Special Relativity: Multi-Particle Systems, Elastic Scattering, and Particle Decays, Study notes of Mechanics

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Mechanics

Physics 151^ Lecture 16Special Relativity

(Chapter 7)

What We Did Last Time „^ Defined covariant form of physical quantities

„^ Collectively called “tensors”

„^ Scalars, 4-vectors, 1-forms, rank-2 tensors, … „^ Found how to Lorentz transform them

„^ Use Lorentz tensor & metric tensor

„^ Covariant form of Newton’s equation with EM force

„^ Equation of motion

Æ

„^ EM potential

Æ^ 4-vector (

φ/c,

A )

„^ EM field

Æ^ Faraday tensor

Å^

I gave you a wrong one… dp^

K μ^ d

μ =τ

Multi-Particle System „^ Consider a system with particles

s^ = 1, 2, …

„^ Total momentum „^ Equation of motion for each particle

„^ EoM for the total momentum is

„^ Not a very clean equation

s s P^

p μ

μ =^ ∑

s

s dp^ s

K

μ^ d

μ

Different time for each particle!

s s^

s^ s

s^

s dp s

dP^

K

dt^

μ d

μ

μ

=^

∑^

∑

Trouble ahead…

Momentum Conservation „^ Imagine a 2-particle system with no external force

„^ But there are internal forces between the particles „^ Law of action and reaction

„^ To conserve the total momentum in all frames,the particles interacting with each other must have thesame velocity

s s^

s^ s

s^

s dp s

dP^

K

dt^

γ d

=^

∑^

∑ 1 2

2 1 K^

K

→^

→ = −

1 2

1 2

1

2 1

2

dP^

K^

K

dt

γ

γ

→^

→

=^

+^

This is zero only if

1

2

γ^

Is this a weird restriction, or what?

Particle Collisions „^ Interactions between particles must be local

„^ Force exchanged when they collide „^ Free motion between collisions

„^ Consider the collision as a black box

„^ We don’t know what happens in the box (not classical) „^ Motion outside the box is easy

Æ^ Relativistic Kinematics

„^ How much can we learn without opening the box?

Center-of-Momentum Frame „^ Local interactions conserve total 4-momentum

„^ i.e., total energy and total 3-momentum are conserved „^ We know how to Lorentz transform it

„^ Define the center-of-momentum frame in which

„^ It’s the frame in which the total 3-momentum is zero „^ Or, the center of mass is at rest „^ Often called center-of-mass frame as well

1

n s s

E

p^

p^

c

μ

⎛μ =

=^

= ⎜^

⎝^

∑^

p

n s^1 s E^

E =

=^ ∑

n = s ∑^1 s = p^

p

i^

i

i^

i

p^

p^

L p

L p

μ

μ

μ^ ν^

μ^ ν ν

ν

′^

=^

=^

∑^

∑^

as usual

s^

s

s^

s

p^

p^

L p

L p

μ

μ

μ^ ν^

μ^ ν ν

ν

′^

=^

=^

∑^

∑

p

Two-Particle Collision „^ Consider collision of particle 1 on particle 2 at rest

„^ Total 4-momentum is „^ Total CoM energy „^ Boost of CoM frame is

1

1

1 (^

,^ )

p^

E^ c =

p^

2

(^ ,^2

p^

m c =

2

1

(^

,^ )

p^

E^ c

m c =^

+^

p

2

2

2

2 4

2

1

2

1 2

(^

)^

E^

p^ p c

m^

m^

c^

E m c

μ^ μ ′^ =^

=^

+^

Fixed-target collision E’^ grows slowly with

E^1

1

1 1

1

1

2

1 1

2 m

E^ c

m c

m

c^

γ m c

=^

+^

p^

v^

Approaches

v /c for large^1

E^1

Creation Threshold „^ Suppose we are trying to create a new particle

„^ In the best scenario, particles 1 and 2 merge to create a newheavy particle 3 „^ Total 4-momentum would be simply „^ Total CoM energy is

