Basic Theory of Relativity

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Basic Theory of RelativityBy Yedidia Neumeier October 2008

Introduction

Why studying relativity?

TBC

Simultaneous events and synchronized clocks

I find it best to start with the following quote from Einstein famous 1905 paper1We have to take into account that all our judgments in which time plays a part are always

judgments ofsimultaneous events. If, for instance, I say, That train arrives here at 7 o'clock, I

mean something like this: The pointing of the small hand of my watch to 7 and the arrival of thetrain are simultaneous events.

And I would like to follow with yet another quote from the same paper.

If at the point A of space there is a clock, an observer at A can determine the time values of events

in the immediate proximity of A by finding the positions of the hands which are simultaneous withthese events. If there is at the point B of space another clock in all respects resembling the one at

A, it is possible for an observer at B to determine the time values of events in the immediate

neighbourhood of B. But it is not possible without further assumption to compare, in respect oftime, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.''

We have not defined a common ``time'' for A and B,

The above two quotes are easily grasped by our intuition. To farther our notion to inclusesimultaneous events in distant location I would like to take a step back and start by first define

what do we mean by time. Time is ultimately counting of periodic events. These periodic events

may be sun rise that count days, a pendulum swings, or vibration in a crystal. For our use we

invent a useful clock. It is made of a solid rod with perfect mirrors at the edges and a short lightpulse bouncing between them. For time keeping there is a counter that counts the bouncing events

on one of the mirrors. We imagine that we can reset the counter to zero at our desire. Consider

now that we have produced two such identical clocks. We assume that the length of the rods isvery small so that we can measure really short time intervals. We now leave Clock A at home

and take clock B to a large distance very large being that the rod length is insignificant with

respect to the distance between the clock. We set clock A to zero and simultaneously we send alight pulse to clock B. When the light arrived at clock B it set the counter of B to zero and bounce

back to A. When the pulse arrived back at clock A the counter reads, say travelt we now reset the

counter of A to 0t such that 0 2travelt t= . Clocks A and B are now said to be synchronized, the a)

clock in the same rate and, b) their counters have synchronized with a known time interval event.If now we have that an airplane passed over A at 12At = and a car passed by B at 12Bt = we saythat the two events are simultaneous. It should be noted that the above outlined process hold trueonly when both point A and B belong to the same Galilean system, meaning that within a frame

system which is in a uniform (non accelerating) motion the two points posses no relative velocity,

namely they are stationary with respect to each other. We will see that simultaneous events are notpossible in two separated points which are moving with respect to each other.

2007 Y. Neumeier and J. Crane 1 4/5/2012


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The two postulations of theory of relativity

1) The lows of physics hold the same form for all systems which are in uniform motion

2) Quoted from Einstein paper,Light is always propagated in empty space with a definite

velocity c which is independent of the state of motion of the emitting body. We shall see that

this leads also to that speed of light in vacuum is independent of the speed of its source andshall be measure as constant c by all observers in uniform motion

The idea behind first postulation was already promoted by physicists before Einstein, particularlyso Ernest Mach which rejected the idea that an absolute rest system of reference exist with respect

to which other systems velocity can thus be determined in an absolute manner. Einstein in his

words just suggested that We will raise this conjecture (the purport of which will hereafter becalled the ``Principle of Relativity'') to the status of a postulate The lack of absolute purports that

all system motion is always relative, hence the source of the name for the theory.

The second postulation is wholly Einstein. It will be useful to discuss in some details. To be in the

spirit of the state of transportation in Einstein time consider two observers along a straight railwaywith synchronized stoppers in a windless day (the air is not moving). Consider now a train

standing in a station ahead of them, the engineer make a sharp blast with its horn. When eachobserver hears the blast he stopped it time counter. Comparing now the readings of the two stop

watches and taking the ratio of the distance between the observers and the time interval measure by

the difference of the reading will, not surprisingly, reveal that the speed of the traveling wave isthat of the speed of sound in air which is about 330 m/sec. Now consider that our train which is a

modern one move forward toward the observers in speed of 100 m/sec (bullet trains actually can

actually come close to this speed of about 340 km/hr). When traveling at this speed the engineer

sound a short blast from the horn and again the two observers take their corresponding timereadings. What do you think will be the propagation speed calculated in this case? You may be

surprised but it come the same as before around 330 m/sec. Why is this? Acoustic waves travel inconstant speed with respect to the medium, in this case air, the speed being a function of the airtemperature which for the discussion is uniform throughout the space of interest. Now assume that

the train which we are discussing is a long one with the horn situated in its midst and many enough

cars ahead and behind it so that I can locate observers behind and in front of the sound sourcewhich will travel along with however belong to the same system of reference. Doing this we will

find out that the observer ahead of the horn measure traveling speed of the sound to be 330-

100=230m/sec whereas the observers behind the horn will measure propagation speed of330+100=430 m/sec. We thus can say that for an observer sitting at the horn, the propagation

speed of sound depends upon the direction. The magnitude of this speed thus depends upon the

velocity vector measured by the relative motion of the medium (air) and the observer. One can

argue in the opposite direction namely, the relative velocity with respect to the medium can bedetermined from the dependence of the propagating speed upon the direction. True enough! This

means that even if I cannot detect the medium I can still measure the relative velocity effect by

measuring the dependence of the propagation speed upon the direction. This was the idea behindthe luminiferous aether . It was hypothesized that the electromagnetic waves which light had

been already known to be part of, propagate in a medium that exist everywhere. It was thus

assumed that as earth travel through this medium that was denoted Aether or Ether, there shouldbe difference in the speed of light depending upon the direction with or against the Aether wind.



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One of the most famous experiments designed to capture this effect was done by Michelson and

Morley2 with the most negative result. There had been many attempt to explain this particular

result as well as results of other experiments all defying the expected effect of the aether. We willtake here a simple stipulation taken by Einstein, namely, there is no significance to the concept of

aether as electromagnetic waves propagate in vacuum with no need for medium. Einstein further

stated that any convention of velocity or direction with respect to vacuum is thus false out of whichit follows that the light travel in the same speed in all direction of the emitter regardless of it state

of motion (assumingly that it is in uniform motion). Later in his book3 Einstein shows how from

the constancy of light speed with respect to the source and the requirements that no differences inthe laws of physics for all inertial systems one must arrive at the constancy of the speed of light

with respect to the source as well as any other observer in uniform motion. Instead of going

through this argument we will take here a bit different approach. Go back to our observers on the

railway, we bring now another train and put those observer on the second train. Clearly now thistrain moves at a certain speed the propagation speed of the sound emitted from the first train is

varied in accordance with the respective speed of the second train to the air that serve as a medium.

