DV Calculations of 4-momentum components

Calculations of 4-momentum components

Initial Conditions

Lab Frame

Figure 1: Definition of variables in the Lab Frame

Begining with the assumption that the incoming electron, p₁, has momentum of 11000 MeV in the positive z direction.

$\vec p_{1(z)}\equiv \vec p_{1}=11000 MeV \widehat {z}$

We can also assume the Moller electron, p₂, is initially at rest

$\vec p_{2}\equiv 0$

This gives the total energy in this frame as

$E\equiv E_1+E_2$

where,

$E\equiv \sqrt{p^2+m^2}$

This gives,

$E\equiv \sqrt{(p_{1})^2+(m_{1})^2}+m_{2}$

$\Longrightarrow E\equiv \sqrt{(11000 MeV)^2+(.511 MeV)^2}+.511MeV\approx 11000.511 MeV$

Center of Mass Frame

4-Momentum Invariants

Figure 2: Definition of variables in the Center of Mass Frame

Starting with the definition for the total relativistic energy:

$E^2\equiv p^2c^2+m^2c^4$

$\Longrightarrow {E^2}-p^2c^2=(mc^2)^2$

Since we can assume that the frame of reference is an inertial frame, it moves at a constant velocity, the mass should remain constant.

$\frac {d\vec p}{dt}=0\Rightarrow \frac{d(m\vec v)}{dt}=\frac{c\ dm}{dt}\Rightarrow \frac{dm}{dt}=0$

$\therefore m=const$

We can use 4-momenta vectors, i.e. ${\mathbf P}\equiv \left(\begin{matrix} E\\ p_x \\ p_y \\ p_z \end{matrix} \right)=\left(\begin{matrix} E\\ \vec p \end{matrix} \right)$ ,with c=1, to describe the variables in the CM Frame.

Using the fact that the scalar product of a 4-momenta with itself,

${\mathbf P_1}\cdot {\mathbf P^1}=P_{\mu}g_{\mu \nu}P^{\nu}=\left(\begin{matrix} E\\ p_x \\ p_y \\ p_z \end{matrix} \right)\cdot \left( \begin{matrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 &0 & 0 &-1\end{matrix} \right)\cdot \left(\begin{matrix} E & p_x & p_y & p_z \end{matrix} \right)$

${\mathbf P_1}\cdot {\mathbf P^1}=E_1E_1-\vec p_1\cdot \vec p_1 =m_{1}^2=s$

is invariant.

Using this notation, the sum of two 4-momenta forms a 4-vector as well

${\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\\vec p_1 +\vec p_2 \end{matrix} \right)= {\mathbf P}$

The length of this four-vector is an invariant as well

${\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 +\vec p_2 )^2=(m_1+m_2)^2=s$

Equal masses

For incoming electrons moving only in the z-direction, we can write

${\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)={\mathbf P}$

We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum

$\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+E_{2}\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)$

Without knowing the values for gamma or beta, we can utalize the fact that lengths of the two 4-momenta are invariant

$s={\mathbf P^*}^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=(m_{1}^*+m_{2}^*)^2$

$s={\mathbf P}^2=(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2=(m_{1}+m_{2})^2$

This gives,

$\Longrightarrow (m_{1}^*+m_{2}^*)^2=(m_{1}+m_{2})^2$

Using the fact that

$\begin{cases} m_{1}=m_{2} \\ m_{1}^*=m_{2}^* \end{cases}$

since the rest mass energy of the electrons remains the same in inertial frames.

Substituting, we find

$(m_{1}^*+m_{1}^*)^2=(m_{1}+m_{1})^2$

$2m_{1}^*=2m_{1}$

$m_{1}^*=m_{1}$

$\Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2}$

This confirms that the mass remains constant between the frames of reference.

Total Energy in CM

Setting the lengths of the 4-momenta equal to each other,

${\mathbf P^*}^2={\mathbf P}^2$

we can use this for the collision of two particles of mass m. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as

$(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$(E^*)^2-(\vec p\ ^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$(E^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$E^*=\sqrt{(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2}$

$E^*=\sqrt{E_{1}^2+2E_{1}E_{2}+E_{2}^2-\vec p_{1} . \vec p_{2} -\vec p_{1} . \vec p_{1} -\vec p_{2} . \vec p_{1} -\vec p_{2} . \vec p_{2} }$

$E^*=\sqrt{(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-\vec p_{1} . \vec p_{2} -\vec p_{2} . \vec p_{1} }$

$E^*=\sqrt{(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) }$

Using the relations $\beta\equiv \vec p/E\Longrightarrow \vec p=\beta E$

$E^*=\sqrt{2m^2+2E_{1}E_{2}(1-\beta_{1}\beta_{2}\cos(\theta))}$

where $\theta$ is the angle between the particles in the Lab frame.

