Register to reply

The velocity of electron near speed of light?

by randybryan
Tags: electron relativity
Share this thread:
randybryan
#1
Apr8-10, 12:55 PM
P: 48
This isn't a homework question, simply one I found in a book that I'm trying to do:

momentum p, of electron at speed v near speed of light increases according to formula

p = [tex]\frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]

if an electron is subject to constant force F, Newton's second law of describing motion is

[tex]\frac{dp}{dt}[/tex] = [tex]\frac{d}{dt}[/tex] [tex]\frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex] = F

This all makes sense to me. It then says, find v(t) and show that v --> c as t --> infinity. Find the distance travelled by the electron in time t if it starts from rest.

Now I could get an expression for v by using the first formula, but I don't understand how I can show that v -->c as t --> infinity as t isn't in the equation. I haven't even attempted the second part, but I'm assuming some integration is involved

Can anyone help?
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
jeppetrost
#2
Apr8-10, 01:43 PM
P: 88
What you want to do to find v(t) is to solve the differential equation:
[tex]
F = \frac{d}{dt} \left( m v(t) \gamma (t)\right)
[/tex]
for v(t).
Now, you can differentiate the product to get
[tex]
F = m a(t) \gamma (t) + m v(t) \left( \frac{v(t)a(t)}{c^2}\gamma^3(t)\right)=m a(t) (\gamma + \beta ^2(t)\gamma ^3(t)), ~~~~ \beta=\frac{v(t)}{c}
[/tex]
Now, solving this (I used Maple) and imposing the condition v(0)=0, one gets the expression
[tex]
v(t)=\frac{Fct}{\sqrt{F^2 t^2 + c^2 m^2}}
[/tex]
As you can see, v approaches c a time goes to infinity.
The expression is easily (with some computer) integrated to give you an expression for the distance travelled over time:
[tex]
d(t)=\frac{c}{F} \sqrt{t^2F^2+c^2m^2}
[/tex]
starthaus
#3
Apr8-10, 05:49 PM
P: 1,568
Quote Quote by randybryan View Post
This isn't a homework question, simply one I found in a book that I'm trying to do:

momentum p, of electron at speed v near speed of light increases according to formula

p = [tex]\frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]

if an electron is subject to constant force F, Newton's second law of describing motion is

[tex]\frac{dp}{dt}[/tex] = [tex]\frac{d}{dt}[/tex] [tex]\frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex] = F

This all makes sense to me. It then says, find v(t) and show that v --> c as t --> infinity. Find the distance travelled by the electron in time t if it starts from rest.

Now I could get an expression for v by using the first formula, but I don't understand how I can show that v -->c as t --> infinity as t isn't in the equation. I haven't even attempted the second part, but I'm assuming some integration is involved

Can anyone help?
F=m*d/dt(v/sqrt(1-(v/c)^2)

so,

F/m=d/dt(v/sqrt(1-(v/c)^2)


Since F/m is constant, the above becomes a very simple differential equation with the solution:

v=at/sqrt(1+(at/c)^2)

For at<<c, you recover the Newtonian equation v=at

If you integrate one more time, you will get x as a function of a and t. Indeed:

dx/dt=at/sqrt(1+(at/c)^2)


x(t)=c^2/a*(sqrt(1+(at/c)^2)-1)

Again, for at<<c, you recover the Newtonian formula x(t)=at^2/2


Register to reply

Related Discussions
Speed of light pulse if you have a velocity of c Advanced Physics Homework 4
Question about light speed (is velocity cumulative?) Special & General Relativity 9
Time travel when velocity> light speed Special & General Relativity 3
Is the speed of light a velocity or acceleration? Special & General Relativity 18
Rotational velocity and the speed of light General Physics 13