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?
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]
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