In the frame where one particle (p₂) is at rest

$\Longrightarrow \beta_{2}=0$

$\Longrightarrow p_{2}=0$

which implies,

$E_{2}=\sqrt{p_{2}^2+m^2}=m$

$E^*=(2m(m+E_{1})^{1/2}=(2(.511MeV)(.511MeV+\sqrt{(11000 MeV)^2+(.511 MeV)^2})^{1/2}\approx 106.030760886 MeV$

where $E_{1}=\sqrt{p_{1}^2+m^2}\approx 11000 MeV$

Scattered and Moller Electron energies in CM

Inspecting the Lorentz transformation to the Center of Mass frame:

$\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+E_{2}\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)$

For the case of a stationary electron, this simplifies to:

$\left( \begin{matrix} E^* \\ p^*_{x} \\ p^*_{y} \\ p^*_{z}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^*\gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+m\\ 0 \\ 0 \\ p_{1(z)}+0\end{matrix} \right)$

which gives,

$\Longrightarrow\begin{cases} E^*=\gamma^* (E_{1}+m)-\beta^* \gamma^* p_{1(z)} \\ p^*_{z}=-\beta^* \gamma^*(E_{1}+m)+\gamma^* p^*_{1(z)} \end{cases}$

Solving for $\beta^*$, with $p^*_{z}=0$

$\Longrightarrow \beta^* \gamma^*(E_{1}+m)=\gamma^* p_{1(z)}$

$\Longrightarrow \beta^*=\frac{p_{1}}{(E_{1}+m)}$

Similarly, solving for $\gamma^*$ by substituting in $\beta^*$

$E^*=\gamma^* (E_{1}+m)-\frac{p_{1}}{(E_{1}+m)} \gamma^* p_{1(z)}$

$E^*=\gamma^* \frac{(E_{1}+m)^2}{(E_{1}+m)}-\gamma^*\frac{(p_{1(z)})^2}{(E_{1}+m)}$

Using the fact that $E^*=[(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2]^{1/2}$

$E^*=\gamma^* \frac{E^*\ ^2}{(E_{1}+m)}$

$\Longrightarrow \gamma^*=\frac{(E_1+m)} {E^*}$

Using the relation

$\left( \begin{matrix} E^*_{1}+E^*_{2} \\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+m\\ 0 \\ 0 \\ p_{1(z)}+0\end{matrix} \right)$

$\Longrightarrow \left( \begin{matrix} E^*_{2} \\ p^*_{2(x)} \\ p^*_{2(y)} \\ p^*_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}m\\ 0 \\ 0 \\ 0\end{matrix} \right)$

$\Longrightarrow\begin{cases} E^*_{2}=\gamma^* (m) \\ p^*_{2(z)}=-\beta^* \gamma^* (m) \end{cases}$

$\Longrightarrow\begin{cases} E^*_{2}=\frac{(E_{1}+m)}{E^*} (m) \\ p^*_{2(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} (m_{2}) \end{cases}$

$\Longrightarrow\begin{cases} E^*_{2}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (.511 MeV) \approx 53.015 MeV\\ p^*_{2(z)}=-\frac{11000 MeV}{106.031 MeV} (.511 MeV) \approx -53.013 MeV \end{cases}$

$\Longrightarrow \left( \begin{matrix} E^*_{1}\\ p^*_{1(x)} \\ p^*_{1(y)} \\ p^*_{1(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}\\ 0 \\ 0 \\ p_{1(z)}\end{matrix} \right)$

$\Longrightarrow\begin{cases} E^*_{1}=\gamma^* (E_{1})-\beta^* \gamma^* p_{1(z)} \\ p^*_{1(z)}=-\beta^* \gamma^*(E_{1})+\gamma^* p_{1(z)} \end{cases}$

$\Longrightarrow\begin{cases} E^*_{1}=\frac{(E_{1}+m)}{E^*} (E_{1})-\frac{p_{1(z)}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*} p_{1(z)} \\ p^*_{1(z)}=-\frac{p_{1}}{(E_{1}+m)} \frac{(E_{1}+m)}{E^*}(E_{1})+\frac{(E_{1}+m)}{E^*} p_{1(z)} \end{cases}$