If no medium is required and moreover no medium exist, such speed does not exist and the

propagation speed is constant for any observer thereof. The reader will benefits from readingAppendix II which deals with waves and the original Lorentz transformation derivation approach.

Einstein postulations, Lorentz transformation and kinematics

Although it is claimed that Lorentz transformation was hypothesized on the ground of maintaining

the integrity of Maxwell equations in moving reference systems, one may suspectthat Lorentz

transformations are closely related to basic postulations of Einstein of the special relativity. In thefollowing we shall derive them based upon those postulations.

The two postulations of Einstein are1. All systems in constant speed experience the same law of physics

2. The speed of light is the same for all observers in those systems

First we examine the impact how these postulations, particularly the second one upon the basicdimensions. For this purpose we consider the so called Michelson Morley configuration1,shown

in Error: Reference source not found, it consists of an L-shaped device comprised of two mirrors at

the ends of equal legs of length L and a light source at the vertex. The L structure is fixed in asystem, which we denote as X, that is moving with speed u (to the right) with respect to system X.

When the origins of the two systems coincide two short light pulses are shot in frame X in the

direction of the two mirrors. Observer seating at the origin of X will see them going horizontally

and vertically and bouncing with the same inclination and arriving simultaneously back to thesource point (which is the origin of X) because the legs are of equal length. Since the arrival of

the two beams to the same point, namely the source point occurs simultaneously in system X, the

two beams should also converge simultaneously for an observer in system X which is local to thepoint of origin at the converging event. This means that, assuming observers distributed along

system X each having a synchronize clock, then at some time t measured by one of these clocks

and at some point, the observer at this point will see the origin of X passing by and

1 After a similar configuration used in Michelson Morley famous experiment.

http://en.wikipedia.org/wiki/Michelson-Morley_experimenthttp://en.wikipedia.org/wiki/Michelson-Morley_experiment


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simultaneously to him the two beams converging to his location. Consider now the traces and

timing of the rays in system X, shown in Error: Reference source not found in the left. According

to Einsteins postulation, the observer X will measure the same propagating speed, c, of the lightpulse. Consider now the two tracks that the two rays take as shown in X system Start with the

(green) ray that travels parallel to the x-axis. Without loosing any generality we can say that the

ray starts at t = 0 and arrives at the horizontal mirror at time 1t by which time, the mirror moved tothe right a distance 1ut . It then reflects and arrives back at the origin after a time interval 2t .

During the back and forth trip the origin point has moved to the right a distance equals to

( )1 2u t t + . Concurrently, the vertical ray travels along the diagonal at speed c, according to the

law that the light always travels the same speed, and reflects from the vertical mirror at time 3t , it

than travel 4t to the origin.

Ln

1x

X

1ut

3ut

( )1 2u t t +

Lp

4ut

Mirror

L

Mirror

L

u

1x

X

2-1Geometry of the moving L structure and ray traces in stationary system, left and body system attached to

the structure, right.

Let the observable horizontal and vertical lengths of the L shape legs measured in system X be Lpand Ln, respectively, (p for parallel and n for normal). For the horizontal path,

1 1pct L ut = + (1)thus,

1

pLt

c u=

(2)

( )2 1 1 2 2p pct L ut u t t L ut = + + = (3)thus,

2

pL

t c u= + (4)From (2) and (4) after simple manipulations we get the timing of the horizontal path,

1 2 2

2

2

1

pL ct t

u

c

+ =

(5)

For the vertical path,



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( ) ( ) ( )2 2 2

3 3nct L ut = + (6)

( )2

3 2 2

nLtc u

=

(7)

Likewise

( ) ( ) ( )2 2 24 4nct L ut = + (8)

( )2

4 2 2

nLtc u

=

(9)

Note that the equality of 3t and 4t implies that the incident angle equals the reflection angle, this is

however not necessarily correct in general in moving mirrors, it is rather a special case when the

mirror surface is parallel to the motion direction.

3 4 2 2

2 /

1

nL ct tu c

+ =

(10)

The process requires that the horizontal and vertical traveling terminate simultaneously thus,

1 2 3 4t t t t + = + (11)and consequently, from (5) and (9)

2 2 2 2

2 2

1 1

p nL c L c

u c u c=

(12)

which yields,2 21p nL L u c= (13)

We thus far got it that in system X the length of the arm parallel to the movement is contracted by

factor 2 21 u c with respect to the normal leg. We recognize that this contraction isindependent of the sign of the velocity and we recognize that the length of the parallel leg is going

to zero unless the normal leg length is going to infinity as u is getting closer and closer to the speed

of light. Note however that we do not know how the length of the normal leg Lp measured in X

relate to L measured in the body-fixed system X, we will thus write.

'n

L kL= (14)2 2' 1pL kL u c= (15)

For brevity we we define which is commonly named Lorentz factor2 21 u c @ (16)

We need to find k. Before we can do it however we need to realize some points. First, it is clear

that the relations in Eq. (12) are general in that a) the origins of the two systems do not need to

coincide at the moment of the light shooting. b) the relations regarding distances expressed in Eq.12 are general and not limited to fixed rigid legs rather, based upon the meaning of spatial

measurements that we introduced above it apply to all distances even if they change in time.

Based on this clarifications, imagine now a point p whose coordinated in system X are x(t) and

y(t), the system X is moving in velocity u (to the right) with respect to the stationary system Xsuch that at t=0 the two origins coincide and at that we also set t=0. After t and t time

corresponding to the time measures in systems X and X the situation is as depicted in Figure 2-2.



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ut

P(x,y,t)

(x,y,t)

YY

X X

Figure 2-2: Point coordinates in two systems

Accordingly we can write.( ) '( ')x t ut k x t= + (17)( ) '( ')y t k y t= (18)

In the above equations we realize something that looks not kosher as we have two time basesnamely t and t in each equation. To allow this one may assume that the reading of the

synchronized clocks of system X are displayed to observers on system X, more mathematically

say that for each t there exist t such that both Eq. 16 and 17 are satisfied, we will find thetransformation of the time later on.

