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Relativistic Force on a Mass
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[QUOTE="six7th, post: 4541732, member: 351050"] [h2]Homework Statement [/h2] a) A mass m starts at rest. Starting at t = 0 (measured in the lab frame), you apply a constant force f to it (lab frame). How long (in the lab frame) does it take for the mass to move a distance x (also measured in the lab frame)? Check that your answer makes sense in the non-relativistic limit. After a very long time, the speed of the mass m will approach c. It turns out that it approaches c sufficiently fast so that after a very long time, the mass will remain (approximately, asymptotically) a constant distance (as measured in the lab frame) behind a photon that was emitted at t = 0 from the starting position of the mass. What is this distance? [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] a) The force on the mass is given by [itex] F = \frac{dP}{dt}[/itex], Integrating w.r.t to time we get: [itex]Ft = P[/itex] Using relativistic momentum this is: [itex]Ft = \gamma mv[/itex] Now solving for t: [itex]t= \frac{\gamma m v}{F}[/itex] I'm not really sure where to go from here. Do I express v as dx/dt and then integrate, which is complicated as [itex]\gamma[/itex] is also a function of v. Edit: Ok I tried solving for v instead and end up with: [itex] v = \frac{Ftc}{\sqrt{c^2m^2+F^2t^2}}[/itex] Using v = dx/dt and solving for x(t): [itex]x(t) = \frac{c\sqrt{c^2m^2+F^2t^2}}{F}[/itex] Squaring both sides gives and rearranging [itex](\frac{Fx(t)}{c})^2 = m^2c^2+F^2t^2[/itex] Not sure this helps with the question but I noticed that the energy of the object is just the work done, Fx(t), and the momentum is F*t. Using this we get [itex](\frac{E}{c})^2 = m^2c^2+p^2[/itex] [itex]E^2 = m^2c^4+p^2c^2[/itex] As this is the invariant does this show that it makes sense in a non relativistic limit? [/QUOTE]
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Relativistic Force on a Mass
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