Information about the kinematics plotter

kinematics calculator and plotter | github repo for kinematics software

Show historical background and plot example

Two-body reaction kinematics

Initial state

Consider a particle with mass $m_{1}$ and kinetic energy $T_{1}$ incident on a stationary particle with mass $m_{2}$ . We can write a reaction of these particles to create two new particles as $m_{1} + m_{2} \to m_{3} + m_{4}$ . The energy of the projectile is $E_{1} = m_{1} + T_{1}$ . The momentum of the projectile is $p_{1} = \sqrt{E_{1}^{2} - m_{1}^{2}} = \sqrt{T_{1} (T_{1} + 2 m_{1})}$ .

A Lorentz invariant for the initial projectile $m_{1}$ and target $m_{2}$ system is $E_{tot}^{2} - p_{tot}^{2}$ , with $E_{tot} = (m_{1} + T_{1} + m_{2})$ and $p_{tot} = p_{1}$ . Algebraic manipulation gives the simplified Lorentz invariant $s = {(m_{1} + m_{2})}^{2} + 2 m_{2} T_{1}$ For identical particles with opposing momenta, the second term would be $2 m T \to 4 (m + T) T$ . So a pair of accelerators gets an additional factor of quadratic gain in energy available for reactions, which is why the largest accelerators like CERN have collided opposing beams of particles.

Center of momentum (CM) reference frame

In the center of momentum (CM) frame, the momenta of $m_{1}$ and $m_{2}$ are equal and opposite. The invariant quantity $s$ does not change. In the CM frame it can be expressed as $s = {(\sqrt{m_{1}^{2} + p_{cm}^{2}} + \sqrt{m_{2}^{2} + p_{cm}^{2}})}^{2}$ Since s is an invariant constant, this equation defines the CM momentum. $p_{cm} = \sqrt{\frac{{(s - m_{1}^{2} - m_{2}^{2})}^{2} - 4 m_{1}^{2} m_{2}^{2}}{4 s}}$ This is the momentum of both particles in the CM frame. To determine the rapidity $χ$ in the CM frame, we consider that a boost to the target with the CM rapidity gives the target the CM momentum: $(m_{2}, 0) \to (E_{2}, p_{2}) = (m_{2} \cosh (χ), m_{2} \sinh (χ))$ A simple fact about hyperbolic trigonometric functions, $\cosh χ + \sinh χ = e^{χ}$ , allows us to calculate the CM rapidity in closed form: $χ = \ln (\frac{p_{cm} + \sqrt{m_{2}^{2} + p_{cm}^{2}}}{m_{2}})$

So far we have found the invariant $s$ and CM rapidity $χ$ from the given initial values of the projectile mass, target mass and projectile kinetic energy. To compute the properties of the reaction products, we must also know the ejectile and recoil masses, $m_{3}$ and $m_{4}$ . For clarity, we will refer to the CM momenta of the products as $q$ , since it will generally be different from $p_{cm}$ . The relationship between the product masses, the Lorentz invariant and the CM momentum is $q = \sqrt{\frac{{(s - m_{3}^{2} - m_{4}^{2})}^{2} - 4 m_{3}^{2} m_{4}^{2}}{4 s}}$

In principle, this is all that is needed to calculate the energies and momenta of the products as a function of either lab angle or CM angle. These relationships are fully determined by conservation of energy and momentum. The quantity of interest in the relevant nuclear and particle physics experiments is the probability for a certain set of reaction products to be produced as a function of the initial energy ( $E_{cm}$ or $\sqrt{s}$ , determined by $T_{1}$ ) and angle (often $θ_{3cm}$ , inferred from the measured angle in the laboratory of $θ_{3}$ ).

Outgoing product states

In the CM frame, the reaction products each have the CM momentum. Their CM energies are $E_{3cm} = \sqrt{q^{2} + m_{3}^{2}}$ and $E_{4cm} = \sqrt{q^{2} + m_{4}^{2}}$ . The corresponding lab energies and momenta are found by applying a boost with the CM rapidity $χ$ to these values as a function of CM angle. The ejectile energy, parallel momentum and perpendicular momentum are given by $\begin{array}{rcl} E_{3} & = & E_{3cm} \cosh χ + q \sinh χ \cos θ_{3cm} \\ p_{3} \cos θ_{3} & = & q \cosh χ \cos θ_{3cm} + E_{3cm} \sinh χ \\ p_{3} \sin θ_{3} & = & q \sin θ_{3cm} \end{array}$ The same values for the recoil are given by the same boost with the sign changed on $q$ : $\begin{array}{rcl} E_{4} & = & E_{4cm} \cosh χ - q \sinh χ \cos θ_{3cm} \\ p_{4} \cos θ_{4} & = & - q \cosh χ \cos θ_{3cm} + E_{4cm} \sinh χ \\ p_{3} \sin θ_{4} & = & q \sin θ_{3cm} \end{array}$