Now we state the fundamental low with respect to the above systems. One may take this law as an

axiom

If observer A sees observer B moving in velocity u with respect to him then observer B sees

observer A moving with velocity u.

This law holds both in classical (Galilean) and relativistic kinematics.

Owing to this we should be able to apply the canonic transformations (16) and (17) in reverse

order implying'( ') ' ( )x t ut k x t= + (19)'( ') ( )y t k y t= (20)

The four equations 16-19 can be solved for the unknowns k and t as function of t or, if one like t

as function of t. Particularly, Eqs 17 and 19 give( ) '( ') ( ) '( ')y t y t k y t y t = (21)

which is satisfied only if,1k = (22)

Using this we have from 16 and 18( ) '( ')x t ut x t= + (23)'( ') ' ( )x t ut x t= + (24)

From 22 we get

( )'( ') ( )x t x t ut = (25)From 23 we get

( )( ) '( ') 'x t x t ut = + (26)22 and 25 gives



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'( ') ''( ')

x t utut x t

+= + (27)

From which we get

2' '( ')

ut t x t

c

= +

(28)

And likewise, from 23 and 24 we get

2' ( )

ut t x t

c

=

(29)

Those are the so called Lorentz transformations lets put them together and frame them and give

them even a special notations

2 21 u c @ Lorentz factor

( )( ) '( ') 'x t x t ut = + L.1

2' '( ')

ut t x t

c

= +

L.2

( )'( ') ( )x t x t ut = L.3

2' ( )

ut t x t

c

=

L.4

'( ') ( )y t y t= L.5We are going to discuss the above transformations in deapth but before hand lets carry one formal

simple step further. Taking the differential of the above equations give

( )' 'dx dx udt = + (30)

2

' 'u

dt dt dxc

= +

(31)

( )'dx dx udt = (32)

2'

udt dt dx

c

=

(33)

'dy dy= (34)Dividing 29 by 30 and dividing both numerator and denominator by dt gives after a simplemanipulation

2

'

'

'1 '

dxu

dx dt

dx udt dt c

+ =

+

(35)

and dividing 33 by 30 gives in the same manner

2

'

''

1'

dydy dt

dx udt

dt c

= +

(36)

By dividing 31 by 32 and 33 by 32 we get



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2

'

'1

dxu

dx dt

dx udt

dt c

=

(37)

2

'

'1

dy

dy dtdx udt

dt c

=

(38)

Denoting by xv , yv and 'xv , 'yv the x and y components of the velocity of point P in the X and X

system respectively, we observe that xdx

vdt

, ydy

vdt

and'

''

x

dxv

dt ,

''

'y

dyv

dt , thus, Eqs (34)-37

give the famous velocity addition formulas provided by Einstein.

( )

2

'

1 '

x

x

x

v uv

u

v c

+=

+ V.1

2

'

1 '

y

y

x

vv

uv

c

= + V.2

( )

2

'

1

x

x

x

v uv

uv

c

=

V.3

2

'

1

y

y

x

vv

uv c

=

V.4

where 2 21 u c @ is the above defined Lorentz factorThe velocity transformation implies that if the relative speed between the systems equals the speed

of light, the point will move in the speed of light in the direction of u and null velocity in the y

direction.Go back now to the L transformations equations L.1-L.5 and ponder the essence of it. In a Galilean

system one would say that if observer X measure the location of p as x at time t then observer X

measure the location as x at the same time t. The L transformation however returns both locationand time as different in the two system, so what is the implication of the transformation? The

answer is as follows. If we put an observer on the point x whose clock is synchronized with the X

system clock and likewise we have such an observer on x synchronized with clock X then whenthey see each other meaning they cross over, they will measure times t and t respectively. It is

very important to keep this so called definition in mind for the future discussion. In view of this,

equations 22 and 23 provide for very useful relations, namely

( )'t x x u= (39)

( )' 't x x u= (40)



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Consider now our two systems as shown in the figure below system X moves in velocity u to the

right, the distance between A and B measure at X is ABl and the distance between a and b

measured in X is abl Denote the events where A and a cross each other and B and b cross

each other, of A,a and B,b, respectively. Using Eqs 38 we get

, ,' 0A a A at t= = (41)

( ),B b AB abt l l u= (42)

( ),'B b AB abt l l u= (43)From 40.a and b we get

,

,

1

' 1

B b AB ab ab AB AB ab

B b AB ab ab AB AB ab

t l l l l l l

t l l l l l l

= = =

(44)

Equation 50 is important; it shows that there is no faster or slower clocks per se, which system will

measure shorter or longer elapse time depend upon the ratio AB abl l of the distances. Particularly,

when 0abl = corresponding to an observer sitting on the origin of X and traveling to pint B, then

,

,

1'B b

B b

t

t = and the case 0ABl = corresponding to an observer in the origin of X moving between a

and b on X

u

X

X

a

A B

b

abl

ABl

Figure 2-3 Geometry for timing of events

The cases where one clock (observer) move between two points is of great inerest. It explains

the phenomena of the muons2 that are created around 10 km high in the atmosphere and travel at

near light speed toward the earth. Even with the speed as high as that of light of about 3x10 8 m/secwe, the stationary observers, on earth, would expect a traveling time of about 3x10-5 sec. The

2The following is from Wikipedia (http://en.wikipedia.org/wiki/Muon)The nuclei that make up cosmic rays are able to travel from their distant sources to the Earth because of the lowdensity of matter in space. Nuclei interact strongly with other matter, so when the cosmic rays approach Earth they

begin to collide with the nuclei of atmospheric gases. These collisions, in a process known as a shower, result in the

production of many pions and kaons, unstable mesons which quickly decay into muons. Because muons do not

interact strongly with the atmosphere and because of the relativistic effect of time dilation many of these muons are

able to reach the surface of the Earth. Muons are ionizing radiation, and may easily be detected by many types of

particle detectors such as bubble chambers or scintillation detectors. If several muons are observed by separated

detectors at the same instant it is clear that they must have been produced in the same shower event.

