# Using relativity: speed of electron in electric field

1. Nov 8, 2013

### skrat

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. Nov 8, 2013

### 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.

3. Nov 8, 2013

### vanhees71

$$u^0=\gamma_0 c \cosh(\alpha \tau), \quad u^2=\gamma_0 c \sinh(\alpha \tau)$$
and also
$$x^0=c t=\frac{\gamma_0 c}{\alpha} \sinh(\alpha \tau).$$
Now just use
$$\cosh(\alpha \tau)=\sqrt{1+\sinh^2(\alpha \tau)}.$$
This gives
$$u^2=c \alpha t, \quad u^0=c \sqrt{\gamma_0^2+\alpha^2 t^2},$$
and finally
$$v^2=c \frac{u^2}{u^0}=\frac{c \alpha t}{\sqrt{\gamma_0^2+\alpha^2 t^2}}.$$
Now you can calculate easily the rest.

4. Nov 8, 2013

### skrat

Ok, that makes sense. =)

One more question... How do I now calculate it's displacement from original direction?

5. Nov 8, 2013

### Staff: Mentor

If you know v(t) in the direction of the electric field, this should be no problem.

6. Nov 8, 2013

### skrat

aaaa, ok! I get it now. :)

THANKS to both of you!

7. Nov 8, 2013

### skrat

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