Subfield: Modern Physics


Question: A.6 Determine the time T that the protons take to pass through this field


![](https://cdn.mathpix.com/cropped/2023_12_21_b9b9ac37d6714962e913g-1.jpg?height=429&width=1262&top_left_y=762&top_left_x=403)

Figure 1: Sketch of an accelerator module.

Solution:
2nd Newton's law

$$
\begin{gathered}
F=\frac{d p}{d t} \quad \text { leads to } \\
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
\end{gathered}
$$

Conservation of energy:

$$
E_{t o t}=m \cdot c^{2}+e \cdot V
$$

Since

$$
\begin{gathered}
E_{t o t}^{2}=\left(m \cdot c^{2}\right)^{2}+\left(p_{f} \cdot c\right)^{2} \\
\rightarrow p_{f}=\frac{1}{c} \cdot \sqrt{\left(m \cdot c^{2}+e \cdot V\right)^{2}-\left(m \cdot c^{2}\right)^{2}}=\sqrt{2 e \cdot m \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
2nd Newton's law

$$
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
$$

velocity from A1 or from conservation of energy

$$
v=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}}
$$

and hence for $\gamma$

$$
\begin{gathered}
\gamma=1 / \sqrt{1-\frac{v^{2}}{c^{2}}}=1+\frac{e \cdot V}{m_{p} \cdot c^{2}} \\
\rightarrow p_{f}=\gamma \cdot m_{p} \cdot v=\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right) \cdot m_{p} \cdot c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d \cdot m_{p} \cdot c}{V \cdot e} \cdot \sqrt{\left(\frac{m_{p} \cdot c^{2}+e \cdot V}{m_{p} \cdot c^{2}}\right)^{2}-1}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
Energy increases linearly with distance $\mathrm{x}$

$$
\begin{gathered}
E(x)=\frac{e \cdot V \cdot x}{d} \\
t=\int d t=\int_{0}^{d} \frac{d x}{v(x)} \\
v(x)=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}}\right)^{2}}=c \cdot \frac{\sqrt{\left(m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}\right)^{2}-\left(m_{p} \cdot c^{2}\right)^{2}}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}} \\
=c \cdot \frac{\sqrt{\left(1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}\right)^{2}-1}}{1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}}
\end{gathered}
$$

Substitution : $\xi=\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}} \quad \frac{d \xi}{d x}=\frac{e \cdot V}{d \cdot m_{p} \cdot c^{2}}$

$$
\rightarrow t=\frac{1}{c} \int_{0}^{b} \frac{1+\xi}{\sqrt{(1+\xi)^{2}-1}} \frac{d \cdot m_{p} \cdot c^{2}}{e \cdot V} d \xi \quad b=\frac{e \cdot V}{m_{p} \cdot c^{2}}
$$

$$
1+\xi:=\cosh (s) \quad \frac{d \xi}{d s}=\sinh (s)
$$

0.1

$$
t=\frac{m_{p} \cdot c \cdot d}{e \cdot V} \int \frac{\cosh (s) \cdot \sinh (s) d s}{\sqrt{\cosh ^{2}(s)-1}}=\frac{m_{p} \cdot c \cdot d}{e \cdot V}[\sinh (s)]_{b_{1}}^{b_{2}}
$$

$$
\text { with } \quad b_{1}=\cosh ^{-1}(1), \quad b_{2}=\cosh ^{-1}\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right)
$$

$$
T=218 \mathrm{~ns}
$$
Alternative: differential equation 

$$
\begin{gathered}
\rightarrow \frac{V \cdot e}{d}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m \cdot v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)=\frac{m \cdot a\left(1-\frac{v^{2}}{c^{2}}\right)+m \cdot a \frac{v^{2}}{c^{2}}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}}=\gamma^{3} \cdot m \cdot a \\
a=\ddot{s}=\frac{V \cdot e}{d \cdot m}\left(1-\frac{\dot{s}^{2}}{c^{2}}\right)^{\frac{3}{2}}
\end{gathered}
$$

$$
\begin{gathered}
\rightarrow s(t)=\frac{c}{V \cdot e}\left(\sqrt{e^{2} \cdot V^{2} \cdot t^{2}+c^{2} \cdot m^{2} \cdot d^{2}}-c \cdot m \cdot d\right) \\
s=d \rightarrow T=\frac{d}{V \cdot e} \sqrt{\left(\frac{V \cdot e}{c}\right)^{2}+2 V \cdot e \cdot m} \\
T=218 \mathrm{~ns}
\end{gathered}
$$

Final answer: 218 ns