Relativistic energy of harmonic oscillator What is the relativistic energy of an harmonic oscillator:
$$\frac{m_0 c^2}{\sqrt{(1-\frac{v^2}{c^2})}}+\frac{1}{2}kx^2$$
Or
$$\frac{{m_0 c^2}+\frac{1}{2}kx^2 }{\sqrt{(1-\frac{v^2}{c^2})}}$$
I think the first one is true but I need an exact logic or derivation.
 A: First of all the speed $v$ is the property of the particle. The points on spring move at different speeds. So you can't write like the 2nd one. Even the 1st one is wrong. Also practically speaking the Hooks law works only for small distances and velocities, and after that it fails. If you assume there exists a mass less spring(technically I should say it has negligible kinetic energy otherwise it should always move at speed $c$) which follows Hooke's law at all distances and speeds, then since relativistic contraction does not induce any stresses, we should divide the spring into infinitesimal parts and find their infinitesimal lengths (measured in the lab frame) at the instant.
We can interpret the dx in Hook's law for an infinitesimal spring($k_i$ is the spring constant of that infinitesimal part)
$$dF=k_idx $$
as the difference between proper lengths of an infinitesimal part at an instant(obtained by multiplying the infinitesimal lengths (measured in the lab frame) by $\gamma$ of that infinitesimal spring at that instant). So it will become somewhat complex. But still the 1st equation is a better approximation than the 2nd one.
A: The energy in special relativity (SR) is defined to be $\gamma m_0c^2$ via the scalar product between 4-momenta. So... I think that $kx^2/2$ would then be encompassed in $\gamma m_0c^2$. You should be able to answer this with an application of math by using Hamilton's action principle and the Euler-Lagrange equations.
A: What you need here is the special relativity version of the work-energy theorem.
The proof is given in many places, including that Wikipedia page, but you start out from the idea that the relativistic force is given by :
$$\vec{F}=\frac{d}{dt}(\gamma m_0 \vec{v})$$
and you will (after some math) get :
$$m_0c^2(\gamma_2-\gamma_1)=\int_{\vec{x_1}}^{\vec{x_2}}\vec{F}\cdot d\vec{x}$$
Where the subscripts refer to initial and final positions.  The expression on the left is the change in kinetic energy, and the expression on the right is simply the work done.
For the simple harmonic oscillator in your case that reduces to :
$$W_{21}=\frac{1}{\gamma_2-\gamma_1}\frac k 2 (x_2^2-x_1^2)k$$
