FYS3500 - spring 2019

Relativistic Kinematics*

Alex ReadUniversity of OsloDepartment of Physics

*Martin&Shaw, Particle Physics, 4th Ed., Appendix A.1,A.2 (Last update 18.02.2019 16:40)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

!2

S

S’

v

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

!2

S

S’

v

β =vc

γ =1

1 − β2

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

❖ Since these are linear transformations they apply equally well to sums of these “4-vectors”.

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

❖ Since these are linear transformations they apply equally well to sums of these “4-vectors”.

❖ Check energy and momentum in S’ when a particle with mass m is at rest in S (energy=mc2):

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

❖ Since these are linear transformations they apply equally well to sums of these “4-vectors”.

❖ Check energy and momentum in S’ when a particle with mass m is at rest in S (energy=mc2):

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

px′� = −γβE/c = − mγvE′�/c = γmc2/c

E′� = mγc2

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

❖ Since these are linear transformations they apply equally well to sums of these “4-vectors”.

❖ Check energy and momentum in S’ when a particle with mass m is at rest in S (energy=mc2):

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

px′� = −γβE/c = − mγvE′�/c = γmc2/c

E′� = mγc2

(particle moves backward in S’)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Lorentz Transformations

❖ Since these are linear transformations they apply equally well to sums of these “4-vectors”.

❖ Check energy and momentum in S’ when a particle with mass m is at rest in S (energy=mc2):

!2

S

S’

v

xyzct

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

xyzct

β =vc

γ =1

1 − β2

pxpypz

E/c

′�

=

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

pxpypz

E/c

px′� = −γβE/c = − mγvE′�/c = γmc2/c

E′� = mγc2

✅

✅

(particle moves backward in S’)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

4-vector dot product

❖ Let’s define two 4-vectors and a 4-vector dot product.

!3

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

4-vector dot product

❖ Let’s define two 4-vectors and a 4-vector dot product.

!3

A =

axayaz

A0

, B =

bx

by

bz

B0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

4-vector dot product

❖ Let’s define two 4-vectors and a 4-vector dot product.

!3

A =

axayaz

A0

, B =

bx

by

bz

B0

A ⋅ B ≡ a ⋅ b − A0B0

≡ ATηB

≡ [ax, ay, az, A0]1 0 0 00 1 0 00 0 1 00 0 0 −1

bx

by

bz

B0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

1 0 0 00 1 0 00 0 1 00 0 0 −1

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

bx

by

bz

B0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

1 0 0 00 1 0 00 0 1 00 0 0 −1

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

γ 0 0 −γβ0 1 0 00 0 1 0γβ 0 0 −γ

bx

by

bz

B0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

1 0 0 00 1 0 00 0 1 00 0 0 −1

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

γ 0 0 −γβ0 1 0 00 0 1 0γβ 0 0 −γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ2(1 − β2) 0 0 −γ2β(1 − 1)

0 1 0 00 0 1 0

−γ2β(1 − 1) 0 0 −γ2(1 − β2)

bx

by

bz

B0

= [ax, ay, az, A0]1 0 0 00 1 0 00 0 1 00 0 0 −1

bx

by

bz

B0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

1 0 0 00 1 0 00 0 1 00 0 0 −1

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

γ 0 0 −γβ0 1 0 00 0 1 0γβ 0 0 −γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ2(1 − β2) 0 0 −γ2β(1 − 1)

0 1 0 00 0 1 0

−γ2β(1 − 1) 0 0 −γ2(1 − β2)

bx

by

bz

B0

= [ax, ay, az, A0]1 0 0 00 1 0 00 0 1 00 0 0 −1

bx

by

bz

B0

A′� ⋅ B′� = A ⋅ B

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Dot product of 4-vectors❖ What if we take the dot product in S’ ?

!4

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

1 0 0 00 1 0 00 0 1 00 0 0 −1

γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ 0 0 −γβ0 1 0 00 0 1 0

−γβ 0 0 γ

γ 0 0 −γβ0 1 0 00 0 1 0γβ 0 0 −γ

bx

by

bz

B0

A′� ⋅ B′� = [ax, ay, az, A0]γ2(1 − β2) 0 0 −γ2β(1 − 1)

0 1 0 00 0 1 0

−γ2β(1 − 1) 0 0 −γ2(1 − β2)

bx

by

bz

B0

= [ax, ay, az, A0]1 0 0 00 1 0 00 0 1 00 0 0 −1

bx

by

bz

B0

A′� ⋅ B′� = A ⋅ B Powerful result!

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

!5

XT = [ x , ct], PT = [ p , E/c]

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

P2 = | p |2 − E2/c2

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

P2 = | p |2 − E2/c2 (Is this a useful Lorentz invariant? Yes!)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

❖ P2 is Lorentz-invariant for a single particle, and also for a system of particles. What is it in the rest frame of a single particle, i.e. when ?

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

P2 = | p |2 − E2/c2

p = 0

(Is this a useful Lorentz invariant? Yes!)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

❖ P2 is Lorentz-invariant for a single particle, and also for a system of particles. What is it in the rest frame of a single particle, i.e. when ?

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

P2 = | p |2 − E2/c2

p = 0

P2 = − m2

(Is this a useful Lorentz invariant? Yes!)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Various dot-products❖ Let

❖ P2 is Lorentz-invariant for a single particle, and also for a system of particles. What is it in the rest frame of a single particle, i.e. when ?

