Commit 4b3d879a authored by Stefan Gehr's avatar Stefan Gehr
Browse files

Korrektes beta in 2

parent b29dddb9
......@@ -81,36 +81,34 @@ Laut Wikipedia lautet die Formel
\gamma
&= \left(1-\beta^2\right)^{-\frac{1}{2}} \\
\Leftrightarrow \beta
&= 1-\frac{1}{\gamma^2}
= 1 - \frac{1}{7461.522473522262^2}
= \num{0.9999999820383968} \\
&\approx \num{0.99999998}
&= \sqrt{1-\frac{1}{\gamma^2}}
= \sqrt{1 - \frac{1}{7461.522473522262^2}}
= \num{0.9999999910191983}
.\end{align*}
\subsection{}
\begin{align*}
s
&= v \, \Delta t \\
&= v \, \Delta t = \beta c \, \Delta t \\
\Leftrightarrow \Delta t
&= \frac{s}{v} = \frac{\SI{27}{\kilo\metre}}{\num{0.9999999820383968}\,c}
= \SI{9.006230732116448e-05}{\second} \\
&\approx \SI{9.0}{\second}
&= \frac{s}{\beta c} = \frac{\SI{27}{\kilo\metre}}{\num{0.9999999910191983}\,c}
= \SI{9.006230651233277e-05}{\second} \\
&\approx \SI{90}{\micro\second}
.\end{align*}
Da sich das Proton auf einer Kreisbahn bewegt, befindet es sich nicht in einem Inertialsystem.
Deshalb müsste man die allgemeine Relativitätstheorie benutzen um die Eigenzeit \(\Delta t'\) zu berechnen.
Geht man jedoch davon aus, dass es sich in einem Inertialsystem befindet, dann erhält man
\begin{align*}
\Delta t' = \frac{\Delta t}{\gamma} = \frac{\SI{9.006230732116448e-05}{\second}}{\num{7461.522473522262}}
= \SI{1.2070231998999792e-08}{\second}
\Delta t' = \frac{\Delta t}{\gamma} = \frac{\SI{9.006230651233277e-05}{\second}}{\num{7461.522473522262}}
= \SI{1.2070231890599434e-08}{\second}
.\end{align*}
\subsection{}
\begin{align*}
p
&= qBR \\
B &= \frac{p}{qR} = \frac{2\pi \,p}{e \, s}
= \frac{2\pi \, \gamma m_{\mathrm{P}}v}{e \, s} \\
&= \frac{2\pi \cdot \num{7461.522473522262} \cdot m_{\mathrm{P}} \cdot \num{0.9999999820383968}\,c}{e \cdot \SI{27}{\kilo\metre}} \\
&= \SI{5.43440050481426}{\tesla}
\approx \SI{5.4}{\tesla}
= \frac{2\pi \, \gamma m_{\mathrm{P}}v}{e \, s}
= \SI{5.4344005536195334}{\tesla} \\
&\approx \SI{5.4}{\tesla}
.\end{align*}
\section{Invariante Masse}
\subsection{}
......
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