Lecture 1: Addendum

Cross Section

• Let’s consider a reaction of the form 𝑎 + 𝑏 𝑐 + 𝑑

• with 2 particles in the initial state • If we regard 𝑎 as the projectile/beam particle and 𝑏 the target

particle, – the Cross Section for the above reaction is defined as the

transition rate 𝑊 per unit incident flux per target particle• Let’s now explicitly calculate what this definition means for an

infinitesimal target area element 𝑑𝑥 𝑑𝑦 transverse to the beam • Calling 𝐹 = 𝐹 𝑥, 𝑦, 𝑡 the flux of beam particles per 𝑐𝑚2 per

𝑠𝑒𝑐 at 𝑥, 𝑦, 𝑡 and 𝐷 = 𝐷 𝑥, 𝑦, 𝑡 the target density i.e. the number of target particles per 𝑐𝑚2 at 𝑥, 𝑦, 𝑡

• we obtain

𝜎 = 𝑊 =𝑑𝑇

𝐷𝑑𝑥𝑑𝑦 𝐹• where 𝑑𝑇 is the transition rate due to the 𝑑𝑥 𝑑𝑦 portion of the

target

Reversing the argument: from 𝜎 to 𝑑𝑇

• Last equation can be inverted to get the transition rate from the Cross Section, i.e.

𝑑𝑇 = 𝜎 𝐷 𝑥, 𝑦, 𝑡 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦• and therefore integrating over the beam flux and

the target density, we can derive the total rate at instant 𝑡

𝑇 𝑡 = 𝜎 𝐷 𝑥, 𝑦, 𝑡 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦• This is exactly what we use to compute the collider

luminosity• If 𝐷 𝑥, 𝑦, 𝑡 is constant for any 𝑥, 𝑦, 𝑡 where 𝐹 𝑥, 𝑦, 𝑡 ≠ 0, as for a typical fixed target experiment, we obtain

𝑇 𝑡 = 𝜎 𝐷 𝐹 𝑥, 𝑦, 𝑡 𝑑𝑥 𝑑𝑦

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 4

Understanding Synchrotron Radiation Shapes

• Actually, it’s pretty immediate.• In an inertial frame, where the particle is at rest at the time t=ti, we can legitimately calculate the

radiated field due to an instantaneous acceleration using the non-relativistic Hertz dipole formula • We can then boost it to the lab frame to obtain our result• Well, the Hertz dipole radiated field features surfaces of equal field intensity, both E and H, which are

toric and result from the rotation around the vector of a circle passing through the position of the particle at the time ti and tangent to

• Since and are normal to the vector and relatively normal too, the energy flux density, i.e. the

Poynting vector , will be directed as and have the same modulus on each point of the toric

surface.

• Therefore, the radiated energy per unit time and solid angle element about will just scale as or equivalently as

• Indeed,

)( itv

r

e

)( itv

)( itv

E

H

HE

r

d

dkdrHEdPrad 22sin

2r

2sin

)( itv

r

r

Z

X

Y

)( itv

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 5

A Further Step

• Now, boosting in a direction, we can understand how the radiated field looks for a relativistically moving particle

• To this extent, we would rather use a more intuitive argument based on the boost of the radiated photons instead of the rigorous and more complex one, which would directly transform the fields (F)

• To do this, we have simply to recall that a Lorentz boost will modify the direction of a photon independently of its energy

• In other words, if you pick a very collimated jet of photons radiated around a certain direction in the space, no matter of the energies of the single photons, it will remain a jet also after the boost: the jet can open up or shrink down by a certain amount, which does not depend from the energies of the single photons

• Even more, also the energy of each photons in the jet will change by a factor which does not depend from their energies

• This means that the radiated energy in one direction, which consists of photons of different energies, will remain energy radiated in one direction also in the new frame: the energy will be multiplied by a factor depending on its original direction only, and its direction transformed into the new one.

• Let’s justify these assertions

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 6

Boosting Photons

• Let’s consider a photon of energy E’=p’c in an inertial frame X’Y’Z’• We choose as Z’-axis that directed as , where is the boost

vector, and as Y’Z’ plane that containing the photon• ’ is the angle between the photon and the Z’-axis

X

Y

Z

zp

yp

c

c

• Let’s now boost this photon to the XYZ frame moving with speed in X’Y’Z’

cos

sin

0

00

0100

0010

00

p

p

p

p

p

p

p

z

y

x

c

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 7

A Few Mathematics Yet

• Let’s see how the photon get transformed in the new frame and it compares with the original one

dd

d

p

p

E

E

pp

p

pp

p

p

p

p

p

p

p

z

y

z

y

x

cos1

1

cos

sin

cos1

coscos

cos1

sinsin

cos

sincos1

cos

sin

0

cos

cos

sin

0

00

0100

0010

00

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 8

The Results

• For large all the photons, but the small fraction of them with , get boosted in the forward direction with a much higher energy ( ) and at an angle of the order of

