relativistic reactions visualized through right triangles in space

7/29/2019 Relativistic Reactions Visualized Through Right Triangles in Space

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arXiv:0905.353

7v1[physics.pop-ph]21May2009

Relativistic reactions visualized through right triangles in space

Theocharis A. ApostolatosSection of Astronomy, Astrophysics, and Mechanics

University of AthensPanepistimiopolis, Zografos,

GR-15783, Athens, Greece

(Dated: May 22, 2009)

In this article we exploit the fact that the special relativistic formula which relates the energy andthe 3-momentum of an elementary particle with its rest mass, resembles the pythagorean theoremfor right triangles. Using such triangles, suitably arranged, we can prove all kind of equalities orinequalities concerning the kinematical properties of elementary particles in a wide variety of casesregarding particles collision, decay or production. Moreover, relations that are somehow hard toproduce by the usual analytic methods arise much more naturally through geometric constructionsbased on right triangles.

PACS numbers:

I. INTRODUCTION

After the notorious Address delivered by H. Minkowski

at the Assembly of German Natural Scientists and Physi-cians in 1908 [1], which was devoted to space and time,all physical quantities which were previously described asvectors in 3-D Euclidean space, were recast accordinglyto form either vectors in a 4-D pseudo-Euclidean space-time (like the 4-momentum), or to tensorial objects (likethe electromagnetic field). The magnitude of all 4-vectorsthat are used in special relativity is defined as

||A||2 = AA = (A0)2 + A2,

where A is the 3-D vector that represents the spatial partof A. This expression resembles the pythagorean theo-

rem relating the sides of a right triangle, with the timecomponent A0 playing the role of triangles hypotenuse.This property has been extensively used in many in-troductory textbooks devoted to Special Relativity [2],in order to visualize relations that apply between vari-ous physical quantities which become interrelated in theframework of Minkowskis space-time. However, there isan intrinsic problem in drawing such right triangles in ageneric situation: one has to draw 4-D objects since oneof the sides of the triangle is already a vector living inthe 3-D Euclidean space, while the other perpendicularside should extend beyond this 3-D space. This practicalproblem arises whenever one deals with more than two

such triangles, since then there is not enough space todraw everything. In all cases explored here we will showthat the most generic situation can always be analyzedwith only two such triangles, the vector sides of whichare defining a plane while the third dimension could beused to develop the other perpendicular sides of the cor-responding triangles.

Electronic address: [email protected]

In this paper we demonstrate the use of such righttriangles in analyzing elementary-particle reactions; thecorresponding sides of each triangle will represent the en-ergy, E, the 3-momentum, p, and the rest mass, m, ofeach particle (we will work with units where the velocityof light is c = 1). This diagrammatic tool has hardlybeen used in textbooks (for a recent article that refers tothis tool see [3]) for these physical quantities, although,as it will be shown later, the quantitative use of such ge-ometric tools is in many cases advantageous compared tothe usual analytic methods. Solving problems concern-ing particles that collide with each other and produce anumber of new particles by the usual algebraic way isquite often a rather complicated procedure, mainly dueto the fact that apart of energy and momentum conser-vation there are also constraints of quadratic form fromthe energy-momentum-mass relation. Thus if one does

not follow the most clever way to analyze the problem,computations may become really meshy. The situationlooks like a labyrinth; if one does not know the right pathconnecting the starting point with the endpoint, variouspaths leading to dead ends will be chosen until the aimis accomplished.

By using right triangles for each particle, arranged inspace in a convenient way, we can get answers about allphysical quantities in a much more straightforward fash-ion. Moreover, these geometric constructions can easilybe used by someone to draw preliminary qualitative con-clusions, like is a reaction allowed or is it forbidden,or could the angle between the lines of motion of twoproducts exceed 90 or not? Also, if one intends toconstruct a new problem, the graphical analysis throughright triangles may help in posing well defined questionsthat have a clear answer. The only difficulty in this geo-metrical approach of solving or sketching problems is todraw a suitable 3-D construction of right triangles. Wewill demonstrate the power of this geometrical methodby solving accordingly a few characteristic problems. Foreach problem we will draw the right triangles in such anarrangement in 3-D space that answers will come out
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quickly and easily.

II. ORTHOGONAL TRIANGLES

CORRESPONDING TO PARTICLES

As was mentioned previously the basic unit in our ge-ometric construction will be a right triangle with hy-

potenuse E, and two perpendicular sides with magni-tude |p| and m respectively, where E, p, m are the en-ergy, the 3-momentum, and the mass of a single particle(see Fig. 1), all of them measured in a specific inertialframe of reference. Since p is a vector in a 3-D Eu-clidean space, the corresponding side will be representedby a vector; hence particular attention should often bepaid in arranging the vectorial side of the triangle alongthe corresponding direction in a 3-D drawing. The ki-netic energy of a particle is defined as

T = E m, (1)

thus it is represented by the difference between the corre-

sponding sides of the right triangle. Also the velocity ofthe particle in the particular frame of reference is givenby the ratio

v =p

E, (2)

that is by the cosine of the right angle subtending themass side of the triangle.

Em

p

cos ||-1v

FIG. 1: This is the basic geometric unit that will be usedthroughout the paper to obtain geometric solutions in a widevariety of problems related to relativistic collisions. It is aright triangle that represents the relativistic relation E2 =

p2

+ m2

. For the two extreme cases (classical particle andhighly relativistic particle) the corresponding triangles are likedegenerate two-sided triangles. For a photon-like particle oneof the sides of the right triangle has exactly zero length.

The diagram corresponding to the two extreme cases,that of a classical particle and that of a highly relativis-tic one, should be considered separately. The former onewill be represented by a degenerate right triangle with its

hypotenuse, E, being almost equal to the m side, whilethe latter particle will be represented again by a degen-erate right triangle with its hypotenuse, E, being almostequal to the |p| side. By Taylor expansion we see that inthe former (classical) case

E =

m2 + p2 = m

1 +

p2

m2= m +

p2

2m, (3)

the last term being the classical Newtonian kinetic en-ergy, while in the latter (ultra-relativistic) case the cor-responding relation is

E = |p| +m2

2|p|(4)

by mere similarity of the corresponding sides (m |p|)in these two extreme cases. Of course photons, or anyother zero rest-mass particle will be represented by a de-generate right triangle with two equal sides (E = |p|).

