) = z(cid:48) = 0) always coincides with the x-axis (the
line y = z = 0); this would not be possible if y(cid:48) and z(cid:48)
depended on x and t. z z’

IMPORTANT: Eq. (8) indicates that it is possible to
have two spacially separated events A and B that are y~ y~’
simultaneous in frame K and, yet, non-simultaneous in
frame K(cid:48), that is
FIG.3: (Coloronline.) “Inverted”framesinrelativemotion.
∆t =0, ∆x (cid:54)=0: ∆t(cid:48) =α ∆x (cid:54)=0. (9)
AB AB AB 41 AB
...which is just a relabeling of coordinate marks, pre-
This is not as obvious as might seem: for example, be-
serves the right-handedness of the coordinate systems
foreEinsteinitwasassumedthatwhenever∆tAB iszero, and is physically equivalent to (inverted) frame K˜(cid:48) mov-
∆t(cid:48) must also be zero. So keeping α (v) in (8) is a
AB 41 ing with velocity v˜ = −v relative to (inverted) frame K˜
significant departure from classical Newtonian mechan-
(see Fig. 3), so that
ics.
Once the standard method of clock synchronization is x˜(cid:48) = α(−v)(x˜−vt), (18)
adopted,itis,however,relativelyeasytogiveanexample
y˜(cid:48) = k(−v)y˜, (19)
of two events satisfying (9). Try that on your own!
z(cid:48) = k(−v)z, (20)
t(cid:48) = δ(−v)x˜+γ(−v)t, (21)
Step 3: Isotropy of space
or,
...alsoimpliesthaty(cid:48) andz(cid:48) arephysicallyequivalent,
so that α (v)=α (v)≡k(v), and thus −x(cid:48) = α(−v)(−x−vt), (22)
22 33
−y(cid:48) = −k(−v)y, (23)
x(cid:48) = α (v)x+α (v)t, (10)
11 14
z(cid:48) = k(−v)z, (24)
y(cid:48) = k(v)y, (11)
t(cid:48) = −δ(−v)x+γ(−v)t, (25)
z(cid:48) = k(v)z, (12)
t(cid:48) = α41(v)x+α44(v)t. (13) which gives
α(−v) = α(v), (26)
Step 4: Motion of O(cid:48) (the origin of frame K(cid:48))
k(−v) = k(v), (27)
δ(−v) = −δ(v), (28)
...asseenfromK givesx =vt , orx −vt =0.
O(cid:48) O(cid:48) O(cid:48) O(cid:48)
On the other hand, as seen from K(cid:48), x(cid:48) = 0. For this γ(−v) = γ(v). (29)
O(cid:48)
to be possible, we must have x(cid:48) ∝(x−vt), and thus
x(cid:48) = α(v)(x−vt), (14) Step 6: Relativity principle and isotropy of space
y(cid:48) = k(v)y, (15)
z(cid:48) = k(v)z, (16) ...tellusthatthevelocityofK relativetoK(cid:48),asmea-
sured by K(cid:48) using primed coordinates (x(cid:48),t(cid:48)), is equal to
t(cid:48) = δ(v)x+γ(v)t, (17)
−v. REMINDER: the velocity of K(cid:48) relative to K, as
where we have re-labeled α ≡δ and α ≡γ. measured by K using unprimed coordinates (x,t), is v.
41 44
NOTE: The γ just introduced will soon become the I justify this by considering two local observers co-
celebrated gamma factor. moving with O and O(cid:48), respectively, and firing identical
4
spring guns in opposite directions at the moment when and
theypasseachother(foramoreformalapproach,see[3]).
If the ball shot in the +x direction by O stays next to x = γ(v)(x(cid:48)+vt(cid:48)), (49)
O(cid:48) then,bytherelativityprincipleandisotropyofspace, y = y(cid:48), (50)
the ball shot in the −x(cid:48) direction by O(cid:48) should stay next z = z(cid:48), (51)
toO. ThismeansthatthevelocityofO relativetoO(cid:48) as
measured by K(cid:48) is negative of the velocity of O(cid:48) relative t = −δ(v)x(cid:48)+γ(v)t(cid:48), (52)
to O as measured by K. Thus,
or, in matrix notation,
x = α(−v)(x(cid:48)+vt(cid:48)), (30) (cid:20)x(cid:48)(cid:21) (cid:20)γ(v) −vγ(v)(cid:21)(cid:20)x(cid:21)
