lecture 12: collisions (scattering) · lecture 11: collisions (scattering) stationary particle hit...
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Lecture 11: Collisions (scattering)
Stationary particle hit by identical particle in an elastic collision
In Newtonian mechanics angle between trajectories is 90° - prove.
But in relativistic mechanics the angle is < 90° due to increase of mass with 𝑣.
(squeezing in forward direction)
Collisions
Elastic: particles have same 𝐸0 = 𝑚0𝑐2 after collision.
Consider the special case, where after collision the particles travel at equal angles to direction of incident particle (viewed in lab frame)
To find 𝜃 in terms of 𝐸0, 𝐾1
Cons. of energy ⟹ 𝐸1 + 𝐸0 = 2𝐸2 (1)
Cons. of mom ⟹ 𝑝1 = 2𝑝2𝐶𝑜𝑠𝜃
2(2)
𝐸1 , 𝑝1 𝐸0 , 0
𝐸2 , 𝑝2
𝐸2 , 𝑝2
𝜃
2
𝜃
2
Collisions
(1) 𝐸1 + 𝐸0 = 2𝐸2
(2) 𝑝1 = 2𝑝2𝐶𝑜𝑠𝜃
2
Also in rel. mechanics: 𝐸2 − 𝑝2𝑐2 = 𝑚02𝑐4 = 𝐸0
2
Thus 𝑝12𝑐2 = 𝐸1
2 − 𝐸02 (3)
𝑝22𝑐2 = 𝐸2
2 − 𝐸02 (3)
It is convenient to introduce 𝐾1 (K.E. of incident particle)
𝐸1 = 𝐸0 + 𝐾1 (4)
Collisions
(1) 𝐸1 + 𝐸0 = 2𝐸2 (2) 𝑝1 = 2𝑝2𝐶𝑜𝑠𝜃
2
(3) 𝑝12𝑐2 = 𝐸1
2 − 𝐸02, 𝑝2
2𝑐2 = 𝐸22 − 𝐸0
2 (4) 𝐸1 = 𝐸0 + 𝐾1
(3),(4) ⇒ 𝑝12𝑐2 = 𝐸0 + 𝐾1
2 − 𝐸02 = 𝐾1(2𝐸0 + 𝐾1)
(3),(4),(1) ⇒ 𝑝22𝑐2 = 𝐸0 +
𝐾1
2
2− 𝐸0
2 = 𝐾1(𝐸0 +𝐾1
4)
Elastic collision ⇒ 𝐾𝑏𝑒𝑓𝑜𝑟𝑒 = 𝐾𝑎𝑓𝑡𝑒𝑟
∴ 2 ⇒ 4𝐶𝑜𝑠2𝜃
2=
2𝐸0 + 𝐾1
𝐸0 +𝐾14
𝐶𝑜𝑠2𝜃
2=
2𝐸0 + 𝐾1
4𝐸0 + 𝐾1
Collisions
𝐶𝑜𝑠2𝜃
2=
2𝐸0 + 𝐾1
4𝐸0 + 𝐾1
We know 𝐶𝑜𝑠 𝜃 = 2𝐶𝑜𝑠2 𝜃
2− 1
∴ 𝐶𝑜𝑠 𝜃 =4𝐸0 + 2𝐾1
4𝐸0 + 𝐾1− 1
𝐶𝑜𝑠 𝜃 =𝐾1
4𝐸0 + 𝐾1
Newtonian case: 𝐾1 ≪ 𝐸0 : 𝐶𝑜𝑠 𝜃 = 0 , θ =𝜋
2
Relativistic case: 𝐾1 𝑛𝑜𝑡 ≪ 𝐸0 ∴ 𝜃 <𝜋
2
Proton rest mass energy: m0c2 = 900𝑀𝑒𝑉
3-Dec-13 6
From Special Relativity by A. P. French
3-Dec-13 7
From Special Relativity by A. P. French
Compton Effect
Collision of photon (𝛾) with effectively free electron (𝑒)
A.H. Compton from 1919 to 1923 used x-ray photons and electrons loosely bound to atoms.
This collision is elastic in the sense that no energy is converted to other forms (rest mass).
But, energy is transferred from 𝛾 to 𝑒
ℎ𝑣 ↓ ∴ 𝜆 ↑
This is a very clear example of the particle nature of photons
Compton Effect
𝜃
𝜑
𝐸2 , 𝑝2
𝐸, 𝑝
𝐸1 , 𝑝1
𝛾𝛾
𝑚0 , 𝐸0
𝑒 stationary rest mass 𝑚0
Remember p = 𝑚𝑣 =𝐸
𝑐2 𝑣 but 𝑣 = 𝑐 for photons
∴ 𝑝 =𝐸
𝑐
𝑝1 =𝐸1
𝑐 𝑛1 𝑝2 =
𝐸2
𝑐 𝑛2
𝑛1 and 𝑛2 are unit vectors
𝑒−
𝑛1
𝑛2
𝜃
Compton Effect
Cons. energy ⇒ 𝐸1 + 𝑚0𝑐2 = 𝐸2 + 𝐸 (1)
Cons. Momentum ⇒𝐸1
𝑐 𝑛1 =
𝐸2
𝑐 𝑛2 + 𝑝 (2)
Also 𝐸2 − 𝑝2𝑐2 = 𝑚02𝑐4 (3)
𝐸1 − 𝐸2 + 𝑚0𝑐2 2 − 𝐸1 𝑛1 − 𝐸2 𝑛2
2 = 𝑚02𝑐4
𝐸12 − 2𝐸1𝐸2 + 𝐸2
2 + 2 𝐸1 − 𝐸2 𝑚0𝑐2 + 𝑚0
2𝑐4 − (𝐸12 − 2𝐸1𝐸2𝐶𝑜𝑠 𝜃 + 𝐸2
2) = 𝑚02𝑐4
2 𝐸1 − 𝐸2 𝑚0𝑐2 − 2𝐸1𝐸2 1 − 𝐶𝑜𝑠 𝜃 = 0
From (1) From(2)
Compton Effect
2 𝐸1 − 𝐸2 𝑚0𝑐2 − 2𝐸1𝐸2 1 − 𝐶𝑜𝑠𝜃 = 0
𝑜𝑟
1
𝐸2−
1
𝐸1=
1
𝑚0𝑐2 (1 − 𝐶𝑜𝑠𝜃)
𝐵𝑢𝑡, 𝐸1 = ℎ𝑣1 =ℎ𝑐
𝜆1𝐸2 = ℎ𝑣2 =
ℎ𝑐
𝜆2
∴ 𝜆2 − 𝜆1 =ℎ
𝑚0𝑐(1 − 𝐶𝑜𝑠𝜃)
Compton Effect
𝜆2 − 𝜆1 =ℎ
𝑚0𝑐(1 − 𝐶𝑜𝑠𝜃)
For 𝑒−, ℎ
𝑚0𝑐= 2.4 × 10−12𝑚 = 0.0024𝑛𝑚
Compton used 𝑥-rays with 𝜆1 = 0.07𝑛𝑚 (p.196 French)
Δ𝜆 ∝ (1 − 𝐶𝑜𝑠𝜃)
Use diffraction from crystal to measure 𝜆 of 𝑥-rays
Later, in 1950, Cross and Ramsey using 𝛾-rays of 2.6𝑀𝑒𝑉 also detected recoiling 𝑒.
• Angle between 𝑒 and scattered photon is correct
• Scattered photon and recoiling 𝑒 detected in coincidence
3-Dec-13 13
From Special Relativity by A. P. French