Derivation of the Lorentz transformation without the use of Einstein's second postulate
DDerivation of the Lorentz transformation without the use of Einstein’s secondpostulate
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 firstdelivered in PHYS 4202 E&M class during the Spring semester of 2014 at the University of Georgia.The main motivation for pursuing this approach was to develop a better understanding of why thefaster-than-light neutrino controversy (OPERA experiment, 2011) was much ado about nothing.
Special relativity as a theory of space and time
All physical phenomena take place in space and time.The theory of space and time (in the absence of gravity)is called the Special Theory of Relativity.We do not get bogged down with the philosophicalproblems related to the concepts of space and time. Wesimply acknowledge the fact that in physics the notionsof space and time are regarded as basic and cannot bereduced to something more elementary or fundamental.We therefore stick to pragmatic operational definitions:
Time is what clocks measure. Space is what measuringrods measure.
In order to study and make conclusions about the prop-erties of space and time we need an observer. A naturalchoice is an observer who moves freely (the one who isfree from any external influences). An observer is not asingle person sitting at the origin of a rectangular coor-dinate grid. Rather, it is a bunch of friends (call it Team K ) equipped with identical clocks distributed through-out the grid who record the events happening at theirrespective locations.How do we know that this bunch of friends is freefrom any external influences? We look around and makesure that nothing is pulling or pushing on any memberof the bunch; no strings, no springs, no ropes are at-tached to them. An even better way is to use a collectionof “floating-ball detectors” (Fig. 1) distributed through-out the grid [1]. When detector balls are released, theyshould remain at rest inside their respective capsules. Ifany ball touches the touch-sensitive surface of the cap-sule, the frame is not inertial.In the reference frame associated with a freely movingobserver (our rectangular coordinate grid), Galileo’s Lawof Inertia is satisfied: a point mass, itself free from anyexternal influences, moves with constant velocity. To beable to say what “constant velocity” really means, andthus to verify the law of inertia, we need to be able tomeasure distances and time intervals between events hap-pening at different grid locations. Floating Ball Capsule Touch-sensitive surface
FIG. 1: (Color online.) A floating-ball inertial detector. After[1].
Definition of clock synchronization
It is pretty clear how to measure distances: the teamsimply uses its rectangular grid of rods.It is also clear how to measure time intervals at a par-ticular location: the team member situated at that loca-tion simply looks at his respective clock. What’s not soclear, however, is how the team measures time intervalsbetween events that are spacially separated.A confusion about measuring this kind of time inter-vals was going on for two hundred years or so, until oneday Einstein said: “We need the notion of synchronizedclocks! Clock synchronization must be operationally de-fined.”The idea that clock synchronization and, consequently,the notion of simultaneity of spacially separated events,has to be defined (and not assumed apriori ) is the singlemost important idea of Einstein’s, the heart of specialrelativity. Einstein proposed to use light pulses. Theprocedure then went like this:In frame K , consider two identical clocks equippedwith light detectors, sitting some distance apart, at A and B . Consider another clock equipped with a lightemitter at location C which is half way between A and B (we can verify that C is indeed half way between A and a r X i v : . [ phy s i c s . c l a ss - ph ] J a n B with the