Relativistic Potential Energy

1. Oct 26, 2004

Hyperreality

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

$$\frac{mc^2}{\sqrt{(1-\frac{v^2}{c^2})}}-mc^2+U=E$$

If we differentiate both side with respect to its velocity,

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

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
$$U=\frac{mc^2}{\sqrt{(1-\frac{v^2}{c^2})}}+C.$$

What is C?

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

Last edited: Oct 26, 2004
2. Oct 27, 2004

blue_sky

The prob. is not what C is.
You should write:
$$\frac{mv}{\sqrt{(1-\frac{v^2}{c^2})^3}}+dU/dv=dE/dv$$

blue

3. Oct 29, 2004

Hyperreality

Yes, I should.

But note that E is the total energy of the system, so it is a contant,

ie, dE/dv = 0.

4. May 20, 2009

llamascience

Be careful, you must be precise with your definitions or you will run into problems.

Energy is more appropriately defined as the functional composition of the Hamiltonian with the coordinates/momenta of the system as a function of time i.e. E(t)=H(q(t),p(t)). This is, by conservation laws, constant, but this is not what you had in your equation. You had the Hamiltonian, which indeed has dependence on momentum (and thus velocity) otherwise the particle (by Hamilton's equations) would be at rest.

In your instance the Hamiltonian is defined as H(q,p) = K(p) + U(q,p) where K(p) and U(q,p) are the kinetic and potential energies for a particle at position q with momentum p.

However, you did something slightly correct and interesting, which is that potential energy can never be found absolutely. You can only ever hope to measure/calculate differences in potential across space, thus all potentials can have a (spatial) constant added to them and the observable physics of the system in question will not change.