At this point we should note that the shape of anytriangle depends on the frame of reference on which the

specific particle is observed (two of its sides E, |p| areframe dependent). Therefore a geometric construction ofmany such right triangles at the same diagram will referto a single frame of reference for all particles. Althoughin algebraic analysis of problems related to relativisticcollisions we usually change frame of reference to makecomputations easier, there is no need to appeal to suchtricks in geometric solutions, since the geometric conclu-sion will be clear, independently of the frame of referenceused to draw all triangles. Actually by avoiding Lorentztransformations to shift frames of reference we avoid alot of algebraic computations.

The geometric solutions are based on conservation of

4-momentum in relativistic reactions; that is simultane-ous conservation of the total energy, and the total 3-momentum of all particles that participate in a reaction.Therefore all geometric constructions that correspond toallowed arrangements for the kinematical characteristicsof the particles involved, share common total length ofE-sides and vectorial sum of p sides.

Next, we proceed to describe some properties regard-ing the extremum of specific quantities for a number ofparticles. These general propositions, that we will provegraphically, will be later used in some of the problemsthat we will analyze by means of our geometrical method.Proposition I: The total 3-momentum of N parti-cles with energies Ei and masses mi, respectively (i =1, 2, . . . , N ), gets maximized if all particles are movingalong the same direction.Proof: This proposition is quite obvious, and there isno need to use any right triangles to prove it. Howeverthe proof is purely geometrical, as in the forthcomingpropositions and problems. Given the magnitudes of allenergies and masses, the magnitudes of all 3-momentaare fixed: |pi| =

E2i m

2i . Therefore if we connect

all these vectors head-to-tail we construct the total 3-momentum of the system of particles (which could be


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either the reacting particles or the products). From thetriangle inequality the total 3-momentum of all particlesptot satisfies the inequality

|ptot| |p1| + |p2| + . . . + |pN|, (5)

where the equality holds ifpi/|pi| = pj/|pj | for all i, j =1, 2, . . . , N , that is when all the 3-momenta are aligned toeach other (common direction of motion for all particles).Intuitively we could just say that the head of the last 3-momentum vector and the tail of the first one are furtherapart when the broken line of the corresponding vectorsis straightened.Proposition II: Two or more particles can be consid-ered as a single particle with respect to geometric repre-sentation by right triangles.Proof: It is obvious that we could always construct aright triangle with one leg equal to the total 3-momentumof the particles

ptot =i

pi (6)

and hypotenuse equal to the sum of the energies of allparticles

Etot =i

Ei (7)

since for each particle it holds Ei |pi| (the equal-ity holds only for massless particles). The latter sum(Eq. (??)) could serve as a hypotenuse since

Etot =i

Ei i

|pi|

i

pi

= |ptot|. (8)

It should be noted though that this representation isnot one-to-one; for a specific value of Etot and ptotthere is too much freedom in choosing the energy and3-momentum of each particle.

The characteristic mass of the representative right tri-angle is not equal to the sum of the corresponding par-ticles masses; it is equal to (E2tot p

2tot)

1/2, which inits turn is equal to the total energy in the center-of-momentum frame for these particles (the frame in whichptot = 0. In propositions III and IV we will actuallyreplace many particles in the corresponding inductiveproofs by a single one. As we shall see in both casesthe minimization/maximization procedure leads to equalvelocities for all particles. This is a distinct representa-tion of many particles by a single composite particle. It isthe only case where the representation is one-to-one sincethen the rest mass in the center-of-momentum frame isexactly equal to the some of all rest masses. From thepoint of view of our geometrical method there is only oneway to construct many similar right triangles with one oftheir sides (the mass sides) given; this corresponds toparticles with equal velocities.Proposition II I: For a given total 3-momentum of Nparticles, the sum of their energies

i Ei assumes its

minimum value when all particles move with the samevelocity v.Proof: Let us begin assuming that there are only twoparticles. Consider the geometric construction of Fig. 2depicting two right triangles with sides E1, m1, p1, andE2, m2, p2. From now on we will call the plane definedby the vector sides p1, p2 momenta screen, since we aregoing to use extensively this plane to move the vectors

of 3-momenta around. On this plane we have also drawnthe total 3-momentum of the two particles, which ac-cording to Proposition III is assumed to be fixed. Anypoint B on momenta screen represents a specific split forthe 3-momenta of the two particles. When two line seg-ments, AA and with lengths equal to the masses ofthe two particles are drawn perpendicular to the plane ofmomenta screen as in Fig. 2, the total length of the hy-potenuses AB, B corresponds to the total energy of theparticles. Clearly the shortest distance between the twopoints A and (which corresponds to the minimum totalenergy) will be the straight distance A. This straightline intersects vector ptot at point K, forming two sim-ilar right triangles AAK and K. From similarity ofthe two triangles the equality of the ratios of masses and3-momenta is apparent.

By induction we could generalize our result to morethan two particles: For N particles we represent the firstN 1 particles by a single right triangle according toProposition II and we minimize the energy of the N-th particle and the rest particles. Then we proceed tominimize the energy of the first N 1 particles and soon. Thus we conclude that the total energy of all particlesis minimized when the total 3-momentum is split in Nsegments proportional to the corresponding masses of theparticles. Then the right triangles that correspond to allparticles are similar triangles and thus all particles should

have the same velocity (cf. Fig. 1).Note that in the case where three or more particles are

involved we cannot draw any momenta screen and thenconstruct all the corresponding right triangles perpendic-ular to such a plane, since three or more arbitrary vectorsdo not lie in the same plane. This explains why we hadto appeal to induction to prove this proposition.

At this point we could add a bit of information toProposition II. According to Proposition III the mass ofthe composite particle described by a single right trian-gle, which represents a number of particles (see Proposi-tion II),

Mcomp = E2tot p2tot, (9)is higher than the total rest mass of the specific particles,since for a given ptot| of such particles their total energyis minimized when all particles are moving with the samevelocity. As it was noticed in the analysis of PropositionII, this optimum case corresponds to Mcomp =

i mi.

