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…mber Version and change in ending for new Survey.
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VERSION = 5.06.01 | ||
VERSION_NUM = 50601 | ||
VERSION_DATE = 2020.09.01 | ||
VERSION = 5.07.00 | ||
VERSION_NUM = 50700 | ||
VERSION_DATE = 2021.05.02 |
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\chapter{Relation between $p_t$ and $\delta _p$} | ||
\label{chap:pt_rel_dp} | ||
In this chapter we establish the exact relation of $p_t$ and $\delta _p$ and show how it can be approximated. | ||
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Note that for simplicity $c$ has been set to 1 in this document. This does not change the outcome of the derivations. | ||
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The definition of $p_t$ is the following: | ||
\begin{equation} | ||
p_t = \frac{E-E_0}{P_0} | ||
\label{eqn:pt} | ||
\end{equation} | ||
, where $E$ is the total energy of the particle, $E_0$ is the energy of the reference particle and $P_0$ is the momentum of the reference particle. | ||
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$\delta_p$ is defined as: | ||
\begin{equation} | ||
\delta_p = \frac{P-P_0}{P_0} | ||
\end{equation} | ||
, where $P$ is the momentum of the particle. | ||
We will use the following relation in several places in the following document: | ||
$E=\frac{P}{\beta} = \frac{P_0(1+\delta _p)}{\beta}$. | ||
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The following part is a derivation of the relation between $p_t$ and $\delta_p$ in the general case. | ||
\begin{equation} | ||
E = \sqrt{P^2+m_0^2}= \sqrt{P_0^2(1+\delta _p)^2+m_0^2} | ||
\label{eqn:e1} | ||
\end{equation} | ||
, where $m_0$ is the rest mass of the particle. | ||
Rearranging equation~\ref{eqn:pt} we can also write the energy as | ||
\begin{equation} | ||
E = P_0 pt + E_0 = P_0 pt + \frac{P_0}{\beta _0} | ||
\label{eqn:e2} | ||
\end{equation} | ||
, where we used the fact that $ E_0 =\frac{P_0}{\beta _0}$ | ||
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Squaring equation~\ref{eqn:e1} and equation~\ref{eqn:e2} we can write: | ||
\begin{eqnarray} | ||
P_0^2(1+\delta _p)^2+m_0^2 = (P_0pt + \frac{P_0}{\beta _0})^2 = P_0 ^2 p_t ^2 + \frac{2P_0^2 p_t}{\beta_0}+\frac{P_0^2}{\beta_0 ^2} | ||
\label{eqn:squared_long} | ||
\end{eqnarray} | ||
We note that we can write $\beta _0 = \frac{P_0}{E_0} = \frac{P_0}{\sqrt{P_0^2+m_0^2}}$ which gives: | ||
\begin{equation} | ||
\frac{P_0^2}{\beta_0^2} = P_0^2 + m_0^2 | ||
\label{eqn:p0b0} | ||
\end{equation} | ||
Substituting equation~\ref{eqn:p0b0} in to equation \ref{eqn:squared_long} gives: | ||
\begin{eqnarray} | ||
P_0^2(1+\delta _p)^2+m_0^2 =P_0 ^2 p_t ^2 + \frac{2P_0^2 p_t}{\beta_0}+P_0^2+m_0 ^2 . | ||
\end{eqnarray} | ||
We can now cancel $m_0$ on both sides, divide by $P_0^2$ and then finally we take the square root and we obtain the relation: | ||
\begin{eqnarray} | ||
1+\delta _p =\sqrt{p_t ^2 + \frac{2p_t}{\beta_0}+1} | ||
\end{eqnarray} | ||
If we now want an approximation to the above formula we again take the square and subtract 1 from both side and we get: | ||
\begin{eqnarray} | ||
2\delta _p+\delta _p^2 =p_t ^2 + \frac{2p_t}{\beta_0} | ||
\end{eqnarray} | ||
In the normal cases $p_t << 1$ so $p_t >> p_t ^2$. We then only use the leading order in $p_t$ and $\delta _p$, which gives us: $\frac{2p_t}{\beta_0} \approx 2 \delta _p$ | ||
\begin{eqnarray} | ||
p_t \approx \beta_0 \delta _p | ||
\end{eqnarray} |
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1-1500 * skip # head * any abs=6e-10 rel=4e-7 | ||
1501-1550 * any abs=6e-9 rel=4e-7 | ||
1-1360 * skip # head * any abs=6e-10 rel=4e-7 | ||
* * any abs=6e-9 rel=4e-7 | ||
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