http://en.wikipedia.org/wiki/Muonhttp://en.wikipedia.org/wiki/Muon


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muons, however, have an average lifetime of 2x10-6 sec well shorter than the expected traveling

time. Yet, because of the above discussed time difference in the time as measured by the muons

single clock and the earth bounded synchronized clocks up high in the atmosphere and in the lab,the muons clock experience much shorter traveling time and hence arrive alive to the earth

bounded detectors. The following is an important conclusion

If two events occur to a single observer and the same two events occur to two separated observerswhich having synchronized clocks, then the duration between the events recorded by the single

observer is always lesser than that recorded by the two synchronized clocks of the two observers

For example assume you are on a spaceship moving between earth and the moon and you observe

the sun spots appearance and disappearance, say a spot appear when you are still close to earth and

disappear when you are near the moon. There is a base on the moon with a clock synchronize to aclock on earth, the clock on earth recording the start of the event when the earth observer see the

appearance of the spot and the observer on moon recording the termination time when he sees the

spot disappearing. The time start and end difference between these clocks are greater than that on

the spaceship. Note that an event is defined in terms of localization meaning the light of theappearing spot arrive at the spaceship and earth when they are collocated, the same with the ending

which is the collocation of the spaceship and the moon.

Einstein took it one step further thereby kindling the famous twin paradox. In his famous 1905

paper On The Electrodynamics of Moving Bodies Einstein stated

From this there ensues the following peculiar consequence. If at the points A and B of K there

are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at

A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks nolonger synchronize, but the clock moved from A to B lags behind the other which has remained at

B by 2 212 tv c (up to magnitudes of fourth and higher order), t being the time occupied in the

journey from A to B.

It is at once apparent that this result still holds good if the clock moves from A to B in anypolygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved

line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with

constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has

remained at rest the travelled clock on its arrival at A will be2 21

2tv c second slow. Thence we

conclude that a balance-clock7at the equator must go more slowly, by a very small amount, than a

precisely similar clock situated at one of the poles under otherwise identical conditions.

In 1911, Paul Langevin made this concept more vivid and comprehensible by his now-iconic

thought experiment of the twins, one an astronaut and the other a homebody. The astronaut brother

undertakes a long space journey in a rocket moving at near light speed, while the other remains onEarth. When the traveling brother finally returns to Earth, it is discovered that he is younger than

his sibling.

http://www.fourmilab.ch/etexts/einstein/specrel/www/http://www.fourmilab.ch/etexts/einstein/specrel/www/#foot172http://www.fourmilab.ch/etexts/einstein/specrel/www/#foot172http://en.wikipedia.org/wiki/Paul_Langevinhttp://www.fourmilab.ch/etexts/einstein/specrel/www/http://www.fourmilab.ch/etexts/einstein/specrel/www/#foot172http://en.wikipedia.org/wiki/Paul_Langevin


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According to the Wikipedia (the framed text in the source): In 1911, Einstein restated in a very

similar way:

If we placed a living organism in a box ... one could arrange that the organism, after any arbitrary

lengthy flight, could be returned to its original spot in a scarcely altered condition, while

corresponding organisms which had remained in their original positions had already long since

given way to new generations. For the moving organism the lengthy time of the journey was a mereinstant, provided the motion took place with approximately the speed of light. (in Resnick and

Halliday, 1992)

The paradox is. Assuming constant speed with the exception of short phases in the takeoff andturn around, one would expect that the relative motion is such that from the point of view of an

observer on the spaceship the homebody is the one that goes away and come back. This of course

leads to a contradiction.

One can find many discussions of this so called paradox, the ones I checked, like the one in theWikipedia was cumbersome and did not get to the crux of the matter. Other discussions, from new

Einsteins like Subhash Kak.4 from the Louisiana State University reveal novel discoveries as

the Daily Reveille Online Edition Issue date: 2/23/07 (the link is not active any more) reads:Professor solves Einstein's twin paradox. the reader could enjoy reading the clip and some of the

responses

Going further yet, Wang Guowen from the College of Physics, Peking University, Beijing, China,in January 25, 2005. In hispaperEinstein's concept of a clock and clock paradox, found that:Regrettably, the concept of the clock that Einstein retained is equivalent toNewtons concept of

absolute time that he rejected. This is a blemish in Einsteinsotherwise perfect special relativity.

Enough now with this, lets get to the crux.

Indeed the fact that the spaceship undergoes velocity transition is the key for resolving the

paradox. We have thus to look in some details what happens when such a transition take place.

Lets start with our two systems X and X both at rest (so to say) having the origins coincide.Lets assume that we have a stick of length L in system X the one end of which is in the origin of

system X. Now let the origin of X which is to say one of the stick edges take a change invelocity. Since we view the stick as rigid we assume that after the velocity transition completed

the stick retains its original qualities. For an observer on X this means that the length measured in

X will be L*. Now, the distance between the ends of the stick measured in system X is simply

( )0

t

Le Tel L v v dt = + (45)Where Le and Te are for leading and trailing edges (in the x direction), respectively. Equation (39)

shows that it is impossible to transform even a rigid body to a new velocity without having some

transient time.

( )0

t

Le TeL L v v dt = + (46)or

( ) ( )0

1

t

Le TeL v v dt = (47)

http://en.wikipedia.org/wiki/Twin_paradoxhttp://en.wikipedia.org/wiki/1911http://en.wikipedia.org/wiki/1911http://en.wikipedia.org/wiki/Twin_paradoxhttp://www.lsureveille.com/home/index.cfm?event=displayArticle&ustory_id=e1c6564f-4be3-44ba-a190-df457cce55e9http://arxiv.org/abs/physics/0501131v1http://arxiv.org/abs/physics/0501131v1http://en.wikipedia.org/wiki/Twin_paradoxhttp://en.wikipedia.org/wiki/1911http://en.wikipedia.org/wiki/Twin_paradoxhttp://www.lsureveille.com/home/index.cfm?event=displayArticle&ustory_id=e1c6564f-4be3-44ba-a190-df457cce55e9http://arxiv.org/abs/physics/0501131v1


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Since the absolute velocity of either end cannot exceed c we can write

( )12

L

c > (48)

Note that in the case where both end move in unison with the same speed or if the leading edgemoves with velocity c then the distance between the points as measured in system X remain L.