„^ How much energy

E^1

do we have to

give particle 1?^ „^ For large

m ,^3

E grows with^1

(^2) m 3

1

2

3

p^

p^

p^

p

=^

+^

2 2

2 2 3 3

3

E^

c^

p^ p

m c μ^ μ ′^

=^

=^

(^23) E^ ′ = m c

CoM energy mustmatch the mass ofthe new particle

2

2

2 4

2

2 4

1

2

1 2

3

(^

)^

E^

m^

m^

c^

E m c

m c

′^ =^

+^

+^

2

2

2 2 3

1

2

1

2 (^

m^

m^

m^ c

E^

−^ − m

Elastic Scattering „^ Particle 1 hits particle 2 and get elastically scattered

„^ Cross section is calculated in CoM frame

„^ By treating it as a central-force problem „^ Experiment is done in the laboratory frame

„^ We need to learn how to translate between the CoMand the laboratory frames

CoM ϑ^

p^1

p^3 p^4

′ p 1

′ p 3

′ p 2 ′ p 4

x y

Elastic Scattering „^ First, what’s the boost?

„^ Total momentum is „^ Let’s get

γ^ as well

1

2

1

2

1

(^

,^ )

p^

p^

p^

E^ c

m c

μ

μ^

μ =^

+^

=^

+^

p

2 2

2 2

2 2

1

2

1

1

2

1 2

(^

)^

p^ p

E^ c

m c

m c

m c

E m

μ^ μ^

=^

+^

−^

=^

+^

p

0

2 1 c^2 p^

E^

m c

=^

=^

p^

p

β

0

2 1

2 2 4

2 4

2

1

2

p^

E^

m c

p^ p

m c

m c

E m c

μ^ μ

γ^

=^

=^

+^

Elastic Scattering „^ What happens to the kinetic energy?

„^ At

Θ^ = 0

„^ With a little bit of work „^ Worst case is

Θ^ =

2

2

3

1

1

(^

cos^

)^

(^

cos^

E^

E^

p c

γ^

⎡^

=^

−^

Θ^

−^

−^

⎣^

2

2

3

1

1

(^

E^

E^

E

γ^

=^

−^

=^

Makes sense

3

(^12)

1

1

2 (

(^

cos^

(^

)^

T T

ρ^ ρ^

=^ −

−^

+^

E +

E

1

2 m^ m ρ^ =

2 1

1

1 T^ m c = E

Kineticenergies

2

3 min

2

1

1

(^ )

(^

(^

)^

T T

ρ− ρ ρ =^

+^

+^

E

Sign wrong in textbook

Elastic Scattering

„^ Non-relativistic limit „^ Ultra-relativistic limit

„^ As

T increases, the energy loss becomes very large^1

2

2 2

3 min

2

1

1

1

2 1

(^ )

(^

)^

(^

T^

m^

m^

c

T^

m T ρ− (^) ρ

=^

2 E

3 min

2

(^ )^1

(^

(^

T T

ρ − ρ =^

If^ m^1

<<^ m

, i.e., the target is heavy, 2 almost no energy is lost in the collision

2

3 min

2

1

1

(^ )

(^

(^

)^

T T

ρ− ρ ρ =^

+^

+^

E

( T )^3 min

is independent of

T^1

Particle Decays „^ A day at the B

A^ B

AR^

experiment at SLAC

„^ Collide

+^ e and

-^ e to generate a few 100,

Υ(4S) particles

„^ … each of which decays into two

(^0) B mesons

„^ … some of which decays into a

J/^ ψ

and a

KS

„^ …

J/^ ψ

decays into

+^ e and

-^ e , or

+^ μand

- μ

„^ …

K^ S^

decays into

+^ πand

- π

„^ Measure 3-momenta of the stable particles

„^ Masses known

Æ^

Calculate 4-momenta

„^ Rebuild the decay chain backwards and calculate invariantmasses of them all

Æ^

Do they match the expected masses?

J/

ψ^

mass

Combine

+ ee

-^ or

• μ^ μ

-^ and to see if they make a

J/^ ψ