- #1
skrat
- 748
- 8
Homework Statement
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?
Homework 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}))##
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!