help of the grid of rods that had already beenput in place when we constructed our frame K ). Then,at some instant, emit two pulses from C in opposite di-rections, and let those pulses arrive at A and B . If theclocks at A and B show same time when the pulses arrivethen the clocks there are synchronized, by definition .The light pulses used in the synchronization procedurecan be replaced with two identical balls initially sitting at C and connected by a compressed spring. The spring isreleased (say, the thread holding the spring is cut in themiddle), the balls fly off in opposite directions towards A and B , respectively.How do we know that the balls are identical? BecauseTeam K made them in accordance with a specific manu-facturing procedure.How do we know that all clocks at K are identical?Because Team K made all of them in accordance with aspecific manufacturing procedure.How do we know that a tic-toc of any clock sittingin frame K corresponds to 1 second? Because Team K called a tic-toc of a clock made in accordance with themanufacturing procedure “a second”.Similarly, clocks in K (cid:48) are regarded as identical andtick-tocking at 1 second intervals because in that frameall of the clocks were made in accordance with the samemanufacturing procedure.Now, how do we know that the manufacturing proce-dures in K and K (cid:48) are the same? (Say, how do we knowthat a Swiss shop in frame K makes watches the sameway as its counterpart in frame K (cid:48) ?) Hmm. . . . That’san interesting question to ponder about.When studying spacetime from the point of view ofinertial frames of reference discussed above, people dis-covered the following. Properties of space and time:
1. At least one inertial reference frame exists. (Geo-centric is OK for crude experiments; geliocentric isbetter; the frame in which microwave backgroundradiation is uniform is probably closest to ideal).2. Space is uniform (translations; 3 parameters).3. Space is isotropic (rotations; 3 parameters).4. Time is uniform (translation; 1 parameter).5. Space is continuous (down to ∼ − [m]).6. Time is continuous (down to ∼ − [s]).7. Space is Euclidean (apart from local distortions,which we ignore; cosmological observations put thelimit at ∼ [m], the size of the visible Universe;this property is what makes rectangular grids ofrods possible).8. Relativity Principle (boosts; 3 parameters). Einstein constructed his theory of relativity on the ba-sis of (1) The Principle of Relativity (laws of nature arethe same in all inertial reference frames), and (2) ThePostulate of the Constancy of the Speed of Light (thespeed of light measured by any inertial observer is in-dependent of the state of motion of the emitting body).[NOTE: This is not the same as saying that the speedof light emitted and measured in K is the same as thespeed of light emitted and measured in K (cid:48) . This lattertype of constancy of the speed of light is already impliedby the principle of relativity.]Here we want to stick to mechanics and push thederivation of the coordinate transformation as far aspossible without the use of the highly counterintuitiveEinstein’s Second Postulate. The method that achievesthis will be presented below and was originally due toVladimir Ignatowsky [2].[ DISCLAIMER:
I never read Ignatowsky’s originalpapers, but the idea is well-known within the community,often mentioned and discussed. Anyone with time toburn can reproduce the steps without much difficulty.The derivation below consists of 14 steps. If you canreduce that number, let me know.]
Step 1: Galileo’s Law of Inertia for freely movingparticles . . . implies the linearity of the coordinate transforma-tion between K and K (cid:48) (see Fig. 2), x (cid:48) = α ( v ) x + α ( v ) y + α ( v ) z + α ( v ) t, (1) y (cid:48) = α ( v ) x + α ( v ) y + α ( v ) z + α ( v ) t, (2) z (cid:48) = α ( v ) x + α ( v ) y + α ( v ) z + α ( v ) t, (3) t (cid:48) = α ( v ) x + α ( v ) y + α ( v ) z + α ( v ) t. (4)Here we assumed that the origins of the two coordinatesystems coincide, that is event (0 , , ,