y = k(−v)y(cid:48), (31) t(cid:48) = δ(v) γ(v) t , (53)
z = k(−v)z(cid:48), (32)
and
t = δ(−v)x(cid:48)+γ(−v)t(cid:48), (33)
(cid:20)x(cid:21) (cid:20) γ(v) +vγ(v)(cid:21)(cid:20)x(cid:48)(cid:21)
= . (54)
and since t −δ(v) γ(v) t(cid:48)
y =k(−v)y(cid:48) =k(−v)k(v)y =k2(v)y, (34)
Step 8: The odd function δ(v)
we get
...can be written as
k(v)=±1. (35)
δ(v)=−vf(v2)γ(v), (55)
Choosing k(v)=+1, which corresponds to parallel rela-
tiveorientationofy andy(cid:48) (aswellasofz andz(cid:48)), gives, since γ(v) is even. [NOTE: The newly introduced func-
for the direct transformation, tion f of v2 will turn out to be a constant! Actually, one
ofthegoalsoftheremainingstepsofthisderivationisto
x(cid:48) = α(v)(x−vt), (36) show that f is a constant. It will later be identifies with
y(cid:48) = y, (37) 1/c2.] Therefore,
z(cid:48) = z, (38) (cid:20)x(cid:48)(cid:21) (cid:20) 1 −v(cid:21)(cid:20)x(cid:21)
t(cid:48) = δ(v)x+γ(v)t, (39) t(cid:48) =γ −vf 1 t , (56)
and, for the inverse transformation, and
x = α(v)(x(cid:48)+vt(cid:48)), (40) (cid:20)x(cid:21) (cid:20) 1 v(cid:21)(cid:20)x(cid:48)(cid:21)
=γ . (57)
y = y(cid:48), (41) t vf 1 t(cid:48)
z = z(cid:48), (42)
t = −δ(v)x(cid:48)+γ(v)t(cid:48). (43) Step 9: Lorentz transformation followed by its
inverse must give the identity transformation
Step 7: Motion of O (the origin of frame K) This seems physically reasonable. We have,
(cid:20)x(cid:48)(cid:21) (cid:20) 1 −v(cid:21)(cid:20)x(cid:21)
...as seen from K gives xO = 0; also, as seen from t(cid:48) = γ −vf 1 t
K(cid:48), it gives x(cid:48) =α(v)(x −vt )=−vα(v)t and t(cid:48) =
δ(v)xO+γ(v)tOO =γ(v)tOO. FromOthis, xt(cid:48)O(cid:48)O =−Ovαγ((vv)). OBut = γ2(cid:20)−1vf −1v(cid:21)(cid:20)v1f v1(cid:21)(cid:20)xt(cid:48)(cid:48)(cid:21)
xt(cid:48)O(cid:48)O =−v, which gives = γ2(cid:20)1−v2f 0 (cid:21)(cid:20)x(cid:48)(cid:21), (58)
0 1−v2f t(cid:48)
α(v)=γ(v). (44)
and thus
As a result,
γ2(1−v2f)=1, (59)
x(cid:48) = γ(v)(x−vt), (45)
y(cid:48) = y, (46) from where
z(cid:48) = z, (47) 1
γ =± . (60)
t(cid:48) = δ(v)x+γ(v)t, (48) (cid:112)1−v2f
5
To preserve the parallel orientation of the x and x(cid:48) axes Squaring and rearranging give
wehavetochoosetheplussign(ascanbeseenbytaking
the v →0 limit), so that (cid:18) v+v(cid:48) (cid:19)2
(v(cid:48)(cid:48))2 = , (68)
1+vv(cid:48)f
1
γ = . (61)
(cid:112)1−v2f or,
Thus, v+v(cid:48)
v(cid:48)(cid:48) =± . (69)
(cid:20)x(cid:48)(cid:21) 1 (cid:20) 1 −v(cid:21)(cid:20)x(cid:21) 1+vv(cid:48)f
= , (62)
t(cid:48) (cid:112)1−v2f −vf 1 t Choosing the plus sign (to make sure that v(cid:48)(cid:48) = v when
v(cid:48) =0), we get
and
(cid:20)x(cid:21) 1 (cid:20) 1 v(cid:21)(cid:20)x(cid:48)(cid:21) v+v(cid:48)
= , (63) v(cid:48)(cid:48) = . (70)
t (cid:112)1−v2f vf 1 t(cid:48) 1+vv(cid:48)f
where, we recall, f =f(v2).
Step 12: The universal constant f cannot be
negative ...
Step 10: Two Lorentz transformations performed in
succession is a Lorentz transformation
...because in that case the conclusions of relativistic
dynamics would violate experimental observations! For
This step is crucial for everything that we’ve been do- example, the force law,
ingsofar,foritshowsthatf isaconstant,whichwillbe
identified with 1/c2, where c is Nature’s limiting speed. d mvp =F, (71)
(cid:113)
