The velocity of electron near speed of light?

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SUMMARY

The discussion focuses on the behavior of an electron's velocity as it approaches the speed of light when subjected to a constant force. The momentum of the electron is described by the formula p = mv / √(1 - v²/c²), and the relationship between force and momentum is expressed through Newton's second law as dp/dt = F. The solution for the velocity as a function of time is derived as v(t) = Fct / √(F²t² + c²m²), demonstrating that v approaches c as t approaches infinity. Additionally, the distance traveled by the electron over time is given by d(t) = (c/F)√(t²F² + c²m²).

PREREQUISITES
  • Understanding of relativistic momentum and the formula p = mv / √(1 - v²/c²)
  • Familiarity with Newton's second law and its application in differential equations
  • Basic knowledge of calculus, specifically integration and differentiation
  • Experience with mathematical software tools like Maple for solving equations
NEXT STEPS
  • Study the derivation of relativistic momentum and its implications in physics
  • Learn how to solve differential equations related to motion under constant force
  • Explore the use of Maple or similar software for solving complex physics equations
  • Investigate the concept of limits in calculus to understand behavior as variables approach infinity
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Physicists, students of advanced mechanics, and anyone interested in the implications of relativistic effects on particle motion.

randybryan
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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 traveled 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?
 
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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 traveled over time:
[tex] d(t)=\frac{c}{F} \sqrt{t^2F^2+c^2m^2}[/tex]
 
Last edited:
randybryan said:
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 traveled 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
 

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