# Relativistic electron in static electric field

• Hybird
In summary, the problem involves an electron placed in a static electric field and its initial velocity is 0. The goal is to calculate V(t) using the equations F_{e}=\frac{d}{dt}p(t) and p(t) = \gamma(t)mv(t), where \gamma is 1/sqrt(1-v^2/c^2). The solution involves multiplying the electric field by the charge of an electron and integrating with respect to time. The resulting formula for V(t) approaches c as t goes to infinity.
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)..

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?

Hybird said:
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

Hybird said:
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.

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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?

Hybird said:
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 ...

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## 1. What is a relativistic electron?

A relativistic electron is an electron that is moving at a significant fraction of the speed of light. This means that its behavior is governed by the principles of relativity, which describe how objects behave at high speeds.

## 2. What is a static electric field?

A static electric field is a type of electric field that does not change over time. In other words, the strength and direction of the field remain constant.

## 3. How does a relativistic electron behave in a static electric field?

A relativistic electron in a static electric field will experience a force due to the electric field. This force will cause the electron to accelerate, and its behavior will be described by the principles of relativity.

## 4. What is the relationship between a relativistic electron and a static electric field?

The behavior of a relativistic electron is influenced by the presence of a static electric field. The strength and direction of the electric field will determine the force experienced by the electron, which in turn affects its motion.

## 5. What are some applications of studying relativistic electrons in static electric fields?

Studying the behavior of relativistic electrons in static electric fields has many practical applications, such as in particle accelerators, plasma physics research, and the development of new technologies like electron microscopes and particle detectors.

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