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The diagrammatic approach to the collision problems in Newtonian mechanics is useful. We show in this article that the same technique can be applied to the case of the special relativity. The two circles play an important role in Newtonian mechanics, while in the special relativity, we need one circle and one ellipse. The circle shows the collision in the center-of-mass system. And the ellipse shows the collision in the laboratory system. These two figures give all information on two dimensional elastic collisions in the special relativity.

Collisions of the interacting particles have fundamental importance in physics. We often use the accelerated particles to investigate the substances. Cosmic ray which is often accelerated up to almost the speed of light collides with other particles in the air. For those particles which have high energy, special relativity has to be considered to investigate the collisions.

Diagrammatic technique gives the powerful tool to investigate the collision in Newtonian mechanics [

This paper is organized in the following way. In Section 2, we recall two dimensional elastic collisions with equations. In Section 3, we show the diagrammatic approach for two dimensional elastic collision in order. First, we draw a circle for the center-of-mass system. Then we add to draw an ellipse to obtain the momentum after the collision in the laboratory system. In Section 4, we investigate the special case in which the two particles are identical. We also compare the cases that the projectile has different speed and we find that the low speed limit recovers the Newtonian case [

Let us take a look the two dimensional elastic collision for later use.

The relation between the laboratory and center-of-mass systems is governed by the Lorentz transformation [

β = p A E A / c + m B c , (1)

where c is the speed of light. The momentum p A is defined by its velocity v A as p A = m A v A / 1 − ( v A / c ) 2 and the energy is given by E A / c = p A 2 + ( m A c ) 2 . The γ-factor is obtained by

γ = 1 1 − β 2 = E A / c + m B c ( m A c ) 2 + ( m B c ) 2 + 2 E A m B . (2)

From the Lorentz transformation, the momentum of the incident particles in the center-of-mass system is given by

p ∗ = p A ∗ = p B ∗ = p A m B c ( m A c ) 2 + ( m B c ) 2 + 2 E A m B , (3)

where note that the momenta in the center-of-mass system are the same in magnitude after the collision: p ∗ = p ′ A ∗ = p ′ B ∗ .

In the same way as the Newtonian mechanics [

p ′ A x = β γ E A ∗ / c + γ p ∗ cos θ ∗ , (4)

p ′ A y = p ∗ sin θ ∗ , (5)

where E A ∗ / c = ( p ∗ ) 2 + ( m A c ) 2 = E ′ A ∗ / c . From these equations and the relation cos 2 θ ∗ + sin 2 θ ∗ = 1 , we obtain

( p ′ A x − β γ E A ∗ / c γ p ∗ ) 2 + ( p ′ A y p ∗ ) 2 = 1. (6)

This equation indicates the ellipse [

minor semiaxis p ∗ = p A m B c ( m A c ) 2 + ( m B c ) 2 + 2 E A m B , (7)

major semiaxis γ p ∗ = β γ E B ∗ / c = p A { ( m B c ) 2 + E A m B } ( m A c ) 2 + ( m B c ) 2 + 2 E A m B , (8)

eccentricity β γ p ∗ = p A 2 m B c ( m A c ) 2 + ( m B c ) 2 + 2 E A m B , (9)

midpoint of foci β γ E A ∗ / c = p A { ( m A c ) 2 + E A m B } ( m A c ) 2 + ( m B c ) 2 + 2 E A m B , (10)

are uniquely determined by the initial conditions of the collision. The energy of the target in the center-of-mass system is defined by E B ∗ / c = ( p ∗ ) 2 + ( m B c ) 2 = E ′ B ∗ / c , which is the same in magnitude before and after the collision.

In this section, we deduce all relations, which we recalled in the former section, from the diagrammatic technique.

Firstly, we draw a dashed circle whose radius is p A ∗ = p B ∗ = p ∗ in Equation (3), as depicted in

center-of-mass system. Now, we draw arrows of momenta into the circle. The momenta before the collision are supposed to be along the x-axis

O A = p A ∗ , O B = p B ∗ = − p A ∗ . (11)

After the collision, the momenta stay the same in magnitude, but change the direction

O C = p ′ A ∗ , O D = p ′ B ∗ = − p ′ A ∗ . (12)

Since the scattering angle θ ∗ = ∠ C O A cannot be determined by the conservations of momentum and energy, the point C lies anywhere on the circle and the point D is opposite side against the point C. It is determined according to what we are asked in the collision problems.

Next, as shown in

O G = p A = β γ E A ∗ / c + β γ E B ∗ / c = O E + E G (13)

is the momentum of the projectile in the laboratory system before the collision.

Next, as depicted in _{x}-axis until the broken line intersects with the ellipse. We call this point of intersection as F. Then, the vector O F = p ′ A becomes the momentum of the projectile A after the collision. The angle ∠ F O G = θ is the scattered angle of the particle A in the laboratory system. We note that the angel θ ∗ in

gives the angle θ in the laboratory system, we first draw the vector O F = p ′ A in the ellipse. Then, we trace from F to C along the broken line. The vector O C = p ′ A ∗ shows the momentum of the projectile A in the center-of-mass system. And the angle ∠ C O A = θ ∗ is the scattered angle of this system.

Next, the vector F G = O H = p ′ B shows the momentum of the target B in the laboratory system after the collision. The angle ∠ F G O = ∠ G O H = ϕ is the scattered angle of the target B. The vector O G = O E + E G = O F + O H shows the momentum conservation law p A = p ′ A + p ′ B of the collision.

The ellipse has or has not intersections with p_{y}-axis, according as m A < m B or m A > m B . It is found from the magnitude of γ p ∗ and β γ E A ∗ / c in Equations (8) and (10). The corresponding diagrams are shown in

The case m A = m B = m becomes quite simple as shown in

minor semiaxis p ∗ = m c p A 2 m c ( E A / c + m c ) , (14)

major semiaxis γ p ∗ = β γ E B ∗ / c = p A 2 , (15)

eccentricity β γ p ∗ = E A / c − m c 2 , (16)

midpoint of foci β γ E A ∗ / c = p A 2 . (17)

In this case, since γ p ∗ = β γ E A ∗ / c , the p_{y}-axis becomes a tangent to the ellipse and the tip of the vector O H = p ′ B is also on the ellipse.

The cases of which the initial speed of the projectile A has v A = 0.6 c and v A = 0.1 c are shown in

minor semiaxis p ∗ → m B m A + m B p A , (18)

major semiaxis γ p ∗ = β γ E B ∗ / c → m B m A + m B p A , (19)

eccentricity β γ p ∗ → 0, (20)

midpoint of foci β γ E A ∗ / c → m A m A + m B p A . (21)

The semiaxes become the same length and the eccentricity tends to zero. The case of the Newtonian collision problems [

We derive the diagrammatic presentation of the two dimensional elastic collision problem in the special relativity. We draw the circle for the center-of-mass system and the ellipse for the laboratory system. Those circle and ellipse show the whole story of the two dimensional elastic collisions. When we use the graph paper for drawing those figures, we are able to measure the length of momentum vectors and the scattered angle by using the ruler and the protractor. This diagrammatic technique can help us understand collision problems qualitatively and quantitatively.

The author thanks the anonymous reviewer for his helpful suggestions.

The authors declare no conflicts of interest regarding the publication of this paper.

Ogura, A. (2018) Diagrammatic Approach for Investigating Two Dimensional Elastic Collisions in Momentum Space II: Special Relativity. World Journal of Mechanics, 8, 353-361. https://doi.org/10.4236/wjm.2018.89026