$\Longrightarrow\begin{cases} E^*_{1}=\frac{(11000 MeV+.511 MeV)}{106.031 MeV} (11000 MeV)-\frac{11000 MeV}{106.031 MeV} 11000 MeV \approx 53.015 MeV\\ p^*_{1(z)}=-\frac{11000 MeV}{106.031 MeV}(11000 MeV)+\frac{(11000 MeV+.511 MeV)}{106.031 MeV} 11000 MeV \approx 53.013 MeV \end{cases}$

$p^*_{1} =\sqrt {(p^*_{1(x)})^2+(p^*_{1(y)})^2+(p^*_{1(z)})^2} \Longrightarrow p^*_{1}=p^*_{1(z)}$

$p^*_{2} =\sqrt {(p^*_{2(x)})^2+(p^*_{2(y)})^2+(p^*_{2(z)})^2} \Longrightarrow p^*_{2}=p^*_{2(z)}$

This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions.

$E^*=E^*_1+E^*_2=\sqrt{2m(m+E_1)}$

Final Conditions

Moller electron Lab Frame

Finding the correct kinematic values starting from knowing the momentum of the Moller electron, $p^'_{2}$ , in the Lab frame,

xz Plane

Figure 3: Definition of Moller electron variables in the Lab Frame in the x-z plane. Using $\theta '_2=\arccos \left(\frac{p^'_{2(z)}}{p^'_{2}}\right)$

$\Longrightarrow {p^'_{2(z)}=p^'_{2}\cos(\theta '_2)}$

Checking on the sign resulting from the cosine function, we are limited to:

$0^\circ \le \theta '_2 \le 60^\circ \equiv 0 \le \theta '_2 \le 1.046\ Radians$

Since,

$\frac{p^'_{2(z)}}{p^'_{2}}=cos(\theta '_2)$

$\Longrightarrow p^'_{2(z)}\ should\ always\ be\ positive$

xy Plane

Figure 4: Definition of Moller electron variables in the Lab Frame in the x-y plane. Similarly, $\phi '_2=\arccos \left( \frac{p^'_{2(x) Lab}}{p^'_{2(xy)}} \right)$

where $p_{2(xy)}^'=\sqrt{(p_{2(x)}^')^2+(p^'_{2(y)})^2}$

$(p^'_{2(xy)})^2=(p^'_{2(x)})^2+(p^'_{2(y)})^2$

and using $p^2=p_{(x)}^2+p_{(y)}^2+p_{(z)}^2$

this gives $(p^'_{2})^2=(p^'_{2(xy)})^2+(p^'_{2(z)})^2$

$\Longrightarrow (p'_{2})^2-(p'_{2(z)})^2 = (p'_{2(xy)})^2$

$\Longrightarrow p_{2(xy)}^'=\sqrt{(p^'_{2})^2-(p^'_{2(z)})^2}$

which gives$\phi '_2 = \arccos \left( \frac{p_{2(x)}'}{\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}}}\right)$

$\Longrightarrow p_{2(x)}'=\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}} \cos(\phi)$

Similarly, using $p_{2}^2=p_{2(x)}^2+p_{2(y)}^2+p_{2(z)}^2$

$\Longrightarrow p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}=p_{2(y)}^{'\ 2}$

$p_{2(y)}'=\sqrt{p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}}$

$p_{x}$ and $p_{y}$ results based on $\phi$

Checking on the sign from the cosine results for $\phi '_2$

We have the limiting range that $\phi$ must fall within:

$-\pi \le \phi '_2 \le \pi\ Radians$

Examining the signs of the components which make up the angle $\phi$ in the 4 quadrants which make up the xy plane:

$For\ 0 \ge \phi '_2 \ge \frac{-\pi}{2}\ Radians$

px=POSITIVE

py=NEGATIVE

$For\ 0 \le \phi '_2 \le \frac{\pi}{2}\ Radians$

px=POSITIVE

py=POSITIVE

$For\ \frac{-\pi}{2} \ge \phi '_2 \ge -\pi\ Radians$

px=NEGATIVE

py=NEGATIVE

$For\ \frac{\pi}{2} \le \phi '_2 \le \pi\ Radians$

px=NEGATIVE

py=POSITIVE

Moller electron Center of Mass Frame

Relativistically, the x and y components remain the same in the conversion from the Lab frame to the Center of Mass frame, since the direction of motion is only in the z direction.

$p^*_{2(x)}\Leftrightarrow p_{2(x)}'$

$p^*_{2(y)}\Leftrightarrow p_{2(y)}'$

$p^*_{2(z)}=\sqrt {(p^*_2)^2-(p^*_{2(x)})^2-(p^*_{2(y)})^2}$

Redefining the components in simpler terms, we use the fact that

$E^*\equiv E^*_1+E^*_2$ $2E^*_2=\sqrt{2m(m+E_1)}$

$E^*_2=\sqrt{\frac{m(m+E_1)}{2}}$

$p^*_{2}=\sqrt{E_{2}^{*2}-m^2}$

Initially, before the collision in the CM frame, p₂ was in the negative z direction. After the collision, the direction should reverse to the positive z direction. This same switching of the momentum direction alters p₁ as well.

