- #1
Confused0ne
- 3
- 1
1. Homework Statement
Express the beam energy loss in the accelerator in terms of the change in the beam momentum.
It is taken from one paper, where
\begin{equation}
\Delta T = \left( \frac{1+\gamma}{\gamma}\right) \frac{T_0 \Delta p}{p_0}
\end{equation}expression is used for transition from one equation to the next one.
Even though it does seem legit, I don't seem to find the way to actually show it mathematically.
Please help!
2. Homework Equations
Kinetic Energy:
\begin{equation}
T= mc^2(\gamma -1)
\end{equation}
Momentum:
\begin{equation}
p= \gamma mV
\end{equation}3. The Attempt at a Solution .
\begin{equation}
\begin{split}
\Delta T = T - T_0 = mc^2 (\gamma - \gamma_0)=\gamma_0 mc^2 (\gamma/ \gamma_0 -1)= \\
=\frac{T_0}{p_0}(\gamma m V_0 - p_0)=\frac{T_0}{p_0} (\gamma m V_0 - \gamma_0 m V_0)=\\
%=\frac{T_0}{p_0} \left( \frac{1+\gamma}{\gamma}\right) \frac{(\gamma^2 m V_0 - \gamma \gamma_0 m V_0)}{1+\gamma}
\end{split}
\end{equation}
I have attempted different manipulations from here, but didn't get anywhere feasible.
I have also tried the other way around, starting with the right part of the equation
\begin{equation}
\begin{split}
\Delta p \left(\frac{\gamma +1}{\gamma} \right)= (p - p_0) \left(\frac{\gamma +1}{\gamma}\right) = \\
=\frac{\gamma^2 m V + \gamma mV -\gamma_0 m V_0 - \gamma \gamma_0 m V_0 }{\gamma}=\\
=\gamma m V +mV - \gamma_0 m V_0/\gamma - \gamma_0 m V_0
\end{split}
\end{equation}
It looks like, I got only one term $\gamma_0 m V_0$, but I am missing something to finalize the other terms.
What AM i Missing?
Express the beam energy loss in the accelerator in terms of the change in the beam momentum.
It is taken from one paper, where
\begin{equation}
\Delta T = \left( \frac{1+\gamma}{\gamma}\right) \frac{T_0 \Delta p}{p_0}
\end{equation}expression is used for transition from one equation to the next one.
Even though it does seem legit, I don't seem to find the way to actually show it mathematically.
Please help!
2. Homework Equations
Kinetic Energy:
\begin{equation}
T= mc^2(\gamma -1)
\end{equation}
Momentum:
\begin{equation}
p= \gamma mV
\end{equation}3. The Attempt at a Solution .
\begin{equation}
\begin{split}
\Delta T = T - T_0 = mc^2 (\gamma - \gamma_0)=\gamma_0 mc^2 (\gamma/ \gamma_0 -1)= \\
=\frac{T_0}{p_0}(\gamma m V_0 - p_0)=\frac{T_0}{p_0} (\gamma m V_0 - \gamma_0 m V_0)=\\
%=\frac{T_0}{p_0} \left( \frac{1+\gamma}{\gamma}\right) \frac{(\gamma^2 m V_0 - \gamma \gamma_0 m V_0)}{1+\gamma}
\end{split}
\end{equation}
I have attempted different manipulations from here, but didn't get anywhere feasible.
I have also tried the other way around, starting with the right part of the equation
\begin{equation}
\begin{split}
\Delta p \left(\frac{\gamma +1}{\gamma} \right)= (p - p_0) \left(\frac{\gamma +1}{\gamma}\right) = \\
=\frac{\gamma^2 m V + \gamma mV -\gamma_0 m V_0 - \gamma \gamma_0 m V_0 }{\gamma}=\\
=\gamma m V +mV - \gamma_0 m V_0/\gamma - \gamma_0 m V_0
\end{split}
\end{equation}
It looks like, I got only one term $\gamma_0 m V_0$, but I am missing something to finalize the other terms.
What AM i Missing?