0) in K has coor-dinates (0 , , ,
0) in K (cid:48) . K K ’ ( x ’ , y ’ , z ’, t ’ ) ( x , y , z , t ) P v O ’ O x ’ y ’ z ’ y x z FIG. 2: (Color online.) Two inertial reference frames (or-thogonal grids of rods equipped with synchronized clocks) inrelative motion along the x -axis. Step 2: Isotropy and homogeneity of space andhomogeneity of time . . . 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 x (cid:48) = α ( v ) x + α ( v ) t, (5) y (cid:48) = α ( v ) y, (6) z (cid:48) = α ( v ) z, (7) t (cid:48) = α ( v ) x + α ( v ) t. (8)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 (theline y (cid:48) = z (cid:48) = 0) always coincides with the x -axis (theline y = z = 0); this would not be possible if y (cid:48) and z (cid:48) depended on x and t .IMPORTANT: Eq. (8) indicates that it is possible tohave two spacially separated events A and B that aresimultaneous in frame K and, yet, non-simultaneous inframe K (cid:48) , that is∆ t AB = 0 , ∆ x AB (cid:54) = 0 : ∆ t (cid:48) AB = α ∆ x AB (cid:54) = 0 . (9)This is not as obvious as might seem: for example, be-fore Einstein it was assumed that whenever ∆ t AB is zero,∆ t (cid:48) AB must also be zero. So keeping α ( v ) in (8) is asignificant departure from classical Newtonian mechan-ics.Once the standard method of clock synchronization isadopted, it is, however, relatively easy to give an exampleof two events satisfying (9). Try that on your own! Step 3: Isotropy of space . . . also implies that y (cid:48) and z (cid:48) are physically equivalent,so that α ( v ) = α ( v ) ≡ k ( v ), and thus x (cid:48) = α ( v ) x + α ( v ) t, (10) y (cid:48) = k ( v ) y, (11) z (cid:48) = k ( v ) z, (12) t (cid:48) = α ( v ) x + α ( v ) t. (13) Step 4: Motion of O (cid:48) (the origin of frame K (cid:48) ) . . . as seen from K gives x O (cid:48) = vt O (cid:48) , or x O (cid:48) − vt O (cid:48) = 0.On the other hand, as seen from K (cid:48) , x (cid:48) O (cid:48) = 0. For thisto be possible, we must have x (cid:48) ∝ ( x − vt ), and thus x (cid:48) = α ( v )( x − vt ) , (14) y (cid:48) = k ( v ) y, (15) z (cid:48) = k ( v ) z, (16) t (cid:48) = δ ( v ) x + γ ( v ) t, (17)where we have re-labeled α ≡ δ and α ≡ γ .NOTE: The γ just introduced will soon become thecelebrated gamma factor. Step 5: Inversion ˜ x = − x , ˜ y = − y , and ˜ x (cid:48) = − x (cid:48) , ˜ y (cid:48) = − y (cid:48) K K ’ ( x ’ , y ’ , z ’, t ’ ) ( x , y , z , t ) P v = − v O ’ O x ’ y ’ z ’ y x z ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ FIG. 3: (Color online.) “Inverted” frames in relative motion. . . . which is just a relabeling of coordinate marks, pre-serves the right-handedness of the coordinate systemsand is physically equivalent to (inverted) frame ˜ K (cid:48) mov-ing with velocity ˜ v = − v relative to (inverted) frame ˜ K (see Fig. 3), so that˜ x (cid:48) = α ( − v )(˜ x − vt ) , (18)˜ y (cid:48) = k ( − v )˜ y, (19) z (cid:48) = k ( − v ) z, (20) t (cid:48) = δ ( − v )˜ x + γ ( − v ) t, (21)or, − x (cid:48) = α ( − v )( − x − vt ) , (22) − y (cid:48) = − k ( − v ) y, (23) z (cid:48) = k ( − v ) z, (24) t (cid:48) = − δ ( − v ) x + γ ( − v ) t, (25)which gives α ( − v ) = α ( v ) , (26) k ( − v ) = k ( v ) , (27) δ ( − v ) = − δ ( v ) , (28) γ ( − v ) = γ ( v ) . (29) Step 6: Relativity principle and isotropy of space . . . tell us that the velocity of K relative to K (cid:48) , as mea-sured by K (cid:48) using primed coordinates ( x (cid:48) , t (cid:48) ), is equal to − v . REMINDER: the velocity of K (cid:48) relative to K , asmeasured by K using unprimed coordinates ( x, t ), is v .I justify this by considering two local observers co-moving with O and O (cid:48) , respectively, and firing identicalspring guns in opposite directions at