What will be the relativistic change in momentum?

1. Apr 6, 2005

gulsen

a falling object has a measured momentum p0 at t0 and p1 at t1 by an observer on ground. if p0, t0 and t1 are known,what will be the relativistic change in momentum?(assuming that only force affecting is gravity force)

2. Apr 6, 2005

Andrew Mason

I am assuming this is some kind of cosmic ray particle with relativistic energy.

I am a little beat tonight, but why would it not just be $\Delta p = p_1 - p_0$? You measure momentum in the same (earth) frame of reference both times:

$$p_0 = \gamma_0m_0v_0$$ and later:

$$p_1 = \gamma_1m_0v_1$$

AM

3. Apr 7, 2005

gulsen

and that's the problem. v1 is unknown. we only know p0, t0 and t1, and a function gravity related to the distance r (say we know g(r) and m -- still GmM/r^2 in GR?). how do we derieve v1 from these?

4. Apr 7, 2005

dextercioby

The problem is nasty,to say the least.It's a standard problem in SR,so i assume (i'm not sure,i don't have the books) it can be encountered in classic texts like Wheeler's SR.

It's a relativistic particle in a Newtonian gravitational potential.It falls freely,so you can say that the accleration is constant and equal to "g"...Solve it or search for uniformly accelerated particle in SR...

Daniel.

5. Apr 7, 2005

Data

Edit: Ignore this post for now... look at those below~

Yes, you can still use the Newtonian model for gravity. Just set up a differential equation

$$\frac{GM}{r^2} = \frac{d^2r}{dt^2}$$

with initial condition

$$\frac{dr}{dt}\biggr |_{t=0} = \frac{p_0}{m_0}$$

and solve for $dr/dt$, ie. the velocity, then use your solution to find the velocity at time $t$.

If you are allowed to make the approximation

$$\frac{d^2r}{dt^2} = g$$

where $g$ is the acceleration at sea level, then the DE is much simpler.

Edit: Though I haven't though about it too much, you may actually have to use the $g$ approximation unless you are also given $r(0)$.

Last edited: Apr 7, 2005
6. Apr 7, 2005

dextercioby

No,no,it's a relativistic particle.It has 4 momentum...

$$\frac{d}{dt}\left(\frac{1}{\sqrt{1-\frac{v^{2}(t)}{c^{2}}}} v(t)\right) = g$$

Daniel.

Last edited: Apr 7, 2005
7. Apr 7, 2005

Data

The derivative of momentum isn't acceleration. You mean

$$\frac{d}{dt} \left(\frac{ v(t)}{\sqrt{1-\frac{v^2(t)}{c^2}}}\right)=g$$

but other than that, you're actually probably right. It might expect you to use that form for the DE. Time to dig up SRT books~