Equation (42) shows that one cannot talk about moving frames without discussing the actualdimension of the frame for example if we have a spaceship with length of 10m to be acceleratedfrom zero to the speed of light the minimum time required to achieve this will be

( ) 8810

1 1 1 1.66 102 3 10

> =

not too long a time, however if a galaxy of many year light size decide to accelerate to this speed,

that will take much longer. Back now to our paradox analysis.Consider the geometry in Figure 2-4 On a stationary system E we put earth at the origin of E, the

planet P at distance L and a reference point E_r distance L behind. Two spaceships S and S-r are

initially at rest with respect to the system E at distance L from each other. All clockssynchronized. At t=0 on the clock at E as well as on their respective clocks S and S_r perform

acceleration each ship as local rigid body so after short time each space ship completed a rigidtransient and move at velocity u to the right. We assume that this phase is relatively short so takethe clock at E to be still zero as well as the clocks at S ans S_r. Note that E_r and P clocks are

synchronized to E so we have nothing further to say about them. The distance between S and S_r

measured from E is L before as well as after the take off. The two spaceships maintain constant

distance between them. Now lets see what we can say about S. It has a local system that wasrigidly transformed. This local system can only measure time because it extend basically nowhere

except in the very small neighborhood of the clock. Therefore, any events measured by the clock in

this system should be in agreement with the earth bound inertial system. However, because thesystem is localized the observer of the system lack the ability to map is outside the way it is done

in the earth bound system with the distributed clocks. As a matter of fact any prediction that we

make with respect to the distances perceived by S hav to be done using the E system. Forexample, we want to know what would be the distance that S measure between him and P

immediately after takeoff. The only tool that S has is a light pulse that he sends toward P and wait

to get back after bouncing on P. The only way to calculate this would be using the synchronized

clocks on E. Solving for the location of S that moves with speed u and the light that move in speedc bounce back and meet S we have

2ct L u t = (49)from which we get

( )2t L c u= + (50)The location at which the light will meet S is

( )2x Lu c u= + (51)Using Eq. (40) we get

( ) ( )' ' 2t x x u L c u = = + (52)Not surprisingly 't t=Now consider the observer S, he sees the light going away with velocity c and come back with the

same velocity. If you see something going and coming back at the same velocity with respect toyou you can rest assure that half time it spend going away and half time coming back. Therefore

from S point of view



13/29

' 2 ' 2 'ut ct L+ = (53)and thus,

( )' 2 't L u c= + (54)Equating (52)and (54)we finally get

'L L= (55)From this prospective one can say that planet P is located at distance 'L L= on the coordinatesystem of S. Is that so indeed? Consider now Figure 2-4. On a global inertial system which we

will denote as E (for earth) we have the planet P and another space base that is stationary of course,

we name it R (for reference). Three spaceships are ready to taking off from R, E and P all clockson R,E,P and the space ships are synchronized. The spaceships take off in the same time indicated

on their respective clock and accelerate fast so they reach the same velocity after short enough

time. Because of the short phase of the acceleration we consider the clocks all to read the sametime (approximately) as before take off. We take this time to be zero for convenience. We note

that since the distance between the space ships is constant as they travel the represent points on an

imaginary moving stick. The distance between the points of the moving stick is L thus implying

that when measured in the local system of S or S_r or S_p it would be L/. You are not convinced?Lets make the calculations then. As before S send a pulse of light toward say, S_p. In system Ethe calculation goes as follows. Denote by 1t and 2t the time the light travel toward and back from

S_p, we have

1 1ct L ut = + (56)thus,

1

Lt

c u=

(57)

( )2 1 1 2 2ct L ut u t t L ut = + + = (58)

thus,

2

Lt

c u=

+(59)

After simple manipulations we get the time for the light to rich S_p bounce back and hit S,

1 2 2

2

2

1

L ct t t

u

c

= + =

(60)

The location at which the light will get back at S is thus,

2

2

2

1

L cx u

u

c

=

(61)

Using Eq. (40)

( ) 22

2 1 2' ' 0

1

u L c L ct x x u

u u

c

= = =

(62)

Finally, since the distance remain constant we have



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2 /' 1/ 2 ' 1/ 2

L c LL c t c

= = = (63)

S

E P

L LS_ r

R

S_p

Figure 2-4 Geometry of stationary system earth E, planet P and earth reference R and three corresponding

space ships

We thus, showed that if right after takeoff the S ping both planet P and his fellowman S_p or

alternatively R and S_r and after getting the pings back he calculate retroactively the distances at

take off he will find two entire different distances one is factor less and the other factor largerthan the nominal rest distance L. This is so even though at take off S_p was very near P (and so

was S_r at R). Which of the distances is relevant to our discussion? It turns out both are relevant.

As far as calculating the trip time measured by S clock, the shorter distance provide the neededinformation. However with respect to how we should judge the relative motion of E with respect

to S we find out that while S has a rendezvous with plant P which is in distance L at the earth

system, his friend on mother earth has a rendezvous with his friend S_r which is L/distance. Wecan say that while S completed a round trip with less time than E, E completed a round trip with

more distance. That much for the twins paradox. It is now time to move forward.

Momentum mass energy and temperature

Put ourselves in Einstein place pondering what the implication of the two postulations on themechanics of bodies. The trick is to start with something fundamental that we believe hold true in

the new circumstances and see where it leads us. We choose to assume that conservation of

momentum is a fundamental law and will hold true under all circumstances of interest Accordinglywe consider the following: Two systems namely the stationary system X and the moving system

X that moves with velocity v to the right. In the moving system, we perform the tests as shown in

Error: Reference source not found, two balls numbered 1 and 2 each having equal rest mass 0m

moving toward each other each with opposing velocity of u measured in system X thus the total

momentum in system X is zero. Before continuing we need of course to define what is the rest

mass 0m . I put in the frame below four definitions that I found, all of the saying the same. Now, in

low velocities the masses are naturally the so called rest mass.



