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Using relativity: speed of electron in electric field

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Electron, accelerated by voltage 100 kV, hits homogeneous electric field 1 kV/m. How fast is the electron after 30 ns and how big is its offset from original direction?


    2. Relevant equations
    ##\vec{E}=(0,E,0)##
    ##\vec{v_0}=(v_0,0,0)## therefore ##u_{0}^{\mu }=(c\gamma _0,\gamma_0 \vec{v})##
    ##\mathfrak{F}^{\mu }=(\frac{e\vec{E}\vec{u}}{c},e(\vec{E}+\vec{v}\times \vec{B}))##


    3. The attempt at a solution
    Firstly, to calculate the initial speed of electron:
    ##T=mc^2(\gamma -1)=eU##
    ##\gamma =\frac{mc^2+eU}{mc^2}##
    ##\sqrt{1-v^2/c^2}=\frac{mc^2}{mc^2+eU}##
    ##v_0=c\sqrt{1-\frac{m^2c^4}{(mc^2+eU)^2}}=0,55c##

    Now let's write Newtons law...:
    ##\frac{\partial u^{(0)}}{\partial \tau }=\frac{eE}{mc}u^{(2)}=\alpha u^{(2)}## for ##\frac{eE}{mc}=\alpha ##
    ##\frac{\partial u^{(2)}}{\partial \tau }=\frac{eE}{mc}u^{(0)}=\alpha u^{(0)}##
    and
    ##\frac{\partial u^{(3)}}{\partial \tau }=\frac{\partial u^{(1)}}{\partial \tau }=0##

    First two give me differential equation ##\frac{\partial^2u^{(0)} }{\partial \tau ^2}=\alpha ^{2}u^{(0)}##. Using ##u^{(0)}=Acosh(\alpha \tau )+Bsinh(\alpha \tau )## and initial conditions:

    ##u^{(0)}=c\gamma _0cosh(\alpha \tau )## and ##u^{(2)}=c\gamma _0sinh(\alpha \tau )##
    Now here is the part I am having troubles with:
    I believe that ##u^{(0)}## is also equal to ##c\gamma ## where ##\gamma ## is different that ##\gamma _0## since the speed is different.
    If that is true, that ##u^{(0)}=c\gamma _0cosh(\alpha \tau )=c\gamma## so ##\gamma _0cosh(\alpha \tau )=\gamma##

    If I integrate ##u^{(0)}=\frac{\partial x^{(0)}}{\partial \tau }## ..... ##x^{(0)}=\int_{0}^{\tau }\gamma _0cosh(\alpha \tau )d\tau =\frac{\gamma _o}{\alpha }sinh(\alpha \tau )## but ##x^{(0)}=ct##.

    This now gives me ##\alpha \tau =arcsinh(\frac{t\alpha }{\gamma _0})## and to go back to equation ##u^{(0)}=c\gamma _0cosh(\alpha \tau )=c\gamma## ... for calculated ##\alpha \tau ## this gives me almost identical gammas ##\gamma _0=\gamma ## which would mean that the velocity after 30 ns is practically the same...


    I don't believe that. Can anybody help me understand where I made a mistake? The velocity can't be the same.... No way! It has to be larger than initial! At least that's what i think...
    Thanks!!
     
  2. jcsd
  3. Nov 8, 2013 #2

    mfb

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    Staff: Mentor

    The speed does not change much. Just consider the extreme case: what is the largest potential difference any particle could pick up in 30 ns? Well, in 30ns, a particle cannot travel more than 9m, which corresponds to 9kV, much less than your 100kV. In addition, your electron is not moving with the speed of light in that direction - its initial velocity in that direction is zero! Its displacement will be smaller by far. A nonrelativistic calculation gives 8cm, or just 80eV. Nothing compared to 100keV.
     
  4. Nov 8, 2013 #3

    vanhees71

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    Hi, you are almost ready! Your solutions look good. I got
    [tex]u^0=\gamma_0 c \cosh(\alpha \tau), \quad u^2=\gamma_0 c \sinh(\alpha \tau)[/tex]
    and also
    [tex]x^0=c t=\frac{\gamma_0 c}{\alpha} \sinh(\alpha \tau).[/tex]
    Now just use
    [tex]\cosh(\alpha \tau)=\sqrt{1+\sinh^2(\alpha \tau)}.[/tex]
    This gives
    [tex]u^2=c \alpha t, \quad u^0=c \sqrt{\gamma_0^2+\alpha^2 t^2},[/tex]
    and finally
    [tex]v^2=c \frac{u^2}{u^0}=\frac{c \alpha t}{\sqrt{\gamma_0^2+\alpha^2 t^2}}.[/tex]
    Now you can calculate easily the rest.
     
  5. Nov 8, 2013 #4
    Ok, that makes sense. =)

    One more question... How do I now calculate it's displacement from original direction?
     
  6. Nov 8, 2013 #5

    mfb

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    Staff: Mentor

    If you know v(t) in the direction of the electric field, this should be no problem.
     
  7. Nov 8, 2013 #6
    aaaa, ok! I get it now. :)

    THANKS to both of you!
     
  8. Nov 8, 2013 #7
    What if I find myself doing this for three dimensional space? For example:
    ##\vec{E}=(Ecos\varphi ,Esin\varphi ,0)##, ##\vec{B}=(0,0,B)##

    Than:

    ##\vec{E}\vec{u}=u^{(1)}Ecos\varphi +u^{(2)}Esin\varphi ## and
    ##\vec{E}+\vec{u}\times \vec{B}=(eEcos\varphi +Bu^{(2)},eEsin\varphi -Bu^{(1)},0)##

    so

    ##\frac{\partial u^{(0)}}{\partial \tau }=\frac{eE}{mc}(cos\varphi +u^{(2)}Esin\varphi )##,

    ##\frac{\partial u^{(1)}}{\partial \tau }=\frac{1}{m}(eEcos\varphi +eBu^{(2)})##,

    ##\frac{\partial u^{(2)}}{\partial \tau }=\frac{1}{m}(eEsin\varphi -eBu^{(1)})## and

    ##\frac{\partial u^{(3)}}{\partial \tau }=0##

    Now.. ##\frac{\partial^2 u^{(0)}}{\partial \tau ^2}=\frac{eE}{m^2c}(eE+eB(cos\varphi u^{(2)}-sin\varphi u^{(1)}))##

    How am I supposed to solve that? :/
     
    Last edited: Nov 8, 2013
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