Thus

Mcomp

(Etot |min)2 p2tot =

i

mi. (10)


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E1

E2

m2

m1

p1 p2

ptot()E+E1 2 minA

B

K

FIG. 2: In order to minimize the total energy of two particleswe should split their total 3-momentum in two parts with ratioequal to the ratio of their masses (point B should be moved topoint K). The mass segments have been drawn perpendicularto the plane of momenta screen.

Proposition IV: For a given total energy ofN particlesthe sum of their 3-momenta

i pi assumes its maximum

magnitude if all particles are moving with the same ve-locity v. This is simply the inverse of Proposition III.

Proof: We shall use induction once again. Referring toFig. 3 that depicts the right triangles of two particles, it isclear that we are free to move the mass segments m1, m2(which stick out of the plane of momenta screen in a per-pendicular fashion) around as long as we keep the sum ofthe two hypotenuses, E1 + E2, represented by the lengthof segments AB and B, fixed. The total 3-momentumof the two particles is the distance between the projec-

tions of points A and B on momenta screen. In order torender this distance maximum we have to move the twomass segments as far as possible from each other. Themaximum distance between the mass segments, compat-ible with the total energy constraint, is achieved whenthe line AB is a straight line. Then the two right tri-angles that form are similar, and the ratio of the twomasses m1/m2 equals the ratio of the magnitudes of thetwo 3-momenta |p1|/|p

2|, while both 3-momenta lie along

the same line. This configuration corresponds again tocommon velocity for both particles. For more than twoparticles, we first maximize the 3-momentum of the firstpair of particles while keeping the sum of their energiesconstant. Then we replace these two particles by a single

one which moves with the common velocity of the twoparticles and has mass equal to the sum of the masses ofthe initial particles and then we proceed further. In everystep we maximize the total 3-momentum, without chang-ing the total energy, by attributing the same velocity toall particles considered up to that point.

As mentioned before, the similarity of triangles thatarises from the graphical solution of the last two proposi-tions corresponds to the case where all particles are mov-ing with the same velocity. This is also the velocity of the

E1

E2

m2m1

p1 p2

ptot

( )E +E1 2A

B

p2

p1

p tot ,max

m2

FIG. 3: The procedure of maximizing the magnitude of thetotal 3-momentum of two particles for a given total energy.Initially the two right triangles are not coplanar. By movingthe mass segments m1, m2 around we maximize the total 3-momentum by putting them as far apart as possible. Thentwo similar right triangles form. This situation correspondsto the same velocity for both particles.

center-of-momentum frame. In this frame all particles arethen at rest. Therefore this minimization/maximizationis succeeded when the particles are still in their center ofmomentum.

The following problems of relativistic reactions will bepresented in a sequence of gradual difficulty with respectto the corresponding analytic solutions for each of them.On the other hand applying the geometrical tools in orderto draw, at least qualitative conclusions, is in all cases ofthe same difficulty and much more direct. Thus, althoughin the first couple of problems one may argue that thegeometric method does not offer an easier solution thanthe usual algebraic solution, for the rest of problems the

directness of the geometrical proof is apparent.

III. PROBLEM 1

We start familiarizing ourselves with the geometricconstructions by treating accordingly the following sim-ple problem: Show that a photon cannot be split spon-taneously into two or more massive particles (like in pairproduction).

In Figure 4 we have drawn on momenta screen theinitial 3-momentum p of the photon the magnitude ofwhich equals the energy of the photon E. The 3-momenta

of two new hypothetic particles have also been drawn onmomenta screen, as well as their energies coming out ofthis plane. However, the following inequality

E1 + E2 = AB + B A > Z = E (11)

is true. Therefore, due to conservation of energy thistype of reaction is forbidden. The same is also true formore than two particles, since all but one particles couldhave been replaced by a single representative triangle ac-cording to Proposition II.


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E1 E2

m2

m1

p1 p2

pEA

B

Z

FIG. 4: The geometric construction for the spontaneous decayof a photon.

The case of a single massive particle produced by aphoton is obviously forbidden since a degenerate triangle(photon) could not be equal to a non-degenerate triangle(massive particle).

IV. PROBLEM 2

Next, we show that particle creation from a single pho-ton is allowed, as long as a massive particle (usually anatomic nucleus), assumed to be initially motionless, isused to interact with the photon, while after the reactionthis particle remains unaltered. Therefore the massiveparticle plays the role of a catalyst for such a reaction.We will also obtain the sufficient condition for this reac-tion to happen.

E12

E

M

m12

p12 p

p

E

A

B

Z

E

FIG. 5: The geometric construction corresponding to the al-lowed production of a pair of massive particles through inter-action of a photon with a massive target particle.

Let us depict the photon with a degenerate two-sidedtriangle (E = |p|), and the massive particle used tointeract with the photon again as a degenerate two-sidedtriangle (since it is motionless) with E = M (see Figure5). If two new particles (#1, #2) are finally produced wedraw on momenta screen their total 3-momentum as ap12 vector and the 3-momentum of the target particle asp. Of course p12 +p

= p, while conservation of energy

rules that

AB + B = E12 + E = E + M = Z + Z, (12)

where the subscript 12 is used to denote the 3-momentumand energy of a composite particle that corresponds tothe two new massive particles that are produced (seeProposition II) and p, E are the 3-momentum and the

energy of the massive target after the collision, respec-tively. The composite particle is depicted by the righttriangle AB. The massive target particle, after the re-action will in general be set in motion, and the corre-sponding right triangle for this particle is BZ. There-fore the reaction is allowed to take place, as long as thegeometric equality of Eq. (12) holds good.

A

m1

m2 m2

p2p1

p1

p

f

FIG. 6: The configuration of right triangles which correspondsto a minimum total energy. This configuration yields thethreshold energy of a photon that could lead to pair produc-tion, if the angle is such that Z + Z = A.

