- #1
soothsayer
- 423
- 5
I'm having some problems trying to figure out how to derive relativistic momentum. The way it was explained to me, classically, p=mv=m(dx/dt),but dx/dt is measured differently in different reference frames. So, if you look at time dilation, t=ɣt' where ɣ= 1/(1-(v/c)^2)^1/2 and t' is time in the moving inertial reference frame. So, dt/dt' = ɣ. m(dx/dt') = m(dx/dt)(dt/dt') = ɣmv. So, p=ɣmv, which is the given formula for special relativistic momentum (though usually u is used instead of v to distinguish velocities of frames and objects). I get all of this.
My question is, don't we have to find p=m(dx'/dt')? if dx/dx' = ɣ and dt/dt' = ɣ, then
m(dx'/dt') = m(dx/dt)(dt/dt')(dx'/dx) = mv(ɣ/ɣ) = mv.
Or, basically, x'=x/ɣ and t'=t/ɣ, so dx/dt = dx'/dt'. Length contracts at the same rate time dilates (dictated by value of ɣ). So where does p=ɣmv come from?
My question is, don't we have to find p=m(dx'/dt')? if dx/dx' = ɣ and dt/dt' = ɣ, then
m(dx'/dt') = m(dx/dt)(dt/dt')(dx'/dx) = mv(ɣ/ɣ) = mv.
Or, basically, x'=x/ɣ and t'=t/ɣ, so dx/dt = dx'/dt'. Length contracts at the same rate time dilates (dictated by value of ɣ). So where does p=ɣmv come from?