the moment whenthey pass each other (for a more formal approach, see [3]).If the ball shot in the + x direction by O stays next to O (cid:48) then, by the relativity principle and isotropy of space,the ball shot in the − x (cid:48) direction by O (cid:48) should stay nextto O . This means that the velocity of O relative to O (cid:48) asmeasured by K (cid:48) is negative of the velocity of O (cid:48) relativeto O as measured by K . Thus, x = α ( − v )( x (cid:48) + vt (cid:48) ) , (30) y = k ( − v ) y (cid:48) , (31) z = k ( − v ) z (cid:48) , (32) t = δ ( − v ) x (cid:48) + γ ( − v ) t (cid:48) , (33)and since y = k ( − v ) y (cid:48) = k ( − v ) k ( v ) y = k ( v ) y, (34)we get k ( v ) = ± . (35)Choosing k ( v ) = +1, which corresponds to parallel rela-tive orientation of y and y (cid:48) (as well as of z and z (cid:48) ), gives,for the direct transformation, x (cid:48) = α ( v )( x − vt ) , (36) y (cid:48) = y, (37) z (cid:48) = z, (38) t (cid:48) = δ ( v ) x + γ ( v ) t, (39)and, for the inverse transformation, x = α ( v )( x (cid:48) + vt (cid:48) ) , (40) y = y (cid:48) , (41) z = z (cid:48) , (42) t = − δ ( v ) x (cid:48) + γ ( v ) t (cid:48) . (43) Step 7: Motion of O (the origin of frame K ) . . . as seen from K gives x O = 0; also, as seen from K (cid:48) , it gives x (cid:48) O = α ( v )( x O − vt O ) = − vα ( v ) t O and t (cid:48) O = δ ( v ) x O + γ ( v ) t O = γ ( v ) t O . From this, x (cid:48) O t (cid:48) O = − v α ( v ) γ ( v ) . But x (cid:48) O t (cid:48) O = − v , which gives α ( v ) = γ ( v ) . (44)As a result, x (cid:48) = γ ( v )( x − vt ) , (45) y (cid:48) = y, (46) z (cid:48) = z, (47) t (cid:48) = δ ( v ) x + γ ( v ) t, (48) and x = γ ( v )( x (cid:48) + vt (cid:48) ) , (49) y = y (cid:48) , (50) z = z (cid:48) , (51) t = − δ ( v ) x (cid:48) + γ ( v ) t (cid:48) , (52)or, in matrix notation, (cid:20) x (cid:48) t (cid:48) (cid:21) = (cid:20) γ ( v ) − vγ ( v ) δ ( v ) γ ( v ) (cid:21) (cid:20) xt (cid:21) , (53)and (cid:20) xt (cid:21) = (cid:20) γ ( v ) + vγ ( v ) − δ ( v ) γ ( v ) (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) . (54) Step 8: The odd function δ ( v ) . . . can be written as δ ( v ) = − vf ( v ) γ ( v ) , (55)since γ ( v ) is even. [NOTE: The newly introduced func-tion f of v will turn out to be a constant! Actually, oneof the goals of the remaining steps of this derivation is toshow that f is a constant. It will later be identifies with1 /c .] Therefore, (cid:20) x (cid:48) t (cid:48) (cid:21) = γ (cid:20) − v − vf (cid:21) (cid:20) xt (cid:21) , (56)and (cid:20) xt (cid:21) = γ (cid:20) vvf (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) . (57) Step 9: Lorentz transformation followed by itsinverse must give the identity transformation
This seems physically reasonable. We have, (cid:20) x (cid:48) t (cid:48) (cid:21) = γ (cid:20) − v − vf (cid:21) (cid:20) xt (cid:21) = γ (cid:20) − v − vf (cid:21) (cid:20) vvf (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) = γ (cid:20) − v f
00 1 − v f (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) , (58)and thus γ (1 − v f ) = 1 , (59)from where γ = ± (cid:112) − v f . (60)To preserve the parallel orientation of the x and x (cid:48) axeswe have to choose the plus sign (as can be seen by takingthe v → γ = 1 (cid:112) − v f . (61)Thus, (cid:20) x (cid:48) t (cid:48) (cid:21) = 1 (cid:112) − v f (cid:20) − v − vf (cid:21) (cid:20) xt (cid:21) , (62)and (cid:20) xt (cid:21) = 1 (cid:112) − v f (cid:20) vvf (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) , (63)where, we recall, f = f ( v ). Step 10: Two Lorentz transformations performed insuccession is a Lorentz transformation