Now let us find the condition for such a reaction to bepossible to happen, that is to find the minimum energyfor the photon in order to be able to produce two massiveparticles. From Proposition III, the minimum total en-ergy of particles #1, #2, and the massive target-particleis achieved when all 3-momenta p1, p2, and p are par-allel to each other and have magnitudes proportional totheir masses, m1, m2, and M, respectively. This extremalsituation, if allowed, sets the threshold condition for thereaction to take place. Therefore, the necessary conditionfor the pair production (via a motionless massive target)to take place is the initial available energy E+ M to beequal to the minimal total energy of all products. Thegeometric construction of triangles that corresponds tothis extremal situation is shown in Figure 6, and the re-quired condition is to find a point along the dashed linewhich lies at a distance m1 + m2 + M from point A sothat the length of A equals the length of Z + Z. Itis intuitively obvious that such a point always exists,since while the point moves along the dashed line thedifference between the two lengths varies continuously,with the former length (A) being longer than the latterone (Z + Z) when the point lies near the projectionof A on the dashed line, while the two lengths are in


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opposite relation to each other when point is moved far away (Z >> m1 + m2 + M) since then

A =

(Z)2 + (m1 + m2 + M)2 Z < Z + M. (13)

Therefore, whatever the mass of the target, there is always a sufficiently energetic photon that could lead to the

production of the two specific massive particles. Let us now quantify this conclusion. From the triangles depicted inFigure 6 we have

(E1 + E2 + E)min = A = E

(thres) + M = A(cos +

M

M + m1 + m2sin ), (14)

where = BA. This leads to the algebraic relation

cos +M

M + m1 + m2sin = 1. (15)

Expressing all trigonometric functions shown up in the above formula in terms of tan , we obtain the solution for

tan =2M(M + m1 + m2)

(M + m1 + m2)2

M2

. (16)

Therefore the photon should have at least energy equal to

E(thres) = |p(thres) | =

M + m1 + m2tan

= (m1 + m2)

1 +

m1 + m22M

, (17)

in order to be able to produce the particular pair of particles.

If the photon has energy higher than this minimumvalue, the conservation of energy is not satisfied by thisoptimum configuration (it would then be Z + Z > Ain Fig. 6). One should then arrange the 3-momenta of all

particles after the reaction in such a way that the totalenergy exceeds this minimum total energy by breakingthe constraint of having all 3-momenta corresponding toequal velocities for all three particles, thus permitting theline A to be a broken one as in Fig. 5.

From Eq. (17) we observe that for M >> m1 + m2,the threshold energy for the photon is approximately

E(thres) m1 + m2. Now we will directly demonstrate

this fact geometrically without going through the generalresult (17). We draw a segment of length Z = M, andwith center Z we draw a circle of radius r = E < M.We also draw another circle of radius R = E + M(which is the initial energy of the system), having its

center at . The broken line ZA (see Fig. 7) withZ = = 90, and A = m1 + m2 constitutes the con-figuration of masses and 3-momenta for the products ofthe reaction which corresponds to the minimum total en-ergy (cf. Fig. 6). In this case the initial 3-momentumof the photon is distributed proportionally to M andm1, m2 after the reaction in such a way that all particle-triangles are similar to each other. From the drawingin Fig. 7 it is clear that the radius of the small circle

E(thres) should be approximately equal to A, that is

equal to m1 + m2 when M >> m1 + m2 with a relativeerror of order r/(2R) = (m1 + m2)/[2(M+ m1 + m2)] =(m1 + m2)/(2M). This error represents the missing seg-ment in order to form an exact square that is shown in

Fig. 7, and it actually leads to the exact expression forthe threshold energy (see Eq. 17).

V. PROBLEM 3

The following problem is something like a generaliza-tion of Problem 2. Let us assume that a particle A hits amotionless target-particle B, and a number of particles Ci(i = 1, 2, . . .) emerge after the collision (the initial parti-cles could be among the products). The question is whatis the minimum energy of particle A in order to renderthis particular reaction possible to happen. Any particle,

apart of B, could be massless.Following the same line of arguments as in Problem

2, we look for the minimum energy required to obtainthe optimal final configuration, which is the one withall particles moving with the same velocity (accordingto proposition III). We now discern two cases: (a) Theproduced particles have total mass less than, or equal to,the total mass of the initial particles. (b) The producedparticles have larger total mass than the initial particles.

In case (a) the optimum configuration cannot comply


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m m1 2+

Rr

r

R

2

/2

FIG. 7: In the case of a very massive target, the configurationthat corresponds to the threshold for a pair production isrepresented by the depicted geometrical structure: Z = M,

Z = r = E(thres) = |p(thres) |, A = m1 + m2, and A =R = M + E(thres) . From this drawing it is obvious that for

R >> r, m1 + m2 = A = r = E(thres) . The error of thisapproximation is of order r2/2R (deviation of a circle from

its tangent).

with the conservation of energy. The argument goes asfollows: if the total mass of the products was equal tothe total mass of the initial particles, the optimal con-figuration for the products would correspond to a righttriangle (see Fig. 8) with vertical sides equal to the to-tal 3-momentum (ZA), and the total mass of them (A),respectively (remember that in the minimum energy con-figuration according to Proposition III all particles aremoving at the same speed; consequently all right trian-

gles are similar and they could all be combined in a sin-gle right triangle with its orthogonal sides equal to thesum of all masses, and the total 3-momenta, respectively,while the hypotenuse is equal to the total energy.) Thusthe energy would be represented by the segment Z .However, from triangular inequality, this is shorter thanZB + B which represents the total energy of the initialparticles. For lower total mass of products, the inequal-ity is even stronger; then the segment Z(a) in Figure8, which represents the energy of the produced particlesin their optimum configuration is shorter than the avail-able energy of the initial particles. Thus in case (a) wecould increase the final energy (in order to comply withenergy conservation), by choosing a non optimal config-uration for the products, without changing the total 3-momentum of the system of particles. Thus the reactioncould happen by following such a non-optimal configura-tion of 3-momenta.

In case (b) we will look for the minimum required 3-momentum of particle A to make conservation of energyhold under the optimum configuration of the products;if |pA| = |p| exceeds this threshold value, a non-optimalarrangement of products 3-momenta could again be fol-lowed to make conservation of energy hold.

