# Relativistic Force on 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?

## 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?

Last edited:

Staff Emeritus
Homework Helper
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.

Staff Emeritus
Homework Helper
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 travelled by a photon cΔt, although surely the distance just goes to infinity as t → ∞.

Staff Emeritus