- #1
AVentura
- 106
- 8
Has anyone ever seen KE derived this way? The only non-classical assumption is that mass = Energy/c^2 (I know, that's a big one) and that it accumulates as mass as the object is accelerated. (sorry for the plain text formatting, I need to learn how to do it right, I may reply with my first try)
KE(v)=∫ dp/dt dx = ∫ v dp
p = mv , m = (mo + KE(v)/c^2)
so: p = mo*v + KE(v)*v/c^2
and dp = mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv
KE(v)=∫ [v*(mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv)]
or KE'(v) = v*mo + KE'(v)*v^2/c^2 + KE(v)*v/c^2
The solution to this DE is k/√(1-v^2/c^2)-mo*c^2
If KE(0)=0 then k=mo*c^2
and KE= mo*c^2(gamma-1), which is correct.
I find this interesting because it does not involve using p=m*dx/dτ as momentum, and instead uses p=mv, with the mass growing. I never understood using a moving object's proper time to find momentum. The observer is the only one claiming motion after all.
KE(v)=∫ dp/dt dx = ∫ v dp
p = mv , m = (mo + KE(v)/c^2)
so: p = mo*v + KE(v)*v/c^2
and dp = mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv
KE(v)=∫ [v*(mo*dv + dKE(v)*v/c^2 + KE(v)/c^2*dv)]
or KE'(v) = v*mo + KE'(v)*v^2/c^2 + KE(v)*v/c^2
The solution to this DE is k/√(1-v^2/c^2)-mo*c^2
If KE(0)=0 then k=mo*c^2
and KE= mo*c^2(gamma-1), which is correct.
I find this interesting because it does not involve using p=m*dx/dτ as momentum, and instead uses p=mv, with the mass growing. I never understood using a moving object's proper time to find momentum. The observer is the only one claiming motion after all.