13-4 Laboratory and center of mass system 1-20-21

1

Laboratory frame and center of mass frame

Masatsugu Suzuki

Department of Physics, SUNY at Binghamton

(Date: February 01, 2021)

The scattering experiment is actually carried out in the laboratory frame, where a beam of

particles is being scattered by some fixed scattering center (target). This is called the

laboratory frame. In fact, the target is always free to move. In order to compare experiment

with theory, it is necessary to convert what is actually observed in the laboratory frame to

what would have been observed in the center of mass frame. Since the observation of

scattered flux is done with macroscopic apparatus, this is a purely classical problem. So it

can be solved using classical notions. We note that the center of mass frame is favorable

compared to the laboratory frame from a theoretical view-point. The separation of the

system from the center of mass leads to the decrease of the degree of freedom from 6 to 3.

In many cases, the scattering theory is discussed in terms of the center of mass frame.

Here the relation between the cross sections in the center of mass frame and the

laboratory frame is discussed.

1. Momentum conservation and the energy conservation

Fig. Laboratory frame. The initial momentum is L

iim 1v . The scattering angle of the

particle 1 with mass m1 is L

1 .

m1 m2

v1 iL

v1 fL

v2 fL

q1L

q2L

2

In the laboratory frame, the velocities of the particles 1 and 2 before and after collision,

are defined by

L

i1v , L

i2v , L

f1v , L

f2v ,

where

02 L

iv

since the particle with mass m2 is at rest before the collision. The center of mass velocity

is given by

L

iCMmm

m1

21

1vv

In the center of mass frame

CM

i1v , CM

i2v , CM

f1v , CM

f2v ,

We have the relation

CM

CM

i

L

i vvv 11 , CM

CM

i

L

i vvv 22 , (1a)

CM

CM

f

L

f vvv 11 , CM

CM

f

L

f vvv 22 , (1b)

where

L

iCM

L

i

CM

imm

m1

21

211 vvvv

, L

iCM

L

i

CM

imm

m1

21

122 vvvv

.

Note that

1 1 2 2 0CM CM

i im m v v or 12 1

2

CM CM

i i

m

m v v

(i) The momentum conservation in the laboratory frame;

3

L

f

L

f

L

i

L

i mmmm 22112211 vvvv .

which means that the velocity of the center of mass remains unchanged before and after

the collision. Using Eqs.(1a) and (1b), we get

)()( 221111 CM

CM

fCM

CM

f

L

i mmm vvvvv

or

0)( 21112211 CM

L

i

CM

f

CM

f mmmmm vvvv

or

02211 CM

f

CM

f mm vv , or CM

f

CM

fm

m1

2

12 vv (2)

(ii) The energy conservation law;

222

2

11

2

22

2

112

1

2

1

2

1

2

1 L

f

L

f

L

i

L

i vmvmvmvm

Using Eqs.(1a) and (1b), we get

211

2

22

2

112

1)(

2

1

2

1 L

iCM

CM

fCM

CM

f mmm vvvvv

or

22 2 2

1 1 2 2 1 1 2 2 1 2 1 1

1 1 1 1( ) ( ) ( ) ( )( )

2 2 2 2

CM CM CM CM L

f f f f CM CM im m m m m m m v v v v v v v

or

4

22 2 2

1 1 2 2 1 1 1 2

221 1

1 1 1 2

1 2

2

1

1 1 1 1( ) ( ) ( )( )

2 2 2 2

1 1( )( )

2 2

1

2

CM CM L

f f i CM

LL ii

L

i

m m m m m

mm m m

m m

v v v v

vv

v

(3)

or

22 2

1 1 2 2 1

1 1 1( ) ( )

2 2 2

CM CM L

f f im m v v v

where 1 1 2 2 0CM CM

f fm m v v , is the reduced mass and is given by

21

21

mm

mm

.

Using Eqs.(2) and (3), we get

21

2

1

2

12

2

112

1)(

2

1)(

2

1 L

i

CM

f

CM

fm

mmm vvv

or

21

21

212

1

2

2

12

112

1)(

2

1)(

2

1 L

i

CM

f

CM

fmm

mm

m

mm vvv

or

2

2 2 11

1 2

( )L

CM if

m vv

m m

, and

22

1 12

1 2

LCM i

f

m vv

m m

5

Fig. The center of mass frame. The scattering angle of the particle 1 with mass m1 is CM . In this plane, the z axis is in the horizontal direction and the y axis is in the

vertical direction.

