# Spring oscillator and time dilatation

Consider the mass M suspended on the (ideal) spring with stiffness D, whose suspension point is in rest in inertial frame K'. I understand how the principle of relativity requires that this harmonic oscillator oscillate at a slower frequency by a factor of gamma (√(1-v^2/c^2)) when measured in inertial frame K (moving with v relative to K'). But I would like to see how the slowing down of this specific type of clock actually happens: Its period is T = 2π√(M/D), so if the oscillation is perpendicular to v ( so that length contraction does not complicate the thing), naively it seems that it slows only by √(gamma), due to M being larger in K. Where is my error? Does Lorentz boost change the stiffness of springs?

• In the period formula, does M refer to the rest mass or the relativistic mass? Also, are you sure the spring's hardness doesn't change under Lorentz boosts? – probably_someone May 22 '18 at 13:43
• We tend to simplify harmonic oscillator models by lumping all the mass into one element with no stiffness and all the stiffness properties into another element with no mass. But the reality is that each element has both properties. So the first question, will the lumped parameter model lead you to the correct solution for relativistic considerations? – docscience May 22 '18 at 13:56
• If the reference frame is moving on the same axis as the spring (e.g., both on the y-axis), then there would be some length contraction for the same spring -- changing its rest length and thus its stiffness. – zhutchens1 May 22 '18 at 14:37
• @zhutchens1 This is why I have specified that the velocity of the frame and the oscillation are perpendicular. – b.Lorenz May 22 '18 at 17:17

An idealized harmonic oscillator is a clock with period $T$ defined as
$T = 2 \pi \sqrt{m/k}$
$m$ rest mass of the material point
$k$ spring constant
An inertial reference frame in relative motion vs. the rest frame of the oscillator measures a period dilated by the $\gamma$ Lorentz factor.
If you want to read the time dilation in the oscillator period definition as measured by the moving frame you have to consider a first $\gamma$ factor attached to the rest mass of the oscillator and a second $\gamma$ factor inversely attached to the spring constant. The latter is due to the definition of the four-force in SR (special relativity). Thus a $\gamma^2$ under a $\sqrt{}$ yields the $\gamma$ you are looking for.