15/29

After colliding we choose the scenario that the two balls bounce elastically vertically each keepingthe same velocity and rest mass, thus maintaining zero momentum as required by conservation of

momentum. How do we know that such an outcome is even possible? Well we dont! we assume

so. We borrow from the classical mechanics the idea of elastic collision in which all mechanicalenergy is conserved, we assume that the kinetic energy is function of the magnitude of the velocity

thus we assume that and 0m thus, the total energy before and after collision in this scenario is

conserved. We choose this scenario as a starting point because in this case we do not need to

address forces, thus enabling us to eliminate the complexity of including yet undefined elements. 3

Also important to note is the total symmetry of our case we start with equal mass having equal

velocity and we end with equal mass each preserving the magnitude of the velocity before the

collision. Under this conditions we can be assured that whatever dependence on the velocitymagnitude that mass would have its equal effects on the two mass should still yield zero

momentum in system X where all mass move with equal velocity. We now transformed the

scenario to system X.

3As it turned out Einstein did not take our approach he instead went through, rather cumbersome electrodynamicalconcepts in which he treated the electron dynamics. As a matter of fact Einstein treatment of the electron motion

under force was nothing else but discussion of mass acceleration and force, so explicitly also stated by Einsteinhimself We remark that these results as to the mass are also valid for ponderable material points, because a

ponderable material point can be made into an electron (in our sense of the word) by the addition of an electriccharge, no matter how small. Also, Einstein himself realized that since the force that he included in his analysis was

not yet defined one could not get definite handle on the mass. He acknowledge this as well in his 1905 paper

With a different definition of force and acceleration we should naturally obtain other values for the masses. This

shows us that in comparing different theories of the motion of the electron we must proceed very cautiously. W.

Pauli noted this disadvantage in his book Theory of relativity (Dover 1981) and dedicated a section (38) to this

subject in which he discuss the contribution of Lewis and Tolman which took an approach similar to the one in

taken this article.


1. The physical mass of a body when it is regarded as being at rest.2. In Special and General Relativity, the observed mass of a body that is not in motion

with respect to the observer. Also called invariant mass.

3. the mass of a body as measured when the body is at rest relative to an observer, aninherent property of the body


16/29

vtYY

XX

u u

u

u

0m 0m

0m

0m

1 2

1

2

3-5 Schematic for momentum calculations for case 1, in system X two balls moving toward each other in

velocity u and bouncing vertically, system X moves with velocity u with respect to system X

Using the law of velocities Eqs. V.1 and V.2 we get the results in the table 1 below. Well, it does

not take a rocket scientist to figure out that if we simply sum up the classic momentum which isthe product of the mass and velocity, we find out that the momentum in the x direction before and

after is not the same. This of course is not surprising given the nonlinear velocity addition. We

can assume that the momentum is no longer a linear function of the velocity but rather somecorrection should be made such that at low velocity it go to the classic linear relations. We can also

assume that the correction will be to the magnitude of the momentum but not to its direction, in

other words we believe that the momentum direction is in the direction of the velocity. I could putforward a pretty strong candidate for this function, however it is much cooler to derive this

function rather than guess it



17/29

velocity Before collision After collision

0

0

1xu

2xu

1yu

2yu

1u

2u

21

u v

uv

c

+

+

v

21

u vuv

c

+

v

2

21

vu

c

2

21

vu

c

21

u v

uv

c

+

+

21

u v

uv

c

+

2 22 2

2

u vv u

c+

2 22 2

2

u vv u

c+

Table 3-1 Velocities for case 1

. For this purpose we will leave for the time being our case and consider the derivation of Lewisand Tolman5 which consider the following case. On a stationary system X a ball A of mass m0 is

pushed upward along the y axis in velocity u. In system X moving to the right in speed v a ball B

of the same mass is pushed downward along the axis y. The balls collide such that the collision

center is along the respective y and y axis. Because of this each ball recoil back along therespective y and y axis. The situation as seen from system X is shown in the diagram below.



18/29

Figure 3-6.Case 2, (Lewis Tolman case),

In the table below we fill all the known quantities. All velocities are known before the collision

and we know that after the collision the balls recoil vertically in their respective system of

coordinates thus, we know the x components of the velocities after collision. Now the symmetryof the case requires that

'yAf yBfw w= 4.1We now make the following assumption about the momentum

( )0J m f w w= r r

4.2

Eq 4.2 states that the momentum direction is that of the velocity, its magnitude linear with themass and a general non linear function of the velocity magnitude. Now, conservation of

momentum in the x direction, according to Eq. 4.2 gives

( ) ( )2 2 2 2xBi xBi yBi xBf xBf yBfw f w w w f w w + = + (4.3)Note that here we assume the mass to keep the same value before and after the collision. This is a

bit of a stretch but can be justified plus remember that we want to get a solution which we will thentest on our case 1, thus we will accept this assumption without further argument. Substitute to 4.3

from the table below we get

( )( ) ( )2 2 2 2 2 21 yBfv f v u v c v f v w + = + (4.4)Which, thus implies

2 21yBfw u v c= (4.5)And thus

'yBf

w u= (4.6)


Ball A

Ball B

v

u


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System X System X

Before collision Ball A 0xAi

w =

yAiw u=

'xAi

w v=

2 2' 1yAiw u v c=

Ball BxBiw v=

2 21yBiw u v c=

' 0xBiw =

'yBiw u=

After collision Ball A 0xAfw =

yAfw u= from Eq. 4.1

'xAfw v= 2 2' 1yAfw u v c=

Ball BxBfw v=

2 21yBf

w u v c= from Eq.4.5

' 0xBfw =

'yBfw u= Eq. 4.6

Table 3-2 velocities for case 2

At this point all entities in the table are known we will now perform momentum balance in the ydirection.

( ) ( )( )2 2 2 2 2 2 20 0 1 1yiJ m uf u m u v c f v u v c= + (4.7a

( ) ( )( )2 2 2 2 2 2 20 0 1 1yfJ m uf u m u v c f v u v c= + + (4.7b)Equating the momentum before and after the collision gives

( ) ( )( )2 2 2 2 2 2 21 1f u v c f v u v c= + (4.8)Here now comes the trick. We know that for low speed, to keep the classical mechanics satisfied

we need to have( )2

01

uf u

(4.9)

Taking thus the limit of Eq. 4.8 as 0u provides

( )2 2 21 1v c f v = (4.10)and thus,

( )2 2 21 1f v v c= (4.11)



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We thus propose that in general

( )2 20 1 1J m w c w= r r r

(4.12)

We can now see if the proposed expression 4.12 make case 1 momentum conservation hold true.