The y and z components of CM

fv1 and CM

fv2 ;

CML

iCMCM

fy

CM

fmm

vmv sinsin

21

1211

v ,

CML

iCMCM

fz

CM

fmm

vmv coscos

21

1211

v ,

CML

iCMCM

fy

CM

fmm

vmv sinsin

21

1122

v ,

CML

iCMCM

fz

CM

fmm

vmv coscos

21

1122

v .

where the z axis is in the horizontal direction and the y axis is in the vertical direction.

Using Eq.(1b), we have

m1 m2

v1 iCM v2 i

CM

v1 fCM

v2 fCM

qCM

6

CM

CM

f

L

f vvv 11 , CM

CM

f

L

f vvv 22 ,

we have

CML

iCMCM

fyCMy

CM

fy

L

fmm

vmv sinsin

21

12111

vvv ,

L

i

CM

L

iCML

i

CM

CMCM

f

zCMz

CM

fz

L

f

vmm

mm

mm

vm

mm

vm

vv

1

21

21

21

11

21

12

1

11

)cos(

cos

cos

vvv

CML

i

yCMy

CM

fy

L

fmm

vmsin

21

1122

vvv ,

L

i

CM

L

iCML

i

zCMz

CM

fz

L

f

vmm

mm

mm

vm

mm

vm

1

21

11

21

11

21

11

22

)cos(

cos

vvv

where

L

iCMmm

m1

21

1vv

.

The scattering angle in the laboratory scheme:

CM

CM

z

L

f

y

L

fL

mm

m

cos

sintan

21

2

1

1

1

v

v

,

CM

CM

CM

CM

z

L

f

y

L

fL

mm

m

cos1

sin

cos

sintan

11

1

2

2

2

v

v

.

7

Center of mass frame. 1 21 1 1 1 1

1 2 1 2

CM L L L L

i i CM i i i

m mv v v v v v

m m m m

,

2 1

1 1

1 2

L

iCM CM

f i

m

m m

v

v v , 1 1

2

1 2

L

iCM

f CM

mv

m m

v

v . 1 1 2

L CM CM

i f f v v v .

12 1

2

CM CM

f f

m

m v v . 1 1

2 1 1

2 1 2

CM CM L

f f i CM

m mv

m m m

v v v .

1 12 1 1

2 1 2

CM CM L

i i i CM

m m

m m m

v v v v .

CM

CM

f

L

f vvv 11 , CM

CM

f

L

f vvv 22 ,

3. Example-1: m2/m1 = 2.

Here we assume that m2/m1 = 2.

(a) 80CM .

(b) 60CM .

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 f

CM

v2 f

L

CM 80 Degreem1 1m2 2

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

8

(c) 40CM

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 f

CM

v2 fL

CM 60 Degreem1 1

m2 2

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

9

(d) 20CM

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 f

CM v2 f

L

CM40 Degree

m1 1

m2 2

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

10

3. Example-2: Identical particles ( 2 1m m )

Here we consider the case of 2 1m m (identical particles)

1

12

L

iCM

f v

v , 1

22

L

iCM

f CMv v

v . 1 1 2

L CM CM

i f f v v v . 2 1

CM CM

f f v v . 2 1

CM CM

i i v v .

CM

CM

f

L

f vvv 11 , CM

CM

f

L

f vvv 22

In the laboratory scheme;

1

L

iEO v��

, 1

L

fOF v��

, 2

L

fOG v��

.

1 1

L CM CM

i i v v v , 1 1

L CM CM

f f v v v , 2 2

L CM CM

f f v v v

In the center of mass scheme:

v1 i

CMv2 i

CM

v1 i

L v1 f

L

v1 fCM

v2 fCM

v2 f

L

CM 20 Degree

m1 1

m2 2

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

11

1

CM

iAO v��

, 2

L

iBO v��

, 1

CM

fOC v��

, 2

CM

fOD v��

CMv : center of mass velocity

CMEA CF DG v

��

((Elastic scattering))

Here we show some examples for the laboratory and center of mass schemes for the elastic

scattering of two particles with the same mass, where 1

L

iv and CM are given and 2 0L

i v . We note

that the points A, B, C, and D are on the same circle with the radius ( 1 / 2L

CM iv v ) centered at

the origin O. The vectors 1

L

fv and 2

L

fv are denoted by the vectors OF��

and OG��

, respectively. The

angle FOG 90 . Because of the momentum conservation ( 1 2 1

L L L

f f i v v v ), we have

OF OG OH ��

. The point H is on the same circle with the radius ( 1

L

iv ) centered at the origin.