This step is crucial for everything that we’ve been do-ing so far, for it shows that f is a constant, which will beidentified with 1 /c , where c is Nature’s limiting speed.We have a sequence of two transformations: from ( x, t )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) ), (cid:20) x (cid:48)(cid:48) t (cid:48)(cid:48) (cid:21) = γ (cid:48) (cid:20) − v (cid:48) − v (cid:48) f (cid:48) (cid:21) (cid:20) x (cid:48) t (cid:48) (cid:21) = γ (cid:48) (cid:20) − v (cid:48) − v (cid:48) f (cid:48) (cid:21) γ (cid:20) − v − vf (cid:21) (cid:20) xt (cid:21) = γ (cid:48) γ (cid:20) vv (cid:48) f − ( v + v (cid:48) ) − ( vf + v (cid:48) f (cid:48) ) 1 + vv (cid:48) f (cid:48) (cid:21) (cid:20) xt (cid:21) , (64)where v is the velocity of K (cid:48) relative to K (as measuredin K using the ( x, t ) coordinates), and v (cid:48) is the velocityof K (cid:48)(cid:48) relative to K (cid:48) (as measured in K (cid:48) using the ( x (cid:48) , t (cid:48) )coordinates). But this could also be written as a singletransformation from ( x, t ) to ( x (cid:48)(cid:48) , t (cid:48)(cid:48) ), (cid:20) x (cid:48)(cid:48) t (cid:48)(cid:48) (cid:21) = γ (cid:48)(cid:48) (cid:20) − v (cid:48)(cid:48) − v (cid:48)(cid:48) f (cid:48)(cid:48) (cid:21) (cid:20) xt (cid:21) , (65)with v (cid:48)(cid:48) being the velocity of K (cid:48)(cid:48) relative to K (as mea-sured in K using the ( x, t ) coordinates). This shows thatthe (1, 1) and (2, 2) elements of the transformation ma-trix (64) must be equal to each other and, thus, f = f (cid:48) , (66)which means that f is a constant that has units of inversespeed squared, [s / m ]. Wow!! Step 11: Velocity addition formula (for referenceframes)
To derive the velocity addition formula (along the x -axis) we use Eqs. (64) and (65) to get γ (cid:48) γ ( v + v (cid:48) ) = γ (cid:48)(cid:48) v (cid:48)(cid:48) . (67) Squaring and rearranging give( v (cid:48)(cid:48) ) = (cid:18) v + v (cid:48) vv (cid:48) f (cid:19) , (68)or, v (cid:48)(cid:48) = ± v + v (cid:48) vv (cid:48) f . (69)Choosing the plus sign (to make sure that v (cid:48)(cid:48) = v when v (cid:48) = 0), we get v (cid:48)(cid:48) = v + v (cid:48) vv (cid:48) f . (70) Step 12: The universal constant f cannot benegative . . . . . . because in that case the conclusions of relativisticdynamics would violate experimental observations! Forexample, the force law, ddt m v p (cid:113) − v p f = F , (71)where v p is the velocity of a particle, would get messedup. In particular, such law would violate the observedfact that it requires an infinite amount of work (and,thus, energy) to accelerate a material particle from restto speeds approaching 3 × [m/s] (this argument is dueto Terletskii [4]). In fact, it would become “easier” to ac-celerate the particle, the faster it is moving. Incidentally,this experimental fact is what “replaces” Einstein’s Sec-ond Postulate in the present derivation! Thus, Eq. (70) isthe limit to which our (actually, Ignatowski’s) derivationcan be pushed.REMARK: Relativistic dynamics has to be discussedseparately. We won’t do that here, but maybe you cansuggest a different reason for f not to be negative? See[3] and [5] for possible approaches. Step 13: Existence of the limiting speed
Denoting f ≡ c , (72)we get the velocity addition formula , v (cid:48)(cid:48) = v + v (cid:48) vv (cid:48) c . (73)If we begin with v (cid:48) < c and attempt to take the limit v (cid:48) → c , we’ll get v (cid:48)(cid:48) → v + c vcc = c, (74)which tells us that c is the limiting speed that a materialobject can attain. (Notice that “material” here means“the one with which an inertial frame can be associated”.The photons do not fall into this category, as will bediscussed shortly!)The possibilities therefore are:1. c = + ∞ (Newtonian mechanics; contradicts (71));2. c > c = 0 (Contradics observations.)So we stick with option 2.What if an object were created to have v (cid:48) > c from thestart (a so-called tachyon), like in the recent superluminalneutrino controversy? We’d get some strange results.For example, if we take v = c/ v (cid:48) = 2 c , we get v (cid:48)(cid:48) = ( c/