A

(b)

p

C(b)

EA

Etot

MBMA

C

(a)

(a)

FIG. 8: Both cases, (a) and (b), are depicted in this diagram.Case (a): when MC MA + MB, in order to comply withenergy conservation, we should have a non optimum config-uration; therefore there is no restriction for EA. Case (b):when MC > MA+ MB, there is a minimum value of |pA|, andthus a minimum value of EA, for which the reaction is allowedto happen by following the optimal energy configuration forthe products.

In Figure 8 we have also drawn the triangle that cor-responds to the optimal configuration for the productsin case (b), which is supposedly compatible with the en-ergy conservation: AZ is the threshold magnitude of 3-momentum of particle A (and thus of all products), ABis the mass of the incident particle, A, B is the massof the target particle B, and A(b) = MC is the sumof masses of all products. From conservation of energyEtot = EA + MB. Thus

(AZ)2 + (A(b))2 =|p|2 + (M(b)C )

2 =

E

2

tot =(EA + MB)

2

,

(18)

and

(AZ)2 + (AB)2 = |p|2 + (MA)2 = E2A. (19)

Subtracting these two relations and solving for EA weobtain the desired threshold energy

EA =M2C M

2A M

2B

2MB, (20)

which is definitely positive since M(b)C > MA + MB. We

should note once again that this final result generalizes

the result of the previous problem since by setting MA =0 (photon), MB = M, and M

(b)C = m1 + m2 + M we

obtain Eq. (17).

VI. PROBLEM 4

What are the possible arrangements of the 3-momentaof two particles as a result of an elastic collision betweenthem? By elastic collision we mean that the two particles


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remain unaltered after the collision, preserving their totalkinetic energy, while no new particles are created.

This is a problem that most clearly demonstrates thepower of geometric constructions in getting qualitativeresults with respect to relativistic particle collisions. Theclassical analytic solution (see [5]) requires first to shiftto the center-of-momentum frame by applying Lorentztransformations on the initial 4-momenta, analyze the

possible arrangements of the particles 4-momenta in thisparticular frame of reference after the collision, and fi-nally go back to laboratory frame by performing aninverse Lorentz transformation of the post-collision 4-momenta.

In our graphical method, we draw on momenta screena hypothetical possible configuration of the two particles3-momenta after collision. From conservation of total 3-momentum these vectors should sum up to the total 3-momentum of the particles before collision. Thus the twovectors on momenta-screen should be the two sides of atriangle with a fixed third side (the total 3-momentum).On the other hand, conservation of total energy means

that the two 3-momentum vectors should have such mag-nitudes that the corresponding energies sum up to a fixedvalue (the total energy of the initial particles). Actually,the specific arrangement of the particles 3-momenta be-fore collision is one of the solutions of all possible con-figurations we are looking for. One should keep in mindthat the orientation of the plane of momenta screen is notknown a priori, since only the total 3-momentum vectoris given; thus the plane on which all configurations of 3-momenta that are compatible with conservation of both3-momentum and energy are drawn, should be rotatedaround the line of total 3-momentum to get all possibleconfigurations of 3-momenta in space. For the momentwe will just fix the orientation of this plane to find thelocus of the heads of the particles 3-momentum vectorsthat correspond to all possible relative arrangements ofthe two particles 3-momenta on this particular plane.

To find these arrangements, we draw two segments(see Fig. 9) of length m1, m2, respectively, perpendic-ular to this plane so that one edge of the first segmentis the tail of the first particles 3-momentum (which isalso the tail of the total 3-momentum), while one edgeof the second segment is the head of the second parti-cles 3-momentum (which is also the head of the total3-momentum). Both segments have been drawn on thesame side of the plane (although they could be drawnon opposite sides without affecting the results). Then,the broken line AKB connecting the free edges of theperpendicular mass-segments, A, B, which lie out of theplane, through point K which lies on moment screen (thispoint is representing a possible arrangement of the two3-momenta vectors) has length equal to the total energyE1 + E2 of the two particles (AK and KB are the hy-potenuses of the two right triangles that correspond tothe two particles). The latter quantity is fixed and equalsthe total energy of the particles before collision. Thus allpossible arrangements of 3-momenta are described by the

A

B

E1

E2

m1

m2

p1

p2

ptot

K

mom

enta

scre

en

FIG. 9: The intersection of the ellipsoid of constant energyby the plane of momenta screen provides all possible configu-rations of particles 3-momenta in an elastic collision.

locus of a point that lies on momenta screen while thesum of its distances from the given points A, B is equal tothe total energy. Obviously the points K that satisfy theabove requirements are described by the intersection ofthe plane of momenta-screen with the axially symmetricellipsoid with focuses A, B and major axis equal to thetotal energy of the particles. The intersection of a planewith an ellipsoid is definitely an ellipse. The explanationis simple. Such an intersection is clearly a closed curve,and since the ellipsoid is described by a quadratic polyno-

mial of cartesian coordinates, its intersection by a planeis described by a quadratic relation as well. However, themost general closed plane curve of quadratic form is anellipse. As mentioned above, we should finally rotate thiselliptical curve around the line of the total 3-momentumto obtain all possible configurations of p1, p2. It is easyto see that this line coincides with the major axis of theconstructed ellipse, by reflection symmetry of the whole3-D construction with respect to the plane defined by thetwo parallel mass segments (see Fig. 9).

