Relativistic electron in static electric field

[Hybird]

Homework Statement

Electron placed in static electric field $$\vec{E}$$ = -$$\Psi$$$$\hat{x}$$ , its initial velocity is 0. Calculate V(t).

Homework Equations

$$F_{e}$$=$$\frac{d}{dt}$$p(t)

p(t) = $$\gamma$$(t)mv(t)

Gamma is 1/sqrt(1-v^2/c^2) of course

The Attempt at a Solution

This is how I go about it and want to know if I'm on the right track.

i) First you multiply the electric field by the charge of an electron to get:

$$F_{e}$$ = $$\frac{d}{dt}$$p(t) = e$$\Psi$$

ii) Then you integrate wrt time to get:

p(t) = e$$\Psi$$t

iii) Then you relate momentum to velocity by:

p(t) = $$\gamma$$mv(t)

iv) Finally you solve for V(t) from the above equation, expressing gamma explicitly I get the following formula:

v(t) = $$\frac{e{\Psi}t}{m}$$*$$\frac{1}{ {\sqrt{1+ {\frac{ e^{2}{\Psi}^{2}t^{2} }{ m^{2}c^{2} }} }} }$$

Does this seem to be the right method? I have to integrate this eventually to get x(t)..

 

[Hybird]

This doesn't seem right because V(t) is unbounded as t goes to infinity and thus we would pass the speed of light correct?

 

[turin]

This doesn't seem right because V(t) is unbounded as t goes to infinity and thus we would pass the speed of light correct?

No. This expression

v(t) = $$\frac{e{\Psi}t}{m}$$*$$\frac{1}{ {\sqrt{1+ {\frac{ e^{2}{\Psi}^{2}t^{2} }{ m^{2}c^{2} }} }} }$$

approaches c as t goes to infinity.

 

[Hybird]

So maybe I am on the right track. Just curious, how do you get that limit? Do you have to binomial expand the square root?

 

[turin]

how do you get that limit? Do you have to binomial expand the square root?

I don't know what you mean by that, but probably not. After you do about 100 million of these kinds of limits, you just start to smell the approach. In this case, I just know that the other stuff under the radical will be much larger than 1, so I ignore the 1. The rest is straightforward. I think the more mathy way to do it is to divide top and bottom by t or something ...