2) + (2 c )1 + ( c/ c ) c = 54 c, (75)so in K the object would move to the right at a slowerspeed than relative to K (cid:48) , while K (cid:48) itself is moving tothe right relative to K . Bizarre, but OK, the two speedsare measured by different observers, so maybe it’s not abig deal . . . . Step 14: Lorentz transformation in standard form
However, if we consider the resulting Lorentz transfor-mation, (cid:20) x (cid:48) t (cid:48) (cid:21) = 1 (cid:113) − v c (cid:20) − v − vc (cid:21) (cid:20) xt (cid:21) , (76)or, t (cid:48) = t − vc x (cid:113) − v c , (77) x (cid:48) = x − vt (cid:113) − v c , (78) y (cid:48) = y, (79) z (cid:48) = z, (80)we notice that in a reference frame K (cid:48) associated withhypothetical tachyons moving with v > c relative to K (imagine a whole fleet of them, forming a grid whichmakes up K (cid:48) ), the spacetime coordinates of any eventwould be imaginary! In order for the spacetime measure-ments to give real values for ( t (cid:48) , x (cid:48) , y (cid:48) , z (cid:48) ), the referenceframe K (cid:48) made of tachyons must be rejected.What about a reference frame made of photons? Inthat case, coordinates would be infinite and should alsobe rejected. So a fleet of photons cannot form a “le-gitimate” reference frame. Nevertheless, we know that photons exist. Similarly, tachyons may also exist and,like photons, (a) should be created instantaneously (thatis, can’t be created at rest, and then accelerated), and(b) should not be allowed to form a “legitimate” inertialreference frame.What about violation of causality? x ’ t ’ t x B A t ’ A t ’ B < t ’ A FIG. 4: (Color online.) Violation of causality by a hypothet-ical tachyon.
Indeed, the Lorentz transformation shows thattachyons violate causality. If we consider two events, A (tachyon creation) and B (tachyon annihilation) with t B > t A such that tachyon’s speed, v p = x B − x A t B − t A > c ,is greater than c as measured in K (see Fig. 4), then inframe K (cid:48) moving with velocity v < c relative to K we’llhave from (77) and (78), t (cid:48) B − t (cid:48) A = γ (cid:104) ( t B − t A ) − vc ( x B − x A ) (cid:105) = γ (cid:20) − vc x B − x A t B − t A (cid:21) ( t B − t A )= γ (cid:16) − vv p c (cid:17) ( t B − t A ) , (81) x (cid:48) B − x (cid:48) A = γ [( x B − x A ) − v ( t B − t A )]= γ (cid:20) − v t B − t A x B − x A (cid:21) ( x B − x A )= γ (cid:18) − vv p (cid:19) ( x B − x A ) , (82)where γ = 1 / (cid:112) − v /c , which shows that it is possibleto find v < c such that t (cid:48) B − t (cid:48) A <
0; that is, in K (cid:48) event B happens before event A . This seems to indicate thattachyons are impossible. However, causality is a conse-quence of the Second Law of Thermodynamics, whichis a statistical law, applicable to macroscopic systems;it does not apply to processes involving individual ele-mentary particles. As a result, the existence of tachyonscannot be so easily ruled out. Step 15: Speed of light is the limiting speed formaterial objects
Finally, returning to Eq. (74), we see that if somethingmoves with c relative to K (cid:48) , it also moves with c rela-tive to any other frame K (cid:48)(cid:48) . That is: the limiting speedis the same in all inertial reference frames. And thereis no mentioning of any emitter. Also, as follows from(73), c is the only speed that has this property (of beingthe same in all inertial frames). We know that light hasthis property (ala Michelson-Morley experiment), so thespeed of light is the limiting speed for material objects.Since neutrinos have mass, they cannot move faster thanlight, and thus superluminal neutrinos are not possible. Immediate consequences of the LorentztransformationA. Length contraction and relativity ofsimultaneity x ’ v x Event B Event A ( = O , for simplicity) t ’ A = 0 t ’ B < t ’ A t A = 0 t B = 0 l l = l /γ < l FIG. 5: (Color online.) Length contraction and relativity ofsimultaneity.
