Ok Janice I will repeat the answer here with more clarification:
If you integrate the first equation above:
[math]
E=\int^x_0{F.dx}=\int^t_0{F.vdt}=\int^t_0{\frac{d(mv)}{dt}.vdt}=\int{(v^2dm+mvdv)}
[/math]
if you integrate it in a regular way you will get the standard equation that gives us the kinetic energy:
[math]E_k=\int{(v^2dm+mvdv)}=m\int^v_0{vdv}=\frac{1}{2}mv^2
[/math]
that is if you assume dm=0 for normal velocities. The reason why we are getting the "half" in this equation is because the velocity increases gradually with time, which makes the integration equals the area of the triangle. Please see the figure in E=mc^2.
Now if you integrate the same first equation above from 0 to c but considering that this happens abruptly without gradual increase on the outer time level because it is not possible to accelerate a mass to the speed of light on the normal time level, but I showed in the article that it is possible on the inner level of time which appear instantaneous on the outer level. So we get:
[math]E=\int{(v^2dm+mvdv)}=m\int^c_0{v.dv}=mc^2[/math]
In this case the integration will give us the area of the square as in the figure in E=mc^2.
Also you can get the same result if you integrate from m to zero or vice versa (and considering dv=0):
[math]E=\int{(v^2dm+mvdv)}=\int^0_m{v^2dm}=mv^2=mc^2
[/math]
Or also from 0 to m, it will give the same result, which corresponds for example to the emission of radiation while the first one corresponds to the absorption.
I hope this clarifies the problem.
Please tell me if you need more clarification, because this is really in the heart of the mathematical formulation of the single monad model.