Beforehand we will do something that will provide much insight and simplify the application ofexpression (4.12) to general cases. First for convenience we write (4.12)

0J m f w=

r(4.13)

were by agreementfis always the operator on the velocity amplitude.2 21 1f w c

r@ (4.14)

Also, if in a system X the operatorfrelate to a velocity vector w such that this vector transform to

w in a system X that moves with respect to X then we denote byfthe operator on w. Consider

now system X moving with velocity v with respect to the stationary system X. Assume now a

velocity vector in X with the components 'xu and 'yu with the corresponding operatorfdenote

the transformation of the velocity vector to system X using the transformation V.1 and V.2 as xu

and yu with the corresponding operatorf, it can be shown, see Appendix I, that

( )2

1 ''

1

xu vf fv

+ =

(4.15

( )2 2

''

1x x

ff u u v

v c = +

(4.16)

' 'y yf u f u = (4.17)

The above two equation are a very powerful tool that simplify our momentum calculationstremendously. Lets use it on case 1. First since in system X all particles move with the same

velocity, the operatorfyields the same quantity for all of them lets denote by A

2 2

( ) '

1

f uA

v c@ (4.18)

We thus have for the momentum before collision

( ) ( ) 2xiJ u v A u v A vA= + + + = (4.19)And after the collision

( ) ( )0 0 2xfJ v A v A vA= + + + = (4.20)Which of course prove that momentum is conserved. Momentum in the y direction is even easier

to prove.

The relations 4.15 - 4.17 can be now used to establish the fundamental low of the momentum.

First, lets define, for convenience at this point a modified mass

0 0 2 2

1

1m m f m

w c =

@ (4.21)

And recalling Eq. 4.12



21/29

0x xJ m f u= (4.22)

0y yJ m f u= (4.23)Multiplying both sides of Eqs (4.15)-(4.17) by 0m

( ) ' ( ) 'x

m F v m G v J = + (4.24

( ) ' ( ) 'x xJ F v J G v m= + (4.25)'y yJ J= (4.26)

Because 2 2( ) 1 1F v v c@ and 2 2( ) 1G v v v c@ are functions ofv only, therefore Eqs(4.24)-(4.26) imply that if in the system X the modified mass and momentum is conserved then

they must conserved in any other system which move in the x direction with respect to system X,

this is expressed in the general statement below.

(4.27)

If in system X

{ {' 'Before Aftercollision collision

m m = and { {' 'Before Aftercollision collision

J J =

Then in system X with respect to which system X is moving or which is moving with respect to

X in any arbitrary velocity v (v


22/29

physics. Alternatively, momentum conservation is a result of mass (energy) conservation and the

first postulation of relativity.

At this point I would like to exploit Eqs. (4.29) and (4.30) for discussing the co-linear collision of

two mass particles

Eq. (4.30) and (4.29) imply

( ) ( ) ( ) ( )

1 21 201 02 01 022 2 2 2

1 2 1 21 1 1 1

f fi ii i f f

i i f f

u uu um m m m

u c u c u c u c

+ = +

(4.31)

( ) ( ) ( ) ( )

01 0201 02

2 2 2 2

1 2 1 21 1 1 1

f fi i

i i f f

m mm m

u c u c u c u c

+ = +

(4.32)

Given known conditions, namely rest masses and velocities prior to collision, we have fourunknowns after the collision, namely two rest masses and velocities. Thus, the two equations

(4.31), (4.32) cannot determine a unique solution. Lets consider only two cases, namely the

elastic and plastic collision. In the former, the rest mass of each of the particles remain unchanged,

in the latter, the two particles move together at the same velocity after colliding.

( ) ( )( )

( )

1 201 02 01 022 2 2

1 21 1 1

fi ii i f f

i i f

uu um m m m

u c u c u c

+ = +

(4.33)

( ) ( )

( )

( )

01 0201 02

2 2 2

1 21 1 1

f fi i

i i f

m mm m

u c u c u c

++ =

(4.34)

( ) ( )

( ) ( )

1 201 022 2

1 2

01 022 2

1 2

1 1

1 11 1

i ii i

i i

f

i i

i i

u um m

u c u cu

m mu c u c

+

=

+

(4.35)

Thus, Eqs (4.33) and (4.35) can be solved to find the change in the sum of the rest mass of the two

particles. Consider now the special case where 01 02 0i i im m m= @ and 2 1i i iu u u = @ we have0fu = and thus

( )

00 2

2

1

if

i

mm

u c=

(4.36)

This shows that in plastic collision the rest mass (not the modified mass m) change. Now lets see

how that fit our conventional concept of mechanics of collisions. First, let expand the function f

for ( ) 2/ 1u c


23/29

we get at 1x


24/29

( )( )

( )

( )( )

2 2

3 22 2

3 22

1 1 ' ' ' ' ' '

1 1

1 ' '

1

mm vJ v m v J J v J

v v v v

v m Jv

= + = + + =

= +

(4.45)

Minimum would occur when the derivative is zero. Equating the derivative (4.45) to zero yields

' 'v J m = (4.46)The sting in this seemingly innocent Eq (4.46) is that it tells us that if ' 0J = then 'm isminimum. Thus in general, an observer in steady motion that measure an overall null momentum

for system of particles is observing the minimum assembled modified mass of the system and at

that, observing the rest mass of the system. We now can defined the absolute zro mass of a systemlike that by

00 0m N m = (4.46)We can thus define a non dimensional temperature as

0

00

mTm

@ (4.46)

where now 0m is the rest mass of the system and 00m is the rest mass of the system at absolute

zero when no random movement exist.The discussion above explains very clearly the connection between the increase of rest mass of a

macro body and the increase it its temperature. We thus know now that temperature is the ratio of

rest mass to absolute zero rest mass. The conclusion from this is that the dynamic of an objectdepend upon its temperature.

It now is apparent that in multi component system the energy possessed by the particles translates

into rest mass following the following definition

( )0 @ 0Jm m == (4.47)It thus time evaluate the energy possessed by a particle as it accumulate speed. First we can now

define force ( ) 021

d d d mF J m u u

dt dt dt u u c

= =

r r r r@ r r (4.48)

( ) 2 2 21

2

ddW F dx m u dx u dm mu du u dm m du

dt = = + = +

r r r r r r r r@ (4.49)

From the definition of m Eq. (4.21), we have2

2 2 01m

u cm

=

r(4.50)

We assume that the rest mass is constant during the process. This is actually by definition if theprocess is reversible and it is obvious given the discussion above.

