- #1
Hyperreality
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If a particle is in a ideal inertial system, with only potential energy and kinetic energy present, then
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??
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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