fys3500 - spring 2019 relativistic · fys3500 - spring 2019 relativistic kinematics* alex read...
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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−