It is remarkable that we have arrived at this conclusion,with respect to all possible arrangements of 3-momenta,by simple geometric arguments without resorting to any

mathematical relations at all.Now let us proceed to quantify our geometric result.We will use the following notation to simplify our for-mulae: E E1 + E2 is the total energy of particles,

p |ptot| = |p1 + p2| is the magnitude of the total 3-momentum, |m1 m2| and m1 + m2. Firstwe write the equation that describes the ellipsoid withfocuses A, B, in cartesian coordinates:

z2

E2+

x2 + y2

E2 p2 2=

1

4, (21)


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z

x

b=a

e

(1

-

)2

1/2ae

a

p

2a E=

A

B

momenta

screen

/2

z

FIG. 10: A 2-D projection (a ground plan) of Fig. 9 on whichthe various lengths related to the ellipsoid are written down.The semi-major axis of the ellipsoid is a = E/2, while the

focal distance AB is 2ae =p

p2 + 2 (e is the eccentricity of

the ellipsoid). Finally the semi-minor axis is b = a

1 e2 =(p

E2 p2 2)/2. All these parameters have been usedto construct the equation describing the plane of momentascreen which is drawn here as a line that intersects both z-and x-axis. The y-axis is perpendicular to the plane of thediagram.

where the z-axis lies along the line AB while the x- and y-axes are perpendicular to the z-axis, with the y-axis beingparallel to the plane of momenta screen. The origin of

this cartesian coordinate system is at the middle point ofthe segment AB, which is also the center of the ellipsoid.On the other hand the plane of momenta screen in thesame coordinate system is described by the equation:

x =

p

z

p2 + 2

2

, (22)

as one could infer from the lengths of various segmentsshown in Fig. 10. Finally if we use a new z-axis definedas the line formed by the intersection of momenta screenby the plane y = 0, with its origin (z = 0) coincidingwith the origin of ptot (empty circle on the z

axis inFig. 10), the relation between the old z- and the newz-values of a point along the z-axis is

z = z

p2 + 2

p+

p2

2p. (23)

Now, if we seek for a simultaneous solution of Eq. (21)

and of Eq. (22) (intersection of the ellipsoid by the planeof momenta screen) and we replace the z-values by itsequivalent z-values we arrive at the general form of theellipse describing the locus of the head points of all pos-sible configurations of 3-momenta vectors after the colli-sion. The corresponding equation is

1

p2

E2

p

p

2

1 +

E2 p2

2+ p2 =

(E2 p2 2)(E2 p2 2)

4(E2 p2), (24)

which is clearly the equation of an ellipse, as it was antici-pated by the aforementioned geometric arguments. Sinceon the plane of momenta screen we draw the 3-momentavectors of the particles after the collision, we have re-placed the z and the y, with p and p, respectively.The and notation corresponds to the components ofeach 3-momentum vector that is parallel or perpendicu-lar to the total 3-momentum vector. Constructing theequation of the ellipse that is describing all possible con-figurations of 3-momenta might be a bit tedious but it isstraightforward compared to the usual analytic method.

We could also draw further conclusions from our ge-ometric construction alone, without any reference toEq. (24). (a) In the non-relativistic limit the two mi-segments (i = 1, 2) are so large with respect to the mag-nitude of the total 3-momentum (mi >> |pi|), that wecould think of both mass segments as lying along thesame line perpendicular to the plane of momenta screen.In this case the ellipsoid of constant energy is an ex-tremely elongated ellipsoid oriented so that its axis ofsymmetry (major axis) is almost perpendicular to theplane of momenta screen. The intersection of such an

ellipsoid with the plane of momenta is the well knowncircular locus of momenta vectors in elastic collision inclassical mechanics (see [4]).

(b) In the relativistic regime the extreme case of a sin-gle point intersection of the ellipsoid with the momentascreen corresponds to a configuration where the plane ofmomenta is tangent to the ellipsoid. Due to reflectionsymmetry with respect to the x z plane (cf. Fig. 9),this intersecting point is along the direction of the to-tal 3-momentum vector (see Fig. 11); therefore the twoparticles can only move along the same direction. More-over, there is a geometric property according to whichthe tangent line to an ellipse forms equal angles withthe focal radii at the point of contact [6]. The tangentsof these angles in our geometric construction are noth-ing but the two particles velocities; thus this singularcase corresponds to two particles moving with exactlythe same speed, one behind the other. Of course suchparticles will never collide; they will always move main-taining their distance. So a single point intersection inour geometrical solution corresponds to this trivial phys-ical configuration.


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10

p1 p2

m1m

2

FIG. 11: When the moment screen (dashed line) is tangentto the ellipsoid of constant energy there is only a single ar-rangement of 3-momenta. The two equal angles correspondto equal velocities for the two particles; thus corresponds toa non-collisional configuration (case b).

m1

m2

(a)

m1

(b)

m2

p

p

p1 min p2 maxp1 min

p1 back

FIG. 12: The relative position of the ellipsoid and the plane ofmomenta screen corresponding to the two cases of comment(c) where p2 = 0: The top drawing (a) is for m1 < m2 andthe bottom (b) for m1 > m2. Both drawings are planar crosssections perpendicular to the momenta-screen (dashed line)as in Figs 10,11.

(c) When one of the particles lets say particle #2is initially at rest, the ellipsoid intersects the momentascreen at one of the edges of the vector p. This hap-pens because the initial (before the collision) configura-tion (p2 = 0, p1 = p) should correspond to one of all pos-sible arrangements of 3-momenta vectors. In this case,that point is the terminal point of p. Now it is clearthat if particle #1 is the lighter one (see Fig. 12(a)) theelliptical locus of the possible arrangements of p1 (thisellipsis is on momenta screen, therefore it is not shown in

Fig. 12) intersects the line of p at the terminal point ofp and at a second point that lies along the opposite di-rection of p. This means that the light particle could bebackscattered. On the other hand for a heavy particle #1(with respect to particle #2) the corresponding situationis depicted in Fig. 12(b). The second intersecting pointof the ellipsoid with the plane of 3-momenta now is an in-termediate point along the vector p. Therefore, a heavy

incident particle cannot be backscattered. Moreover themaximum angle at which particle #1 could be deflectedcorresponds to the tangent line from the starting point ofp to the ellipse on momenta screen that describes all 3-momenta arrangements. Algebraic manipulations of thestandard form of the equation for the tangent line of anellipse from a given point [6] lead to the following valuefor this angle:

tan 1,max =P

(P1 + P2/2)2 (P2/2)2, (25)

where P1 is the minimum magnitude of p1 (it is equal

P1

m1

m2

P2

1,max P

p

FIG. 13: Top: A ground plan of the ellipsoid of constant totalenergy with respect to momenta screen (dashed line). Bot-tom: The elliptical locus of all possible arrangements for thetwo 3-momenta which arises from the intersection of the ellip-soid with the plane of momenta screen. This is the portrait ofmomenta screen corresponding to the above relative positionof the ellipsoid and the momenta screen. The P, P1 and P2lengths used in Eq. 25 are shown on this diagram.

