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Added images to impact parameter article.

• Loading branch information...  Roshan Sawhil authored and Roshan Sawhil committed May 3, 2019
1 parent 4a0b9f9 commit f368a95370c8e9e83c24d4cff05db496974e6efc
Showing with 12 additions and 2 deletions.
1. +12 −2 content/docs/nuclear-physics/impact-parameter.md
 @@ -14,9 +14,12 @@ Firstly, collision as in what we discuss here is slightly different from our int There are broadly two types of collisions: **elastic** and **inelastic**. The collision is elastic if initially the target was at rest and the projectile started off with some kinetic energy, and we find that after the collision, the total kinetic energy of the deflected projectile and the recoiling target equal the initial kinetic energy of the projectile. In short, the kinetic energy is conserved in such collisions. The other type of collision is the inelastic collision - the projectile loses some or all of its kinetic energy in the form of heat generated during the collision. The total energy is still conserved but not the kinetic energy alone. # The definition We suppose it should be obvious that the angle by which the projectile deflects on collision with the target, will be large if the projectile was aimed more directly towards the target (a 180 degree bounce back if it was aimed at exactly the centre of the target). We will now see how we can measure how well a projectile is aimed at the target. Firstly, we assume that the projectile is initially very far from the target and hence the target has almost no influence on it. The projectile drifts freely towards the target along a straight line. It begins curving only when it is considerably near the target. We can trace out the path the projectile will take. Let us now draw a straight line parallel to the initial straight line path of the projectile, passing through the centre of the target. The shortest distance between these two lines is what we call the **impact parameter**. We suppose it should be obvious that the angle by which the projectile deflects on collision with the target, will be large if the projectile was aimed more directly towards the target (a 180 degree bounce back if it was aimed at exactly the centre of the target). We will now see how we can measure how well a projectile is aimed at the target. Firstly, we assume that the projectile is initially very far from the target and hence the target has almost no influence on it. The projectile here drifts freely towards the target along a straight line. It may begin curving due to the force from the target only when it is considerably near the target (in case of billiard balls the deflection happens only on contact). We can trace out the path the projectile will take. Let us now draw a straight line parallel to the initial straight line path of the projectile, passing through the centre of the target. The shortest distance between these two lines is what we call the **impact parameter**. Another way to visualise this is to think in terms of what path the projectile would take if the forces on it due to the target are completely switched off. It would continue to the drift along a straight line and if it missed the target, it would remain undeflected. Then we could say the impact parameter is the distance of closest approach of the projectile. ![Impact parameter] (https://www.dropbox.com/s/z6b8stbyncr2znp/Impact%20parameter.001.jpeg?raw=1) Another way to visualise this is to think in terms of what path the projectile would take if the forces on it, if any, due to the target are completely switched off. It would continue to the drift along a straight line and if it missed the target, it would remain undeflected. Then we could say the impact parameter is the distance of closest approach of the projectile. So it must be clear that a larger impact parameter will mean that the projectile will probably miss the target and if it gets deflected due to a force field, the angle by which it does so will be small. Whereas, if the impact parameter is zero, it means the projectile is headed straight towards the centre of the target and we must expect a head-on collision. It should be noted that the impact parameter must be measured where the projectile is almost free from the influence of the target and moves in a straight line, otherwise we won't have a single value for it - for repulsive deflections, the distance between the projectile path and the line through the centre of the target will constantly increase and vice versa for attractive deflections. @@ -25,18 +28,25 @@ Let us now consider the collision between a point projectile and a hard sphere ( If the projectile itself is a sphere with finite radius \$r\$, the situation would be slightly different. If we track the centre of the projectile and trace out its path, we would call the shortest distance between this path and the line parallel to it drawn through the centre of the target, the impact parameter. Then a collision would be inevitable if the impact parameter was less than the sum of the two radii (\$r+R\$). Whereas if the impact parameter was larger than this, the projectile would miss the target and there would be no collision. Therefore the impact parameter in such a case is determined by both the size of the target and that of the projectile. ![Hard sphere scattering] (https://www.dropbox.com/s/iw5tk4htk2z1j8y/Hard%20sphere%20scattering.001.jpeg?raw=1) # Relation with scattering angle In general when a projectile collides elastically with the target, the projectile gets deflected by some angle and the target recoils. For simplicity, let us assume the target is very massive and hence its recoil is negligible. The question we would like to answer is: how is the angle of deflection, also called the **scattering angle**, of the projectile related to the impact parameter? Obviously, the smaller the impact parameter, the larger is the scattering angle. But we'd like to obtain a precise mathematical relation between the two. Let us denote the scattering angle by \$\theta\$. This is the angle between the straight line path of the projectile before the collision and its new deflected path after collision. Let \$b\$ be the impact parameter. Now, the projectile bounces off the target in such a way that the angle that the incident path makes with a line stretching through the centre of the sphere through the point of impact, equals the angle made by the deflected path with the same line through the centre. Let us call this angle \$\alpha\$. ![Scattering angle] (https://www.dropbox.com/s/fekyn4tfx8qwnir/Scattering%20angle.001.jpeg?raw=1) From geometry it is evident that \$b=R\sin\alpha\$. Also, supplementarity requires that \$\alpha+\alpha+\theta=180^o\$ or \$\alpha=90^o-{\theta\over2}\$. Hence,
\begin{align*} b&=R\sin\left(90^o-{\theta\over 2}\right)\\ b&=R\cos\left({\theta\over 2}\right) \end{align*}
This is the relation between the impact parameter \$b\$ and the scattering angle \$\theta\$ that we were looking for. Let's test it. In case of a head-on collision, we have \$\theta=180^o\$. Then the impact parameter must be \$b=R\cos(180)\$. This means, \$b=R\cos90^o=0\$. Which was expected. \$b=0\$ means the path of the projectile coincides with the line connecting with the centre of the target, leading to a head-on collision.

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