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Energy loss in terms of beam momentum change

  1. Oct 8, 2015 #1
    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
    \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. Relevant 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?
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2015 #2

    mfb

    User Avatar
    2016 Award

    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##).
     
  4. Oct 9, 2015 #3
    Thanks for your reply.
    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...
     
  5. Oct 9, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Hmm okay, the velocity makes it messy. What about ##T^2 = p^2 + m^2## (ignoring factors of c)?
     
  6. Oct 14, 2015 #5
    I finally figured it out!

    The key is to apply logarithmic differentiation! \\
    1)
    \begin{equation}
    \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}
    \end{equation}\\
    2)
    Momentum-energy relation can be written as
    \begin{equation}
    pc= \beta E
    \end{equation}

    from here
    \begin{equation}
    \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}
    \end{equation}\\
    3)
    \begin{equation}
    \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}
    \end{equation}
     
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