Here we have a rod of (proper) length (cid:96) ≡ x (cid:48) B − x (cid:48) A > K (cid:48) , see Fig. 5. Its speed relativeto frame K is v . The two events, A and B , represent themeetings of the two clocks at the ends of the rod withthe corresponding clocks in the K frame at t A = t B . Wehave from (77) and (78), t (cid:48) B − t (cid:48) A = γ (cid:16) − vc (cid:17) ( x B − x A ) , (83) x (cid:48) B − x (cid:48) A = γ ( x B − x A ) , (84)or t (cid:48) B − t (cid:48) A = (cid:16) − vc (cid:17) ( x (cid:48) B − x (cid:48) A ) , (85) x B − x A = x (cid:48) B − x (cid:48) A γ . (86) Eq. (85) says that t (cid:48) B − t (cid:48) A <
0, that is, the meeting eventsare not simultaneous in K (cid:48) ( relativity of simultaneity ).Eq. (86) says that the length of the rod, (cid:96) ≡ x B − x A , asmeasured in K is smaller than its proper length by thegamma factor, (cid:96) = (cid:96) /γ, (87)the phenomenon of length contraction . B. Time dilation x ’ v x Event B Event A ( = O , for simplicity) t ’ A = 0 t ’ B < t B t A = 0 t B > 0 x ’ Same moving clock
FIG. 6: (Color online.) Time dilation.
This time a single clock belonging to K (cid:48) , Fig. 6, passesby two different clocks in K . The corresponding twoevents, A and B , have x (cid:48) A = x (cid:48) B , and are related to eachother by t (cid:48) B − t (cid:48) A = γ (cid:104) ( t B − t A ) − vc ( x B − x A ) (cid:105) = γ (cid:20) − vc x B − x A t B − t A (cid:21) ( t B − t A )= γ (cid:18) − v c (cid:19) ( t B − t A )= t B − t A γ . (88)This means that upon arrival at B the moving clock willread less time than the K -clock sitting at that location.This phenomenon is called time dilation (moving clocksrun slower). Acknowledgments
I thank Todd Baker, Amara Katabarwa, and LorisMagnani for helpful discussions. [1] T. A. Moore,
Six Ideas That Shaped Physics. Unit R: TheLaws of Physics Are Frame Independent , Secon Edition (McGraw Hill, 2003).[2] W. von Ignatowsky, Das Relativit¨atsprinzip, Archiv der
Mathematik und Physik , 124 (1910); ibid. , 17-40(1911).[3] V. Berzi and V. Gorini, “Reciprocity Principle andthe Lorentz Transformations,” J. Math. Phys. , 1518(1969). [4] Y. P. Terletskii, Paradoxes in the Theory of Relativity (Plenum, New York, 1968).[5] N. D. Mermin, “Relativity without light,” Am. J. Phys.52