Alternatively, we can also use the relation,

$sin \left( \theta_{2}'\right)=\left(\frac{p_{2(x)}'}{p_{2}'}\right)$

$\Longrightarrow p_{2}'\ sin \left( \theta_{2}'\right)=p_{2(x)}'$

$\Longrightarrow p_{2}^*\ sin \left( \theta_{2}^*\right)=p_{2(x)}^*$

However, $p_{2(x)}'\equiv p_{2(x)}^*$ Since only parallel components are altered in a Lorentz shift.

$\Longrightarrow p_{2}^*\ sin \left( \theta_{2}^*\right)=p_{2}'\ sin \left( \theta_{2}'\right)$

$\Longrightarrow \theta_{2}^*=arcsin \left(\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\right)$

This function puts limits on the angles or momentum we can use, since

Therefore, we need $\frac{p_{2}'}{p_{2}^*}sin \theta_2' \le 1$

Electron Center of Mass Frame

Relativistically, the x, y, and z components have the same magnitude, but opposite direction, in the conversion from the Moller electron's Center of Mass frame to the electron's Center of Mass frame.

$p^*_{2(x)}= -p^*_{1(x)}$

$p^*_{2(y)}= -p^*_{1(y)}$

$p^*_{2(z)}=-p^*_{1(z)}$

where previously it was shown

$p^*_{1}\equiv -p^*_{2}$

$E^*_{1}\equiv E^*_{2}$

$\theta_{1}^*=\theta_{1}^*+\pi$

Electron Lab Frame

We can perform a Lorentz transformation from the Center of Mass frame, with zero total momentum, to the Lab frame.

$\left( \begin{matrix}E'_{1}+E'_{2}\\ p'_{1(x)}+p'_{2(x)} \\p'_{1(y)}+ p'_{2(y)} \\ p'_{1(z)}+p'_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & \beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ \beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E^*_{1}+E^*_{2}\\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{(1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)$

$p^*_{1(x)}= p'_{1(x)}$

$p^*_{1(y)}= p'_{1(y)}$

$\Longrightarrow\begin{cases} E'=\gamma E^*+\beta \gamma p^*_{z} \\ p'_{z}=\beta \gamma E^*+ \gamma p^*_{z} \end{cases}$

$\Longrightarrow\begin{cases} \gamma = \frac{E'}{E^*} \\ \beta =\frac{p'_{z}}{E'} \end{cases}$

Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant

$s={\mathbf P^*} ^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2$

$s={\mathbf P}^2=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2$

Setting these equal to each other, we can use this for the collision of two particles of mass m₁ and m₂. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as

$(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E'_{1}+E'_{2})^2-(\vec p\ '_{1}+\vec p\ '_{2})^2$

$(E^*)^2-(\vec p\ ^*)^2=(E'_{1}+E'_{2})^2-(\vec p\ ')^2$

$(E^*)^2=(E')^2-(\vec p\ ')^2$

$(E^*)^2+(\vec{p'})^2=(E')^2$

$E'=\sqrt{(E^*)^2+(\vec p\ ')^2}$

Since momentum is conserved, the initial momentum from the incident electron and stationary electron is still the same in the Lab frame, therefore $p\ '_{Lab}=11000 MeV$

$E'=\sqrt{(106.031 MeV)^2+(11000 MeV)^2}\approx 11000.511 MeV$

This is the same as the initial total energy in the Lab frame, which should be expected since scattering is considered to be an elastic collision.

Since,

$E'\equiv E'_{1}+E'_{2}$

$\Longrightarrow E'_{1}=E'-E'_{2}$

Using the relation

$E'^2\equiv p^2+m^2$

$\Longrightarrow p_{1}^{'\ 2}=E_{1}^{'\ 2}-m_{1}^2=E_{1}^{'\ 2}-(.511 MeV)^2$

Using

$p^2 \equiv p_x^2+p_y^2+p_z^2$

$\Longrightarrow p_{1(z)}'=\sqrt {p_{1}^{'\ 2}-p_{1(x)}^{'\ 2}-p_{1(y)}^{'\ 2}}$

DV_RunGroupC_Moller#Momentum distributions in the Center of Mass Frame

DV_MollerTrackRecon#Calculations_of_4-momentum_components