to that part of p that is left out of the ellipsoid), P2 isthe maximum magnitude of p2 (it is the part of p thatlies inside the ellipsoid), and P is the semi-minor axisof the ellipse (the maximum magnitude of p1 and p2that is perpendicular to p). All these elements are eas-ily computed from expression (24). The correspondingexpressions yield the following values:


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11

P = max[p] =

(E2 p2 2)(E2 p2 2)

4(E2 p2)

P1 = min[p|p=0] =p

2

1 +

E2 p2

(E2 p2 2)(E2 p2 2)E2

4(E2 p2)2

P2 = max[p|p=0] min[p|p=0] = (E2 p2 2)(E2 p2 2)E2(E2 p2)2 .(26)

After quite some algebra we obtain the following simple value for this angle:

tan 1,max =m2

m21 m22

. (27)

Oddly enough, this happens to be exactly the answer one gets for the analogous classical non-relativistic elasticcollision (see [4]).

(d) In a relativistic billiard game (the analogue of theclassical non-relativistic one), the major axis of the el-

lipse describing all 3-momenta configurations is equal tothe magnitude of the total 3-momentum (one of the twoidentical balls is assumed to be initially at rest). Hencethe angle between the lines of motion of the two ballsafter their elastic collision is the inscribed angle whichsubtends a diameter of the elliptical locus of 3-momenta(cf. Figure 14) and with its vertex at the starting pointof the total p (the leftmost point of the ellipse). Thisangle lies between 90 and min = 2tan1(b/a), whereb, a are the semi-minor and the semi-major axis of the el-lipse, respectively. The former happens when one of theballs is almost still after the collision (the angle formedby the dotted lines in Fig. 14), while the latter corre-

sponds to the case of equal 3-momenta after collision.The non-relativistic locus of momenta is a circle insteadof an ellipse (cf. case (a)); all inscribed angles then thatsubtend to a diameter are equal to 90, and the balls al-ways move perpendicular to each other after the collision,as is well known by billiard players.

VII. PROBLEM 5

In beta decay, what is the range of kinetic energy of theelectron produced, and what is the distribution functionof that energy?

This is a reaction where a single particle (in the case ofbeta decay this particle is a neutron) spontaneously dis-integrate into three particles; one of them (an electronantineutrino) being a very light one. It was first foundby L. Meitner and O. Hahn that the electrons producedin beta decay have a continuous spectrum although onlytwo particles (an electron and a proton) were apparentlyproduced. As we will demonstrate through geometricalarguments, production of only two particles is not com-patible with a continuous spectrum. Pauli suggested that

p

p1 p2

p2

min

FIG. 14: The 3-momenta p1, p2 of two billiard balls after anelastic collision are represented by two vectors that sum upto the major axis of an ellipse, the total 3-momentum p. Theangle between them is the inscribed angle from the leftmostpoint of the ellipse that subtends one of the diameters of theellipse. This angle ranges from 90 (approximately the angleformed by the dotted chords), to min (dashed chords).

production of a third light particle, which could not bedetected then, could resolve the mystery. Fermi bap-tized that particle neutrino, and it was actually detecteda quarter of a century later.

Lets first assume that a neutron, which is initially atrest, spontaneously decays into two particles; a protonand an electron. On momenta screen we draw (cf. Figure15) the two 3-momenta vectors so that they add up tozero (since the parent particle is assumed to be still inthe lab frame). Two segments of length mp and me aredrawn perpendicular to the momenta-screen plane at oneof the common edges of the two opposite 3-momenta, oneach side of the plane. The length of the broken line AKBconnecting the second common edges of 3-momenta, K,with the free edges A, B of the mass segments (me andmp respectively), equals the total energy of the system,that is the rest mass mn of the neutron. It is intuitivelyobvious that there is only one solution for the magnitudeof 3-momenta of the daughter particles (apart from their


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12

mp

me

Ep..

Ee..pe

pp

A

B

K

FIG. 15: There is only one way to distribute the total en-ergy in two particles that are produced from the spontaneousdecay of a neutron. The two particles should have oppo-site 3-momenta with the same magnitude. These magnitudesshould be compatible with the aforementioned conservationof energy. The mutual line of motion, though, could have anydirection.

direction). This solution is represented by the radius of

the circular intersection of an ellipsoid with its foci atA, B, and with major axis equal to mn with the plane ofmomenta screen. Therefore, if there was no extra particleproduced, the electron should be monoenergetic.

Now if we allow for one more particle (more than oneextra particles could be considered equivalent to a singlecomposite particle; cf. Proposition II), there is a con-tinuous sequence of arrangements for the 3-momenta ofproton and electron, and correspondingly energies, thatcould accommodate an extra particle. More specificallywe will focus our attention on a zero-rest-mass parti-cle, like what was assumed for many years for the neu-trinos (practically it could be considered as such, com-pared to the other two heavy-mass particles). The three

3-momenta drawn on momenta screen should still sumup to zero, forming a generic triangle KM. Since thenew particle has no rest mass, the three-segment bro-ken line AKB (K representing the new particles 3-momentum), should have total length equal to mn. It iseasy to see that we could continuously deform the triangleof 3-momenta KM while keeping the total energy (rep-resented by the length of AKB) fixed. This is a simplepictorial argument that demonstrates why the electronsin beta decay should come out with a continuous spec-trum.