Wehaveasequenceoftwotransformations: from(x,t) dt 1−v2f
to (x(cid:48),t(cid:48)), and then from (x(cid:48),t(cid:48)) to (x(cid:48)(cid:48),t(cid:48)(cid:48)), p
(cid:20)x(cid:48)(cid:48)(cid:21) (cid:20) 1 −v(cid:48)(cid:21)(cid:20)x(cid:48)(cid:21) where vp is the velocity of a particle, would get messed
t(cid:48)(cid:48) = γ(cid:48) −v(cid:48)f(cid:48) 1 t(cid:48) up. In particular, such law would violate the observed
fact that it requires an infinite amount of work (and,
= γ(cid:48)(cid:20) 1 −v(cid:48)(cid:21)γ(cid:20) 1 −v(cid:21)(cid:20)x(cid:21) thus, energy) to accelerate a material particle from rest
−v(cid:48)f(cid:48) 1 −vf 1 t tospeedsapproaching3×108 [m/s](thisargumentisdue
(cid:20) 1+vv(cid:48)f −(v+v(cid:48))(cid:21)(cid:20)x(cid:21) toTerletskii[4]). Infact,itwouldbecome“easier”toac-
= γ(cid:48)γ , (64)
−(vf +v(cid:48)f(cid:48)) 1+vv(cid:48)f(cid:48) t celeratetheparticle,thefasteritismoving. Incidentally,
this experimental fact is what “replaces” Einstein’s Sec-
where v is the velocity of K(cid:48) relative to K (as measured ondPostulateinthepresentderivation! Thus,Eq.(70)is
in K using the (x,t) coordinates), and v(cid:48) is the velocity thelimittowhichour(actually, Ignatowski’s)derivation
ofK(cid:48)(cid:48) relativetoK(cid:48) (asmeasuredinK(cid:48) usingthe(x(cid:48),t(cid:48)) can be pushed.
coordinates). But this could also be written as a single REMARK: Relativistic dynamics has to be discussed
transformation from (x,t) to (x(cid:48)(cid:48),t(cid:48)(cid:48)), separately. We won’t do that here, but maybe you can
(cid:20)x(cid:48)(cid:48)(cid:21) (cid:20) 1 −v(cid:48)(cid:48)(cid:21)(cid:20)x(cid:21) suggest a different reason for f not to be negative? See
= γ(cid:48)(cid:48) , (65) [3] and [5] for possible approaches.
t(cid:48)(cid:48) −v(cid:48)(cid:48)f(cid:48)(cid:48) 1 t
with v(cid:48)(cid:48) being the velocity of K(cid:48)(cid:48) relative to K (as mea-
suredinK usingthe(x,t)coordinates). Thisshowsthat Step 13: Existence of the limiting speed
the (1, 1) and (2, 2) elements of the transformation ma-
trix (64) must be equal to each other and, thus, Denoting
f =f(cid:48), (66) 1
f ≡ , (72)
whichmeansthatf isaconstantthathasunitsofinverse c2
speed squared, [s2/m2]. Wow!! we get the velocity addition formula,
v+v(cid:48)
v(cid:48)(cid:48) = . (73)
Step 11: Velocity addition formula (for reference 1+ vv(cid:48)
frames) c2
If we begin with v(cid:48) < c and attempt to take the limit
To derive the velocity addition formula (along the x- v(cid:48) →c, we’ll get
axis) we use Eqs. (64) and (65) to get
v+c
v(cid:48)(cid:48) → =c, (74)
γ(cid:48)γ(v+v(cid:48))=γ(cid:48)(cid:48)v(cid:48)(cid:48). (67) 1+ vc
c2
6
whichtellsusthatcisthelimitingspeedthatamaterial photons exist. Similarly, tachyons may also exist and,
object can attain. (Notice that “material” here means likephotons,(a)shouldbecreatedinstantaneously(that
“theonewithwhichaninertialframecanbeassociated”. is, can’t be created at rest, and then accelerated), and
The photons do not fall into this category, as will be (b) should not be allowed to form a “legitimate” inertial
discussed shortly!) reference frame.
The possibilities therefore are: What about violation of causality?
1. c=+∞ (Newtonian mechanics; contradicts (71));
t
t’
2. c>0 and finite (Special Relativity); B