Accordingly2

2 2 0

32

mdu c dm

m=

r(4.51)

Substituting (4.50) and (4.51) into (4.49) gives



25/29

2 22 2 20 0

31

m mdW c dm m c dm c dm

m m

= + = (4.52)

If we just carry out the integral of both sides and set the arbitrary constant to zero and denote the

workWas the energy Ewe get the iconic famous formula2E mc=

(4.53)Some discussion:



26/29

Appendix I Transformation of the momentum operator

Consider system X in which we have a velocity vector with the components 'xu and 'yu the

system X moves with velocity v with respect to stationary system X, the velocity components in

the stationary system are xu and yu . The momentum operator is given by:

0' ' 'x xJ m u f= I.1-a

0' ' 'y yJ m u f= I.1-bwhere

( ) ( )22

2

' '' 1 1

x yu uf

c

+@ I.2

Likewise

( ) ( )22

21 1

x yu uf

c

+@ I.3

The product of the velocity components and the functionfprovide the momentum operator. Weshall see now how this operator transform from system X to system XStart with the velocity transformation, see Eqs V.1 and V.2 in the main text

2

'

1 '

xx

x

u vu

u v c

+=

+ (I.4)

2 2

2

'1

1 '

y

y

x

uu v c

u v c=

+ (I.5)

( ) ( ) ( )222 2 2 2

2

1' ' 1

1 'x y x y

x

u u u u v u v cu v c

= + = + + + (I.6)

Now

( ) ( )

( ) ( ) ( )

( )

22

2 22 2 2

2 22

11 1

' ' 1 11

1 '

x y

x y

x

u uf

cu v u v c

cu v c

+ =

+ +

+

@(I.7)

Denoting for brevity

' 'x xu u c@ ; ' 'y yu u c@ ; v v c@ , (I.8)

we can write (I.7)

( ) ( ) ( )

( ) ( ) ( ) ( )( )

22

2 22 2 2

1 '1 1

1 ' ' ' 1

x y x

x x y

D

u u u vf

cu v u v u v

+ + =

+ + + @

1 4 4 4 4 4 4 42 4 4 4 4 4 4 43

(I.9)

Manipulating the expression in the denominator inside the radical



27/29

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )

( )( ) ( )( ) ( ) ( )

( )( ) ( ) ( ) ( )

22 2 2

2 22 2 2 2

2 22 22 2 2

22 2 2 2

22 2 2

1 ' ' ' 1

1 ' ' 2 ' ' '

1 2 ' ' ' 2 ' ' '

1 ' 1 ' 1 '

1 ' 1 1 '

1

x x y

x x x y y

x x x x y y

x x y

x y

D u v u v u v

u v u u v v u u v

u v u v u u v v u u v

u u v v u

u v v u

v

+ + + =

= + + + + =

= + + + =

= =

= =

=

@

( ) ( ) ( )( )( )222 1 ' 'x yu u +

(I.10)

Substitute forD From (I.10) into (I.9) we get

( ) ( ) ( )

( ) ( )

( )

22

2222

1 '1 1

1 1 ' '

x y x

x y

u u u vf

cv u u

+ + =

+@ (I.11)

Using definition (I.2.) in (I.11) gives

( )2

1 ''

1

xu vf fv

+ =

(I.12)

Eq. (I.11) gives us an easy way to transform the otherwise messy formula forfbut it get evenbetter. Using I.4 and I.5 and the definitions (I.8) we get

( )( )

2

''

1x x

fu f u v

v c = +

(I.13)

' 'y y

u f u f =

(I.14)

Eqs. (I.13) and (I.14) are extremely useful in transforming the momentum from system to system

Appendix II Elastic and Electromagnetic Waves2

2

2 2

10E

c t

=

22

2 2

10B

c t

=



28/29



29/29

1 ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905

This edition of Einstein's On the Electrodynamics of Moving Bodiesis based on the English translation of his

original 1905 German-language paper (published as Zur Elektrodynamik bewegter Krper, in Annalen der

Physik. 17:891, 1905) which appeared in the book The Principle of Relativity, published in 1923 by Methuen and

Company, Ltd. of London. Most of the papers in that collection are English translations by W. Perrett and G.B.

Jeffery from the German Das Relativatsprinzip, 4th ed., published by in 1922 by Tuebner. All of these sources

are now in the public domain; this document, derived from them, remains in the public domain and may be

reproduced in any manner or medium without permission, restriction, attribution, or compensation.

2 http://en.wikipedia.org/wiki/Michelson-Morley_experiment3Relativity: The Special and General Theory by Albert Einstein 1916, book can be downloaded fromhttp://www.marxists.org/reference/archive/einstein/works/1910s/relative/index.htm

4Subhash Kak Moving Observers in an Isotropic Universe

JournalInternational Journal of Theoretical PhysicsPublisherSpringer NetherlandsISSN0020-7748 (Print) 1572-9575 (Online)IssueVolume 46, Number 5 / May, 2007DOI10.1007/s10773-006-9281-2Pages1424-1430SubjectCollectionPhysics and AstronomySpringerLink DateWednesday, January 17, 20075 G. N. Lewis and R.C. Tolman, Phil. Mag., 18 (1909) 510.
http://www.amazon.com/exec/obidos/ASIN/0486600815/fourmilabwwwfourhttp://www.springerlink.com/content/101594/?p=824e40cab4ff467ca37ed7d3557a5ca3&pi=0http://www.springerlink.com/content/m2ul882w6357/?p=824e40cab4ff467ca37ed7d3557a5ca3&pi=0http://www.springerlink.com/physics-and-astronomy/http://www.amazon.com/exec/obidos/ASIN/0486600815/fourmilabwwwfourhttp://www.springerlink.com/content/101594/?p=824e40cab4ff467ca37ed7d3557a5ca3&pi=0http://www.springerlink.com/content/m2ul882w6357/?p=824e40cab4ff467ca37ed7d3557a5ca3&pi=0http://www.springerlink.com/physics-and-astronomy/

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