Since FOG 90 (which is independent of CM ), the points O, E, and F lie on the same circle

with the radius CMv .

12

Fig. Scattering of two identical particles (the same mass) in the laboratory and center of mass

schemes. 90CM . 45L .



13


schemes. 60CM . . 30L

14



15



16


schemes, CM is changed as a parameter. / 2L CM .

4. Differential cross sections in the laboratory frame and the center of mass

frame

The relation between the cross sections in the center of mass frame and the laboratory

frame can be obtained from the fact that the same particles which go into the solid angle

CMd at CM in the center of mass frame go into the solid angle

Ld at L in the

laboratory frame. Therefore, we have

v1 f

CM

v1 i

CMv2 i

CM

v1 i

L

v1 f

Lv1 f

CM

v2 f

CM v2 fL

E A

O

B

C

D F

F

H

G

vCM

vCM

LCM

x

y

v1 f

CM

v1 i

CMv2 i

CM

v1 i

Lv1 f

Lv1 f

CM

v2 fCM v

2 f

LE A

O

B

C

DF

F

H

G

vCM

vCM

LCM

x

y

v1 f

CM

v1 i

CMv2 i

CM

v1 i

Lv1 f

Lv1 f

CM

v2 f

CM v2 f

L

E A

O

B

C

DF

F

H

G

vCM

vCM

L

CM

x

y

v1 f

CM

v1 i

CMv2 i

CM

v1 i

L

v1fLv

1 f

CM

v2 f

CM v2 f

L

E A

O

B

C

DF

F

H

G

vCM

vCM

L

CM

x

y

17

CMCMCM

CM

LLL

L dd sin)(sin)( 111 ,

or

)(sin

sin)(

11

1

CM

CMLL

CMCML

Ld

d

.

Using the relation

CM

CML

mm

m

cos

sintan

21

21 ,

we have

)(cos1

)cos21()(

2/32

1

CM

CMCM

CML

L

,

where

2

1

m

m .

The kinetic energy of the laboratory frame is

2112

1 L

iL mK v .

In the center of mass frame, it is

21

21

21

2

1 L

iCMmm

mmK v

.

Then we have

LCMK

mm

mK

21

2

.

18

((Mathematica))

_______________________________________________________________________

REFERENCES

R.H. Dicke and J.P. Wittke, Introduction to Quantum Mechanics (Addison-Wesley, Reading MA,

1966).

A. Das, Lectures on Quantum Mechanics, 2nd edition (World Scientific Publishing, 2012).

J.R. Taylor, Scattering Theory: The Quantum Theory on Nonrelativistic Collisions (John Wiley &

Sons, 1972).

P.T. Martin, Introduction to Quantum Mechanics, second edition (McGraw-Hill Book Company,

New York, 1968).

A.G. Sitenko, Theory of Scattering (Pergamon Press, Oxford, 1971).

N. Zettili, Quantum Mechanics: Concepts and Applications, 2nd edition (Wiley).

APPENDIX-I: Examples.

Clear@"Global`"D; θL = ArcTanB m2 Sin@θDm1 + m2 Cos@θD F;

f1 =1

D@θL, θDSin@θDSin@θLD êê Simplify@�, Hm1 + m2 Cos@θDL > 0D & ;

f2 = f1 ê. 8m1 → γ m2< êê Simplify@�, m2 > 0D &

I1 + γ2 + 2 γ Cos@θDM3ê2

1 + γ Cos@θD

19

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 fCM

v2 f

CM

v2 f

L

CM 60 Degree

m1 1

m2 1

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

20

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 f

CM

v2 fL

CM 60 Degreem1 1

m2 2

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

21

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 f

CM

v2 fL

CM60 Degree

m1 1

m2 3

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

22

APPENDIX-II Mathematica

v1 i

CMv2 i

CM

v1 i

L

v1 f

L

v1 f

CM

v2 fCM

v2 f

L

CM 60 Degree

m1 1

m2 4

E A O B

C

D

F

G

vCM

vCM

vCM

L

CM

x

y

25

APPENDIX-III Mathemtica (identical mass)

Identical mass; 1 2m m

13-4 Laboratory and center of mass system 1-20-21

Documents

Transcript of 13-4 Laboratory and center of mass system 1-20-21