# Relativistic Force and Momentum

1. Oct 5, 2005

### Jibobo

I've been having a lot of trouble with this problem. There's definitely something I'm missing and it most likely has to do with the force.

"A mass $$m$$ is thrown from the origin at $$t = 0$$ with initial three-momentum $$p_0$$ in the y direction. If it is subject to a constant force $$F_0$$ in the x direction, find its velocity $$v$$ as a function of $$t$$ and by integrating $$v$$, find its trajectory. You will need to integrate functions such as $$t/\sqrt{a + bt^2}$$ and $$1/\sqrt{a + bt^2}$$.

In addition, check that in the non-relativistic limit, ($$c \rightarrow \infty$$), $$x(t)$$ is what you expect for motion in a constant field and check that $$y(t)$$ is what you expect for motion in a constant field when the force is orthogonal to the y direction.

HINT: You will need the Taylor expansion of the functions $$\sqrt{1 + x}$$ and $$\ln(1 + x)$$."

The 2nd part seems easy, but I'm simply not sure how to find $$x(t)$$ or $$y(t)$$ in the first place.

My work so far:
$$\gamma = 1/\sqrt(1 - v^2/c^2)\\ \gamma_v_0 = 1/\sqrt(1 - {v_0}^2/c^2)$$
$$p_0 = (0, p_0, 0) = m*\gamma_v0*(0, v_0, 0)$$
$$F_0 = (F_0, 0, 0)$$
$$F = dP/dt, \mbox{so } P - p_0 = F*t$$
$$P = F_0*t + p_0 = m*\gamma*v$$
$$v = (v_x, v_y, v_z), v_z = 0$$
$$m*\gamma*v_x = F_0*t$$
$$m*\gamma*v_y = p_0 = m*\gamma_v0*v_0$$

I'm not exactly sure how to proceed from here since I can't really isolate $$v_x$$ or $$v_y$$ because the gamma term contains only the magnitude of $$v$$. Should I use $$\|v\| = \sqrt{v_x^2 + v_y^2}$$ and then work through some really terrible algebra? Or is this even the right way to approach this problem?

Edit: I've actually done the terrible alegbra using $$\|v\| = \sqrt{v_x^2 + v_y^2}$$, but the equations I end up with are ridiculous. Can anyone suggest a different method?

Last edited: Oct 6, 2005
2. Oct 12, 2005

### pervect

Staff Emeritus
The 4-momentum of the particle, which I'll call P, is supposed to be a 4-vector.

Write down the four components of P:

E (the energy)
Px (the momentum in the x direction)
Py (the momentum in the y direction)
Pz (the momentum in the z direction, which is very easy, it's zero)

You have something that you call P, but it doesn't appear to be a 4-vector.