How Does Relativistic Force Affect the Movement and Speed of a Mass?

[six7th]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

a) The force on the mass is given by $F = \frac{dP}{dt}$, Integrating w.r.t to time we get:

$Ft = P$

Using relativistic momentum this is:

$Ft = \gamma mv$

Now solving for t:

$t= \frac{\gamma m v}{F}$

I'm not really sure where to go from here. Do I express v as dx/dt and then integrate, which is complicated as $\gamma$ is also a function of v.

Edit: Ok I tried solving for v instead and end up with:

$v = \frac{Ftc}{\sqrt{c^2m^2+F^2t^2}}$

Using v = dx/dt and solving for x(t):

$x(t) = \frac{c\sqrt{c^2m^2+F^2t^2}}{F}$

Squaring both sides gives and rearranging

$(\frac{Fx(t)}{c})^2 = m^2c^2+F^2t^2$

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

$(\frac{E}{c})^2 = m^2c^2+p^2$
$E^2 = m^2c^4+p^2c^2$

As this is the invariant does this show that it makes sense in a non relativistic limit?

 

[vela]

That looks good. What do you get when you integrate that?

 

[six7th]

Sorry I accidentally clicked post before I finished typing everything up. The main post has been updated with how far I've got.

 

[vela]

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.

 

[six7th]

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:

$x(t) = \frac{mc^2}{F}\sqrt{1+(\frac{Ft}{mc})^2} - \frac{mc^2}{F} = \frac{mc^2}{F}(\sqrt{1+(\frac{Ft}{mc})^2} -1)$

Rearranging for t:

$t = ±\sqrt{\frac{v^2}{c^2}+\frac{2mx}{F}}$

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

So for part b),

$x(t) = \frac{mc^2}{F}(\sqrt{1+(\frac{Ft}{mc})^2} -1)$

I'm thinking I should take the limit t → ∞ and then subtract this from the distance traveled by a photon cΔt, although surely the distance just goes to infinity as t → ∞.

 

[vela]

Just assume $t$ is large, not infinite.