If a particle is in a ideal inertial system, with only potential energy and kinetic energy present, then(adsbygoogle = window.adsbygoogle || []).push({});

K + U = E

If we take in account of the relativistic effect, we get

[tex]\frac{mc^2}{\sqrt{(1-\frac{v^2}{c^2})}}-mc^2+U=E[/tex]

If we differentiate both side with respect to its velocity,

[tex]\frac{mv}{\sqrt{(1-\frac{v^2}{c^2})^3}}+dU/dv=0[/tex]

So far, I'm fairly sure my derivations are correct, for I use the last result to derive the "relativistc" force F=ma x gamma^3.

Now, for the next bit, if I solve for U using indefinite integral I ended with

[tex]U=\frac{mc^2}{\sqrt{(1-\frac{v^2}{c^2})}}+C.[/tex]

What is C?

If I did it by definite integral from 0 to v, I ended with the relativistic kinetic energy. Which is right??

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# Relativistic Potential Energy

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