3. c=0 (Contradics observations.) A
t’
So we stick with option 2. A
x’
Whatifanobjectwerecreatedtohavev(cid:48) >cfromthe

start(aso-calledtachyon),likeintherecentsuperlumina l t’ < t’
B A
neutrino controversy? We’d get some strange results.
For example, if we take v =c/2 and v(cid:48) =2c, we get
x
(c/2)+(2c) 5
v(cid:48)(cid:48) = = c, (75)
1+ (c/2)(2c) 4
c2

so in K the object would move to the right at a slower
speed than relative to K(cid:48), while K(cid:48) itself is moving to
the right relative to K. Bizarre, but OK, the two speeds
are measured by different observers, so maybe it’s not a
FIG. 4: (Color online.) Violation of causality by a hypothet-
big deal ....
ical tachyon.
Step 14: Lorentz transformation in standard form Indeed, the Lorentz transformation shows that
tachyons violate causality. If we consider two events,
A (tachyon creation) and B (tachyon annihilation) with
However, ifweconsidertheresultingLorentztransfor-
mation, tB > tA such that tachyon’s speed, vp = xtBB−−txAA > c,
is greater than c as measured in K (see Fig. 4), then in
(cid:20)x(cid:48)(cid:21) 1 (cid:20) 1 −v(cid:21)(cid:20)x(cid:21) frame K(cid:48) moving with velocity v <c relative to K we’ll
= , (76)
t(cid:48) (cid:113) −v 1 t have from (77) and (78),
1− v2 c2
c2
(cid:104) v (cid:105)
or, t(cid:48) −t(cid:48) = γ (t −t )− (x −x )
B A B A c2 B A
(cid:20) (cid:21)
t(cid:48) = (cid:113)t− cv2x , (77) = γ 1− cv2xtB −−xtA (tB −tA)
1− v2 B A
c2 = γ(cid:16)1− vvp(cid:17)(t −t ), (81)
x−vt c2 B A
x(cid:48) = , (78)
(cid:113)1− v2 x(cid:48)B −x(cid:48)A = γ[(xB −xA)−v(tB −tA)]
c2 (cid:20) (cid:21)
t −t
y(cid:48) = y, (79) = γ 1−vxB −xA (xB −xA)
B A
z(cid:48) = z, (80) (cid:18) (cid:19)
v
= γ 1− (x −x ), (82)
we notice that in a reference frame K(cid:48) associated with vp B A
hypothetical tachyons moving with v > c relative to K
(cid:112)
(imagine a whole fleet of them, forming a grid which where γ =1/ 1−v2/c2, which shows thatit is possible
makes up K(cid:48)), the spacetime coordinates of any event to find v <c such that t(cid:48) −t(cid:48) <0; that is, in K(cid:48) event
B A
wouldbeimaginary! Inorderforthespacetimemeasure- B happens before event A. This seems to indicate that
ments to give real values for (t(cid:48),x(cid:48),y(cid:48),z(cid:48)), the reference tachyons are impossible. However, causality is a conse-
frame K(cid:48) made of tachyons must be rejected. quence of the Second Law of Thermodynamics, which
What about a reference frame made of photons? In is a statistical law, applicable to macroscopic systems;
that case, coordinates would be infinite and should also it does not apply to processes involving individual ele-
be rejected. So a fleet of photons cannot form a “le- mentary particles. As a result, the existence of tachyons
gitimate” reference frame. Nevertheless, we know that cannot be so easily ruled out.
7
Step 15: Speed of light is the limiting speed for Eq.(85)saysthatt(cid:48) −t(cid:48) <0,thatis,themeetingevents
B A
material objects are not simultaneous in K(cid:48) (relativity of simultaneity).
Eq. (86) says that the length of the rod, (cid:96)≡x −x , as
B A
Finally,returningtoEq.(74),weseethatifsomething measured in K is smaller than its proper length by the
moves with c relative to K(cid:48), it also moves with c rela- gamma factor,
tive to any other frame K(cid:48)(cid:48). That is: the limiting speed
is the same in all inertial reference frames. And there (cid:96)=(cid:96)0/γ, (87)
is no mentioning of any emitter. Also, as follows from
(73), c is the only speed that has this property (of being the phenomenon of length contraction.
the same in all inertial frames). We know that light has
this property (ala Michelson-Morley experiment), so the
speed of light is the limiting speed for material objects. B. Time dilation
Sinceneutrinoshavemass,theycannotmovefasterthan
light, and thus superluminal neutrinos are not possible.