❖ We identify as the invariant mass of a single particle, but also the invariant mass of a system of particles that we can calculate in the Lorentz frame with , i.e. the rest frame of the system of particles.

!5

XT = [ x , ct], PT = [ p , E/c]

X ⋅ P = p ⋅ x − Et (The factor in the exponential of the free-particle wavefunction is Lorentz-invariant)

P2 = | p |2 − E2/c2

p = 0

P2 = − m2

∑ p i = 0

W2 = − P2

(Is this a useful Lorentz invariant? Yes!)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

!6

Ebeam ⟶ ⟵ Ebeam

f

f

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

!6

e+e− → γ* → ff (e.g. μ+μ−)

Ebeam ⟶ ⟵ Ebeam

f

f

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

❖ What is the invariant mass, or to say it another way, the center of mass energy of the virtual photon?

!6

e+e− → γ* → ff (e.g. μ+μ−)

Ebeam ⟶ ⟵ Ebeam

f

f

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

❖ What is the invariant mass, or to say it another way, the center of mass energy of the virtual photon?

❖ This laboratory frame is also the center of mass frame

!6

e+e− → γ* → ff (e.g. μ+μ−)

Ebeam ⟶ ⟵ Ebeam

f

f

(∑i

p i = 0)

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

❖ What is the invariant mass, or to say it another way, the center of mass energy of the virtual photon?

❖ This laboratory frame is also the center of mass frame

!6

e+e− → γ* → ff (e.g. μ+μ−)

Ebeam ⟶ ⟵ Ebeam

f

f

(∑i

p i = 0)′�′�m′�′� ≡ s = 2Ebeam, or s = 4E2

beam

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Symmetric collider kinematics

❖ What is the invariant mass, or to say it another way, the center of mass energy of the virtual photon?

❖ This laboratory frame is also the center of mass frame

❖ With enough beam energy and the right couplings we can make a heavy particle at rest and observe its decays (e.g. the Z0 boson)

!6

e+e− → γ* → ff (e.g. μ+μ−)

Ebeam ⟶ ⟵ Ebeam

f

f

(∑i

p i = 0)′�′�m′�′� ≡ s = 2Ebeam, or s = 4E2

beam

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Fixed-target kinematics

!7

mbeam, p beam mtarget ≠ 0, p target = 0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Fixed-target kinematics

❖ What is the energy in the center of mass?

!7

mbeam, p beam mtarget ≠ 0, p target = 0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Fixed-target kinematics

❖ What is the energy in the center of mass?

!7

mbeam, p beam mtarget ≠ 0, p target = 0

W2 = (∑i

Ei)2

− (∑i

p i)2

= (Ebeam + mtarget)2

− p 2beam

= E2beam + m2

target + 2Ebeammtarget − p2beam

= m2beam + p2

beam + m2target + 2Ebeammtarget − p2

beam

= m2beam + m2

target + 2Ebeammtarget

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Fixed-target kinematics

❖ What is the energy in the center of mass?

!7

mbeam, p beam mtarget ≠ 0, p target = 0

ECM = W = m2beam + m2

target + 2mtargetEbeam

W2 = (∑i

Ei)2

− (∑i

p i)2

= (Ebeam + mtarget)2

− p 2beam

= E2beam + m2

target + 2Ebeammtarget − p2beam

= m2beam + p2

beam + m2target + 2Ebeammtarget − p2

beam

= m2beam + m2

target + 2Ebeammtarget

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Antiprotons from proton beam and target

!8

pp → ppppmbeam, p beam mtarget ≠ 0, p target = 0

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Antiprotons from proton beam and target

❖ Must have minimum energy of 4 proton masses in the center of mass:

!8

pp → ppppmbeam, p beam mtarget ≠ 0, p target = 0

ECM ≥ 4mp

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Antiprotons from proton beam and target

❖ Must have minimum energy of 4 proton masses in the center of mass:

!8

pp → ppppmbeam, p beam mtarget ≠ 0, p target = 0

ECM ≥ 4mp

ECM = m2beam + m2

target + 2mtargetEbeam

16m2p = m2

p + m2p + 2mpEbeam

7mp = Ebeam

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Antiprotons from proton beam and target

❖ Must have minimum energy of 4 proton masses in the center of mass:

❖ Need a linear accelerator with proton beam energy above ~7 GeV

!8

pp → ppppmbeam, p beam mtarget ≠ 0, p target = 0

ECM ≥ 4mp

ECM = m2beam + m2

target + 2mtargetEbeam

16m2p = m2

p + m2p + 2mpEbeam

7mp = Ebeam

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Invariant masses of unstable particles

!9

W2 = (∑i

Ei)2

− (∑i

p i)2

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Invariant masses of unstable particles

!9

W2 = (∑i

Ei)2

− (∑i

p i)2

B0s → μ+μ−

FYS3500 Spring 2019 Alex Read, U. Oslo, Dept. Physics

Invariant masses of unstable particles

!9

W2 = (∑i

Ei)2

− (∑i

p i)2

B0s → μ+μ−

H → Z0Z0* → e+e−e+e−

+μ+μ−μ+μ−

+e+e−μ+μ−