# Energy loss in terms of beam momentum change

1. Oct 8, 2015

### Confused0ne

1. The problem statement, all variables and given/known data

Express the beam energy loss in the accelerator in terms of the change in the beam momentum.

It is taken from one paper, where

\Delta T = \left( \frac{1+\gamma}{\gamma}\right) \frac{T_0 \Delta p}{p_0}

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.

2. Relevant equations

Kinetic Energy:

T= mc^2(\gamma -1)

Momentum:

p= \gamma mV

3. The attempt at a solution.

\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}

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{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}

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?

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2. Oct 8, 2015

### Staff: Mentor

You can write the last line as $\gamma_0 m V_0 (\dots )$. I guess it needs some approximation at some point (small $\Delta p$).

3. Oct 9, 2015

### Confused0ne

Since I need to express \Delta T in \Delta p, I don't see how small \Delta p approximation helps me. As soon as I try something like that (e.g. use p_0 in place of p), it all goes to hell...

γ0mV0(…) doesn't bring me anywhere either...

4. Oct 9, 2015

### Staff: Mentor

Hmm okay, the velocity makes it messy. What about $T^2 = p^2 + m^2$ (ignoring factors of c)?

5. Oct 14, 2015

### Confused0ne

I finally figured it out!

The key is to apply logarithmic differentiation! \\
1)

\frac{dp}{p} = \frac{d\gamma}{\gamma} + \frac{d\beta}{\beta} = \left(\frac{\beta^2}{1-\beta^2} +1\right)\frac{d\beta}{\beta}= \gamma^2 \frac{d\beta}{\beta}
\\
2)
Momentum-energy relation can be written as

pc= \beta E

from here

\frac{dE}{E} = \frac{dp}{p} - \frac{d\beta}{\beta} = \frac{dp}{p} - \frac{1}{\gamma^2}\frac{dp}{p}= \left( 1-\frac{1}{\gamma^2} \right) \frac{dp}{p}
\\
3)

\begin{split}
\frac{dT}{T} = \frac{dE}{mc^2 (\gamma -1)}
=\frac{dE}{E} \frac{E}{mc^2 (\gamma -1)}
= \left( 1-\frac{1}{\gamma^2} \right) \frac{dp}{p} \frac{mc^2\gamma}{mc^2 (\gamma -1)}=\\
= \left(\frac{\gamma^2-1}{\gamma^2} \right)\frac{\gamma}{(\gamma -1)}\frac{dp}{p}
=\left(\frac{\gamma+1}{\gamma} \right) \frac{dp}{p}
\end{split}