Next we will turn into a more quantitative analysis ofall possible arrangements for the energies of the three

particles. First of all we can convince ourselves thatan electron could be produced with no kinetic energyat all. This arises when the KM side of the triangle of3-momenta has zero length (K and M points coincide),while the third vertex is such that

AM + M + B = me + M +

(M)2 + m2p

= mn. (28)

This situation is unique with respect to the magnitude

mp

me

Ep

Ee.

pe

pp

A

B

K

p M

FIG. 16: A third massless particle among the products allowsfor a wide range of energies for the electron. Any choice for thepoints K, on momenta screen, such that AK + K + B =mn, represents a distinct configuration of 3-momenta for theproducts of beta decay.

ofE = |p| = M, as with the decay into two particles.On the other hand, if the electron has a specific non-zero

kinetic energy, there is a whole family of arrangements forthe 3-momenta of the rest two particles. These arrange-ments are described by the intersection of the ellipsoidwith major axis equal to the rest of energy shared by theother two particles (mn me Te) and B, K as foci, withthe momenta screen. The maximum allowed energy forthe electron is such that there is but a unique arrange-ment for the other two particles momenta. In order tohave a single intersection point (unique solution) of anellipsoid with a plane on which one of the ellipsoids fociis lying, the ellipsoid should be a degenerate one withmajor axis equal to the focal distance. This is the casewhere no energy is left for the antineutrino; therefore itis exactly the situation with only two particles, an elec-tron and a proton, which was analyzed above. The wholerange of energies for the electron within this interval ofkinetic energies are actually observed in beta-decay ex-periments.

Besides predicting the range of the spectrum itself,we can also predict the distribution function for the en-ergy of the electron, at least with respect to kinemat-ics. For a given kinetic energy of the electron, whichcorresponds to a magnitude of |pe|, there is a wholefamily of vector arrangements for the 3-momenta of therest two particles. In order to visualize these arrange-ments, we should draw the ellipsoid of constant energyEp + E = mn Ee = mn me Te (this is the majoraxis of the ellipsoid), with the ending point ofpe, K, andthe free edge of the vertical segment mp, B, as its foci,and then find its intersection with the plane of momentascreen. The ellipse C (see Fig. 17) that will arise fromsuch an intersection is the locus of all possible p that arecompatible with the particular value ofTe. Therefore thepossible arrangements of the three particles 3-momentathat correspond to Te for the electron, are described bythe surface of a sphere of radius |pe| (all possible direc-tions for the electron), times the surface of the ellipsoid


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13

that comes about when the ellipse C is rotated aroundthe direction of pe (all possible inclinations of the planeof momenta around the axis of pe). Hence, the distribu-tion function of Te will be given by the product of thetwo surfaces. The surface of the corresponding sphere is

Ssph = 4|pe|2 (29)

while the surface of an ellipsoidal surface that arises fromthe revolution of C around pe is

Sell = 2ab

b

a+

sin1

1 (b/a)21 (b/a)2

(30)

where a, b are the semi-major and the semi-minor axes ofthe ellipse respectively. Finally, the derivative d|pe|/dTeis needed to transform the distribution into a distributionover Te:

P(Te) =d|pe|

dTeSsphSell = 4(Te + me)|pe|S

ell. (31)

Computing the dimensions a, b of the ellipse is straight-forward, though quite tedious. It can be performedthrough algebraic computations similar to that used inproblem 4 to obtain the intersection of an ellipsoid witha plane. In Fig. 18 we have plotted the spectrum of theelectron as a function of its kinetic energy Te (Eq. (31))for a neutron undergoing beta decay, assuming it wasinitially at rest.

It should be noted though, that the usual experimentalcurves for beta decay correspond to a metastable nucleusundergoing beta decay, instead of a single neutron. Thisaffects the value of Te,max (the maximum value of Te)and the general shape of the curve. Also, our descriptionis accurate with respect only to the kinematics implied

by special relativity, without taking into account the in-ternal dynamics of the transformation.

me

mp

Ee

Ep

pe p

pp

A

K

C

FIG. 17: The arrangements ofp and pp are described by theintersection of an ellipsoid with the plane of momenta screen,as with problem 4.

0.2 0.4 0.6 0.8 1.0TeTe,max

Te

FIG. 18: The probability distribution ofTe in beta decay dueto kinematics only.

VIII. CONCLUSION

The graphic tools we have used in this paper are sim-ply right triangles, suitably drawn in order to describethe kinematics of either simple or more complicated rel-ativistic reactions.

FIG. 19: A mechanical construction to draw right trianglescorresponding to relativistic particles that take part in rela-tivistic reactions.

Using these tools when teaching relativistic reactions,not only is fun, but also renders complicated analyticcomputations unnecessary. The student can easily visu-alize the kinematics allowed by energy and momentumconservation. The teacher can benefit from using 3-D di-agrams with right triangles, when she or he attempts toconstruct new problems with reactions. After a quickdrawing she or he could make sure that the problemhas the desired solution. Furthermore, the teacher couldmake teaching more vivid by using a simple mechani-cal construction: a simple metal planar board, rods ofadjustable length to use as mass segments with suitablemagnetic bases so that one could stick them perpendicu-lar to the board, an adjustable string and a board marker


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14

(see Fig. 19) to build 3-D diagrams that correspond toany relativistic reaction one may think of. We bet thatsuch a construction will make the analysis of relativistickinematics even more fun.

Acknowledgments

This work was supported by the research funding pro-gram Kapodistrias with Grant No 70/4/7672.

[1] H. Minkowski, Address delivered at the 80th Assemblyof German Natural Scientists and Physicians, at Cologne,1905, shown up and translated in The Principle of Rela-tivity, (Dover Publications, New York, NY 1952).

[2] E. F. Taylor and J. A. Wheeler Spacetime Physics,(W. H. Freeman and Company, New York, NY 1966).

[3] During the submission process I came across some verybeautiful recent articles by Pr. L. B. Okun which are

closely related to this topic. Thus some of the introductorynotes of the present paper are presented and explained inthe paper of Pr. Okun as well: L. B. Okun Am. J. Phys.,

77 430 (2009), and L. B. Okun Physics-Uspekhi 51 622(2008).

[4] L. D. Landau and E. M. Lifshitz Mechanics Course of The-oretical Physics vol. 1, (Butterworth-Heinemann, Oxford,1976), 3rd. ed.

[5] L. D. Landau and E. M. Lifshitz The Classical TheoryofFields, Course of Theoretical Physics vol. 2, (Butterworth-Heinemann, Oxford, 1979), 4th. ed.

[6] C. O. OakleyAn Outline of Analytic Geometry

, (Barnes& Noble, Inc., New York, NY, 1950).

relativistic reactions visualized through right triangles in space

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