Same moving clock

t’ = 0 t’ < t
A B B
Immediate consequences of the Lorentz
transformation x’ x’ v

A. Length contraction and relativity of
simultaneity
t = 0 t > 0 x
A B
E vent A E vent B
( = O, for simplicity)
t’ = 0 t’ < t’
A B A

l0 x’ v FIG. 6: (Color online.) Time dilation.

l = l0 /γ < l0 ThistimeasingleclockbelongingtoK(cid:48),Fig.6,passes
by two different clocks in K. The corresponding two
t = 0 t = 0 x
A B events, A and B, have x(cid:48) =x(cid:48) , and are related to each
A B
E vent A E vent B other by
( = O, for simplicity)
(cid:104) v (cid:105)
t(cid:48) −t(cid:48) = γ (t −t )− (x −x )
FIG. 5: (Color online.) Length contraction and relativity of B A B A c2 B A
(cid:20) (cid:21)
simultaneity. = γ 1− v xB −xA (t −t )
c2 t −t B A
B A
Herewehavearodof(proper)length(cid:96)0 ≡x(cid:48)B−x(cid:48)A >0 (cid:18) v2(cid:19)
sitting at rest in frame K(cid:48), see Fig. 5. Its speed relative = γ 1− c2 (tB −tA)
to frame K is v. The two events, A and B, represent the
t −t
meetings of the two clocks at the ends of the rod with = B A. (88)
γ
the corresponding clocks in the K frame at t =t . We
A B
have from (77) and (78),
This means that upon arrival at B the moving clock will
(cid:16) v (cid:17) read less time than the K-clock sitting at that location.
t(cid:48) −t(cid:48) = γ − (x −x ), (83)
B A c2 B A This phenomenon is called time dilation (moving clocks
x(cid:48) −x(cid:48) = γ(x −x ), (84) run slower).
B A B A
or
Acknowledgments
(cid:16) v (cid:17)
t(cid:48) −t(cid:48) = − (x(cid:48) −x(cid:48) ), (85)
B A c2 B A
I thank Todd Baker, Amara Katabarwa, and Loris
x(cid:48) −x(cid:48)
x −x = B A. (86) Magnani for helpful discussions.
B A γ
[1] T.A.Moore,Six Ideas That Shaped Physics. Unit R: The (McGraw Hill, 2003).
Laws of Physics Are Frame Independent, Secon Edition [2] W. von Ignatowsky, Das Relativita¨tsprinzip, Archiv der
8
Mathematik und Physik 17, 124 (1910); ibid. 18, 17-40 [4] Y. P. Terletskii, Paradoxes in the Theory of Relativity
(1911). (Plenum, New York, 1968).
[3] V. Berzi and V. Gorini, “Reciprocity Principle and [5] N. D. Mermin, “Relativity without light,” Am. J. Phys.
the Lorentz Transformations,” J. Math. Phys. 10, 1518 52 (2), 119 (1984).
(1969).