• Now we can easily reproduce the shapes we start with• Have just to boost each ray/photon of the toric distribution in the inertial

frame to the lab frame moving with -c relative velocity • A little caveat yet: since the radiated energy flux are per unit solid angle and

time, we have to put all the transformed radiated photons in the corresponding solid angle and time

• i.e. gets shrank by whereas E amplified by • Therefore a total factor• Putting everything together we obtain

cos

1EE

tdd

Ed

dtd

Ed

232

2

cos1

2cos1cos1cos1

sinsin

tddtdd

ddtdddtd

tdd 2cos1

32 cos1

cos1

Acceleration parallel to the boost

• The boost is along the axis of the torus, therefore

• and the Y’Z’ section is symmetric • In this case we have

2

2232

2

sincos1

tdd

Ed

dtd

Ed

Z

r

e

Y

Calculating Equal Field Intensity Lines

• Since

• and is by definition constant on the Equal Field Intensity Lines (EFIL) , we can infer that

• This means that the EFIL are described by the vector having polar angle and azimuthal angle equal to that of the particular section of the torus we start with

• Thus, calculating and we can easily draw the lines

HE

2

22322

2

sincos1

tdd

EdrHE

dtd

Ed

2322sincos1r

r

r

EFIL for acceleration parallel to the boost

• symmetric• Radiation will be in the forward direction, no

matter about the sign of the acceleration

• Maximum radiated power at about =1/• EFIL scales longitudinally with -boost

– This is a common feature for all the EFIL

0

0,2

0,4

0,6

0,8

1

1,2

-20 0 20 40 60 80 100 120 140 160 180

boost = 10

boost =100

1/14/2021 Experimental Methods in High Energy Physics - An Introductory Course (L. Moroni) 12

Acceleration normal to the boost

• Here we do not have any more the azimuthal symmetry, thus have to distinguish between the two extreme cases for =/2 and =0

• Thus, for =/2, we have

• and for =0

Z

r

e

Y

Z

r

e

X =/2 =0

2

232

2

cos1

tdd

Ed

dtd

Ed

2

2232

2

coscos1

tdd

Ed

dtd

Ed

EFIL for acceleration normal to the boost

• For =/2

• For =0

0

0,2

0,4

0,6

0,8

1

1,2

-50 0 50 100 150 200 250 300

boost = 100

0

0,2

0,4

0,6

0,8

-50 0 50 100 150 200 250 300

bost = 100

and what about the field intensity on these EFIL?

• All the previous EFIL are normalized to maximum value of the Hertz dipole radiation, i.e.

• Therefore, the fields on them are for the same or modulus of the force in the primed reference frame where the charged particle is instantaneously at rest

• This is why the field intensity on these lines resultsabout of the same order, , and also the total radiated energy

• If we instead require and be equal in

the lab, the picture dramatically changes!

116

2

32

00

2

2

td

pd

cm

eHE

d

Pd

2

td

pd

2

dt

pd T

2

dt

pd L

HE

2

Let’s see how much it changes

• The ratio of force or to acceleration is when the force is normal to

the acceleration and in case they are parallel

• Therefore, , and for we get

• Transforming the acceleration to the particle rest frame, we obtain

and

• Since the ratio , and thus the radiated field for

acceleration normal to the boost acquire an extra factor 2

• In conclusion, for the same force acting on a charged particle moving with

velocity c in the lab, the radiated energy is a factor 2 larger when the

force is normal to the velocity

dt

pdm

m3

(see for instance Landau,The Classical Theory of Fields, paragraph 1-7 and 2-3)

cT ma

dt

pd L

L madt

pd 3dt

pd

dt

pd LT 2c

L

aa

cc aca 22c

cL ac

aca 2

2

23

L

c

a

a

td

pd

td

pd LT

Space Charge Effect in the BeamSP

AC

E C

HA

RG

E, b

y K

. Sch

ind

l, C

ERN

SPACE CHARGE, by K. Schindl, CERN

SPACE CHARGE, by K. Schindl, CERN

SPACE CHARGE, by K. Schindl, CERN

SPACE CHARGE, by K. Schindl, CERN

SPACE CHARGE, by K. Schindl, CERN

The frequent student’s question

• How could it possibly be that in the beam particle CoM frame still the particles repel each other because of the Coulomb force?

• The trick is due to the fact that the time intervals get boosted by , which approachingthe speed of light diverges

• Thus, if in t sec a particle moves away from another one by 1 cm in the CoM transversal plane, the same 1 cm movement will be seen in the lab to happen in t sec– Just for 1 TeV proton, 1 sec will be boosted to 103

sec, about 16 minutes.