These equations fully determine the outgoing energies as a function of angle. It is also possible to calculate the lab energy as a function of the lab angle for the ejectile and recoil. To do this, isolate the terms with the CM angle and CM momentum, then square and add them to eliminate the CM angle by using the trigonometric identity $\sin^{2} θ + \cos^{2} θ = 1$ . Further rearrangement and application of the hyperbolic trigonometric identity $\cosh^{2} χ - \sinh^{2} χ = 1$ yield quadratic equations for both the ejectile and recoil momomenta as a function of their lab angles: $\begin{matrix} p_{3}^{2} (1 + \sinh^{2} χ \sin^{2} θ_{3}) - 2 p_{3} E_{3cm} \cosh χ \cos θ_{3} + m_{3}^{2} \sinh^{2} χ - q^{2} & = 0 \\ p_{4}^{2} (1 + \sinh^{2} χ \sin^{2} θ_{4}) - 2 p_{4} E_{4cm} \cosh χ \cos θ_{4} + m_{4}^{2} \sinh^{2} χ - q^{2} & = 0 \end{matrix}$

The solution to the quadratic equation gives $p_{3} = \frac{E_{3cm} \sinh χ \cos θ_{3} \pm \cosh χ \sqrt{q^{2} - m_{3}^{2} \sinh^{2} χ \sin^{2} θ_{3}}}{1 + \sinh^{2} χ \sin^{2} θ_{3}}$ The same solution applies with the 3 subscripts changed to 4. The quantity under the radical in the second term of the numerator must be greater than or equal to zero, since the momentum must be real. If $m_{3} \sinh χ > q$ , then this implies a maximum lab angle for the ejectile of $θ_{3} = \arcsin (\frac{q}{m_{3} \sinh χ})$ The same condition and solution apply to the recoil as well, using its mass instead.

If there is a maximum scattering angle, then both signs apply in the equation for momentum as a function of angle. If there is no maximum angle, then the positive sign is correct. To see this, consider the case where one of the final state partcles is a photon. The numerator in the equation for momentum as a function of angle in the massless limit becomes $q \sinh χ \cos θ \pm q \cosh χ$ . The magnitude of the momentum must not be negative, so the positive sign is correct.

Limit for elastic scattering of target

Consider the limiting form of the equations for the recoil at very small values of $θ_{3cm}$ . For elastic scattering, $E_{4cm} = m_{2} \cosh χ$ . For infinitesimal values of the CM angle, we write its cosine as $1 - ε^{2} / 2$ and its sine as $ε$ . To leading order, we find $\begin{array}{rcl} E_{4} & = & m_{2} (1 + \sinh^{2} χ ε^{2} / 2) \\ p_{4} \cos θ_{4} & = & m_{2} \sinh χ \cosh χ ε^{2} / 2 \\ p_{4} \sin θ_{4} & = & m_{2} \sinh χ ε \end{array}$ To leading order, the kinetic energy takes the form of the non-relativistic limit and it is completely explained by the perpendicular component of the momentum. The component of momentum parallel to the incident momentum is only proportional to the infinitesimal quantity squared and we can neglect it. The target's momentum is perpendicular to the incident projectile and its limiting angle is 90 degrees, in the case of elastic scattering.

Limiting angle for elastic scattering of a heavy particle from a light particle

Consider one more elastic scattering limit. Assume that the projectile is a heavy particle and the target is a light particle. Just to be specific, let the heavy particle be a proton and the light particle be an electron. In the case of elastic scattering, the initial particles are the same as the final particles, so for our example we have $q = m_{e} \sinh χ$ . The formula for the maximum scattering angle becomes $θ_{3} = \arcsin (\frac{m_{4}}{m_{3}}) \to θ_{p} = \frac{m_{e}}{m_{p}}$ The portion written in terms of the electron and proton masses has used the small angle approximation, but this is quite valid for the mass ratio of approximately $1 / 2000$ . This result is completely general for any case of a heavy particle scattering off a light particle. Just to give one more concrete example, the maximum scattering angle of a deuteron incident on a proton is the inverse sine of 1/2, or 30 degrees.

Conversion between laboratory and CM frames

For planning and analyzing experiments, it is often useful to convert between CM and laboratory cross sections. To do this, one must know how the solid angles vary with respect to each other. Using the relativistic transformations to write the laboratory angle as a function of the CM angle, one possible expression is $\tan θ = \frac{q \sin θ_{cm}}{q \cos θ_{cm} \cosh χ + E_{cm} \sinh χ}$ Take the implicit derivative of both sides with respect to angle to find $\frac{d θ}{d θ_{cm}} = \frac{q}{p} (\cos θ_{cm} \cos θ + \sin θ_{cm} \sin θ \cosh χ)$ Finally, multiply by the ratio of sines to find $\frac{d Ω}{d Ω_{cm}} = \frac{q^{2}}{p^{2}} (\cos θ_{cm} \cos θ + \sin θ_{cm} \sin θ \cosh χ)$ The above equation was derived for the ejectile. The only difference for the recoil is that the first term in parentheses has a minus sign.