help: how to derive relativistic kinetic energy----------------
ChN92
one can obtain the equation for the non-relativits kinetic energy by defining
kindetic energy to be the work it required to accelerate teh mass from zero
velocity to the final speed. when i tried to do this for the relativisitc
case
i obtain
K ===== /
relativistic | d/dt (gamma m v) . dx
/
is this right? the time derivative d/dt, is this the proper time or what?
do i do this?
K ===== /
relativistic | d/dt (gamma m v) . dx
/
/
== | d(gamma) m v . dx + gamma m dv/dt . dx
| -------------
/ dt
is there a better way to derive the relativistic kinetic energy equation?
Agustín Sánchez
ChN92 wrote:
I've got your formulae on my mail in a very bad format, which make hard to see
any mistake could be in your reasoning. I'll try to show to you a simple
derivation for kinetic energy of a relativisic particle, that you can easily find
(f.e. Alonso-Finn "Fundamental Unversity Physics", vol I , sect. 11,6
Addison-Wesley) .
I hope you'll get the formulae in a good format. First of all, since we
calculate the kinetic energy of a traveling particle at a speed v (regarding to
us), the time derivative is our proper time. In next formulae "S" satates for the
integral symbol applicated to boundary limits 0 and v
Ek = Wtot = S Ftot . ds = S d/t(m v) ds = S v d(mv)
integrating by parts, we have:
Ek = mv^2 - S mv dv
where m is the relativistic mass of the particle at speed v. At speed 0 the mass
is m0. So, putting gamma = (1- v^2/c^2)^1/2:
Ek = m0v^2/gamma - S m0 v dv /gamma =
= m0 v^2/gamma + m0 c^2 (1 - v^2/c^2)1/2 - m0 c^2 =
Combining the two first terms in one:
Ek = m0 c^2/gamma - m0 c^2
In the first term of the above formula, m0 /gamma is the same as de
relativistic mass m of the particle. So:
Ek = m c^2 - m0 c^2 = (m - m0) c^2
that shows the equivalence between kinetic energy and the mass increment. This m
is the value of mass mesasured by us. An observer traveling at same speed that the
particle could not see any changes in the initial value m0.
Bye