lecture 12: collisions (scattering) · lecture 11: collisions (scattering) stationary particle hit...

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

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From Special Relativity by A. P. French