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Relativistic Force on a Mass

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations




    3. The attempt at a solution

    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?
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2

    vela

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    That looks good. What do you get when you integrate that?
     
  4. Oct 17, 2013 #3
    Sorry I accidentally clicked post before I finished typing everything up. The main post has been updated with how far I've got.
     
  5. Oct 17, 2013 #4

    vela

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    You're getting there, but you need to include the constant of integration and apply initial conditions or, equivalently, use definite integrals. Note that with your result, when t=0, x isn't 0.

    I think it makes things a bit clearer if you pull out the factor of (mc)^2 to get everything in terms of the dimensionless quantity Ft/mc:
    $$\int_0^x dx = c\int_0^t \frac{\frac{Ft}{mc}}{\sqrt{1+\left(\frac{Ft}{mc}\right)^2}}\,dt = \frac{mc^2}{F}\left.\sqrt{1+\left(\frac{Ft}{mc}\right)^2}\right|_0^t = \dots$$ In the non-relativistic limit, the imparted impulse Ft is much less than mc.
     
  6. Oct 17, 2013 #5
    Thanks, I wrongly assumed the constants would be 0 because the object started at rest and I didn't think to check my result matched the initial conditions!

    Ok, so after finding the constant I get:

    [itex]x(t) = \frac{mc^2}{F}\sqrt{1+(\frac{Ft}{mc})^2} - \frac{mc^2}{F} = \frac{mc^2}{F}(\sqrt{1+(\frac{Ft}{mc})^2} -1)[/itex]

    Rearranging for t:

    [itex]t = ±\sqrt{\frac{v^2}{c^2}+\frac{2mx}{F}}[/itex]

    Which reduces to what we expect in the non-relativistic limit!

    So for part b),

    [itex]x(t) = \frac{mc^2}{F}(\sqrt{1+(\frac{Ft}{mc})^2} -1)[/itex]

    I'm thinking I should take the limit t → ∞ and then subtract this from the distance travelled by a photon cΔt, although surely the distance just goes to infinity as t → ∞.
     
  7. Oct 17, 2013 #6

    vela

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    Just assume ##t## is large, not infinite.
     
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