Is there a relativistic version of the work-energy theorem?

jaketodd

The work-energy theorem is stated here on Wikipedia. On the same page it says "regardless of the choice of reference frame, the work energy theorem remains valid and the work done on the object is equal to the change in kinetic energy."

I am wondering if there is a relativistic version of the work-energy theorem.

Thanks,

Jake

DrStupid

No. The classical version applies to relativity too. But the result is different.

elfmotat

[itex]W=\Delta E_k=(\gamma -1)mc^2[/itex]

Just integrate force over some distance, starting from rest:

[tex]W=\int Fdx=\int \frac{dp~dx}{dt}=\int vdp=vp-\int pdv=\gamma mv^2-\int \gamma mvdv[/tex]

What you end up with is:

[itex]W=\gamma mc^2+C[/itex]

with some constant of integration C. If you set v=0, you take W=0 and γ=1, therefore C=-mc². So what you get is:

[itex]W=\gamma mc^2-mc^2=(\gamma -1)mc^2=E_k[/itex]

Pengwuino

In general, [itex]W=\int F dx[/itex], but the Lorentz transformations tell you the relationship between dx in the two different inertial frames as [itex]dx = {{dx'}\over{\sqrt{1-{{v^2}\over{c^2}}}}}[/itex].

DrStupid

Two different inertial frames?